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When a characteristic-zero conclusion is intended, assume additionally that \\(\\operatorname{char}K=0\\); the general counting statements do not require that restriction.  Let \\(\\mathcal M\\) be a moduli functor on complete local Noetherian \\(\\Ocal\\)-algebras with residue field \\(k\\), equipped with a well-defined reduction map\n\\[\n\\red_A:\\mathcal M(A)\\longrightarrow \\mathcal M(k).\n\\]\nFor a Galois representation this means either that a framed integral model has been fixed or that the residual semisimplified class is used.","discrete valuation ring with uniformizer \\(\\varpi\\), maximal ideal \\((\\varpi)\\), residue field\n\\[\nk=\\Ocal/(\\varpi),\\qquad |k|=q,\n\\]\nand fraction field \\(K\\).  When a characteristic-zero conclusion is intended, assume additionally that \\(\\operatorname{char}K=0\\); the general counting statements do not require that restriction.  Let \\(\\mathcal M\\) be a moduli functor on complete local Noetherian \\(\\Ocal\\)-algebras with residue field \\(k\\), equipped with a well-defined reduction map\n\\[\n\\red_A:\\mathcal M(A)\\longrightarrow \\mathcal M(k).\n\\]\nFor a Galois representation this means either that a framed integral model has been fixed or that the residual semisimplified class is used.","zer \\(\\varpi\\), maximal ideal \\((\\varpi)\\), residue field\n\\[\nk=\\Ocal/(\\varpi),\\qquad |k|=q,\n\\]\nand fraction field \\(K\\).  When a characteristic-zero conclusion is intended, assume additionally that \\(\\operatorname{char}K=0\\); the general counting statements do not require that restriction.  Let \\(\\mathcal M\\) be a moduli functor on complete local Noetherian \\(\\Ocal\\)-algebras with residue field \\(k\\), equipped with a well-defined reduction map\n\\[\n\\red_A:\\mathcal M(A)\\longrightarrow \\mathcal M(k).\n\\]\nFor a Galois representation this means either that a framed integral model has been fixed or that the residual semisimplified class is used.","e field\n\\[\nk=\\Ocal/(\\varpi),\\qquad |k|=q,\n\\]\nand fraction field \\(K\\).  When a characteristic-zero conclusion is intended, assume additionally that \\(\\operatorname{char}K=0\\); the general counting statements do not require that restriction.  Let \\(\\mathcal M\\) be a moduli functor on complete local Noetherian \\(\\Ocal\\)-algebras with residue field \\(k\\), equipped with a well-defined reduction map\n\\[\n\\red_A:\\mathcal M(A)\\longrightarrow \\mathcal M(k).\n\\]\nFor a Galois representation this means either that a framed integral model has been fixed or that the residual semisimplified class is used.","\\begin{definition}[Residual seed system]\\label{px:1.1}\nFor \\(\\bar x\\in\\mathcal M(k)\\), define the residual fiber\n\\[\n\\mathcal M_{\\bar x}(A)\n=\n\\{x\\in\\mathcal M(A):\\red_A(x)=\\bar x\\}.\n\\]\nThe point \\(\\bar x\\) is the residual root.  The family of all nonempty fibers is the residual special fiber, and the formal deformation problems above them form the residual forest.\n\\end{definition}","For a Galois problem the principal example is\n\\[\n\\bar\\rho:G_{F,S}\\longrightarrow \\GL_n(k),\n\\]\nwith a deformation functor \\(\\Def_{\\bar\\rho}\\) cut out by a determinant and selected local conditions.","\\begin{theorem}[Residual partition]\\label{px:2.1}\nFor every coefficient algebra \\(A\\),\n\\[\n\\boxed{\n\\mathcal M(A)\n=\n\\bigsqcup_{\\bar x\\in\\mathcal M(k)}\n\\mathcal M_{\\bar x}(A).\n}\n\\]\n\\end{theorem}","\\begin{proof}\nEvery \\(A\\)-valued object has one reduced object \\(\\red_A(x)\\in\\mathcal M(k)\\), so the union is exhaustive.  Two distinct fibers cannot intersect because an object cannot have two different reductions.  Hence the union is disjoint.\n\\end{proof}","\\begin{theorem}[No crossing under positive-depth perturbation]\\label{px:2.2}\nSuppose \\(x,y\\in\\mathcal M(\\Ocal)\\) are related by a deformation operation that is congruent to the identity modulo \\(\\varpi\\).  Then\n\\[\n\\red(x)=\\red(y).\n\\]\nEvery finite composition remains in one residual fiber.  If an infinite perturbation stream converges in the \\(\\varpi\\)-adic topology, its limit remains in the same fiber.  The existence of that limit is an explicit hypothesis, not a consequence of positive depth alone.\n\\end{theorem}","\\begin{proof}\nA correction in \\(\\varpi\\Ocal\\) vanishes after reduction modulo \\(\\varpi\\).  Thus one operation preserves the residual root, and induction proves the finite statement.  For a convergent stream, convergence at the first finite precision makes its reduction eventually equal to that of its limit.  Every term already has the initial reduction, so the limit has that reduction too.  Completeness applies to Cauchy streams; it does not make every root-preserving stream Cauchy.  For example, \\(0,3,0,3,\\ldots\\) has constant reduction modulo \\(3\\) but no limit modulo \\(9\\).\n\\end{proof}","ation preserves the residual root, and induction proves the finite statement.  For a convergent stream, convergence at the first finite precision makes its reduction eventually equal to that of its limit.  Every term already has the initial reduction, so the limit has that reduction too.  Completeness applies to Cauchy streams; it does not make every root-preserving stream Cauchy.  For example, \\(0,3,0,3,\\ldots\\) has constant reduction modulo \\(3\\) but no limit modulo \\(9\\).\n\\end{proof}","proves the finite statement.  For a convergent stream, convergence at the first finite precision makes its reduction eventually equal to that of its limit.  Every term already has the initial reduction, so the limit has that reduction too.  Completeness applies to Cauchy streams; it does not make every root-preserving stream Cauchy.  For example, \\(0,3,0,3,\\ldots\\) has constant reduction modulo \\(3\\) but no limit modulo \\(9\\).\n\\end{proof}","t.  For a convergent stream, convergence at the first finite precision makes its reduction eventually equal to that of its limit.  Every term already has the initial reduction, so the limit has that reduction too.  Completeness applies to Cauchy streams; it does not make every root-preserving stream Cauchy.  For example, \\(0,3,0,3,\\ldots\\) has constant reduction modulo \\(3\\) but no limit modulo \\(9\\).\n\\end{proof}","\\begin{numericalcertificate}[Exact residual partition]\\label{px:2.4}\nThe finite implementation partitions the integers \\(0,\\ldots,8\\) into the three classes modulo \\(3\\):\n\\[\n\\{0,3,6\\},\\qquad \\{1,4,7\\},\\qquad \\{2,5,8\\}.\n\\]\nEvery tested correction of the form \\(x\\mapsto x+p h\\) preserves the residual class exactly.\n\\end{numericalcertificate}","Fix \\(\\bar x\\in\\mathcal M(k)\\).  For \\(m\\ge1\\), set\n\\[\nX_m(\\bar x)=\\mathcal M_{\\bar x}(\\Ocal/\\varpi^m).\n\\]\nReduction gives transition maps\n\\[\nr_{m+1,m}:X_{m+1}(\\bar x)\\longrightarrow X_m(\\bar x).\n\\]\nA compatible branch is a sequence \\((x_m)_{m\\ge1}\\) satisfying\n\\[\nr_{m+1,m}(x_{m+1})=x_m.\n\\]","\\begin{classicalinput}[Represented complete deformation problems]\\label{px:3.1}\nWhen a deformation functor is represented by a complete local ring \\(R_{\\bar x}\\), its \\(\\Ocal\\)-valued points are continuous local maps \\(R_{\\bar x}\\to\\Ocal\\).  Completeness identifies such maps with compatible maps to every quotient \\(\\Ocal/\\varpi^m\\).\n\\end{classicalinput}","\\begin{theorem}[Formal inverse-limit reconstruction]\\label{px:3.2}\nAssume \\(\\mathcal M_{\\bar x}\\) is represented by a complete local \\(\\Ocal\\)-algebra \\(R_{\\bar x}\\).  Then reduction induces a natural bijection\n\\[\n\\boxed{\n\\mathcal M_{\\bar x}(\\Ocal)\n\\cong\n\\varprojlim_m\n\\mathcal M_{\\bar x}(\\Ocal/\\varpi^m).\n}\n\\]\n\\end{theorem}","\\begin{proof}\nA continuous local homomorphism \\(R_{\\bar x}\\to\\Ocal\\) reduces to a compatible family of homomorphisms into \\(\\Ocal/\\varpi^m\\).  Conversely, compatible finite-level maps determine a map\n\\[\nR_{\\bar x}\\longrightarrow\\varprojlim_m\\Ocal/\\varpi^m\\cong\\Ocal.\n\\]\nThe two constructions are inverse because equality in a separated complete ring is detected modulo every power of its maximal ideal.\n\\end{proof}","\\begin{theorem}[Finite-branch compactness]\\label{px:3.3}\nSuppose every \\(X_m(\\bar x)\\) is finite and nonempty and every node has finitely many children.  Then the lift tree above \\(\\bar x\\) contains an infinite compatible branch.\n\\end{theorem}","\\begin{proposition}[Smooth formal branch count]\\label{px:3.4}\nIf\n\\[\nR_{\\bar x}\\cong\\Ocal[[t_1,\\ldots,t_d]]\\quad\\text{as complete local }\\Ocal\\text{-algebras},\n\\]\nthen\n\\[\n\\boxed{\n\\#X_m(\\bar x)=q^{d(m-1)}.\n}\n\\]\nFix a uniformizer and a section \\(s:k\\to\\Ocal\\) of reduction, with \\(s(0)=0\\). Every \\(\\Ocal\\)-valued branch is encoded by \\(d\\) positive-depth digit streams\n\\[\nt_i=s(a_{i,1})\\varpi+s(a_{i,2})\\varpi^2+\\cdots,\n\\qquad a_{i,j}\\in k,\n\\]\n\\end{proposition}","\\begin{proof}\nA local map sends each parameter into the image of \\((\\varpi)\\) in \\(\\Ocal/\\varpi^m\\). Conversely, every tuple in that image is nilpotent and defines a coefficient-preserving evaluation homomorphism. This homomorphism is local. To prove uniqueness without assuming continuity of an arbitrary competing homomorphism, observe in one variable that a homomorphism killing \\(X^N\\) only sees the finite polynomial truncation modulo \\(X^N\\). Induction over the finite number of variables gives uniqueness in all dimensions. The actual local maps are therefore in bijection with the full parameter tuple set. Its cardinality is \\(q^{d(m-1)}\\), including the zero-dimensional chart.","eter into the image of \\((\\varpi)\\) in \\(\\Ocal/\\varpi^m\\). Conversely, every tuple in that image is nilpotent and defines a coefficient-preserving evaluation homomorphism. This homomorphism is local. To prove uniqueness without assuming continuity of an arbitrary competing homomorphism, observe in one variable that a homomorphism killing \\(X^N\\) only sees the finite polynomial truncation modulo \\(X^N\\). Induction over the finite number of variables gives uniqueness in all dimensions. The actual local maps are therefore in bijection with the full parameter tuple set. Its cardinality is \\(q^{d(m-1)}\\), including the zero-dimensional chart.","ocal. To prove uniqueness without assuming continuity of an arbitrary competing homomorphism, observe in one variable that a homomorphism killing \\(X^N\\) only sees the finite polynomial truncation modulo \\(X^N\\). Induction over the finite number of variables gives uniqueness in all dimensions. The actual local maps are therefore in bijection with the full parameter tuple set. Its cardinality is \\(q^{d(m-1)}\\), including the zero-dimensional chart.","For integral maps, reduce an arbitrary tuple in \\((\\varpi)\\Ocal\\) at every precision, evaluate there, and use the represented-map reconstruction theorem. The finite evaluations are compatible, the reconstructed algebra map is local, and every competing local map has the same finite reductions. Separatedness gives uniqueness. Local maps are continuous in the maximal-ideal adic topologies. This establishes the integral parameter classification rather than assuming it from the finite count.","For the digit claim, subtract the chosen representative of the residue and divide the remainder by the uniformizer. Iterating gives unique digits. Conversely, the partial sums of any digit stream form a compatible Cauchy family and reconstruct uniquely by completeness and separatedness. More precisely, for each parameter there is a unique stream \\((a_{i,j})_{j\\ge1}\\) such that\n\\[\n\\varpi^{n+1}\\mid t_i-\\sum_{j=1}^{n}s(a_{i,j})\\varpi^j\\qquad(n\\ge0).\n\\]\nThe section is a set-theoretic section of reduction, not an asserted embedding of residue fields into the coefficient ring. Finally, a supplied local \\(\\Ocal\\)-algebra isomorphism from \\(R_{\\bar x}\\) transports both the map classification and the count.\n\\end{proof}","patible Cauchy family and reconstruct uniquely by completeness and separatedness. More precisely, for each parameter there is a unique stream \\((a_{i,j})_{j\\ge1}\\) such that\n\\[\n\\varpi^{n+1}\\mid t_i-\\sum_{j=1}^{n}s(a_{i,j})\\varpi^j\\qquad(n\\ge0).\n\\]\nThe section is a set-theoretic section of reduction, not an asserted embedding of residue fields into the coefficient ring. Finally, a supplied local \\(\\Ocal\\)-algebra isomorphism from \\(R_{\\bar x}\\) transports both the map classification and the count.\n\\end{proof}","ct uniquely by completeness and separatedness. More precisely, for each parameter there is a unique stream \\((a_{i,j})_{j\\ge1}\\) such that\n\\[\n\\varpi^{n+1}\\mid t_i-\\sum_{j=1}^{n}s(a_{i,j})\\varpi^j\\qquad(n\\ge0).\n\\]\nThe section is a set-theoretic section of reduction, not an asserted embedding of residue fields into the coefficient ring. Finally, a supplied local \\(\\Ocal\\)-algebra isomorphism from \\(R_{\\bar x}\\) transports both the map classification and the count.\n\\end{proof}","\\begin{figure}[H]\n\\centering\n\\includegraphics[width=0.94\\linewidth]{02_residual_digit_forest.pdf}\n\\caption{The first four levels of a full \\(3\\)-adic digit forest.  A path is a compatible finite-depth perturbation stream.}\n\\end{figure}","\\begin{figure}[H]\n\\centering\n\\includegraphics[width=0.82\\linewidth]{04_inverse_limit_digits.pdf}\n\\caption{A compatible base-five residue tower reconstructs the digit stream \\((2,4,1,3,0,2)\\).}\n\\end{figure}","For a residual representation \\(\\bar\\rho\\), first-order deformations and lifting obstructions are classically controlled by cohomology of the adjoint module.  The projector calculus does not replace that theory.  It resolves the controlling complex into invariant packets when suitable projectors exist.","\\begin{classicalinput}[Tangent and obstruction spaces]\\label{px:4.1}\nFor a framed representation problem, first-order lifts are described by \\(Z^1(G,\\ad\\bar\\rho)\\).  For an unframed Schur problem, tangent directions are described by \\(H^1(G,\\ad\\bar\\rho)\\).  Relations in a presentation of a deformation ring map into \\(H^2(G,\\ad\\bar\\rho)\\), which contains the obstruction classes.\n\\end{classicalinput}","Let\n\\[\nC^0\\xrightarrow{d_0}C^1\\xrightarrow{d_1}C^2\\xrightarrow{}\\cdots\n\\]\nbe a cochain complex over a field.  Suppose a finite family of projectors \\(P_j^r:C^r\\to C^r\\) satisfies\n\\[\n\\sum_jP_j^r=I,\n\\qquad\nP_i^rP_j^r=\\delta_{ij}P_j^r,\n\\qquad\nd_rP_j^r=P_j^{r+1}d_r.\n\\]","\\begin{theorem}[Cohomology packet decomposition]\\label{px:4.2}\nUnder these hypotheses, each packet is a subcomplex and\n\\[\n\\boxed{\nH^r(C^\\bullet)\n\\cong\n\\bigoplus_jH^r(P_jC^\\bullet).\n}\n\\]\nIn particular,\n\\[\nH^1=\\bigoplus_jH^1_j,\n\\qquad\nH^2=\\bigoplus_jH^2_j,\n\\]\nso tangent and obstruction spaces reconstruct exactly from their packet components.\n\\end{theorem}","\\begin{proof}\nFor each packet, use the actual image submodules\n\\(C_j^r=\\operatorname{im}P_j^r\\). Commutation with the differential\nmakes these a subcomplex. Write\n\\[\nZ^r=\\ker d_r,\\qquad B^r=\\operatorname{im}d_{r-1}\\subseteq Z^r,\n\\qquad H^r=Z^r/B^r,\n\\]\nand define \\(Z_j^r,B_j^r,H_j^r\\) using the restricted differentials of\n\\(C_j^\\bullet\\). The inclusion \\(i_j:C_j^\\bullet\\to C^\\bullet\\)\nand the image-valued projection \\(p_j:C^\\bullet\\to C_j^\\bullet\\)\nare cochain maps, so they induce maps on these genuine quotients.","h packet, use the actual image submodules\n\\(C_j^r=\\operatorname{im}P_j^r\\). Commutation with the differential\nmakes these a subcomplex. Write\n\\[\nZ^r=\\ker d_r,\\qquad B^r=\\operatorname{im}d_{r-1}\\subseteq Z^r,\n\\qquad H^r=Z^r/B^r,\n\\]\nand define \\(Z_j^r,B_j^r,H_j^r\\) using the restricted differentials of\n\\(C_j^\\bullet\\). The inclusion \\(i_j:C_j^\\bullet\\to C^\\bullet\\)\nand the image-valued projection \\(p_j:C^\\bullet\\to C_j^\\bullet\\)\nare cochain maps, so they induce maps on these genuine quotients.","For \\(i\\ne j\\), orthogonality gives\n\\(H(p_i)H(i_j)=0\\). Idempotence gives\n\\(H(p_j)H(i_j)=I\\), and completeness gives\n\\(\\sum_j H(i_j)H(p_j)=I\\). Thus the maps\n\\[\n[z]\\longmapsto([P_j^r z])_j,\n\\qquad\n([z_j])_j\\longmapsto\\left[\\sum_j i_j z_j\\right]\n\\]\nare inverse linear maps. The finite product equals the direct sum, proving\nthe assertion. In the formalization, both sides are also identified with\nMathlib's standard homology objects of the genuine graded complexes.\n\\end{proof}","\\paragraph{Scope of the splitting.}\nThe algebraic proof works over any commutative ring and does not divide by\n\\(2\\), so it applies in characteristic two when the stated integral\nchain projectors exist. It does not construct those projectors. The\nidentification of a particular Galois tangent or obstruction problem with\nthis cohomology remains the preceding classical input.\nThe splitting detects vanishing:\n\\[\nH^r(C^\\bullet)=0\\quad\\Longleftrightarrow\\quad\nH^r(C_j^\\bullet)=0\\text{ for every }j.\n\\]\nIt does not force either side to vanish or decouple nonlinear deformation\nequations. Moreover, commutation with the differential is essential:\ncomplementary idempotents on the underlying vector spaces need not\npreserve cocycles or coboundaries.","Scope of the splitting.}\nThe algebraic proof works over any commutative ring and does not divide by\n\\(2\\), so it applies in characteristic two when the stated integral\nchain projectors exist. It does not construct those projectors. The\nidentification of a particular Galois tangent or obstruction problem with\nthis cohomology remains the preceding classical input.\nThe splitting detects vanishing:\n\\[\nH^r(C^\\bullet)=0\\quad\\Longleftrightarrow\\quad\nH^r(C_j^\\bullet)=0\\text{ for every }j.\n\\]\nIt does not force either side to vanish or decouple nonlinear deformation\nequations. Moreover, commutation with the differential is essential:\ncomplementary idempotents on the underlying vector spaces need not\npreserve cocycles or coboundaries.","\\paragraph{Naturality.}\nIf a cochain map \\(F:C^\\bullet\\to D^\\bullet\\) intertwines the supplied\nprojectors, it restricts to maps \\(F_j:C_j^\\bullet\\to D_j^\\bullet\\).\nThe direct-sum decomposition commutes with \\(H^r(F)\\) and the family\n\\(H^r(F_j)\\). This is checked using representatives in the actual\nquotients, independently of any dimension computation.","\\begin{remark}[Odd and dyadic nonlinear equations]\\label{px:4.3}\nAt odd residue characteristic, a differential graded Lie model may be written in Maurer--Cartan form\n\\[\nd\\eta+\\frac12[\\eta,\\eta]=0,\n\\]\nand a cyclic grading converts this into exact packet convolutions.  At residue characteristic two, division by \\(2\\) is not integral.  A quadratic or divided-square refinement must be retained.  This is one reason the dyadic residual forest is weighted rather than obtained by formal substitution into the odd-prime formulas.\n\\end{remark}","\\begin{numericalcertificate}[Finite packet complex]\\label{px:4.4}\nThe exact modular test uses two invariant subcomplexes over \\(\\F_5\\).  Their cohomology dimensions are\n\\[\n(0,0,0),\n\\qquad\n(1,2,1),\n\\]\nand the reconstructed total is\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nThe packet-sum residual is zero.\n\\end{numericalcertificate}","\\paragraph{New checked realization of the finite pattern.}\nThe Block 04 implementation separately uses\n\\[\n\\mathbf F_5^2\\xrightarrow{d_0}\\mathbf F_5^4\\xrightarrow{d_1}\\mathbf F_5^2,\n\\quad d_0(a,b)=(a,0,0,0),\\quad d_1(a,b,c,d)=(b,0).\n\\]\nThe first packet is the exact complex with dimensions \\((1,2,1)\\),\nand the second packet has zero differentials and dimensions \\((1,2,1)\\).\nThe actual middle cohomology is linearly isomorphic to \\(\\mathbf F_5^2\\),\nwhich is checked in Lean. Independent exhaustive coset tests check all\nthree cohomology degrees after changes of basis. This is a new specified\nrealization of the displayed dimension pattern, not a claimed rerun of\nthe original companion implementation.","on of the finite pattern.}\nThe Block 04 implementation separately uses\n\\[\n\\mathbf F_5^2\\xrightarrow{d_0}\\mathbf F_5^4\\xrightarrow{d_1}\\mathbf F_5^2,\n\\quad d_0(a,b)=(a,0,0,0),\\quad d_1(a,b,c,d)=(b,0).\n\\]\nThe first packet is the exact complex with dimensions \\((1,2,1)\\),\nand the second packet has zero differentials and dimensions \\((1,2,1)\\).\nThe actual middle cohomology is linearly isomorphic to \\(\\mathbf F_5^2\\),\nwhich is checked in Lean. Independent exhaustive coset tests check all\nthree cohomology degrees after changes of basis. This is a new specified\nrealization of the displayed dimension pattern, not a claimed rerun of\nthe original companion implementation.","At an odd prime, distinct sign characters remain distinct in the residue field.  At the coefficient prime two,\n\\[\n1\\equiv-1\\pmod2.\n\\]\nThus distinct characteristic-zero packets can share one residual root and separate only after more precision is retained.","\\begin{definition}[Visibility weight]\\label{px:5.1}\nFor two integral packet coordinates \\(a,b\\in\\Ocal\\), define their separation depth\n\\[\nw(a,b)=\\min\\{m\\ge1:a\\not\\equiv b\\pmod{\\varpi^m}\\},\n\\]\nwhen such an \\(m\\) exists.  A dyadic packet decomposition is weighted by these first visibility depths and by the valuations of any denominators required by its rational projectors.\n\\end{definition}","\\begin{example}[The two sign branches]\\label{px:5.2}\nThe characters \\(\\chi_+(g)=1\\) and \\(\\chi_-(g)=-1\\) have the same reduction modulo \\(2\\), but\n\\[\n1\\not\\equiv-1\\pmod4.\n\\]\nTheir first separation depth is therefore \\(2\\).  One mod-two root contains two branches that become distinct at the next level.\n\\end{example}","A simple weighted formal model already captures delayed visibility.  Let \\(w_i\\ge1\\) and constrain\n\\[\nt_i\\in(\\varpi^{w_i}\\Ocal+\\varpi^m\\Ocal)/\\varpi^m\\Ocal.\n\\]\nThen coordinate \\(i\\) is invisible until \\(m>w_i\\).","\\begin{proposition}[Weighted smooth-chart count]\\label{px:5.3}\nFor weights \\(\\bm w=(w_1,\\ldots,w_d)\\),\n\\[\n\\boxed{\nN_{\\bm w}(m)\n=\nq^{\\sum_{i=1}^d\\max(m-w_i,0)}.\n}\n\\]\n\\end{proposition}","\\begin{proof}\nThe image of \\(\\varpi^{w_i}\\Ocal\\) in \\(\\Ocal/\\varpi^m\\) has one element when \\(m\\le w_i\\), and \\(q^{m-w_i}\\) elements when \\(m>w_i\\).  Multiplying the independent coordinate counts gives the formula.\n\\end{proof}","\\begin{definition}[Seed zeta function]\\label{px:6.1}\nFor a residual root \\(\\bar x\\) with finite levels, define\n\\[\n\\boxed{\nZ_{\\bar x}(T)\n=\n\\sum_{m\\ge1}\\#X_m(\\bar x)T^m.\n}\n\\]\nWhen a weighted chart is specified, write \\(Z_{\\bm w}(T)\\).\n\\end{definition}","\\begin{theorem}[Smooth seed zeta function]\\label{px:6.2}\nFor a smooth formal component of dimension \\(d\\),\n\\[\n\\boxed{\nZ_d(T)=\\frac{T}{1-q^dT}.\n}\n\\]\n\\end{theorem}","\\begin{proof}\nBy the proved formal-model map-to-parameter equivalence, \\(\\#X_m=q^{d(m-1)}\\) counts all the actual local coefficient-algebra maps. Summing the geometric series gives the identity. The checked statement is an equality of infinite formal power series, with zero constant coefficient.\n\\end{proof}","\\begin{theorem}[Weighted seed zeta function]\\label{px:6.3}\nLet \\(1\\le w_1\\le\\cdots\\le w_d\\), put \\(W=\\sum_iw_i\\), and let\n\\[\na_m=q^{\\sum_i\\max(m-w_i,0)}.\n\\]\nThen\n\\[\n\\boxed{\nZ_{\\bm w}(T)\n=\n\\sum_{m=1}^{w_d}a_mT^m\n+\nq^{-W}\\frac{(q^dT)^{w_d+1}}{1-q^dT}.\n}\n\\]\nIn particular, every weighted smooth chart has a rational seed zeta function.\n\\end{theorem}","\\begin{proof}\nThe finite sum records the depths before all coordinates are active.  For \\(m\\ge w_d+1\\),\n\\[\na_m=q^{dm-W},\n\\]\nso the remaining terms form one geometric series.\n\\end{proof}","\\begin{theorem}[Singular power seed]\\label{px:6.4}\nFor integers \\(e\\ge1\\) and \\(m\\ge1\\), in \\(\\Ocal/\\varpi^m\\) the equation\n\\[\nx^e=0\n\\]\nhas exactly\n\\[\n\\boxed{\nq^{m-\\lceil m/e\\rceil}\n}\n\\]\nsolutions.  Its seed zeta function is\n\\[\n\\boxed{\nZ_e(T)\n=\n\\frac{\\sum_{r=1}^{e}q^{r-1}T^r}\n{1-q^{e-1}T^e}.\n}\n\\]\n\\end{theorem}","\\begin{proof}\nThe congruence is equivalent to\n\\[\ne\\,v_\\varpi(x)\\ge m,\n\\]\nso \\(x\\) must lie in \\(\\varpi^{\\lceil m/e\\rceil}\\Ocal/\\varpi^m\\), which has the stated cardinality.  Write \\(m=ek+r\\), \\(1\\le r\\le e\\).  The exponent becomes \\((e-1)k+r-1\\).  Summing over \\(k\\ge0\\) and \\(1\\le r\\le e\\) gives the rational expression.\n\\end{proof}","\\begin{definition}[Branching dimension]\\label{px:6.5}\nWhen the limit exists, define the finite-level counting exponent\n\\[\n\\dim_{\\mathrm{seed}}(\\bar x)\n=\n\\lim_{m\\to\\infty}\n\\frac{\\log_q\\#X_m(\\bar x)}{m}.\n\\]\nA limsup may be used in general.\n\\end{definition}","\\begin{corollary}[Fractional singular dimension]\\label{px:6.6}\nFor \\(x^e=0\\),\n\\[\n\\boxed{\n\\dim_{\\mathrm{seed}}=1-\\frac1e.\n}\n\\]\nIn particular, \\(x^2=0\\) has branching dimension \\(1/2\\).","\\paragraph{Counting exponent versus compatible branches.}\nThe preceding exponent counts all finite-level solutions, including nodes which cannot lift indefinitely. It is not the cardinality or dimension of the space of infinite branches. In fact every compatible tower of solutions to \\(x^e=0\\), for \\(e\\ge1\\), is the zero tower. For each \\(m\\), a solution at depth \\(em\\) lies in \\(\\varpi^m\\Ocal/\\varpi^{em}\\); its reduction at depth \\(m\\) is therefore zero. Compatibility proves the assertion at every depth. This observation and the real logarithmic limit are both formalized in Block 02.","ompatible branches.}\nThe preceding exponent counts all finite-level solutions, including nodes which cannot lift indefinitely. It is not the cardinality or dimension of the space of infinite branches. In fact every compatible tower of solutions to \\(x^e=0\\), for \\(e\\ge1\\), is the zero tower. For each \\(m\\), a solution at depth \\(em\\) lies in \\(\\varpi^m\\Ocal/\\varpi^{em}\\); its reduction at depth \\(m\\) is therefore zero. Compatibility proves the assertion at every depth. This observation and the real logarithmic limit are both formalized in Block 02.","\\begin{figure}[H]\n\\centering\n\\includegraphics[width=0.83\\linewidth]{06_smooth_singular_counts.pdf}\n\\caption{A smooth two-dimensional formal chart and the singular equation \\(x^2=0\\) have visibly different exponential growth.}\n\\end{figure}","\\begin{figure}[H]\n\\centering\n\\includegraphics[width=0.78\\linewidth]{08_fractional_seed_dimension.pdf}\n\\caption{The finite-depth branching dimension of \\(x^e=0\\) is \\(1-1/e\\), giving a family of fractional local growth exponents.}\n\\end{figure}","\\begin{definition}[Admissible residual transport]\\label{px:7.1}\nAn admissible transport from \\(\\bar\\rho\\) to \\(\\bar\\rho'\\) consists of\n\\[\n\\mathcal T:\\bar\\rho\\longmapsto\\bar\\rho'\n\\]\ntogether with natural maps\n\\[\n\\mathcal T_A:\\Def_{\\bar\\rho}(A)\\longrightarrow\\Def_{\\bar\\rho'}(A)\n\\]\nfor every coefficient algebra \\(A\\), compatible with base change and reduction.  When the transport is reversible, it is an isomorphism in the residual transport groupoid.  One-way operations give a directed category.\n\\end{definition}","For a finite collection of roots, record a directed edge \\(u\\to v\\) whenever occupancy or the relevant property is known to propagate from \\(u\\) to \\(v\\).","\\begin{definition}[Directed seed rank]\\label{px:7.2}\nFor a directed transport graph \\(G\\), define\n\\[\n\\srank_G\n=\n\\min\\{|S|:\\Cl_G(S)=V(G)\\},\n\\]\nwhere \\(\\Cl_G(S)\\) is the set of vertices reachable from \\(S\\).\n\\end{definition}","\\begin{theorem}[Directed minimum-seed theorem]\\label{px:7.3}\nLet \\(G\\) be a finite directed graph.  Collapse every strongly connected component to one vertex, obtaining the condensation DAG.  Then\n\\[\n\\boxed{\n\\srank_G\n=\n\\#\\{\\text{source strongly connected components of }G\\}.\n}\n\\]\nOne representative from each source component is a minimum seed set.\n\\end{theorem}","If a property $P$ propagates along every supplied edge and holds at all selected\nseeds, it holds at every reached vertex. These are explicit hypotheses. The\nminimum-seed theorem does not construct valid arithmetic edges or supply\ninitial occupancy.","\\paragraph{Why the finite hypothesis cannot be dropped.}\nTake vertices $\\{*\\}\\sqcup\\{v_n:n\\in\\mathbf N\\}$, with $*$ isolated and\nedges $v_{n+1}\\to v_n$. Every strongly connected component is a singleton,\nand $\\{*\\}$ is the unique source component. No finite seed set covers all\nvertices: it can reach only indices no larger than the largest chosen index.\nBoth the unique-source statement and the failure of every finite cover are\nchecked in Lean. Hence one source component in an infinite condensation does\nnot imply that one seed reaches all objects.","\\begin{figure}[H]\n\\centering\n\\includegraphics[width=0.91\\linewidth]{09_directed_transport_graph.pdf}\n\\caption{A finite directed transport model.  The components \\(\\{A,B\\}\\), \\(\\{D,E\\}\\), and \\(\\{G\\}\\) are sources; \\(C\\) and \\(F\\) are reached downstream.}\n\\label{fig:transport-graph}\n\\end{figure}","\\paragraph{New explicit checked realization.}\nSeparately from the unretrieved original companion implementation, Block 05\nspecifies vertices $A,B,C,D,E,F,G$ and the eight edges\n\\[\n A\\to B,\\ B\\to A,\\ B\\to C,\\ C\\to F,\\ D\\to E,\\ E\\to D,\\ E\\to F,\\ G\\to F.\n\\]\nThe source components are $\\{A,B\\}$, $\\{D,E\\}$, and $\\{G\\}$; the other\ncomponents are $\\{C\\}$ and $\\{F\\}$. A bounded path search to length three\nis proved closed under every original edge, which certifies equality with\nunbounded reachability. The resulting label certificate is identified with\nthe actual quotient by mutual reachability. Lean checks seed rank three and\nexactly the four minimum sets\n\\[\n \\{A,D,G\\},\\quad\\{A,E,G\\},\\quad\\{B,D,G\\},\\quad\\{B,E,G\\}.\n\\]\nIndependent exact tests enumerate every directed graph on zero through four\nvertices, including loops, and every seed subset. These finit","\\paragraph{New explicit checked realization.}\nSeparately from the unretrieved original companion implementation, Block 05\nspecifies vertices $A,B,C,D,E,F,G$ and the eight edges\n\\[\n A\\to B,\\ B\\to A,\\ B\\to C,\\ C\\to F,\\ D\\to E,\\ E\\to D,\\ E\\to F,\\ G\\to F.\n\\]\nThe source components are $\\{A,B\\}$, $\\{D,E\\}$, and $\\{G\\}$; the other\ncomponents are $\\{C\\}$ and $\\{F\\}$. A bounded path search to length three\nis proved closed under every original edge, which certifies equality with\nunbounded reachability. The resulting label certificate is identified with\nthe actual quotient by mutual reachability. Lean checks seed rank three and\nexactly the four minimum sets\n\\[\n \\{A,D,G\\},\\quad\\{A,E,G\\},\\quad\\{B,D,G\\},\\quad\\{B,E,G\\}.\n\\]\nIndependent exact tests enumerate every directed graph on zero through four\nvertices, including loops, and every seed subset. These finite tests audit the\nimplementation; the universal theorem is a separate formal proof.","e source components are $\\{A,B\\}$, $\\{D,E\\}$, and $\\{G\\}$; the other\ncomponents are $\\{C\\}$ and $\\{F\\}$. A bounded path search to length three\nis proved closed under every original edge, which certifies equality with\nunbounded reachability. The resulting label certificate is identified with\nthe actual quotient by mutual reachability. Lean checks seed rank three and\nexactly the four minimum sets\n\\[\n \\{A,D,G\\},\\quad\\{A,E,G\\},\\quad\\{B,D,G\\},\\quad\\{B,E,G\\}.\n\\]\nIndependent exact tests enumerate every directed graph on zero through four\nvertices, including loops, and every seed subset. These finite tests audit the\nimplementation; the universal theorem is a separate formal proof.","A compatible system should not be treated as an unrelated collection of \\(\\lambda\\)-adic roots.  Its members share algebraic Frobenius polynomials, Hodge--Tate data, determinant information, and local Weil--Deligne packets.","\\begin{definition}[Common arithmetic spine]\\label{px:8.1}\nLet \\(M\\) be a number field.  A common spine \\(\\Compat\\) consists of:\n\\begin{itemize}\n\\item characteristic polynomials \\(P_v(X)\\in M[X]\\) at almost all places;\n\\item coefficient-prime independent weight and determinant data;\n\\item prescribed local Weil--Deligne or inertial packet labels at the ramified places.\n\\end{itemize}\nFor each finite place \\(\\lambda\\) of \\(M\\), let\n\\[\nX_{\\lambda,m}(\\Compat)\n\\]\nbe the finite set of admissible objects modulo \\(\\varpi_\\lambda^m\\) realizing the reduction of this spine.\n\\end{definition}","efinition}[Common arithmetic spine]\\label{px:8.1}\nLet \\(M\\) be a number field.  A common spine \\(\\Compat\\) consists of:\n\\begin{itemize}\n\\item characteristic polynomials \\(P_v(X)\\in M[X]\\) at almost all places;\n\\item coefficient-prime independent weight and determinant data;\n\\item prescribed local Weil--Deligne or inertial packet labels at the ramified places.\n\\end{itemize}\nFor each finite place \\(\\lambda\\) of \\(M\\), let\n\\[\nX_{\\lambda,m}(\\Compat)\n\\]\nbe the finite set of admissible objects modulo \\(\\varpi_\\lambda^m\\) realizing the reduction of this spine.\n\\end{definition}","on arithmetic spine]\\label{px:8.1}\nLet \\(M\\) be a number field.  A common spine \\(\\Compat\\) consists of:\n\\begin{itemize}\n\\item characteristic polynomials \\(P_v(X)\\in M[X]\\) at almost all places;\n\\item coefficient-prime independent weight and determinant data;\n\\item prescribed local Weil--Deligne or inertial packet labels at the ramified places.\n\\end{itemize}\nFor each finite place \\(\\lambda\\) of \\(M\\), let\n\\[\nX_{\\lambda,m}(\\Compat)\n\\]\nbe the finite set of admissible objects modulo \\(\\varpi_\\lambda^m\\) realizing the reduction of this spine.\n\\end{definition}","ic spine]\\label{px:8.1}\nLet \\(M\\) be a number field.  A common spine \\(\\Compat\\) consists of:\n\\begin{itemize}\n\\item characteristic polynomials \\(P_v(X)\\in M[X]\\) at almost all places;\n\\item coefficient-prime independent weight and determinant data;\n\\item prescribed local Weil--Deligne or inertial packet labels at the ramified places.\n\\end{itemize}\nFor each finite place \\(\\lambda\\) of \\(M\\), let\n\\[\nX_{\\lambda,m}(\\Compat)\n\\]\nbe the finite set of admissible objects modulo \\(\\varpi_\\lambda^m\\) realizing the reduction of this spine.\n\\end{definition}","at\\) consists of:\n\\begin{itemize}\n\\item characteristic polynomials \\(P_v(X)\\in M[X]\\) at almost all places;\n\\item coefficient-prime independent weight and determinant data;\n\\item prescribed local Weil--Deligne or inertial packet labels at the ramified places.\n\\end{itemize}\nFor each finite place \\(\\lambda\\) of \\(M\\), let\n\\[\nX_{\\lambda,m}(\\Compat)\n\\]\nbe the finite set of admissible objects modulo \\(\\varpi_\\lambda^m\\) realizing the reduction of this spine.\n\\end{definition}","The system seed is the fibered object\n\\[\n\\boxed{\n\\Sys_{\\mathrm{sys}}\n=\n\\Sys_2^{\\mathrm{wt}}\n\\times_{\\Compat}\n\\prod_{\\lambda\\nmid2}'\\Sys_\\lambda.\n}\n\\]\nThe fiber product notation states a compatibility requirement.  It does not prove that every fiber is nonempty.","\\begin{theorem}[Adelic finite-depth completion]\\label{px:8.2}\nAssume:\n\\begin{enumerate}\n\\item an algebraic common spine \\(\\Compat\\) over a number field \\(M\\) is fixed;\n\\item for every finite place \\(\\lambda\\) of \\(M\\) and every \\(m\\ge1\\), the companion level \\(X_{\\lambda,m}(\\Compat)\\) is finite and nonempty;\n\\item the reduction forests are finitely branching;\n\\item the selected local conditions are closed under inverse limits.\n\\end{enumerate}\nThen for every \\(\\lambda\\) there is an infinite compatible branch\n\\[\n\\rho_\\lambda\\in\\varprojlim_mX_{\\lambda,m}(\\Compat),\n\\]\nand the collection \\(\\{\\rho_\\lambda\\}_\\lambda\\) realizes the prescribed strictly compatible spine.\n\\end{theorem}","sume:\n\\begin{enumerate}\n\\item an algebraic common spine \\(\\Compat\\) over a number field \\(M\\) is fixed;\n\\item for every finite place \\(\\lambda\\) of \\(M\\) and every \\(m\\ge1\\), the companion level \\(X_{\\lambda,m}(\\Compat)\\) is finite and nonempty;\n\\item the reduction forests are finitely branching;\n\\item the selected local conditions are closed under inverse limits.\n\\end{enumerate}\nThen for every \\(\\lambda\\) there is an infinite compatible branch\n\\[\n\\rho_\\lambda\\in\\varprojlim_mX_{\\lambda,m}(\\Compat),\n\\]\nand the collection \\(\\{\\rho_\\lambda\\}_\\lambda\\) realizes the prescribed strictly compatible spine.\n\\end{theorem}","pine \\(\\Compat\\) over a number field \\(M\\) is fixed;\n\\item for every finite place \\(\\lambda\\) of \\(M\\) and every \\(m\\ge1\\), the companion level \\(X_{\\lambda,m}(\\Compat)\\) is finite and nonempty;\n\\item the reduction forests are finitely branching;\n\\item the selected local conditions are closed under inverse limits.\n\\end{enumerate}\nThen for every \\(\\lambda\\) there is an infinite compatible branch\n\\[\n\\rho_\\lambda\\in\\varprojlim_mX_{\\lambda,m}(\\Compat),\n\\]\nand the collection \\(\\{\\rho_\\lambda\\}_\\lambda\\) realizes the prescribed strictly compatible spine.\n\\end{theorem}","lace \\(\\lambda\\) of \\(M\\) and every \\(m\\ge1\\), the companion level \\(X_{\\lambda,m}(\\Compat)\\) is finite and nonempty;\n\\item the reduction forests are finitely branching;\n\\item the selected local conditions are closed under inverse limits.\n\\end{enumerate}\nThen for every \\(\\lambda\\) there is an infinite compatible branch\n\\[\n\\rho_\\lambda\\in\\varprojlim_mX_{\\lambda,m}(\\Compat),\n\\]\nand the collection \\(\\{\\rho_\\lambda\\}_\\lambda\\) realizes the prescribed strictly compatible spine.\n\\end{theorem}","\\begin{proof}\nFinite-branch compactness gives an infinite branch in every coefficient-prime forest.  Inverse-limit closure gives an admissible \\(\\lambda\\)-adic object.  By definition each finite level has the prescribed reductions of the common Frobenius and local packets.  Equality at every finite precision implies equality in the complete coefficient ring.  Thus all members realize the same spine.\n\\end{proof}","\\begin{corollary}[Residual companion obstruction]\\label{px:8.3}\nIf\n\\[\nX_{\\lambda,1}(\\Compat)=\\varnothing\n\\]\nfor one coefficient prime \\(\\lambda\\), no compatible completion with spine \\(\\Compat\\) exists.  Deeper perturbations at other primes cannot create the missing residual companion.\n\\end{corollary}","\\begin{example}[A scalar companion failure]\\label{px:8.4}\nFor\n\\[\nf(x)=x^2+1,\n\\]\nthere are two simple branches at \\(p=5\\) and \\(p=13\\), no residual roots at \\(p=3\\) or \\(p=7\\), and one root modulo \\(2\\) that has no child modulo \\(4\\).  A complete \\(5\\)-adic branch therefore does not imply an adelic solution.\n\\end{example}","\\begin{figure}[H]\n\\centering\n\\includegraphics[width=0.79\\linewidth]{12_companion_forest_occupancy.pdf}\n\\caption{Finite companion levels for \\(x^2+1=0\\).  The obstruction at \\(2,3,7\\) is visible before any deep lifting argument begins.}\n\\end{figure}","\\paragraph{Checked representation-level completion and its arithmetic boundary.}\nThe eighth formalization block works with actual matrix representations of a\nfixed group, not only with underlying sets. For a complete separated ring $R$\nand ideal $I$, compatible maps\n$\\rho_m:G\\to\\operatorname{GL}_d(R/I^m)$ reconstruct a unique integral\nrepresentation for that fixed compatible family. Multiplication and inverses\nare verified after every reduction. If all finite representations are continuous\nfor the discrete quotient topology, the integral representation is continuous\nfor the ideal-adic topology. This does not assert uniqueness among all possible\nbranches of the allowed forest.","All coefficients of the characteristic polynomial commute with reduction.\nConsequently, prescribed integral characteristic polynomials at every precision\nhold exactly in the reconstructed representation. For a common coefficient field\n$E$, local integral polynomials lie in $\\Ocal_\\lambda[X]$ and are compared with\n$P_v\\in E[X]$ only after mapping both into the local coefficient field. No field\nembedding $E\\to\\Ocal_\\lambda$ is assumed. The allowed-place predicate is explicit\nand may depend on $\\lambda$. Trivial action on a supplied subgroup is likewise\ndetected at every precision; identifying that subgroup with arithmetic inertia\nis a separate input.","fficients of the characteristic polynomial commute with reduction.\nConsequently, prescribed integral characteristic polynomials at every precision\nhold exactly in the reconstructed representation. For a common coefficient field\n$E$, local integral polynomials lie in $\\Ocal_\\lambda[X]$ and are compared with\n$P_v\\in E[X]$ only after mapping both into the local coefficient field. No field\nembedding $E\\to\\Ocal_\\lambda$ is assumed. The allowed-place predicate is explicit\nand may depend on $\\lambda$. Trivial action on a supplied subgroup is likewise\ndetected at every precision; identifying that subgroup with arithmetic inertia\nis a separate input.","d integral characteristic polynomials at every precision\nhold exactly in the reconstructed representation. For a common coefficient field\n$E$, local integral polynomials lie in $\\Ocal_\\lambda[X]$ and are compared with\n$P_v\\in E[X]$ only after mapping both into the local coefficient field. No field\nembedding $E\\to\\Ocal_\\lambda$ is assumed. The allowed-place predicate is explicit\nand may depend on $\\lambda$. Trivial action on a supplied subgroup is likewise\ndetected at every precision; identifying that subgroup with arithmetic inertia\nis a separate input.","\\paragraph{Completion from conjugacy classes (Block 14).}\nOver an $I$-adically complete separated commutative coefficient ring $R$,\nfor a fixed finite rank, the formalization now takes finite-level objects to be\nactual matrix representations modulo conjugation by an invertible matrix.\nAssume the coefficient quotients $R/I^k$ are finite.\nConjugation commutes with coefficient reduction, and each conjugacy class has\nfinitely many frames over a finite coefficient ring. Thus finite allowed class\nlevels give finite full preimages of allowed matrix representations. Downward\nadmissibility and positive-depth nonemptiness yield an exactly compatible\nmatrix branch and an actual integral representation. More strongly, every\nprescribed compatible class branch has such re","\\paragraph{Completion from conjugacy classes (Block 14).}\nOver an $I$-adically complete separated commutative coefficient ring $R$,\nfor a fixed finite rank, the formalization now takes finite-level objects to be\nactual matrix representations modulo conjugation by an invertible matrix.\nAssume the coefficient quotients $R/I^k$ are finite.\nConjugation commutes with coefficient reduction, and each conjugacy class has\nfinitely many frames over a finite coefficient ring. Thus finite allowed class\nlevels give finite full preimages of allowed matrix representations. Downward\nadmissibility and positive-depth nonemptiness yield an exactly compatible\nmatrix branch and an actual integral representation. More strongly, every\nprescribed compatible class branch has such representatives, using finite\nnonempty fibers over that branch.","\\paragraph{Completion from conjugacy classes (Block 14).}\nOver an $I$-adically complete separated commutative coefficient ring $R$,\nfor a fixed finite rank, the formalization now takes finite-level objects to be\nactual matrix representations modulo conjugation by an invertible matrix.\nAssume the coefficient quotients $R/I^k$ are finite.\nConjugation commutes with coefficient reduction, and each conjugacy class has\nfinitely many frames over a finite coefficient ring. Thus finite allowed class\nlevels give finite full preimages of allowed matrix representations. Downward\nadmissibility and positive-depth nonemptiness yield an exactly compatible\nmatrix branch and an actual integral representation. More strongly, every\nprescribed compatible class branch has such representatives, using finite\nnonempty fibers over that branch. No surjectivity of reductions or lifting of\nprescribed change-of-basis matrices is assumed. This does not assert uniqueness\nup to integral conjugacy.","For finitely generated $I$ with finite $R/I$, finiteness of $R/I^k$ is derived.\nOne continuous representative per allowed class suffices, because continuity\nis proved invariant under conjugation. Characteristic polynomials, determinants\nand trivial action on a supplied subgroup are likewise actual invariants.\nThe unframed common-spine theorem constructs continuous coefficient-field\nrepresentations with the prescribed whole polynomials and the explicitly\nsupplied inverse-limit-closed local predicate.","An actual $\\mathbb Z_3$ example applies the class-level completion theorem\nfor the cyclic group of order two and polynomial $X^2-1$, producing a nonidentity generator. Another\nchecked example has compatible classes represented by\n$\\operatorname{diag}(1,-1)$ modulo $3$ and\n$\\operatorname{diag}(-1,1)$ modulo $9$: these chosen matrices disagree under\nreduction, and an actual swap of basis repairs them. Arbitrary independent\nrepresentatives therefore do not replace the framing argument.","\\paragraph{Actual local comparisons from fixed-depth certificates (Block 15).}\nLet $R$ be an $I$-adically complete separated commutative ring with finite\ncoefficient quotients and an injective map into a field $K$. Fix a finite matrix\nrank. For finitely many\nlocal group maps $\\phi_s:H_s\\to G$ and integral template representations\n$\\tau_s$, fix detecting depths $d_s$. At precision $m$ retain matrices $P_{s,m}$\nas part of the admissible object and require\n\\[\n\\rho_m(\\phi_s(h))P_{s,m}=P_{s,m}\\tau_{s,m}(h).\n\\]\nFor $m\\ge d_s$ require their determinants to remain nonzero after reduction to\n$R/I^{d_s}$. Earlier levels omit that test. Finite occupied augmented levels\ngive a compatible branch of both representations and comparisons. The integral\nlimits satisfy the actual equations and","\\paragraph{Actual local comparisons from fixed-depth certificates (Block 15).}\nLet $R$ be an $I$-adically complete separated commutative ring with finite\ncoefficient quotients and an injective map into a field $K$. Fix a finite matrix\nrank. For finitely many\nlocal group maps $\\phi_s:H_s\\to G$ and integral template representations\n$\\tau_s$, fix detecting depths $d_s$. At precision $m$ retain matrices $P_{s,m}$\nas part of the admissible object and require\n\\[\n\\rho_m(\\phi_s(h))P_{s,m}=P_{s,m}\\tau_{s,m}(h).\n\\]\nFor $m\\ge d_s$ require their determinants to remain nonzero after reduction to\n$R/I^{d_s}$. Earlier levels omit that test. Finite occupied augmented levels\ngive a compatible branch of both representations and comparisons. The integral\nlimits satisfy the actual equations and have nonz","\\paragraph{Actual local comparisons from fixed-depth certificates (Block 15).}\nLet $R$ be an $I$-adically complete separated commutative ring with finite\ncoefficient quotients and an injective map into a field $K$. Fix a finite matrix\nrank. For finitely many\nlocal group maps $\\phi_s:H_s\\to G$ and integral template representations\n$\\tau_s$, fix detecting depths $d_s$. At precision $m$ retain matrices $P_{s,m}$\nas part of the admissible object and require\n\\[\n\\rho_m(\\phi_s(h))P_{s,m}=P_{s,m}\\tau_{s,m}(h).\n\\]\nFor $m\\ge d_s$ require their determinants to remain nonzero after reduction to\n$R/I^{d_s}$. Earlier levels omit that test. Finite occupied augmented levels\ngive a compatible branch of both representations and comparisons. The integral\nlimits satisfy the actual equations and have nonzero determinants, so the\ncomparisons are units over $K$. A converse supplies all finite nodes from any\nintegral reali","\\paragraph{Actual local comparisons from fixed-depth certificates (Block 15).}\nLet $R$ be an $I$-adically complete separated commutative ring with finite\ncoefficient quotients and an injective map into a field $K$. Fix a finite matrix\nrank. For finitely many\nlocal group maps $\\phi_s:H_s\\to G$ and integral template representations\n$\\tau_s$, fix detecting depths $d_s$. At precision $m$ retain matrices $P_{s,m}$\nas part of the admissible object and require\n\\[\n\\rho_m(\\phi_s(h))P_{s,m}=P_{s,m}\\tau_{s,m}(h).\n\\]\nFor $m\\ge d_s$ require their determinants to remain nonzero after reduction to\n$R/I^{d_s}$. Earlier levels omit that test. Finite occupied augmented levels\ngive a compatible branch of both representations and comparisons. The integral\nlimits satisfy the actual equations and have nonzero determinants, so the\ncomparisons are units over $K$. A converse supplies all finite nodes from any\nintegral realization with these detectors. Finite-level invertibility and\ncompatible prior cho","\\paragraph{Actual local comparisons from fixed-depth certificates (Block 15).}\nLet $R$ be an $I$-adically complete separated commutative ring with finite\ncoefficient quotients and an injective map into a field $K$. Fix a finite matrix\nrank. For finitely many\nlocal group maps $\\phi_s:H_s\\to G$ and integral template representations\n$\\tau_s$, fix detecting depths $d_s$. At precision $m$ retain matrices $P_{s,m}$\nas part of the admissible object and require\n\\[\n\\rho_m(\\phi_s(h))P_{s,m}=P_{s,m}\\tau_{s,m}(h).\n\\]\nFor $m\\ge d_s$ require their determinants to remain nonzero after reduction to\n$R/I^{d_s}$. Earlier levels omit that test. Finite occupied augmented levels\ngive a compatible branch of both representations and comparisons. The integral\nlimits satisfy the actual equations and have nonzero determinants, so the\ncomparisons are units over $K$. A converse supplies all finite nodes from any\nintegral realization with these detectors. Finite-level invertibility and\ncompatible prior choices of comparisons are not assumed.","agraph{Actual local comparisons from fixed-depth certificates (Block 15).}\nLet $R$ be an $I$-adically complete separated commutative ring with finite\ncoefficient quotients and an injective map into a field $K$. Fix a finite matrix\nrank. For finitely many\nlocal group maps $\\phi_s:H_s\\to G$ and integral template representations\n$\\tau_s$, fix detecting depths $d_s$. At precision $m$ retain matrices $P_{s,m}$\nas part of the admissible object and require\n\\[\n\\rho_m(\\phi_s(h))P_{s,m}=P_{s,m}\\tau_{s,m}(h).\n\\]\nFor $m\\ge d_s$ require their determinants to remain nonzero after reduction to\n$R/I^{d_s}$. Earlier levels omit that test. Finite occupied augmented levels\ngive a compatible branch of both representations and comparisons. The integral\nlimits satisfy the actual equations and have nonzero determinants, so the\ncomparisons are units over $K$. A converse supplies all finite nodes from any\nintegral realization with these detectors. Finite-level invertibility and\ncompatible prior choices of comparisons are not assumed.","cates (Block 15).}\nLet $R$ be an $I$-adically complete separated commutative ring with finite\ncoefficient quotients and an injective map into a field $K$. Fix a finite matrix\nrank. For finitely many\nlocal group maps $\\phi_s:H_s\\to G$ and integral template representations\n$\\tau_s$, fix detecting depths $d_s$. At precision $m$ retain matrices $P_{s,m}$\nas part of the admissible object and require\n\\[\n\\rho_m(\\phi_s(h))P_{s,m}=P_{s,m}\\tau_{s,m}(h).\n\\]\nFor $m\\ge d_s$ require their determinants to remain nonzero after reduction to\n$R/I^{d_s}$. Earlier levels omit that test. Finite occupied augmented levels\ngive a compatible branch of both representations and comparisons. The integral\nlimits satisfy the actual equations and have nonzero determinants, so the\ncomparisons are units over $K$. A converse supplies all finite nodes from any\nintegral realization with these detectors. Finite-level invertibility and\ncompatible prior choices of comparisons are not assumed.","eparated commutative ring with finite\ncoefficient quotients and an injective map into a field $K$. Fix a finite matrix\nrank. For finitely many\nlocal group maps $\\phi_s:H_s\\to G$ and integral template representations\n$\\tau_s$, fix detecting depths $d_s$. At precision $m$ retain matrices $P_{s,m}$\nas part of the admissible object and require\n\\[\n\\rho_m(\\phi_s(h))P_{s,m}=P_{s,m}\\tau_{s,m}(h).\n\\]\nFor $m\\ge d_s$ require their determinants to remain nonzero after reduction to\n$R/I^{d_s}$. Earlier levels omit that test. Finite occupied augmented levels\ngive a compatible branch of both representations and comparisons. The integral\nlimits satisfy the actual equations and have nonzero determinants, so the\ncomparisons are units over $K$. A converse supplies all finite nodes from any\nintegral realization with these detectors. Finite-level invertibility and\ncompatible prior choices of comparisons are not assumed.","e map into a field $K$. Fix a finite matrix\nrank. For finitely many\nlocal group maps $\\phi_s:H_s\\to G$ and integral template representations\n$\\tau_s$, fix detecting depths $d_s$. At precision $m$ retain matrices $P_{s,m}$\nas part of the admissible object and require\n\\[\n\\rho_m(\\phi_s(h))P_{s,m}=P_{s,m}\\tau_{s,m}(h).\n\\]\nFor $m\\ge d_s$ require their determinants to remain nonzero after reduction to\n$R/I^{d_s}$. Earlier levels omit that test. Finite occupied augmented levels\ngive a compatible branch of both representations and comparisons. The integral\nlimits satisfy the actual equations and have nonzero determinants, so the\ncomparisons are units over $K$. A converse supplies all finite nodes from any\nintegral realization with these detectors. Finite-level invertibility and\ncompatible prior choices of comparisons are not assumed.","d require\n\\[\n\\rho_m(\\phi_s(h))P_{s,m}=P_{s,m}\\tau_{s,m}(h).\n\\]\nFor $m\\ge d_s$ require their determinants to remain nonzero after reduction to\n$R/I^{d_s}$. Earlier levels omit that test. Finite occupied augmented levels\ngive a compatible branch of both representations and comparisons. The integral\nlimits satisfy the actual equations and have nonzero determinants, so the\ncomparisons are units over $K$. A converse supplies all finite nodes from any\nintegral realization with these detectors. Finite-level invertibility and\ncompatible prior choices of comparisons are not assumed.","The formalization specializes to the actual maps\n$G_{\\mathbb Q_p}\\to G_{\\mathbb Q}$ induced by the field embeddings and choices\nof algebraic closures. A nontrivial continuous quadratic character of\n$G_{\\mathbb Q}$, constructed from its action on $i^2=-1$, gives the integral\n$2$-adic actions\n\\[\n\\tau(\\sigma)=\\operatorname{diag}(1,-1),\\quad\n\\rho(\\sigma)=\\begin{pmatrix}1&-1","The formalization specializes to the actual maps\n$G_{\\mathbb Q_p}\\to G_{\\mathbb Q}$ induced by the field embeddings and choices\nof algebraic closures. A nontrivial continuous quadratic character of\n$G_{\\mathbb Q}$, constructed from its action on $i^2=-1$, gives the integral\n$2$-adic actions\n\\[\n\\tau(\\sigma)=\\operatorname{diag}(1,-1),\\quad\n\\rho(\\sigma)=\\begin{pmatrix}1&-1\\\\0&-1\\end{pmatrix},\\quad\nP=\\begin{pmatrix}1&1\\\\0&2\\end{pmatrix}\n\\]\nat a nonidentity quadratic element. The determinant $2$ is detected modulo $4$.\nThe representations are conjugate over $\\mathbb Q_2$ but not integrally\nconjugate. A separate $\\mathbb Z_3$ example invokes simultaneous completion\nand reconstructs a comparison which is an integral nonunit and a field unit.","0&2\\end{pmatrix}\n\\]\nat a nonidentity quadratic element. The determinant $2$ is detected modulo $4$.\nThe representations are conjugate over $\\mathbb Q_2$ but not integrally\nconjugate. A separate $\\mathbb Z_3$ example invokes simultaneous completion\nand reconstructs a comparison which is an integral nonunit and a field unit.","The bound is substantive: for the unipotent actions\n$\\rho_a(g)=\\begin{pmatrix}1&ag\\\\0&1\\end{pmatrix}$, every integral intertwiner\nwith $\\rho_1$ has determinant $aP_{22}^2$. Taking $a=\\pi^n$ for a nonzero\n$\\pi\\in I$ rules out every\nfixed detecting depth, even though all the field-valued approximants are\nisomorphic and their integral limit is trivial. This does not refute Claim 8.2,\nwhich already assumes inverse-limit closure of its chosen local conditions.","0&1\\end{pmatrix}$, every integral intertwiner\nwith $\\rho_1$ has determinant $aP_{22}^2$. Taking $a=\\pi^n$ for a nonzero\n$\\pi\\in I$ rules out every\nfixed detecting depth, even though all the field-valued approximants are\nisomorphic and their integral limit is trivial. This does not refute Claim 8.2,\nwhich already assumes inverse-limit closure of its chosen local conditions.","\\paragraph{Derived comparison bounds (Block 16).}\nFor a finite group $\\Gamma$ whose order is a unit in a commutative local\ncoefficient ring $R$, any prescribed equivariant residual isomorphism lifts to\nan integral isomorphism. Lift its matrix to $A$ and define\n\\[\nP=|\\Gamma|^{-1}\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nReindexing proves equivariance, and reduction recovers the prescribed matrix.\nIts determinant is a unit. Complete local coefficients with finite quotients,\noccupied compatible finite conditions and this residual type therefore give\nactual integral completion with detector depth one. A fixed continuous map\n$G\\to\\Gamma$ permits pullback to infinite source groups. The finite-level\nrepresentations must factor through the same map, and the residual type is","\\paragraph{Derived comparison bounds (Block 16).}\nFor a finite group $\\Gamma$ whose order is a unit in a commutative local\ncoefficient ring $R$, any prescribed equivariant residual isomorphism lifts to\nan integral isomorphism. Lift its matrix to $A$ and define\n\\[\nP=|\\Gamma|^{-1}\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nReindexing proves equivariance, and reduction recovers the prescribed matrix.\nIts determinant is a unit. Complete local coefficients with finite quotients,\noccupied compatible finite conditions and this residual type therefore give\nactual integral completion with detector depth one. A fixed continuous map\n$G\\to\\Gamma$ permits pullback to infinite source groups. The finite-level\nrepresentations must factor through the same map, and the residual type is\nimposed on their $\\Gamma$-actions. No uniform determinant bound o","\\paragraph{Derived comparison bounds (Block 16).}\nFor a finite group $\\Gamma$ whose order is a unit in a commutative local\ncoefficient ring $R$, any prescribed equivariant residual isomorphism lifts to\nan integral isomorphism. Lift its matrix to $A$ and define\n\\[\nP=|\\Gamma|^{-1}\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nReindexing proves equivariance, and reduction recovers the prescribed matrix.\nIts determinant is a unit. Complete local coefficients with finite quotients,\noccupied compatible finite conditions and this residual type therefore give\nactual integral completion with detector depth one. A fixed continuous map\n$G\\to\\Gamma$ permits pullback to infinite source groups. The finite-level\nrepresentations must factor through the same map, and the residual type is\nimposed on their $\\Gamma$-actions. No uniform determinant bound or augmented\ncertificate occupancy is assumed in this sector.","ft its matrix to $A$ and define\n\\[\nP=|\\Gamma|^{-1}\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nReindexing proves equivariance, and reduction recovers the prescribed matrix.\nIts determinant is a unit. Complete local coefficients with finite quotients,\noccupied compatible finite conditions and this residual type therefore give\nactual integral completion with detector depth one. A fixed continuous map\n$G\\to\\Gamma$ permits pullback to infinite source groups. The finite-level\nrepresentations must factor through the same map, and the residual type is\nimposed on their $\\Gamma$-actions. No uniform determinant bound or augmented\ncertificate occupancy is assumed in this sector.","ix.\nIts determinant is a unit. Complete local coefficients with finite quotients,\noccupied compatible finite conditions and this residual type therefore give\nactual integral completion with detector depth one. A fixed continuous map\n$G\\to\\Gamma$ permits pullback to infinite source groups. The finite-level\nrepresentations must factor through the same map, and the residual type is\nimposed on their $\\Gamma$-actions. No uniform determinant bound or augmented\ncertificate occupancy is assumed in this sector.","For a second sector let $R$ be a DVR with uniformizer $\\pi$, fraction field $K$\nand matrix rank $n$. If a fixed template $\\tau$ has full $K$-linear action span,\nfinite generation and denominator clearing give a fixed $c$ such that\n\\[\n\\pi^c M_n(R)\\subseteq\\operatorname{span}_R\\{\\tau(g):g\\in G\\}.\n\\]\nNormalize a field comparison to an integral $P$ with a unit entry. The displayed\nspan constructs an integral $B$ with $BP=\\pi^cI$, so\n\\[\n\\det P\\mid\\pi^{cn},\\qquad \\det P\\not\\equiv0\\pmod{\\pi^{cn+1}}.\n\\]\nOne $c$ works for all integral representations in the fixed field-isomorphism\nclass. Jacobson density derives full span from actual irreducibility and a\nscalar commutant. The formalization states these hypotheses directly and","For a second sector let $R$ be a DVR with uniformizer $\\pi$, fraction field $K$\nand matrix rank $n$. If a fixed template $\\tau$ has full $K$-linear action span,\nfinite generation and denominator clearing give a fixed $c$ such that\n\\[\n\\pi^c M_n(R)\\subseteq\\operatorname{span}_R\\{\\tau(g):g\\in G\\}.\n\\]\nNormalize a field comparison to an integral $P$ with a unit entry. The displayed\nspan constructs an integral $B$ with $BP=\\pi^cI$, so\n\\[\n\\det P\\mid\\pi^{cn},\\qquad \\det P\\not\\equiv0\\pmod{\\pi^{cn+1}}.\n\\]\nOne $c$ works for all integral representations in the fixed field-isomorphism\nclass. Jacobson density derives full span from actual irreducibility and a\nscalar commutant. The formalization states these hypotheses directly and does\nnot assume the DVR fractio","For a second sector let $R$ be a DVR with uniformizer $\\pi$, fraction field $K$\nand matrix rank $n$. If a fixed template $\\tau$ has full $K$-linear action span,\nfinite generation and denominator clearing give a fixed $c$ such that\n\\[\n\\pi^c M_n(R)\\subseteq\\operatorname{span}_R\\{\\tau(g):g\\in G\\}.\n\\]\nNormalize a field comparison to an integral $P$ with a unit entry. The displayed\nspan constructs an integral $B$ with $BP=\\pi^cI$, so\n\\[\n\\det P\\mid\\pi^{cn},\\qquad \\det P\\not\\equiv0\\pmod{\\pi^{cn+1}}.\n\\]\nOne $c$ works for all integral representations in the fixed field-isomorphism\nclass. Jacobson density derives full span from actual irreducibility and a\nscalar commutant. The formalization states these hypotheses directly and does\nnot assume the DVR fraction field is algebraic","For a second sector let $R$ be a DVR with uniformizer $\\pi$, fraction field $K$\nand matrix rank $n$. If a fixed template $\\tau$ has full $K$-linear action span,\nfinite generation and denominator clearing give a fixed $c$ such that\n\\[\n\\pi^c M_n(R)\\subseteq\\operatorname{span}_R\\{\\tau(g):g\\in G\\}.\n\\]\nNormalize a field comparison to an integral $P$ with a unit entry. The displayed\nspan constructs an integral $B$ with $BP=\\pi^cI$, so\n\\[\n\\det P\\mid\\pi^{cn},\\qquad \\det P\\not\\equiv0\\pmod{\\pi^{cn+1}}.\n\\]\nOne $c$ works for all integral representations in the fixed field-isomorphism\nclass. Jacobson density derives full span from actual irreducibility and a\nscalar commutant. The formalization states these hypotheses directly and does\nnot assume the DVR fraction field is algebraically closed.","ond sector let $R$ be a DVR with uniformizer $\\pi$, fraction field $K$\nand matrix rank $n$. If a fixed template $\\tau$ has full $K$-linear action span,\nfinite generation and denominator clearing give a fixed $c$ such that\n\\[\n\\pi^c M_n(R)\\subseteq\\operatorname{span}_R\\{\\tau(g):g\\in G\\}.\n\\]\nNormalize a field comparison to an integral $P$ with a unit entry. The displayed\nspan constructs an integral $B$ with $BP=\\pi^cI$, so\n\\[\n\\det P\\mid\\pi^{cn},\\qquad \\det P\\not\\equiv0\\pmod{\\pi^{cn+1}}.\n\\]\nOne $c$ works for all integral representations in the fixed field-isomorphism\nclass. Jacobson density derives full span from actual irreducibility and a\nscalar commutant. The formalization states these hypotheses directly and does\nnot assume the DVR fraction field is algebraically closed.","r let $R$ be a DVR with uniformizer $\\pi$, fraction field $K$\nand matrix rank $n$. If a fixed template $\\tau$ has full $K$-linear action span,\nfinite generation and denominator clearing give a fixed $c$ such that\n\\[\n\\pi^c M_n(R)\\subseteq\\operatorname{span}_R\\{\\tau(g):g\\in G\\}.\n\\]\nNormalize a field comparison to an integral $P$ with a unit entry. The displayed\nspan constructs an integral $B$ with $BP=\\pi^cI$, so\n\\[\n\\det P\\mid\\pi^{cn},\\qquad \\det P\\not\\equiv0\\pmod{\\pi^{cn+1}}.\n\\]\nOne $c$ works for all integral representations in the fixed field-isomorphism\nclass. Jacobson density derives full span from actual irreducibility and a\nscalar commutant. The formalization states these hypotheses directly and does\nnot assume the DVR fraction field is algebraically closed.","R with uniformizer $\\pi$, fraction field $K$\nand matrix rank $n$. If a fixed template $\\tau$ has full $K$-linear action span,\nfinite generation and denominator clearing give a fixed $c$ such that\n\\[\n\\pi^c M_n(R)\\subseteq\\operatorname{span}_R\\{\\tau(g):g\\in G\\}.\n\\]\nNormalize a field comparison to an integral $P$ with a unit entry. The displayed\nspan constructs an integral $B$ with $BP=\\pi^cI$, so\n\\[\n\\det P\\mid\\pi^{cn},\\qquad \\det P\\not\\equiv0\\pmod{\\pi^{cn+1}}.\n\\]\nOne $c$ works for all integral representations in the fixed field-isomorphism\nclass. Jacobson density derives full span from actual irreducibility and a\nscalar commutant. The formalization states these hypotheses directly and does\nnot assume the DVR fraction field is algebraically closed.","fixed template $\\tau$ has full $K$-linear action span,\nfinite generation and denominator clearing give a fixed $c$ such that\n\\[\n\\pi^c M_n(R)\\subseteq\\operatorname{span}_R\\{\\tau(g):g\\in G\\}.\n\\]\nNormalize a field comparison to an integral $P$ with a unit entry. The displayed\nspan constructs an integral $B$ with $BP=\\pi^cI$, so\n\\[\n\\det P\\mid\\pi^{cn},\\qquad \\det P\\not\\equiv0\\pmod{\\pi^{cn+1}}.\n\\]\nOne $c$ works for all integral representations in the fixed field-isomorphism\nclass. Jacobson density derives full span from actual irreducibility and a\nscalar commutant. The formalization states these hypotheses directly and does\nnot assume the DVR fraction field is algebraically closed.","Over $\\mathbb Z_3$, the actual average corrects\n$A=\\left(\\begin{smallmatrix}1&1","Over $\\mathbb Z_3$, the actual average corrects\n$A=\\left(\\begin{smallmatrix}1&1\\\\0&1\\end{smallmatrix}\\right)$ to\n$P=\\left(\\begin{smallmatrix}1&-1/2","0&1\\end{smallmatrix}\\right)$ to\n$P=\\left(\\begin{smallmatrix}1&-1/2\\\\0&1\\end{smallmatrix}\\right)$ for the\ninvolutions $\\rho(g)=\\left(\\begin{smallmatrix}1&1","0&1\\end{smallmatrix}\\right)$ for the\ninvolutions $\\rho(g)=\\left(\\begin{smallmatrix}1&1\\\\0&-1\\end{smallmatrix}\\right)$\nand $\\tau(g)=\\operatorname{diag}(1,-1)$. It preserves the residual comparison\nand has determinant one. Pullback through the quadratic character of the\nactual $G_{\\mathbb Q}$ gives continuous arithmetic examples. In contrast, the\nidentity and $\\operatorname{diag}(1,\\mathrm{sign})$ over $\\mathbb Z/4$ agree\nmodulo two but are not integrally conjugate. Their group order is not a unit.\nThese restricted results do not supply general arithmetic packet attachment,\nall finite-level existence statements, or cross-coefficient compatibility.","0&-1\\end{smallmatrix}\\right)$\nand $\\tau(g)=\\operatorname{diag}(1,-1)$. It preserves the residual comparison\nand has determinant one. Pullback through the quadratic character of the\nactual $G_{\\mathbb Q}$ gives continuous arithmetic examples. In contrast, the\nidentity and $\\operatorname{diag}(1,\\mathrm{sign})$ over $\\mathbb Z/4$ agree\nmodulo two but are not integrally conjugate. Their group order is not a unit.\nThese restricted results do not supply general arithmetic packet attachment,\nall finite-level existence statements, or cross-coefficient compatibility.","\\begin{theorem}[Finite Frobenius separation]\\label{px:9.1}\nLet $\\mathscr W_1,\\ldots,\\mathscr W_r$ be pairwise nonisomorphic strictly\ncompatible systems whose members are absolutely irreducible, with common\nalgebraic unramified trace values $t_i(v)$ in a number field $E$.\nThere are $r$ distinct permitted unramified places $v_1,\\ldots,v_r$ for which\n\\[\n C=(t_i(v_j))_{j,i}\\in M_r(E),\\qquad \\det C\\ne0.\n\\]\nEvery coefficient-field embedding $\\sigma_\\lambda:E\\hookrightarrow E_\\lambda$\ntherefore sends $C$ to an invertible matrix. Its entries agree with the\nordinary representation traces\n$\\operatorname{Tr}(W_{i,\\lambda}(\\operatorname{Frob}_{v_j}))$\nwhen the selected places satisfy the unramified compatibility convention\nfor that coefficient prime. In particular, places above the residue\ncharact","\\begin{theorem}[Finite Frobenius separation]\\label{px:9.1}\nLet $\\mathscr W_1,\\ldots,\\mathscr W_r$ be pairwise nonisomorphic strictly\ncompatible systems whose members are absolutely irreducible, with common\nalgebraic unramified trace values $t_i(v)$ in a number field $E$.\nThere are $r$ distinct permitted unramified places $v_1,\\ldots,v_r$ for which\n\\[\n C=(t_i(v_j))_{j,i}\\in M_r(E),\\qquad \\det C\\ne0.\n\\]\nEvery coefficient-field embedding $\\sigma_\\lambda:E\\hookrightarrow E_\\lambda$\ntherefore sends $C$ to an invertible matrix. Its entries agree with the\nordinary representation traces\n$\\operatorname{Tr}(W_{i,\\lambda}(\\operatorname{Frob}_{v_j}))$\nwhen the selected places satisfy the unramified compatibility convention\nfor that coefficient prime. In particular, places above the residue\ncharacteristic of $\\lambda$ are excluded from this ordinary trace comparison.\n\\end{theorem}","ion]\\label{px:9.1}\nLet $\\mathscr W_1,\\ldots,\\mathscr W_r$ be pairwise nonisomorphic strictly\ncompatible systems whose members are absolutely irreducible, with common\nalgebraic unramified trace values $t_i(v)$ in a number field $E$.\nThere are $r$ distinct permitted unramified places $v_1,\\ldots,v_r$ for which\n\\[\n C=(t_i(v_j))_{j,i}\\in M_r(E),\\qquad \\det C\\ne0.\n\\]\nEvery coefficient-field embedding $\\sigma_\\lambda:E\\hookrightarrow E_\\lambda$\ntherefore sends $C$ to an invertible matrix. Its entries agree with the\nordinary representation traces\n$\\operatorname{Tr}(W_{i,\\lambda}(\\operatorname{Frob}_{v_j}))$\nwhen the selected places satisfy the unramified compatibility convention\nfor that coefficient prime. In particular, places above the residue\ncharacteristic of $\\lambda$ are excluded from this ordinary trace comparison.\n\\end{theorem}","pairwise nonisomorphic strictly\ncompatible systems whose members are absolutely irreducible, with common\nalgebraic unramified trace values $t_i(v)$ in a number field $E$.\nThere are $r$ distinct permitted unramified places $v_1,\\ldots,v_r$ for which\n\\[\n C=(t_i(v_j))_{j,i}\\in M_r(E),\\qquad \\det C\\ne0.\n\\]\nEvery coefficient-field embedding $\\sigma_\\lambda:E\\hookrightarrow E_\\lambda$\ntherefore sends $C$ to an invertible matrix. Its entries agree with the\nordinary representation traces\n$\\operatorname{Tr}(W_{i,\\lambda}(\\operatorname{Frob}_{v_j}))$\nwhen the selected places satisfy the unramified compatibility convention\nfor that coefficient prime. In particular, places above the residue\ncharacteristic of $\\lambda$ are excluded from this ordinary trace comparison.\n\\end{theorem}","cible, with common\nalgebraic unramified trace values $t_i(v)$ in a number field $E$.\nThere are $r$ distinct permitted unramified places $v_1,\\ldots,v_r$ for which\n\\[\n C=(t_i(v_j))_{j,i}\\in M_r(E),\\qquad \\det C\\ne0.\n\\]\nEvery coefficient-field embedding $\\sigma_\\lambda:E\\hookrightarrow E_\\lambda$\ntherefore sends $C$ to an invertible matrix. Its entries agree with the\nordinary representation traces\n$\\operatorname{Tr}(W_{i,\\lambda}(\\operatorname{Frob}_{v_j}))$\nwhen the selected places satisfy the unramified compatibility convention\nfor that coefficient prime. In particular, places above the residue\ncharacteristic of $\\lambda$ are excluded from this ordinary trace comparison.\n\\end{theorem}","ry coefficient-field embedding $\\sigma_\\lambda:E\\hookrightarrow E_\\lambda$\ntherefore sends $C$ to an invertible matrix. Its entries agree with the\nordinary representation traces\n$\\operatorname{Tr}(W_{i,\\lambda}(\\operatorname{Frob}_{v_j}))$\nwhen the selected places satisfy the unramified compatibility convention\nfor that coefficient prime. In particular, places above the residue\ncharacteristic of $\\lambda$ are excluded from this ordinary trace comparison.\n\\end{theorem}","For $r$ independent functions, their evaluation linear forms span the\n$r$-dimensional coefficient dual: a proper span would leave a nonzero\ncoefficient vector annihilated by every evaluation, contradicting independence.\nChoose a basis from the evaluations. Its $r$ distinct sample points give an\ninvertible square matrix. The common algebraic traces put that matrix over\n$E$; an injective field embedding reflects the nonvanishing of its determinant.\nAll further field embeddings preserve the same nonvanishing.\n\\end{proof}","onjugacy-saturated samples and hence the finite separator.\nGeneral arithmetic Frobenius coverage remains unformalized. Block 25 now proves\nthe bridge from componentwise nonisomorphic systems to nonisomorphic members\nat the selected coefficient prime, under density on the required good-place\noverlaps. It also proves the complete restricted separator for a fixed finite\ncyclotomic Galois group over $\\Q$, with arithmetic coverage derived from the\nproved Dirichlet theorem. The full arithmetic theorem remains partial. Selection\nis existential and gives no effective upper bound on the selected primes.\nA common algebraic matrix is still distinct from an ordinary unramified trace\ncomparison at all coefficient primes simultaneously.","\\paragraph{System isomorphism and good-place overlaps (Block 25).}\nThe system equivalence used here is componentwise representation isomorphism:\n\\[\n \\mathscr W_i\\simeq\\mathscr W_j\n \\quad\\Longleftrightarrow\\quad\n \\forall\\lambda,\\ W_{i,\\lambda}\\simeq W_{j,\\lambda}.\n\\]\nFor finite-dimensional irreducible representations of any monoid over a\ncharacteristic-zero field, the new character-rigidity theorem proves that\nequality of characters is equivalent to an actual equivariant linear isomorphism.\nNeither a finite group nor an algebraically closed coefficient field is needed.","Suppose two members are isomorphic at $\\lambda_0$. Injectivity of the common\ncoefficient embedding gives $t_i(v)=t_j(v)$ at every place good for $\\lambda_0$.\nAt another coefficient prime $\\lambda$, this equality compares actual member\ntraces on the intersection of the two good-place sets. Conjugacy density of\nthat intersection and continuity extend the equality to the whole group;\ncharacter rigidity then gives an actual isomorphism at $\\lambda$.\nThus the nonisomorphism of systems supplies the required nonisomorphism of\nmembers. Equality of optional labels or record fields is not used as a\nsubstitute for representation equivalence.","lambda_0$. Injectivity of the common\ncoefficient embedding gives $t_i(v)=t_j(v)$ at every place good for $\\lambda_0$.\nAt another coefficient prime $\\lambda$, this equality compares actual member\ntraces on the intersection of the two good-place sets. Conjugacy density of\nthat intersection and continuity extend the equality to the whole group;\ncharacter rigidity then gives an actual isomorphism at $\\lambda$.\nThus the nonisomorphism of systems supplies the required nonisomorphism of\nmembers. Equality of optional labels or record fields is not used as a\nsubstitute for representation equivalence.","\\paragraph{Arithmetic coverage for a fixed cyclotomic extension (Block 25).}\nLet $n>0$ and let $L=\\Q(\\zeta_n)$ be a cyclotomic number field. The canonical\nisomorphism\n\\[\n \\operatorname{Gal}(L/\\Q)\\simeq(\\mathbb Z/n\\mathbb Z)^\\times\n\\]\ndefines $\\sigma_p$ by $\\sigma_p(\\zeta_n)=\\zeta_n^p$ when $p\\nmid n$.\nFor every automorphism $\\sigma$, finite set $B\\subset\\mathbb N$ and bound $b$,\nLean proves that there is a rational prime $p>b$, with $p\\notin B$ and\n$p\\nmid n$, such that $\\sigma_p=\\sigma$. The prime-existence step uses the\nproved Dirichlet theorem in the pinned Mathlib dependency.","rage for a fixed cyclotomic extension (Block 25).}\nLet $n>0$ and let $L=\\Q(\\zeta_n)$ be a cyclotomic number field. The canonical\nisomorphism\n\\[\n \\operatorname{Gal}(L/\\Q)\\simeq(\\mathbb Z/n\\mathbb Z)^\\times\n\\]\ndefines $\\sigma_p$ by $\\sigma_p(\\zeta_n)=\\zeta_n^p$ when $p\\nmid n$.\nFor every automorphism $\\sigma$, finite set $B\\subset\\mathbb N$ and bound $b$,\nLean proves that there is a rational prime $p>b$, with $p\\notin B$ and\n$p\\nmid n$, such that $\\sigma_p=\\sigma$. The prime-existence step uses the\nproved Dirichlet theorem in the pinned Mathlib dependency.","ixed cyclotomic extension (Block 25).}\nLet $n>0$ and let $L=\\Q(\\zeta_n)$ be a cyclotomic number field. The canonical\nisomorphism\n\\[\n \\operatorname{Gal}(L/\\Q)\\simeq(\\mathbb Z/n\\mathbb Z)^\\times\n\\]\ndefines $\\sigma_p$ by $\\sigma_p(\\zeta_n)=\\zeta_n^p$ when $p\\nmid n$.\nFor every automorphism $\\sigma$, finite set $B\\subset\\mathbb N$ and bound $b$,\nLean proves that there is a rational prime $p>b$, with $p\\notin B$ and\n$p\\nmid n$, such that $\\sigma_p=\\sigma$. The prime-existence step uses the\nproved Dirichlet theorem in the pinned Mathlib dependency.","extension (Block 25).}\nLet $n>0$ and let $L=\\Q(\\zeta_n)$ be a cyclotomic number field. The canonical\nisomorphism\n\\[\n \\operatorname{Gal}(L/\\Q)\\simeq(\\mathbb Z/n\\mathbb Z)^\\times\n\\]\ndefines $\\sigma_p$ by $\\sigma_p(\\zeta_n)=\\zeta_n^p$ when $p\\nmid n$.\nFor every automorphism $\\sigma$, finite set $B\\subset\\mathbb N$ and bound $b$,\nLean proves that there is a rational prime $p>b$, with $p\\notin B$ and\n$p\\nmid n$, such that $\\sigma_p=\\sigma$. The prime-existence step uses the\nproved Dirichlet theorem in the pinned Mathlib dependency.","Let $n>0$ and let $L=\\Q(\\zeta_n)$ be a cyclotomic number field. The canonical\nisomorphism\n\\[\n \\operatorname{Gal}(L/\\Q)\\simeq(\\mathbb Z/n\\mathbb Z)^\\times\n\\]\ndefines $\\sigma_p$ by $\\sigma_p(\\zeta_n)=\\zeta_n^p$ when $p\\nmid n$.\nFor every automorphism $\\sigma$, finite set $B\\subset\\mathbb N$ and bound $b$,\nLean proves that there is a rational prime $p>b$, with $p\\notin B$ and\n$p\\nmid n$, such that $\\sigma_p=\\sigma$. The prime-existence step uses the\nproved Dirichlet theorem in the pinned Mathlib dependency.","The arithmetic attachment is checked in this finite cyclotomic case. There is\nan actual maximal prime $\\mathfrak P$ of $\\mathcal O_L$ above $(p)$, its\nramification index is one, and $\\sigma_p$ belongs to its decomposition group.\nFor every integral $x\\in\\mathcal O_L$,\n\\[\n \\overline{\\sigma_p(x)}=\\overline{x}^{\\,p}\n \\quad\\text{in }\\mathcal O_L/\\mathfrak P.\n\\]\nThe residue identity is proved on all integral elements using\n$\\mathcal O_L=\\mathbb Z[\\zeta_n]$. Existence of $\\mathfrak P$ is obtained by\nlying over, rather than assumed. This specifies arithmetic Frobenius, with\n$p$-power residue action. The action and unramifiedness statements also hold\nfor every maximal prime above the selected $p$.","ent is checked in this finite cyclotomic case. There is\nan actual maximal prime $\\mathfrak P$ of $\\mathcal O_L$ above $(p)$, its\nramification index is one, and $\\sigma_p$ belongs to its decomposition group.\nFor every integral $x\\in\\mathcal O_L$,\n\\[\n \\overline{\\sigma_p(x)}=\\overline{x}^{\\,p}\n \\quad\\text{in }\\mathcal O_L/\\mathfrak P.\n\\]\nThe residue identity is proved on all integral elements using\n$\\mathcal O_L=\\mathbb Z[\\zeta_n]$. Existence of $\\mathfrak P$ is obtained by\nlying over, rather than assumed. This specifies arithmetic Frobenius, with\n$p$-power residue action. The action and unramifiedness statements also hold\nfor every maximal prime above the selected $p$.","here is\nan actual maximal prime $\\mathfrak P$ of $\\mathcal O_L$ above $(p)$, its\nramification index is one, and $\\sigma_p$ belongs to its decomposition group.\nFor every integral $x\\in\\mathcal O_L$,\n\\[\n \\overline{\\sigma_p(x)}=\\overline{x}^{\\,p}\n \\quad\\text{in }\\mathcal O_L/\\mathfrak P.\n\\]\nThe residue identity is proved on all integral elements using\n$\\mathcal O_L=\\mathbb Z[\\zeta_n]$. Existence of $\\mathfrak P$ is obtained by\nlying over, rather than assumed. This specifies arithmetic Frobenius, with\n$p$-power residue action. The action and unramifiedness statements also hold\nfor every maximal prime above the selected $p$.","ne, and $\\sigma_p$ belongs to its decomposition group.\nFor every integral $x\\in\\mathcal O_L$,\n\\[\n \\overline{\\sigma_p(x)}=\\overline{x}^{\\,p}\n \\quad\\text{in }\\mathcal O_L/\\mathfrak P.\n\\]\nThe residue identity is proved on all integral elements using\n$\\mathcal O_L=\\mathbb Z[\\zeta_n]$. Existence of $\\mathfrak P$ is obtained by\nlying over, rather than assumed. This specifies arithmetic Frobenius, with\n$p$-power residue action. The action and unramifiedness statements also hold\nfor every maximal prime above the selected $p$.","=\\overline{x}^{\\,p}\n \\quad\\text{in }\\mathcal O_L/\\mathfrak P.\n\\]\nThe residue identity is proved on all integral elements using\n$\\mathcal O_L=\\mathbb Z[\\zeta_n]$. Existence of $\\mathfrak P$ is obtained by\nlying over, rather than assumed. This specifies arithmetic Frobenius, with\n$p$-power residue action. The action and unramifiedness statements also hold\nfor every maximal prime above the selected $p$.","ite set, with nonzero determinant. Every coefficient embedding preserves its\nnonvanishing; actual member trace comparisons retain that member's good-prime\ncondition. No coverage or character-continuity hypothesis is added in this\ncyclotomic specialization: the prime coverage is proved, and the finite\nextension's Krull topology is discrete. A concrete conductor-four instance uses\nthe actual field $\\Q(\\zeta_4)$.","\\paragraph{Absolute Galois restriction and arithmetic witnesses (Block 26).}\nFix a cyclotomic number field $L=\\Q(\\zeta_n)$ with $n>0$ and an embedding\n$j:L\\hookrightarrow\\overline{\\Q}$. The new restriction map\n\\[\n q_L:G_{\\Q}\\longrightarrow\\operatorname{Gal}(L/\\Q),\\qquad\n j(q_L(\\gamma)x)=\\gamma(j(x)),\n\\]\nis an actual continuous surjective homomorphism with open kernel. Surjectivity\nis derived by extending automorphisms to the algebraic closure. Pullback along\na surjective homomorphism preserves and reflects equivariant linear\nisomorphisms, invariant subspaces, and irreducibility. These assertions keep\nthe underlying linear maps and submodules and do not assume a finite acting\ngroup. In particular, the equivalence relation for systems is preserved and\nreflected componentwise.","The arithmetic witnesses use the integral closure\n$\\mathcal A$ of $\\mathbb Z$ in $\\overline{\\Q}$. For each rational prime $p$,\nLean constructs a maximal ideal $\\mathfrak Q$ of $\\mathcal A$ above $(p)$ and\nan actual element $\\gamma_p\\in G_{\\Q}$ such that\n\\[\n \\gamma_p\\in\\operatorname{Stab}_{G_{\\Q}}(\\mathfrak Q),\\qquad\n \\overline{\\gamma_p(a)}=\\overline a^{\\,p}\n \\quad(a\\in\\mathcal A).\n\\]\nThe existence argument applies Mathlib's proved profinite stabilizer-surjectivity\ntheorem to this actual invariant ring. The residue $p$-power map is proved\nbijective by taking $p$-th roots in $\\overline{\\Q}$ and showing that they\nremain integral when their $p$-th powers are integral. This constructs an\narithmetic Frobenius witness in the prime-ideal decomposition group;","The arithmetic witnesses use the integral closure\n$\\mathcal A$ of $\\mathbb Z$ in $\\overline{\\Q}$. For each rational prime $p$,\nLean constructs a maximal ideal $\\mathfrak Q$ of $\\mathcal A$ above $(p)$ and\nan actual element $\\gamma_p\\in G_{\\Q}$ such that\n\\[\n \\gamma_p\\in\\operatorname{Stab}_{G_{\\Q}}(\\mathfrak Q),\\qquad\n \\overline{\\gamma_p(a)}=\\overline a^{\\,p}\n \\quad(a\\in\\mathcal A).\n\\]\nThe existence argument applies Mathlib's proved profinite stabilizer-surjectivity\ntheorem to this actual invariant ring. The residue $p$-power map is proved\nbijective by taking $p$-th roots in $\\overline{\\Q}$ and showing that they\nremain integral when their $p$-th powers are integral. This constructs an\narithmetic Frobenius witness in the prime-ideal decomposition group; it does\nnot ass","The arithmetic witnesses use the integral closure\n$\\mathcal A$ of $\\mathbb Z$ in $\\overline{\\Q}$. For each rational prime $p$,\nLean constructs a maximal ideal $\\mathfrak Q$ of $\\mathcal A$ above $(p)$ and\nan actual element $\\gamma_p\\in G_{\\Q}$ such that\n\\[\n \\gamma_p\\in\\operatorname{Stab}_{G_{\\Q}}(\\mathfrak Q),\\qquad\n \\overline{\\gamma_p(a)}=\\overline a^{\\,p}\n \\quad(a\\in\\mathcal A).\n\\]\nThe existence argument applies Mathlib's proved profinite stabilizer-surjectivity\ntheorem to this actual invariant ring. The residue $p$-power map is proved\nbijective by taking $p$-th roots in $\\overline{\\Q}$ and showing that they\nremain integral when their $p$-th powers are integral. This constructs an\narithmetic Frobenius witness in the prime-ideal decomposition group; it does\nnot assert that an arbitra","The arithmetic witnesses use the integral closure\n$\\mathcal A$ of $\\mathbb Z$ in $\\overline{\\Q}$. For each rational prime $p$,\nLean constructs a maximal ideal $\\mathfrak Q$ of $\\mathcal A$ above $(p)$ and\nan actual element $\\gamma_p\\in G_{\\Q}$ such that\n\\[\n \\gamma_p\\in\\operatorname{Stab}_{G_{\\Q}}(\\mathfrak Q),\\qquad\n \\overline{\\gamma_p(a)}=\\overline a^{\\,p}\n \\quad(a\\in\\mathcal A).\n\\]\nThe existence argument applies Mathlib's proved profinite stabilizer-surjectivity\ntheorem to this actual invariant ring. The residue $p$-power map is proved\nbijective by taking $p$-th roots in $\\overline{\\Q}$ and showing that they\nremain integral when their $p$-th powers are integral. This constructs an\narithmetic Frobenius witness in the prime-ideal decomposition group; it does\nnot assert that an arbitrary global lift belongs to tha","The arithmetic witnesses use the integral closure\n$\\mathcal A$ of $\\mathbb Z$ in $\\overline{\\Q}$. For each rational prime $p$,\nLean constructs a maximal ideal $\\mathfrak Q$ of $\\mathcal A$ above $(p)$ and\nan actual element $\\gamma_p\\in G_{\\Q}$ such that\n\\[\n \\gamma_p\\in\\operatorname{Stab}_{G_{\\Q}}(\\mathfrak Q),\\qquad\n \\overline{\\gamma_p(a)}=\\overline a^{\\,p}\n \\quad(a\\in\\mathcal A).\n\\]\nThe existence argument applies Mathlib's proved profinite stabilizer-surjectivity\ntheorem to this actual invariant ring. The residue $p$-power map is proved\nbijective by taking $p$-th roots in $\\overline{\\Q}$ and showing that they\nremain integral when their $p$-th powers are integral. This constructs an\narithmetic Frobenius witness in the prime-ideal decomposition group; it does\nnot assert that an arbitrary global lift belongs to that group.","$p$,\nLean constructs a maximal ideal $\\mathfrak Q$ of $\\mathcal A$ above $(p)$ and\nan actual element $\\gamma_p\\in G_{\\Q}$ such that\n\\[\n \\gamma_p\\in\\operatorname{Stab}_{G_{\\Q}}(\\mathfrak Q),\\qquad\n \\overline{\\gamma_p(a)}=\\overline a^{\\,p}\n \\quad(a\\in\\mathcal A).\n\\]\nThe existence argument applies Mathlib's proved profinite stabilizer-surjectivity\ntheorem to this actual invariant ring. The residue $p$-power map is proved\nbijective by taking $p$-th roots in $\\overline{\\Q}$ and showing that they\nremain integral when their $p$-th powers are integral. This constructs an\narithmetic Frobenius witness in the prime-ideal decomposition group; it does\nnot assert that an arbitrary global lift belongs to that group.","frak Q$ of $\\mathcal A$ above $(p)$ and\nan actual element $\\gamma_p\\in G_{\\Q}$ such that\n\\[\n \\gamma_p\\in\\operatorname{Stab}_{G_{\\Q}}(\\mathfrak Q),\\qquad\n \\overline{\\gamma_p(a)}=\\overline a^{\\,p}\n \\quad(a\\in\\mathcal A).\n\\]\nThe existence argument applies Mathlib's proved profinite stabilizer-surjectivity\ntheorem to this actual invariant ring. The residue $p$-power map is proved\nbijective by taking $p$-th roots in $\\overline{\\Q}$ and showing that they\nremain integral when their $p$-th powers are integral. This constructs an\narithmetic Frobenius witness in the prime-ideal decomposition group; it does\nnot assert that an arbitrary global lift belongs to that group.","hcal A$ above $(p)$ and\nan actual element $\\gamma_p\\in G_{\\Q}$ such that\n\\[\n \\gamma_p\\in\\operatorname{Stab}_{G_{\\Q}}(\\mathfrak Q),\\qquad\n \\overline{\\gamma_p(a)}=\\overline a^{\\,p}\n \\quad(a\\in\\mathcal A).\n\\]\nThe existence argument applies Mathlib's proved profinite stabilizer-surjectivity\ntheorem to this actual invariant ring. The residue $p$-power map is proved\nbijective by taking $p$-th roots in $\\overline{\\Q}$ and showing that they\nremain integral when their $p$-th powers are integral. This constructs an\narithmetic Frobenius witness in the prime-ideal decomposition group; it does\nnot assert that an arbitrary global lift belongs to that group.","ch that\n\\[\n \\gamma_p\\in\\operatorname{Stab}_{G_{\\Q}}(\\mathfrak Q),\\qquad\n \\overline{\\gamma_p(a)}=\\overline a^{\\,p}\n \\quad(a\\in\\mathcal A).\n\\]\nThe existence argument applies Mathlib's proved profinite stabilizer-surjectivity\ntheorem to this actual invariant ring. The residue $p$-power map is proved\nbijective by taking $p$-th roots in $\\overline{\\Q}$ and showing that they\nremain integral when their $p$-th powers are integral. This constructs an\narithmetic Frobenius witness in the prime-ideal decomposition group; it does\nnot assert that an arbitrary global lift belongs to that group.","When $p\\nmid n$, the comparison with the finite cyclotomic sample is proved:\n\\[\n q_L(\\gamma_p)=\\sigma_p,\n \\qquad \\sigma_p(\\zeta_n)=\\zeta_n^p.\n\\]\nReduction distinguishes $n$-th roots of unity at such a prime. Indeed, if\n$x^n=y^n=1$ and their reductions agree, the factorization of $x^n-y^n$ gives\n\\[\n (x-y)\\sum_{i=0}^{n-1}x^i y^{n-1-i}=0.\n\\]\nThe second factor reduces to $n\\overline y^{\\,n-1}\\ne0$, so the domain\nproperty gives $x=y$. This proves both Frobenius recognition on the\ncyclotomic generator and triviality of its inertia action. Consequently,\nevery representation pulled back through $q_L$ is unramified there.","son with the finite cyclotomic sample is proved:\n\\[\n q_L(\\gamma_p)=\\sigma_p,\n \\qquad \\sigma_p(\\zeta_n)=\\zeta_n^p.\n\\]\nReduction distinguishes $n$-th roots of unity at such a prime. Indeed, if\n$x^n=y^n=1$ and their reductions agree, the factorization of $x^n-y^n$ gives\n\\[\n (x-y)\\sum_{i=0}^{n-1}x^i y^{n-1-i}=0.\n\\]\nThe second factor reduces to $n\\overline y^{\\,n-1}\\ne0$, so the domain\nproperty gives $x=y$. This proves both Frobenius recognition on the\ncyclotomic generator and triviality of its inertia action. Consequently,\nevery representation pulled back through $q_L$ is unramified there.","uch a prime. Indeed, if\n$x^n=y^n=1$ and their reductions agree, the factorization of $x^n-y^n$ gives\n\\[\n (x-y)\\sum_{i=0}^{n-1}x^i y^{n-1-i}=0.\n\\]\nThe second factor reduces to $n\\overline y^{\\,n-1}\\ne0$, so the domain\nproperty gives $x=y$. This proves both Frobenius recognition on the\ncyclotomic generator and triviality of its inertia action. Consequently,\nevery representation pulled back through $q_L$ is unramified there.","The valuation formulation is checked separately. For any valuation subring\n$A\\subset\\overline{\\Q}$ with residue characteristic $p\\nmid n$, its actual\nresidue-action kernel fixes every $n$-th root of unity. A supplied element of\nits decomposition subgroup with $p$-power residue action restricts to\n$\\sigma_p$. These statements use Mathlib's valuation inertia and decomposition\nsubgroups. No identification of the valuation and integral-closure prime\nmodels is assumed in the formal comparison.","For a compatible family $\\rho_{i,\\lambda}$ of representations of the fixed\nfinite group, set $V_{i,\\lambda}=\\rho_{i,\\lambda}\\circ q_L$. Suppose the\nactual absolute-Galois members are irreducible and the systems are pairwise\ncomponentwise nonisomorphic. The finite prime separator now yields distinct\npermitted rational primes $p_j$, actual prime ideals $\\mathfrak Q_j$ and\narithmetic Frobenius witnesses $\\gamma_j$ as above, outside any prescribed\nfinite set, such that the common algebraic trace matrix has nonzero\ndeterminant. Every coefficient embedding preserves this nonvanishing, and\n\\[\n \\iota_\\lambda(t_i(p_j))=\n \\operatorname{tr}(V_{i,\\lambda}(\\gamma_j))\n\\]\nwhenever $p_j$ is good for that coefficient member. The operator and trace","For a compatible family $\\rho_{i,\\lambda}$ of representations of the fixed\nfinite group, set $V_{i,\\lambda}=\\rho_{i,\\lambda}\\circ q_L$. Suppose the\nactual absolute-Galois members are irreducible and the systems are pairwise\ncomponentwise nonisomorphic. The finite prime separator now yields distinct\npermitted rational primes $p_j$, actual prime ideals $\\mathfrak Q_j$ and\narithmetic Frobenius witnesses $\\gamma_j$ as above, outside any prescribed\nfinite set, such that the common algebraic trace matrix has nonzero\ndeterminant. Every coefficient embedding preserves this nonvanishing, and\n\\[\n \\iota_\\lambda(t_i(p_j))=\n \\operatorname{tr}(V_{i,\\lambda}(\\gamma_j))\n\\]\nwhenever $p_j$ is good for that coefficient member. The operator and trace\nvalues agree for every global lift with the same cyclotomic restriction.\nNeither Frobenius-w","For a compatible family $\\rho_{i,\\lambda}$ of representations of the fixed\nfinite group, set $V_{i,\\lambda}=\\rho_{i,\\lambda}\\circ q_L$. Suppose the\nactual absolute-Galois members are irreducible and the systems are pairwise\ncomponentwise nonisomorphic. The finite prime separator now yields distinct\npermitted rational primes $p_j$, actual prime ideals $\\mathfrak Q_j$ and\narithmetic Frobenius witnesses $\\gamma_j$ as above, outside any prescribed\nfinite set, such that the common algebraic trace matrix has nonzero\ndeterminant. Every coefficient embedding preserves this nonvanishing, and\n\\[\n \\iota_\\lambda(t_i(p_j))=\n \\operatorname{tr}(V_{i,\\lambda}(\\gamma_j))\n\\]\nwhenever $p_j$ is good for that coefficient member. The operator and trace\nvalues agree for every global lift with the same cyclotomic restriction.\nNeither Frobenius-witness existence nor quotient coverage is a supplied\nhypothesis in this restricted arithmetic theorem. The common cyclotomic\nfactorization and the compatible trace data remain explicit hypotheses.","compatible family $\\rho_{i,\\lambda}$ of representations of the fixed\nfinite group, set $V_{i,\\lambda}=\\rho_{i,\\lambda}\\circ q_L$. Suppose the\nactual absolute-Galois members are irreducible and the systems are pairwise\ncomponentwise nonisomorphic. The finite prime separator now yields distinct\npermitted rational primes $p_j$, actual prime ideals $\\mathfrak Q_j$ and\narithmetic Frobenius witnesses $\\gamma_j$ as above, outside any prescribed\nfinite set, such that the common algebraic trace matrix has nonzero\ndeterminant. Every coefficient embedding preserves this nonvanishing, and\n\\[\n \\iota_\\lambda(t_i(p_j))=\n \\operatorname{tr}(V_{i,\\lambda}(\\gamma_j))\n\\]\nwhenever $p_j$ is good for that coefficient member. The operator and trace\nvalues agree for every global lift with the same cyclotomic restriction.\nNeither Frobenius-witness existence nor quotient coverage is a supplied\nhypothesis in this restricted arithmetic theorem. The common cyclotomic\nfactorization and the compatible trace data remain explicit hypotheses.","le and the systems are pairwise\ncomponentwise nonisomorphic. The finite prime separator now yields distinct\npermitted rational primes $p_j$, actual prime ideals $\\mathfrak Q_j$ and\narithmetic Frobenius witnesses $\\gamma_j$ as above, outside any prescribed\nfinite set, such that the common algebraic trace matrix has nonzero\ndeterminant. Every coefficient embedding preserves this nonvanishing, and\n\\[\n \\iota_\\lambda(t_i(p_j))=\n \\operatorname{tr}(V_{i,\\lambda}(\\gamma_j))\n\\]\nwhenever $p_j$ is good for that coefficient member. The operator and trace\nvalues agree for every global lift with the same cyclotomic restriction.\nNeither Frobenius-witness existence nor quotient coverage is a supplied\nhypothesis in this restricted arithmetic theorem. The common cyclotomic\nfactorization and the compatible trace data remain explicit hypotheses.","tor now yields distinct\npermitted rational primes $p_j$, actual prime ideals $\\mathfrak Q_j$ and\narithmetic Frobenius witnesses $\\gamma_j$ as above, outside any prescribed\nfinite set, such that the common algebraic trace matrix has nonzero\ndeterminant. Every coefficient embedding preserves this nonvanishing, and\n\\[\n \\iota_\\lambda(t_i(p_j))=\n \\operatorname{tr}(V_{i,\\lambda}(\\gamma_j))\n\\]\nwhenever $p_j$ is good for that coefficient member. The operator and trace\nvalues agree for every global lift with the same cyclotomic restriction.\nNeither Frobenius-witness existence nor quotient coverage is a supplied\nhypothesis in this restricted arithmetic theorem. The common cyclotomic\nfactorization and the compatible trace data remain explicit hypotheses.","The boundaries are formalized as well. Pullback along a nonsurjective\nhomomorphism can make inequivalent irreducible representations isomorphic.\nSamples can cover one quotient while missing an open quotient of the full\ngroup. Finally, in $\\mathbb Z_2$, the distinct square roots $1$ and $-1$ have\nthe same residue, so the condition that the root order be a unit cannot be\ndiscarded.","\\paragraph{General finite-field Frobenius comparison (Block 27).}\nLet $L/\\Q$ now be any finite normal number field, with a chosen embedding\n$j:L\\hookrightarrow\\overline{\\Q}$. Write $\\mathcal A$ for the integral\nclosure of $\\mathbb Z$ in $\\overline{\\Q}$. The embedding restricts to an\ninjective map $\\mathcal O_L\\to\\mathcal A$ which intertwines the actual\nGalois actions. If $\\mathfrak Q\\subset\\mathcal A$ is maximal above $(p)$,\nits contraction $\\mathfrak P\\subset\\mathcal O_L$ is a nonzero maximal\nideal above $(p)$, and the induced residue map is injective.","te-field Frobenius comparison (Block 27).}\nLet $L/\\Q$ now be any finite normal number field, with a chosen embedding\n$j:L\\hookrightarrow\\overline{\\Q}$. Write $\\mathcal A$ for the integral\nclosure of $\\mathbb Z$ in $\\overline{\\Q}$. The embedding restricts to an\ninjective map $\\mathcal O_L\\to\\mathcal A$ which intertwines the actual\nGalois actions. If $\\mathfrak Q\\subset\\mathcal A$ is maximal above $(p)$,\nits contraction $\\mathfrak P\\subset\\mathcal O_L$ is a nonzero maximal\nideal above $(p)$, and the induced residue map is injective.","arison (Block 27).}\nLet $L/\\Q$ now be any finite normal number field, with a chosen embedding\n$j:L\\hookrightarrow\\overline{\\Q}$. Write $\\mathcal A$ for the integral\nclosure of $\\mathbb Z$ in $\\overline{\\Q}$. The embedding restricts to an\ninjective map $\\mathcal O_L\\to\\mathcal A$ which intertwines the actual\nGalois actions. If $\\mathfrak Q\\subset\\mathcal A$ is maximal above $(p)$,\nits contraction $\\mathfrak P\\subset\\mathcal O_L$ is a nonzero maximal\nideal above $(p)$, and the induced residue map is injective.","ld, with a chosen embedding\n$j:L\\hookrightarrow\\overline{\\Q}$. Write $\\mathcal A$ for the integral\nclosure of $\\mathbb Z$ in $\\overline{\\Q}$. The embedding restricts to an\ninjective map $\\mathcal O_L\\to\\mathcal A$ which intertwines the actual\nGalois actions. If $\\mathfrak Q\\subset\\mathcal A$ is maximal above $(p)$,\nits contraction $\\mathfrak P\\subset\\mathcal O_L$ is a nonzero maximal\nideal above $(p)$, and the induced residue map is injective.","Restriction carries the actual decomposition group at $\\mathfrak Q$ into\nthat at $\\mathfrak P$, carries ideal inertia into ideal inertia, and\npreserves the arithmetic $p$-power residue action. Thus the absolute\nFrobenius constructed in Block 26 gives a simultaneous finite Frobenius\nat the contracted prime. This assertion applies to arbitrary finite normal\nnumber fields and does not require a cyclotomic generator.","The exceptional primes are controlled by the actual field discriminant:\n\\[\n S_L=\\{p\\text{ rational prime}:p\\mid\\operatorname{disc}(L)\\}.\n\\]\nThis set is finite. For $p\\notin S_L$, every prime of $\\mathcal O_L$\nabove $(p)$ is unramified. Lean proves that the actual inertia cardinality\nis the ramification index. It follows that inertia is trivial and the\narithmetic Frobenius at a fixed finite prime is unique as a field\nautomorphism. Neither trivial inertia nor uniqueness is assumed in place\nof the arithmetic unramifiedness condition.","The choice of absolute prime and Frobenius lift is also controlled. Let\n$(\\mathfrak Q_u,\\gamma_u)$ and $(\\mathfrak Q_v,\\gamma_v)$ be any two\nactual arithmetic witnesses at the same rational prime. Transitivity on\nprimes of $\\mathcal A$ above $(p)$ gives $a\\in G_{\\Q}$ with\n\\[\n \\mathfrak Q_v=a\\mathfrak Q_u,\\qquad\n \\gamma_v^{-1}a\\gamma_u a^{-1}\\in I_{\\mathfrak Q_v}.\n\\]\nConsequently, any representation trivial on $I_{\\mathfrak Q_v}$ has the\nsame trace on the two witnesses. For representations factoring through\n$L$, triviality on this actual absolute inertia is derived whenever\n$p\\notin S_L$. The restrictions of all absolute witnesses at such a prime\ntherefore form a single conjugacy class. This conclusion requires only\nthe indicated finite factorization and the discriminant exclusion.","absolute prime and Frobenius lift is also controlled. Let\n$(\\mathfrak Q_u,\\gamma_u)$ and $(\\mathfrak Q_v,\\gamma_v)$ be any two\nactual arithmetic witnesses at the same rational prime. Transitivity on\nprimes of $\\mathcal A$ above $(p)$ gives $a\\in G_{\\Q}$ with\n\\[\n \\mathfrak Q_v=a\\mathfrak Q_u,\\qquad\n \\gamma_v^{-1}a\\gamma_u a^{-1}\\in I_{\\mathfrak Q_v}.\n\\]\nConsequently, any representation trivial on $I_{\\mathfrak Q_v}$ has the\nsame trace on the two witnesses. For representations factoring through\n$L$, triviality on this actual absolute inertia is derived whenever\n$p\\notin S_L$. The restrictions of all absolute witnesses at such a prime\ntherefore form a single conjugacy class. This conclusion requires only\nthe indicated finite factorization and the discriminant exclusion.","k Q_v,\\gamma_v)$ be any two\nactual arithmetic witnesses at the same rational prime. Transitivity on\nprimes of $\\mathcal A$ above $(p)$ gives $a\\in G_{\\Q}$ with\n\\[\n \\mathfrak Q_v=a\\mathfrak Q_u,\\qquad\n \\gamma_v^{-1}a\\gamma_u a^{-1}\\in I_{\\mathfrak Q_v}.\n\\]\nConsequently, any representation trivial on $I_{\\mathfrak Q_v}$ has the\nsame trace on the two witnesses. For representations factoring through\n$L$, triviality on this actual absolute inertia is derived whenever\n$p\\notin S_L$. The restrictions of all absolute witnesses at such a prime\ntherefore form a single conjugacy class. This conclusion requires only\nthe indicated finite factorization and the discriminant exclusion.","rational prime. Transitivity on\nprimes of $\\mathcal A$ above $(p)$ gives $a\\in G_{\\Q}$ with\n\\[\n \\mathfrak Q_v=a\\mathfrak Q_u,\\qquad\n \\gamma_v^{-1}a\\gamma_u a^{-1}\\in I_{\\mathfrak Q_v}.\n\\]\nConsequently, any representation trivial on $I_{\\mathfrak Q_v}$ has the\nsame trace on the two witnesses. For representations factoring through\n$L$, triviality on this actual absolute inertia is derived whenever\n$p\\notin S_L$. The restrictions of all absolute witnesses at such a prime\ntherefore form a single conjugacy class. This conclusion requires only\nthe indicated finite factorization and the discriminant exclusion.","There is a final compatibility between embeddings. For a normal subfield\n$L\\subset\\overline{\\Q}$, restriction defined through a chosen embedding\nis conjugate to canonical restriction by one fixed automorphism of $L$.\nThis identifies their conjugacy classes and lets the arithmetic comparison\napply directly to the fixed fields of open normal subgroups of $G_{\\Q}$.\nThose fixed fields are proved to be finite Galois number fields, with the\noriginal open normal subgroup as the kernel of restriction.","The remaining prime-existence obligation now has a precise formulation.\nFor every finite Galois subfield $L$, every finite set $B$ of rational\nprimes, and every $\\sigma\\in\\operatorname{Gal}(L/\\Q)$, require a prime\n$p\\notin B\\cup S_L$ and an actual absolute arithmetic witness $\\gamma_p$\nsuch that\n\\[\n q_L(\\gamma_p)\\sim\\sigma.\n\\]\nLean proves that this assertion is equivalent to the cofinite quotient\ncoverage required by the general separator, for any fixed choice of actual\nFrobenius witnesses. The equivalence derives independence from the witness\nchoices; it does not supply the displayed prime-existence assertion.\nArbitrary finite exclusions absorb $S_L$, and the resulting conditional\ndensity theorem preserves the separate good-prime sets of coefficient\nmembers. General prime coverage remains unprov","The remaining prime-existence obligation now has a precise formulation.\nFor every finite Galois subfield $L$, every finite set $B$ of rational\nprimes, and every $\\sigma\\in\\operatorname{Gal}(L/\\Q)$, require a prime\n$p\\notin B\\cup S_L$ and an actual absolute arithmetic witness $\\gamma_p$\nsuch that\n\\[\n q_L(\\gamma_p)\\sim\\sigma.\n\\]\nLean proves that this assertion is equivalent to the cofinite quotient\ncoverage required by the general separator, for any fixed choice of actual\nFrobenius witnesses. The equivalence derives independence from the witness\nchoices; it does not supply the displayed prime-existence assertion.\nArbitrary finite exclusions absorb $S_L$, and the resulting conditional\ndensity theorem preserves the separate good-prime sets of coefficient\nmembers. General prime coverage remains unproved, so Claim 9.1 remai","The remaining prime-existence obligation now has a precise formulation.\nFor every finite Galois subfield $L$, every finite set $B$ of rational\nprimes, and every $\\sigma\\in\\operatorname{Gal}(L/\\Q)$, require a prime\n$p\\notin B\\cup S_L$ and an actual absolute arithmetic witness $\\gamma_p$\nsuch that\n\\[\n q_L(\\gamma_p)\\sim\\sigma.\n\\]\nLean proves that this assertion is equivalent to the cofinite quotient\ncoverage required by the general separator, for any fixed choice of actual\nFrobenius witnesses. The equivalence derives independence from the witness\nchoices; it does not supply the displayed prime-existence assertion.\nArbitrary finite exclusions absorb $S_L$, and the resulting conditional\ndensity theorem preserves the separate good-prime sets of coefficient\nmembers. General prime coverage remains unproved, so Claim 9.1 remains\npartial. The source ledger is still seventeen checked and three","The remaining prime-existence obligation now has a precise formulation.\nFor every finite Galois subfield $L$, every finite set $B$ of rational\nprimes, and every $\\sigma\\in\\operatorname{Gal}(L/\\Q)$, require a prime\n$p\\notin B\\cup S_L$ and an actual absolute arithmetic witness $\\gamma_p$\nsuch that\n\\[\n q_L(\\gamma_p)\\sim\\sigma.\n\\]\nLean proves that this assertion is equivalent to the cofinite quotient\ncoverage required by the general separator, for any fixed choice of actual\nFrobenius witnesses. The equivalence derives independence from the witness\nchoices; it does not supply the displayed prime-existence assertion.\nArbitrary finite exclusions absorb $S_L$, and the resulting conditional\ndensity theorem preserves the separate good-prime sets of coefficient\nmembers. General prime coverage remains unproved, so Claim 9.1 remains\npartial. The source ledger is still seventeen checked and three partial.\nThis new arithmetic interfa","The remaining prime-existence obligation now has a precise formulation.\nFor every finite Galois subfield $L$, every finite set $B$ of rational\nprimes, and every $\\sigma\\in\\operatorname{Gal}(L/\\Q)$, require a prime\n$p\\notin B\\cup S_L$ and an actual absolute arithmetic witness $\\gamma_p$\nsuch that\n\\[\n q_L(\\gamma_p)\\sim\\sigma.\n\\]\nLean proves that this assertion is equivalent to the cofinite quotient\ncoverage required by the general separator, for any fixed choice of actual\nFrobenius witnesses. The equivalence derives independence from the witness\nchoices; it does not supply the displayed prime-existence assertion.\nArbitrary finite exclusions absorb $S_L$, and the resulting conditional\ndensity theorem preserves the separate good-prime sets of coefficient\nmembers. General prime coverage remains unproved, so Claim 9.1 remains\npartial. The source ledger is still seventeen checked and three partial.\nThis new arithmetic interface has base $\\Q$. For the source's general-base\ninsta","The remaining prime-existence obligation now has a precise formulation.\nFor every finite Galois subfield $L$, every finite set $B$ of rational\nprimes, and every $\\sigma\\in\\operatorname{Gal}(L/\\Q)$, require a prime\n$p\\notin B\\cup S_L$ and an actual absolute arithmetic witness $\\gamma_p$\nsuch that\n\\[\n q_L(\\gamma_p)\\sim\\sigma.\n\\]\nLean proves that this assertion is equivalent to the cofinite quotient\ncoverage required by the general separator, for any fixed choice of actual\nFrobenius witnesses. The equivalence derives independence from the witness\nchoices; it does not supply the displayed prime-existence assertion.\nArbitrary finite exclusions absorb $S_L$, and the resulting conditional\ndensity theorem preserves the separate good-prime sets of coefficient\nmembers. General prime coverage remains unproved, so Claim 9.1 remains\npartial. The source ledger is still seventeen checked and three partial.\nThis new arithmetic interface has base $\\Q$. For the source's general-base\ninstances involving $G_{F,S}$, the analogous a","e arithmetic witness $\\gamma_p$\nsuch that\n\\[\n q_L(\\gamma_p)\\sim\\sigma.\n\\]\nLean proves that this assertion is equivalent to the cofinite quotient\ncoverage required by the general separator, for any fixed choice of actual\nFrobenius witnesses. The equivalence derives independence from the witness\nchoices; it does not supply the displayed prime-existence assertion.\nArbitrary finite exclusions absorb $S_L$, and the resulting conditional\ndensity theorem preserves the separate good-prime sets of coefficient\nmembers. General prime coverage remains unproved, so Claim 9.1 remains\npartial. The source ledger is still seventeen checked and three partial.\nThis new arithmetic interface has base $\\Q$. For the source's general-base\ninstances involving $G_{F,S}$, the analogous actual-prime and norm-power\nFrobenius attachment and coverage over $F$, or a proved transfer to the\nrational-base result, also remain necessary.","ness\nchoices; it does not supply the displayed prime-existence assertion.\nArbitrary finite exclusions absorb $S_L$, and the resulting conditional\ndensity theorem preserves the separate good-prime sets of coefficient\nmembers. General prime coverage remains unproved, so Claim 9.1 remains\npartial. The source ledger is still seventeen checked and three partial.\nThis new arithmetic interface has base $\\Q$. For the source's general-base\ninstances involving $G_{F,S}$, the analogous actual-prime and norm-power\nFrobenius attachment and coverage over $F$, or a proved transfer to the\nrational-base result, also remain necessary.","stence assertion.\nArbitrary finite exclusions absorb $S_L$, and the resulting conditional\ndensity theorem preserves the separate good-prime sets of coefficient\nmembers. General prime coverage remains unproved, so Claim 9.1 remains\npartial. The source ledger is still seventeen checked and three partial.\nThis new arithmetic interface has base $\\Q$. For the source's general-base\ninstances involving $G_{F,S}$, the analogous actual-prime and norm-power\nFrobenius attachment and coverage over $F$, or a proved transfer to the\nrational-base result, also remain necessary.","nsity theorem preserves the separate good-prime sets of coefficient\nmembers. General prime coverage remains unproved, so Claim 9.1 remains\npartial. The source ledger is still seventeen checked and three partial.\nThis new arithmetic interface has base $\\Q$. For the source's general-base\ninstances involving $G_{F,S}$, the analogous actual-prime and norm-power\nFrobenius attachment and coverage over $F$, or a proved transfer to the\nrational-base result, also remain necessary.","\\paragraph{Finite deletion can destroy density.}\nIn the actual discrete group $C_2=\\{1,a\\}$, set $s(0)=a$ and $s(m+1)=1$.\nThe original samples are surjective and satisfy quotient coverage. Removing\nindex zero leaves only the identity, so both density and quotient coverage fail.\nThe actual rational sign and trivial characters agree on the remaining samples\nbut differ globally. This checked example explains why ordinary coverage alone\ncannot justify deleting the bad places in the system comparison.","\\begin{theorem}[Compatible-carrier summand completion]\\label{px:9.2}\nSuppose a semisimple strictly compatible carrier decomposes as\n\\[\n\\mathscr W\n\\cong\n\\bigoplus_{i=1}^{r}\n\\mathscr W_i^{\\oplus c_i},\n\\]\nwith the \\(\\mathscr W_i\\) as above.  At one coefficient prime \\(\\lambda_0\\), suppose a semisimple summand has the form\n\\[\nV_{\\lambda_0}\n\\cong\n\\bigoplus_{i=1}^{r}\nW_{i,\\lambda_0}^{\\oplus m_i},\n\\qquad\n0\\le m_i\\le c_i.\n\\]\nThen\n\\[\n\\boxed{\nV_\\lambda\n=\n\\bigoplus_{i=1}^{r}\nW_{i,\\lambda}^{\\oplus m_i}\n}\n\\]\ndefines a strictly compatible summand at every coefficient prime.  The vector \\((m_i)\\) is uniquely recoverable from finitely many Frobenius traces supplied by the separation theorem.\n\\end{theorem}","arrier summand completion]\\label{px:9.2}\nSuppose a semisimple strictly compatible carrier decomposes as\n\\[\n\\mathscr W\n\\cong\n\\bigoplus_{i=1}^{r}\n\\mathscr W_i^{\\oplus c_i},\n\\]\nwith the \\(\\mathscr W_i\\) as above.  At one coefficient prime \\(\\lambda_0\\), suppose a semisimple summand has the form\n\\[\nV_{\\lambda_0}\n\\cong\n\\bigoplus_{i=1}^{r}\nW_{i,\\lambda_0}^{\\oplus m_i},\n\\qquad\n0\\le m_i\\le c_i.\n\\]\nThen\n\\[\n\\boxed{\nV_\\lambda\n=\n\\bigoplus_{i=1}^{r}\nW_{i,\\lambda}^{\\oplus m_i}\n}\n\\]\ndefines a strictly compatible summand at every coefficient prime.  The vector \\((m_i)\\) is uniquely recoverable from finitely many Frobenius traces supplied by the separation theorem.\n\\end{theorem}","W_i^{\\oplus c_i},\n\\]\nwith the \\(\\mathscr W_i\\) as above.  At one coefficient prime \\(\\lambda_0\\), suppose a semisimple summand has the form\n\\[\nV_{\\lambda_0}\n\\cong\n\\bigoplus_{i=1}^{r}\nW_{i,\\lambda_0}^{\\oplus m_i},\n\\qquad\n0\\le m_i\\le c_i.\n\\]\nThen\n\\[\n\\boxed{\nV_\\lambda\n=\n\\bigoplus_{i=1}^{r}\nW_{i,\\lambda}^{\\oplus m_i}\n}\n\\]\ndefines a strictly compatible summand at every coefficient prime.  The vector \\((m_i)\\) is uniquely recoverable from finitely many Frobenius traces supplied by the separation theorem.\n\\end{theorem}","\\begin{proof}\nUse the same multiplicities in every member of the compatible carrier.  Direct sums preserve strict compatibility, and \\(m_i\\le c_i\\) makes \\(V_\\lambda\\) a summand of \\(W_\\lambda\\).  At the separating places, the target trace vector is \\(C\\bm m\\).  Full column rank makes \\(\\bm m\\) the unique solution.\n\\end{proof}","\\paragraph{Recovery and realization are different steps.}\nFor a certified square separator the coefficient vector is $C^{-1}t$.\nThis is a valid carrier multiplicity vector only when every entry is a\nnonnegative integer and satisfies its capacity bound $m_i\\le c_i$.\nThose tests and uniqueness of natural multiplicities in characteristic zero\nare separately Lean-checked, as is transport through a field embedding.\nReduction modulo a residue characteristic is not a field embedding and may\nlose rank. The representation construction and common-polynomial part are now\nLean-checked as described below. The full strictly compatible arithmetic\nstatement, including its local conditions, remains partially formalized.","\\paragraph{Explicit summands inside a supplied decomposition.}\nLet $K$ be a field, let $G$ be a monoid, and let\n$\\rho_i:G\\to\\operatorname{End}_K(W_i)$ be actual representations on\nfinite-dimensional spaces, indexed by a finite set $I$. For natural numbers\n$m_i\\le c_i$, set\n\\[\n U_m=\\prod_{i\\in I}\\prod_{k<m_i}W_i,\n \\qquad U_c=\\prod_{i\\in I}\\prod_{k<c_i}W_i.\n\\]\nThese finite products realize the corresponding finite direct sums. The\nrepresentation on $U_m$ acts componentwise. Define $j:U_m\\to U_c$ by retaining\nthe first $m_i$ copies and filling every remaining coordinate with zero.\nDefine $r:U_c\\to U_m$ by restriction to the selected copies. Lean proves\n\\[\n rj=1_{U_m},\\qquad\n j\\rho_m(g)=\\rho_c(g)j,\\qquad\n r\\rho_c(g)=\\rho_m(g)r.\n\\]\nThe image of $j$ is exactly the","\\paragraph{Explicit summands inside a supplied decomposition.}\nLet $K$ be a field, let $G$ be a monoid, and let\n$\\rho_i:G\\to\\operatorname{End}_K(W_i)$ be actual representations on\nfinite-dimensional spaces, indexed by a finite set $I$. For natural numbers\n$m_i\\le c_i$, set\n\\[\n U_m=\\prod_{i\\in I}\\prod_{k<m_i}W_i,\n \\qquad U_c=\\prod_{i\\in I}\\prod_{k<c_i}W_i.\n\\]\nThese finite products realize the corresponding finite direct sums. The\nrepresentation on $U_m$ acts componentwise. Define $j:U_m\\to U_c$ by retaining\nthe first $m_i$ copies and filling every remaining coordinate with zero.\nDefine $r:U_c\\to U_m$ by restriction to the selected copies. Lean proves\n\\[\n rj=1_{U_m},\\qquad\n j\\rho_m(g)=\\rho_c(g)j,\\qquad\n r\\rho_c(g)=\\rho_m(g)r.\n\\]\nThe image of $j$ is exactly the subspace whose unuse","\\paragraph{Explicit summands inside a supplied decomposition.}\nLet $K$ be a field, let $G$ be a monoid, and let\n$\\rho_i:G\\to\\operatorname{End}_K(W_i)$ be actual representations on\nfinite-dimensional spaces, indexed by a finite set $I$. For natural numbers\n$m_i\\le c_i$, set\n\\[\n U_m=\\prod_{i\\in I}\\prod_{k<m_i}W_i,\n \\qquad U_c=\\prod_{i\\in I}\\prod_{k<c_i}W_i.\n\\]\nThese finite products realize the corresponding finite direct sums. The\nrepresentation on $U_m$ acts componentwise. Define $j:U_m\\to U_c$ by retaining\nthe first $m_i$ copies and filling every remaining coordinate with zero.\nDefine $r:U_c\\to U_m$ by restriction to the selected copies. Lean proves\n\\[\n rj=1_{U_m},\\qquad\n j\\rho_m(g)=\\rho_c(g)j,\\qquad\n r\\rho_c(g)=\\rho_m(g)r.\n\\]\nThe image of $j$ is exactly the subspace whose unused coordinates vanish.\nConsequently $jr$ is an equivariant id","\\paragraph{Explicit summands inside a supplied decomposition.}\nLet $K$ be a field, let $G$ be a monoid, and let\n$\\rho_i:G\\to\\operatorname{End}_K(W_i)$ be actual representations on\nfinite-dimensional spaces, indexed by a finite set $I$. For natural numbers\n$m_i\\le c_i$, set\n\\[\n U_m=\\prod_{i\\in I}\\prod_{k<m_i}W_i,\n \\qquad U_c=\\prod_{i\\in I}\\prod_{k<c_i}W_i.\n\\]\nThese finite products realize the corresponding finite direct sums. The\nrepresentation on $U_m$ acts componentwise. Define $j:U_m\\to U_c$ by retaining\nthe first $m_i$ copies and filling every remaining coordinate with zero.\nDefine $r:U_c\\to U_m$ by restriction to the selected copies. Lean proves\n\\[\n rj=1_{U_m},\\qquad\n j\\rho_m(g)=\\rho_c(g)j,\\qquad\n r\\rho_c(g)=\\rho_m(g)r.\n\\]\nThe image of $j$ is exactly the subspace whose unused coordinates vanish.\nConsequently $jr$ is an equivariant idempotent. A supplied equivariant\nisomorphism $D:W\\xrightarrow{\\sim}U_c$ gives the sp","\\paragraph{Explicit summands inside a supplied decomposition.}\nLet $K$ be a field, let $G$ be a monoid, and let\n$\\rho_i:G\\to\\operatorname{End}_K(W_i)$ be actual representations on\nfinite-dimensional spaces, indexed by a finite set $I$. For natural numbers\n$m_i\\le c_i$, set\n\\[\n U_m=\\prod_{i\\in I}\\prod_{k<m_i}W_i,\n \\qquad U_c=\\prod_{i\\in I}\\prod_{k<c_i}W_i.\n\\]\nThese finite products realize the corresponding finite direct sums. The\nrepresentation on $U_m$ acts componentwise. Define $j:U_m\\to U_c$ by retaining\nthe first $m_i$ copies and filling every remaining coordinate with zero.\nDefine $r:U_c\\to U_m$ by restriction to the selected copies. Lean proves\n\\[\n rj=1_{U_m},\\qquad\n j\\rho_m(g)=\\rho_c(g)j,\\qquad\n r\\rho_c(g)=\\rho_m(g)r.\n\\]\nThe image of $j$ is exactly the subspace whose unused coordinates vanish.\nConsequently $jr$ is an equivariant idempotent. A supplied equivariant\nisomorphism $D:W\\xrightarrow{\\sim}U_c$ gives the split maps\n$D^{-1}j$ and $rD$ inside","\\paragraph{Explicit summands inside a supplied decomposition.}\nLet $K$ be a field, let $G$ be a monoid, and let\n$\\rho_i:G\\to\\operatorname{End}_K(W_i)$ be actual representations on\nfinite-dimensional spaces, indexed by a finite set $I$. For natural numbers\n$m_i\\le c_i$, set\n\\[\n U_m=\\prod_{i\\in I}\\prod_{k<m_i}W_i,\n \\qquad U_c=\\prod_{i\\in I}\\prod_{k<c_i}W_i.\n\\]\nThese finite products realize the corresponding finite direct sums. The\nrepresentation on $U_m$ acts componentwise. Define $j:U_m\\to U_c$ by retaining\nthe first $m_i$ copies and filling every remaining coordinate with zero.\nDefine $r:U_c\\to U_m$ by restriction to the selected copies. Lean proves\n\\[\n rj=1_{U_m},\\qquad\n j\\rho_m(g)=\\rho_c(g)j,\\qquad\n r\\rho_c(g)=\\rho_m(g)r.\n\\]\nThe image of $j$ is exactly the subspace whose unused coordinates vanish.\nConsequently $jr$ is an equivariant idempotent. A supplied equivariant\nisomorphism $D:W\\xrightarrow{\\sim}U_c$ gives the split maps\n$D^{-1}j$ and $rD$ inside the original carrier. The decomposition $D$ is an\ninput; these statements do not construct","position.}\nLet $K$ be a field, let $G$ be a monoid, and let\n$\\rho_i:G\\to\\operatorname{End}_K(W_i)$ be actual representations on\nfinite-dimensional spaces, indexed by a finite set $I$. For natural numbers\n$m_i\\le c_i$, set\n\\[\n U_m=\\prod_{i\\in I}\\prod_{k<m_i}W_i,\n \\qquad U_c=\\prod_{i\\in I}\\prod_{k<c_i}W_i.\n\\]\nThese finite products realize the corresponding finite direct sums. The\nrepresentation on $U_m$ acts componentwise. Define $j:U_m\\to U_c$ by retaining\nthe first $m_i$ copies and filling every remaining coordinate with zero.\nDefine $r:U_c\\to U_m$ by restriction to the selected copies. Lean proves\n\\[\n rj=1_{U_m},\\qquad\n j\\rho_m(g)=\\rho_c(g)j,\\qquad\n r\\rho_c(g)=\\rho_m(g)r.\n\\]\nThe image of $j$ is exactly the subspace whose unused coordinates vanish.\nConsequently $jr$ is an equivariant idempotent. A supplied equivariant\nisomorphism $D:W\\xrightarrow{\\sim}U_c$ gives the split maps\n$D^{-1}j$ and $rD$ inside the original carrier. The decomposition $D$ is an\ninput; these statements do not construct a decomposition of an unknown\narithmetic carrier.","t $G$ be a monoid, and let\n$\\rho_i:G\\to\\operatorname{End}_K(W_i)$ be actual representations on\nfinite-dimensional spaces, indexed by a finite set $I$. For natural numbers\n$m_i\\le c_i$, set\n\\[\n U_m=\\prod_{i\\in I}\\prod_{k<m_i}W_i,\n \\qquad U_c=\\prod_{i\\in I}\\prod_{k<c_i}W_i.\n\\]\nThese finite products realize the corresponding finite direct sums. The\nrepresentation on $U_m$ acts componentwise. Define $j:U_m\\to U_c$ by retaining\nthe first $m_i$ copies and filling every remaining coordinate with zero.\nDefine $r:U_c\\to U_m$ by restriction to the selected copies. Lean proves\n\\[\n rj=1_{U_m},\\qquad\n j\\rho_m(g)=\\rho_c(g)j,\\qquad\n r\\rho_c(g)=\\rho_m(g)r.\n\\]\nThe image of $j$ is exactly the subspace whose unused coordinates vanish.\nConsequently $jr$ is an equivariant idempotent. A supplied equivariant\nisomorphism $D:W\\xrightarrow{\\sim}U_c$ gives the split maps\n$D^{-1}j$ and $rD$ inside the original carrier. The decomposition $D$ is an\ninput; these statements do not construct a decomposition of an unknown\narithmetic carrier.","o\\operatorname{End}_K(W_i)$ be actual representations on\nfinite-dimensional spaces, indexed by a finite set $I$. For natural numbers\n$m_i\\le c_i$, set\n\\[\n U_m=\\prod_{i\\in I}\\prod_{k<m_i}W_i,\n \\qquad U_c=\\prod_{i\\in I}\\prod_{k<c_i}W_i.\n\\]\nThese finite products realize the corresponding finite direct sums. The\nrepresentation on $U_m$ acts componentwise. Define $j:U_m\\to U_c$ by retaining\nthe first $m_i$ copies and filling every remaining coordinate with zero.\nDefine $r:U_c\\to U_m$ by restriction to the selected copies. Lean proves\n\\[\n rj=1_{U_m},\\qquad\n j\\rho_m(g)=\\rho_c(g)j,\\qquad\n r\\rho_c(g)=\\rho_m(g)r.\n\\]\nThe image of $j$ is exactly the subspace whose unused coordinates vanish.\nConsequently $jr$ is an equivariant idempotent. A supplied equivariant\nisomorphism $D:W\\xrightarrow{\\sim}U_c$ gives the split maps\n$D^{-1}j$ and $rD$ inside the original carrier. The decomposition $D$ is an\ninput; these statements do not construct a decomposition of an unknown\narithmetic carrier.","ional spaces, indexed by a finite set $I$. For natural numbers\n$m_i\\le c_i$, set\n\\[\n U_m=\\prod_{i\\in I}\\prod_{k<m_i}W_i,\n \\qquad U_c=\\prod_{i\\in I}\\prod_{k<c_i}W_i.\n\\]\nThese finite products realize the corresponding finite direct sums. The\nrepresentation on $U_m$ acts componentwise. Define $j:U_m\\to U_c$ by retaining\nthe first $m_i$ copies and filling every remaining coordinate with zero.\nDefine $r:U_c\\to U_m$ by restriction to the selected copies. Lean proves\n\\[\n rj=1_{U_m},\\qquad\n j\\rho_m(g)=\\rho_c(g)j,\\qquad\n r\\rho_c(g)=\\rho_m(g)r.\n\\]\nThe image of $j$ is exactly the subspace whose unused coordinates vanish.\nConsequently $jr$ is an equivariant idempotent. A supplied equivariant\nisomorphism $D:W\\xrightarrow{\\sim}U_c$ gives the split maps\n$D^{-1}j$ and $rD$ inside the original carrier. The decomposition $D$ is an\ninput; these statements do not construct a decomposition of an unknown\narithmetic carrier.","\\le c_i$, set\n\\[\n U_m=\\prod_{i\\in I}\\prod_{k<m_i}W_i,\n \\qquad U_c=\\prod_{i\\in I}\\prod_{k<c_i}W_i.\n\\]\nThese finite products realize the corresponding finite direct sums. The\nrepresentation on $U_m$ acts componentwise. Define $j:U_m\\to U_c$ by retaining\nthe first $m_i$ copies and filling every remaining coordinate with zero.\nDefine $r:U_c\\to U_m$ by restriction to the selected copies. Lean proves\n\\[\n rj=1_{U_m},\\qquad\n j\\rho_m(g)=\\rho_c(g)j,\\qquad\n r\\rho_c(g)=\\rho_m(g)r.\n\\]\nThe image of $j$ is exactly the subspace whose unused coordinates vanish.\nConsequently $jr$ is an equivariant idempotent. A supplied equivariant\nisomorphism $D:W\\xrightarrow{\\sim}U_c$ gives the split maps\n$D^{-1}j$ and $rD$ inside the original carrier. The decomposition $D$ is an\ninput; these statements do not construct a decomposition of an unknown\narithmetic carrier.","i}W_i.\n\\]\nThese finite products realize the corresponding finite direct sums. The\nrepresentation on $U_m$ acts componentwise. Define $j:U_m\\to U_c$ by retaining\nthe first $m_i$ copies and filling every remaining coordinate with zero.\nDefine $r:U_c\\to U_m$ by restriction to the selected copies. Lean proves\n\\[\n rj=1_{U_m},\\qquad\n j\\rho_m(g)=\\rho_c(g)j,\\qquad\n r\\rho_c(g)=\\rho_m(g)r.\n\\]\nThe image of $j$ is exactly the subspace whose unused coordinates vanish.\nConsequently $jr$ is an equivariant idempotent. A supplied equivariant\nisomorphism $D:W\\xrightarrow{\\sim}U_c$ gives the split maps\n$D^{-1}j$ and $rD$ inside the original carrier. The decomposition $D$ is an\ninput; these statements do not construct a decomposition of an unknown\narithmetic carrier.","ms. The\nrepresentation on $U_m$ acts componentwise. Define $j:U_m\\to U_c$ by retaining\nthe first $m_i$ copies and filling every remaining coordinate with zero.\nDefine $r:U_c\\to U_m$ by restriction to the selected copies. Lean proves\n\\[\n rj=1_{U_m},\\qquad\n j\\rho_m(g)=\\rho_c(g)j,\\qquad\n r\\rho_c(g)=\\rho_m(g)r.\n\\]\nThe image of $j$ is exactly the subspace whose unused coordinates vanish.\nConsequently $jr$ is an equivariant idempotent. A supplied equivariant\nisomorphism $D:W\\xrightarrow{\\sim}U_c$ gives the split maps\n$D^{-1}j$ and $rD$ inside the original carrier. The decomposition $D$ is an\ninput; these statements do not construct a decomposition of an unknown\narithmetic carrier.","_c$ by retaining\nthe first $m_i$ copies and filling every remaining coordinate with zero.\nDefine $r:U_c\\to U_m$ by restriction to the selected copies. Lean proves\n\\[\n rj=1_{U_m},\\qquad\n j\\rho_m(g)=\\rho_c(g)j,\\qquad\n r\\rho_c(g)=\\rho_m(g)r.\n\\]\nThe image of $j$ is exactly the subspace whose unused coordinates vanish.\nConsequently $jr$ is an equivariant idempotent. A supplied equivariant\nisomorphism $D:W\\xrightarrow{\\sim}U_c$ gives the split maps\n$D^{-1}j$ and $rD$ inside the original carrier. The decomposition $D$ is an\ninput; these statements do not construct a decomposition of an unknown\narithmetic carrier.","ing every remaining coordinate with zero.\nDefine $r:U_c\\to U_m$ by restriction to the selected copies. Lean proves\n\\[\n rj=1_{U_m},\\qquad\n j\\rho_m(g)=\\rho_c(g)j,\\qquad\n r\\rho_c(g)=\\rho_m(g)r.\n\\]\nThe image of $j$ is exactly the subspace whose unused coordinates vanish.\nConsequently $jr$ is an equivariant idempotent. A supplied equivariant\nisomorphism $D:W\\xrightarrow{\\sim}U_c$ gives the split maps\n$D^{-1}j$ and $rD$ inside the original carrier. The decomposition $D$ is an\ninput; these statements do not construct a decomposition of an unknown\narithmetic carrier.","ining coordinate with zero.\nDefine $r:U_c\\to U_m$ by restriction to the selected copies. Lean proves\n\\[\n rj=1_{U_m},\\qquad\n j\\rho_m(g)=\\rho_c(g)j,\\qquad\n r\\rho_c(g)=\\rho_m(g)r.\n\\]\nThe image of $j$ is exactly the subspace whose unused coordinates vanish.\nConsequently $jr$ is an equivariant idempotent. A supplied equivariant\nisomorphism $D:W\\xrightarrow{\\sim}U_c$ gives the split maps\n$D^{-1}j$ and $rD$ inside the original carrier. The decomposition $D$ is an\ninput; these statements do not construct a decomposition of an unknown\narithmetic carrier.","restriction to the selected copies. Lean proves\n\\[\n rj=1_{U_m},\\qquad\n j\\rho_m(g)=\\rho_c(g)j,\\qquad\n r\\rho_c(g)=\\rho_m(g)r.\n\\]\nThe image of $j$ is exactly the subspace whose unused coordinates vanish.\nConsequently $jr$ is an equivariant idempotent. A supplied equivariant\nisomorphism $D:W\\xrightarrow{\\sim}U_c$ gives the split maps\n$D^{-1}j$ and $rD$ inside the original carrier. The decomposition $D$ is an\ninput; these statements do not construct a decomposition of an unknown\narithmetic carrier.","The same formalization proves, for each $g\\in G$,\n\\[\n \\det\\bigl(XI-\\rho_m(g)\\bigr)\n =\\prod_{i\\in I}\\det\\bigl(XI-\\rho_i(g)\\bigr)^{m_i},\n \\qquad\n \\operatorname{Tr}\\bigl(\\rho_m(g)\\bigr)\n =\\sum_{i\\in I}(m_i:K)\\operatorname{Tr}\\bigl(\\rho_i(g)\\bigr).\n\\]\nConstituents may have different dimensions. Zero multiplicities, zero-dimensional\nconstituents, and the empty index set are included. If at a permitted sample\n$g$ the constituent polynomials are the images of $P_{i,g}\\in E[X]$ under a\nspecified map $E\\to K$, the summand polynomial is the image of\n$\\prod_iP_{i,g}^{m_i}$. Applying this independently at each coefficient field\npreserves a supplied common polynomial family. This does not establish the\nHodge--Tate or Weil--Deligne conditions in stric","The same formalization proves, for each $g\\in G$,\n\\[\n \\det\\bigl(XI-\\rho_m(g)\\bigr)\n =\\prod_{i\\in I}\\det\\bigl(XI-\\rho_i(g)\\bigr)^{m_i},\n \\qquad\n \\operatorname{Tr}\\bigl(\\rho_m(g)\\bigr)\n =\\sum_{i\\in I}(m_i:K)\\operatorname{Tr}\\bigl(\\rho_i(g)\\bigr).\n\\]\nConstituents may have different dimensions. Zero multiplicities, zero-dimensional\nconstituents, and the empty index set are included. If at a permitted sample\n$g$ the constituent polynomials are the images of $P_{i,g}\\in E[X]$ under a\nspecified map $E\\to K$, the summand polynomial is the image of\n$\\prod_iP_{i,g}^{m_i}$. Applying this independently at each coefficient field\npreserves a supplied common polynomial family. This does not establish the\nHodge--Tate or Weil--Deligne conditions in strict compatibility.","formalization proves, for each $g\\in G$,\n\\[\n \\det\\bigl(XI-\\rho_m(g)\\bigr)\n =\\prod_{i\\in I}\\det\\bigl(XI-\\rho_i(g)\\bigr)^{m_i},\n \\qquad\n \\operatorname{Tr}\\bigl(\\rho_m(g)\\bigr)\n =\\sum_{i\\in I}(m_i:K)\\operatorname{Tr}\\bigl(\\rho_i(g)\\bigr).\n\\]\nConstituents may have different dimensions. Zero multiplicities, zero-dimensional\nconstituents, and the empty index set are included. If at a permitted sample\n$g$ the constituent polynomials are the images of $P_{i,g}\\in E[X]$ under a\nspecified map $E\\to K$, the summand polynomial is the image of\n$\\prod_iP_{i,g}^{m_i}$. Applying this independently at each coefficient field\npreserves a supplied common polynomial family. This does not establish the\nHodge--Tate or Weil--Deligne conditions in strict compatibility.","\\bigl(XI-\\rho_m(g)\\bigr)\n =\\prod_{i\\in I}\\det\\bigl(XI-\\rho_i(g)\\bigr)^{m_i},\n \\qquad\n \\operatorname{Tr}\\bigl(\\rho_m(g)\\bigr)\n =\\sum_{i\\in I}(m_i:K)\\operatorname{Tr}\\bigl(\\rho_i(g)\\bigr).\n\\]\nConstituents may have different dimensions. Zero multiplicities, zero-dimensional\nconstituents, and the empty index set are included. If at a permitted sample\n$g$ the constituent polynomials are the images of $P_{i,g}\\in E[X]$ under a\nspecified map $E\\to K$, the summand polynomial is the image of\n$\\prod_iP_{i,g}^{m_i}$. Applying this independently at each coefficient field\npreserves a supplied common polynomial family. This does not establish the\nHodge--Tate or Weil--Deligne conditions in strict compatibility.","}\\det\\bigl(XI-\\rho_i(g)\\bigr)^{m_i},\n \\qquad\n \\operatorname{Tr}\\bigl(\\rho_m(g)\\bigr)\n =\\sum_{i\\in I}(m_i:K)\\operatorname{Tr}\\bigl(\\rho_i(g)\\bigr).\n\\]\nConstituents may have different dimensions. Zero multiplicities, zero-dimensional\nconstituents, and the empty index set are included. If at a permitted sample\n$g$ the constituent polynomials are the images of $P_{i,g}\\in E[X]$ under a\nspecified map $E\\to K$, the summand polynomial is the image of\n$\\prod_iP_{i,g}^{m_i}$. Applying this independently at each coefficient field\npreserves a supplied common polynomial family. This does not establish the\nHodge--Tate or Weil--Deligne conditions in strict compatibility.","eratorname{Tr}\\bigl(\\rho_m(g)\\bigr)\n =\\sum_{i\\in I}(m_i:K)\\operatorname{Tr}\\bigl(\\rho_i(g)\\bigr).\n\\]\nConstituents may have different dimensions. Zero multiplicities, zero-dimensional\nconstituents, and the empty index set are included. If at a permitted sample\n$g$ the constituent polynomials are the images of $P_{i,g}\\in E[X]$ under a\nspecified map $E\\to K$, the summand polynomial is the image of\n$\\prod_iP_{i,g}^{m_i}$. Applying this independently at each coefficient field\npreserves a supplied common polynomial family. This does not establish the\nHodge--Tate or Weil--Deligne conditions in strict compatibility.","Joint continuity of the constituent actions implies joint continuity of the\nconstructed action. The inclusion and retraction are continuous for the\nproduct topologies, and triviality on a supplied subgroup is preserved.\nNeither continuity nor triviality is inferred for a carrier whose corresponding\nhypothesis has not been supplied. The maps transported by $D$ are proved\nalgebraically; their topological interpretation additionally requires the\nappropriate continuity of $D$ and its inverse.","e\nconstructed action. The inclusion and retraction are continuous for the\nproduct topologies, and triviality on a supplied subgroup is preserved.\nNeither continuity nor triviality is inferred for a carrier whose corresponding\nhypothesis has not been supplied. The maps transported by $D$ are proved\nalgebraically; their topological interpretation additionally requires the\nappropriate continuity of $D$ and its inverse.","For the three-character example below, the construction produces an actual\nthree-dimensional summand inside the carrier of dimension seven. At the sample\n$(0,1)$ its characteristic polynomial is $(X-1)^2(X+1)$. A separate checked\ncounterexample uses the coordinate inclusion $\\mathbf Q\\to\\mathbf Q^2$, its\ncoordinate retraction, and the swap involution: the split law holds, but\ncompression sends the swap to zero and therefore fails to preserve its square.\nThus selecting whole constituent copies, with the proved equivariance, is\nessential to this construction.","\\paragraph{Algebraic monodromy data on the selected copies.}\nFix a field $K$ and a nonzero scalar $q\\in K^\\times$. On each constituent\n$W_i$, supply an automorphism $F_i$ and a nilpotent endomorphism $N_i$ with\n$F_iN_iF_i^{-1}=qN_i$. The scalar $q$ is common to this finite family.\nThe componentwise operators $F_m,N_m$ on $U_m$ satisfy the same relation;\n$F_m$ is invertible and $N_m$ is nilpotent. The split maps constructed above\nintertwine both operators. The proof first identifies the independently defined\nkernel through an equivariant linear isomorphism\n\\[\n \\ker N_m\\simeq\\prod_{i\\in I}\\prod_{k<m_i}\\ker N_i.\n\\]\nThe relation, with $q\\ne0$, proves that both $F_i$ and its inverse preserve\n$\\ker N_i$. Thus $F_i|_{\\ker N_i}$ is an actual automorphism, and Lean proves\n\\","\\paragraph{Algebraic monodromy data on the selected copies.}\nFix a field $K$ and a nonzero scalar $q\\in K^\\times$. On each constituent\n$W_i$, supply an automorphism $F_i$ and a nilpotent endomorphism $N_i$ with\n$F_iN_iF_i^{-1}=qN_i$. The scalar $q$ is common to this finite family.\nThe componentwise operators $F_m,N_m$ on $U_m$ satisfy the same relation;\n$F_m$ is invertible and $N_m$ is nilpotent. The split maps constructed above\nintertwine both operators. The proof first identifies the independently defined\nkernel through an equivariant linear isomorphism\n\\[\n \\ker N_m\\simeq\\prod_{i\\in I}\\prod_{k<m_i}\\ker N_i.\n\\]\nThe relation, with $q\\ne0$, proves that both $F_i$ and its inverse preserve\n$\\ker N_i$. Thus $F_i|_{\\ker N_i}$ is an actual automorphism, and Lean proves\n\\[\n \\det(XI-F_m|_{\\ker N_m})\n   =\\prod","\\paragraph{Algebraic monodromy data on the selected copies.}\nFix a field $K$ and a nonzero scalar $q\\in K^\\times$. On each constituent\n$W_i$, supply an automorphism $F_i$ and a nilpotent endomorphism $N_i$ with\n$F_iN_iF_i^{-1}=qN_i$. The scalar $q$ is common to this finite family.\nThe componentwise operators $F_m,N_m$ on $U_m$ satisfy the same relation;\n$F_m$ is invertible and $N_m$ is nilpotent. The split maps constructed above\nintertwine both operators. The proof first identifies the independently defined\nkernel through an equivariant linear isomorphism\n\\[\n \\ker N_m\\simeq\\prod_{i\\in I}\\prod_{k<m_i}\\ker N_i.\n\\]\nThe relation, with $q\\ne0$, proves that both $F_i$ and its inverse preserve\n$\\ker N_i$. Thus $F_i|_{\\ker N_i}$ is an actual automorphism, and Lean proves\n\\[\n \\det(XI-F_m|_{\\ker N_m})\n   =\\prod_{i\\in I}\\det(XI-F_i|_{\\ker","\\paragraph{Algebraic monodromy data on the selected copies.}\nFix a field $K$ and a nonzero scalar $q\\in K^\\times$. On each constituent\n$W_i$, supply an automorphism $F_i$ and a nilpotent endomorphism $N_i$ with\n$F_iN_iF_i^{-1}=qN_i$. The scalar $q$ is common to this finite family.\nThe componentwise operators $F_m,N_m$ on $U_m$ satisfy the same relation;\n$F_m$ is invertible and $N_m$ is nilpotent. The split maps constructed above\nintertwine both operators. The proof first identifies the independently defined\nkernel through an equivariant linear isomorphism\n\\[\n \\ker N_m\\simeq\\prod_{i\\in I}\\prod_{k<m_i}\\ker N_i.\n\\]\nThe relation, with $q\\ne0$, proves that both $F_i$ and its inverse preserve\n$\\ker N_i$. Thus $F_i|_{\\ker N_i}$ is an actual automorphism, and Lean proves\n\\[\n \\det(XI-F_m|_{\\ker N_m})\n   =\\prod_{i\\in I}\\det(XI-F_i|_{\\ker N_i})^{m_i}.\n\\]\nThe correspon","\\paragraph{Algebraic monodromy data on the selected copies.}\nFix a field $K$ and a nonzero scalar $q\\in K^\\times$. On each constituent\n$W_i$, supply an automorphism $F_i$ and a nilpotent endomorphism $N_i$ with\n$F_iN_iF_i^{-1}=qN_i$. The scalar $q$ is common to this finite family.\nThe componentwise operators $F_m,N_m$ on $U_m$ satisfy the same relation;\n$F_m$ is invertible and $N_m$ is nilpotent. The split maps constructed above\nintertwine both operators. The proof first identifies the independently defined\nkernel through an equivariant linear isomorphism\n\\[\n \\ker N_m\\simeq\\prod_{i\\in I}\\prod_{k<m_i}\\ker N_i.\n\\]\nThe relation, with $q\\ne0$, proves that both $F_i$ and its inverse preserve\n$\\ker N_i$. Thus $F_i|_{\\ker N_i}$ is an actual automorphism, and Lean proves\n\\[\n \\det(XI-F_m|_{\\ker N_m})\n   =\\prod_{i\\in I}\\det(XI-F_i|_{\\ker N_i})^{m_i}.\n\\]\nThe corresponding kernel dimensions and traces a","\\paragraph{Algebraic monodromy data on the selected copies.}\nFix a field $K$ and a nonzero scalar $q\\in K^\\times$. On each constituent\n$W_i$, supply an automorphism $F_i$ and a nilpotent endomorphism $N_i$ with\n$F_iN_iF_i^{-1}=qN_i$. The scalar $q$ is common to this finite family.\nThe componentwise operators $F_m,N_m$ on $U_m$ satisfy the same relation;\n$F_m$ is invertible and $N_m$ is nilpotent. The split maps constructed above\nintertwine both operators. The proof first identifies the independently defined\nkernel through an equivariant linear isomorphism\n\\[\n \\ker N_m\\simeq\\prod_{i\\in I}\\prod_{k<m_i}\\ker N_i.\n\\]\nThe relation, with $q\\ne0$, proves that both $F_i$ and its inverse preserve\n$\\ker N_i$. Thus $F_i|_{\\ker N_i}$ is an actual automorphism, and Lean proves\n\\[\n \\det(XI-F_m|_{\\ker N_m})\n   =\\prod_{i\\in I}\\det(XI-F_i|_{\\ker N_i})^{m_i}.\n\\]\nThe corresponding kernel dimensions and traces are additive with multiplici","\\paragraph{Algebraic monodromy data on the selected copies.}\nFix a field $K$ and a nonzero scalar $q\\in K^\\times$. On each constituent\n$W_i$, supply an automorphism $F_i$ and a nilpotent endomorphism $N_i$ with\n$F_iN_iF_i^{-1}=qN_i$. The scalar $q$ is common to this finite family.\nThe componentwise operators $F_m,N_m$ on $U_m$ satisfy the same relation;\n$F_m$ is invertible and $N_m$ is nilpotent. The split maps constructed above\nintertwine both operators. The proof first identifies the independently defined\nkernel through an equivariant linear isomorphism\n\\[\n \\ker N_m\\simeq\\prod_{i\\in I}\\prod_{k<m_i}\\ker N_i.\n\\]\nThe relation, with $q\\ne0$, proves that both $F_i$ and its inverse preserve\n$\\ker N_i$. Thus $F_i|_{\\ker N_i}$ is an actual automorphism, and Lean proves\n\\[\n \\det(XI-F_m|_{\\ker N_m})\n   =\\prod_{i\\in I}\\det(XI-F_i|_{\\ker N_i})^{m_i}.\n\\]\nThe corresponding kernel dimensions and traces are additive with multiplicity.\nCommon const","\\paragraph{Algebraic monodromy data on the selected copies.}\nFix a field $K$ and a nonzero scalar $q\\in K^\\times$. On each constituent\n$W_i$, supply an automorphism $F_i$ and a nilpotent endomorphism $N_i$ with\n$F_iN_iF_i^{-1}=qN_i$. The scalar $q$ is common to this finite family.\nThe componentwise operators $F_m,N_m$ on $U_m$ satisfy the same relation;\n$F_m$ is invertible and $N_m$ is nilpotent. The split maps constructed above\nintertwine both operators. The proof first identifies the independently defined\nkernel through an equivariant linear isomorphism\n\\[\n \\ker N_m\\simeq\\prod_{i\\in I}\\prod_{k<m_i}\\ker N_i.\n\\]\nThe relation, with $q\\ne0$, proves that both $F_i$ and its inverse preserve\n$\\ker N_i$. Thus $F_i|_{\\ker N_i}$ is an actual automorphism, and Lean proves\n\\[\n \\det(XI-F_m|_{\\ker N_m})\n   =\\prod_{i\\in I}\\det(XI-F_i|_{\\ker N_i})^{m_i}.\n\\]\nThe corresponding kernel dimensions and traces are additive with multiplicity.\nCommon constituent kernel polynomials, when supplied through a coefficient-ring\nmap","\\paragraph{Algebraic monodromy data on the selected copies.}\nFix a field $K$ and a nonzero scalar $q\\in K^\\times$. On each constituent\n$W_i$, supply an automorphism $F_i$ and a nilpotent endomorphism $N_i$ with\n$F_iN_iF_i^{-1}=qN_i$. The scalar $q$ is common to this finite family.\nThe componentwise operators $F_m,N_m$ on $U_m$ satisfy the same relation;\n$F_m$ is invertible and $N_m$ is nilpotent. The split maps constructed above\nintertwine both operators. The proof first identifies the independently defined\nkernel through an equivariant linear isomorphism\n\\[\n \\ker N_m\\simeq\\prod_{i\\in I}\\prod_{k<m_i}\\ker N_i.\n\\]\nThe relation, with $q\\ne0$, proves that both $F_i$ and its inverse preserve\n$\\ker N_i$. Thus $F_i|_{\\ker N_i}$ is an actual automorphism, and Lean proves\n\\[\n \\det(XI-F_m|_{\\ker N_m})\n   =\\prod_{i\\in I}\\det(XI-F_i|_{\\ker N_i})^{m_i}.\n\\]\nThe corresponding kernel dimensions and traces are additive with multiplicity.\nCommon constituent kernel polynomials, when supplied through a coefficient-ring\nmap, give th","\\paragraph{Algebraic monodromy data on the selected copies.}\nFix a field $K$ and a nonzero scalar $q\\in K^\\times$. On each constituent\n$W_i$, supply an automorphism $F_i$ and a nilpotent endomorphism $N_i$ with\n$F_iN_iF_i^{-1}=qN_i$. The scalar $q$ is common to this finite family.\nThe componentwise operators $F_m,N_m$ on $U_m$ satisfy the same relation;\n$F_m$ is invertible and $N_m$ is nilpotent. The split maps constructed above\nintertwine both operators. The proof first identifies the independently defined\nkernel through an equivariant linear isomorphism\n\\[\n \\ker N_m\\simeq\\prod_{i\\in I}\\prod_{k<m_i}\\ker N_i.\n\\]\nThe relation, with $q\\ne0$, proves that both $F_i$ and its inverse preserve\n$\\ker N_i$. Thus $F_i|_{\\ker N_i}$ is an actual automorphism, and Lean proves\n\\[\n \\det(XI-F_m|_{\\ker N_m})\n   =\\prod_{i\\in I}\\det(XI-F_i|_{\\ker N_i})^{m_i}.\n\\]\nThe corresponding kernel dimensions and traces are additive with multiplicity.\nCommon constituent kernel polynomials, when supplied through a coefficient-ring\nmap, give the mapped product polynomial for t","\\paragraph{Algebraic monodromy data on the selected copies.}\nFix a field $K$ and a nonzero scalar $q\\in K^\\times$. On each constituent\n$W_i$, supply an automorphism $F_i$ and a nilpotent endomorphism $N_i$ with\n$F_iN_iF_i^{-1}=qN_i$. The scalar $q$ is common to this finite family.\nThe componentwise operators $F_m,N_m$ on $U_m$ satisfy the same relation;\n$F_m$ is invertible and $N_m$ is nilpotent. The split maps constructed above\nintertwine both operators. The proof first identifies the independently defined\nkernel through an equivariant linear isomorphism\n\\[\n \\ker N_m\\simeq\\prod_{i\\in I}\\prod_{k<m_i}\\ker N_i.\n\\]\nThe relation, with $q\\ne0$, proves that both $F_i$ and its inverse preserve\n$\\ker N_i$. Thus $F_i|_{\\ker N_i}$ is an actual automorphism, and Lean proves\n\\[\n \\det(XI-F_m|_{\\ker N_m})\n   =\\prod_{i\\in I}\\det(XI-F_i|_{\\ker N_i})^{m_i}.\n\\]\nThe corresponding kernel dimensions and traces are additive with multiplicity.\nCommon constituent kernel polynomials, when supplied through a coefficient-ring\nmap, give the mapped product polynomial for the summand. Whole-space","$F_i$ and a nilpotent endomorphism $N_i$ with\n$F_iN_iF_i^{-1}=qN_i$. The scalar $q$ is common to this finite family.\nThe componentwise operators $F_m,N_m$ on $U_m$ satisfy the same relation;\n$F_m$ is invertible and $N_m$ is nilpotent. The split maps constructed above\nintertwine both operators. The proof first identifies the independently defined\nkernel through an equivariant linear isomorphism\n\\[\n \\ker N_m\\simeq\\prod_{i\\in I}\\prod_{k<m_i}\\ker N_i.\n\\]\nThe relation, with $q\\ne0$, proves that both $F_i$ and its inverse preserve\n$\\ker N_i$. Thus $F_i|_{\\ker N_i}$ is an actual automorphism, and Lean proves\n\\[\n \\det(XI-F_m|_{\\ker N_m})\n   =\\prod_{i\\in I}\\det(XI-F_i|_{\\ker N_i})^{m_i}.\n\\]\nThe corresponding kernel dimensions and traces are additive with multiplicity.\nCommon constituent kernel polynomials, when supplied through a coefficient-ring\nmap, give the mapped product polynomial for the summand. Whole-space Frobenius\npolynomials are not substituted for these kernel polynomials.","alar $q$ is common to this finite family.\nThe componentwise operators $F_m,N_m$ on $U_m$ satisfy the same relation;\n$F_m$ is invertible and $N_m$ is nilpotent. The split maps constructed above\nintertwine both operators. The proof first identifies the independently defined\nkernel through an equivariant linear isomorphism\n\\[\n \\ker N_m\\simeq\\prod_{i\\in I}\\prod_{k<m_i}\\ker N_i.\n\\]\nThe relation, with $q\\ne0$, proves that both $F_i$ and its inverse preserve\n$\\ker N_i$. Thus $F_i|_{\\ker N_i}$ is an actual automorphism, and Lean proves\n\\[\n \\det(XI-F_m|_{\\ker N_m})\n   =\\prod_{i\\in I}\\det(XI-F_i|_{\\ker N_i})^{m_i}.\n\\]\nThe corresponding kernel dimensions and traces are additive with multiplicity.\nCommon constituent kernel polynomials, when supplied through a coefficient-ring\nmap, give the mapped product polynomial for the summand. Whole-space Frobenius\npolynomials are not substituted for these kernel polynomials.","finite family.\nThe componentwise operators $F_m,N_m$ on $U_m$ satisfy the same relation;\n$F_m$ is invertible and $N_m$ is nilpotent. The split maps constructed above\nintertwine both operators. The proof first identifies the independently defined\nkernel through an equivariant linear isomorphism\n\\[\n \\ker N_m\\simeq\\prod_{i\\in I}\\prod_{k<m_i}\\ker N_i.\n\\]\nThe relation, with $q\\ne0$, proves that both $F_i$ and its inverse preserve\n$\\ker N_i$. Thus $F_i|_{\\ker N_i}$ is an actual automorphism, and Lean proves\n\\[\n \\det(XI-F_m|_{\\ker N_m})\n   =\\prod_{i\\in I}\\det(XI-F_i|_{\\ker N_i})^{m_i}.\n\\]\nThe corresponding kernel dimensions and traces are additive with multiplicity.\nCommon constituent kernel polynomials, when supplied through a coefficient-ring\nmap, give the mapped product polynomial for the summand. Whole-space Frobenius\npolynomials are not substituted for these kernel polynomials.","se operators $F_m,N_m$ on $U_m$ satisfy the same relation;\n$F_m$ is invertible and $N_m$ is nilpotent. The split maps constructed above\nintertwine both operators. The proof first identifies the independently defined\nkernel through an equivariant linear isomorphism\n\\[\n \\ker N_m\\simeq\\prod_{i\\in I}\\prod_{k<m_i}\\ker N_i.\n\\]\nThe relation, with $q\\ne0$, proves that both $F_i$ and its inverse preserve\n$\\ker N_i$. Thus $F_i|_{\\ker N_i}$ is an actual automorphism, and Lean proves\n\\[\n \\det(XI-F_m|_{\\ker N_m})\n   =\\prod_{i\\in I}\\det(XI-F_i|_{\\ker N_i})^{m_i}.\n\\]\nThe corresponding kernel dimensions and traces are additive with multiplicity.\nCommon constituent kernel polynomials, when supplied through a coefficient-ring\nmap, give the mapped product polynomial for the summand. Whole-space Frobenius\npolynomials are not substituted for these kernel polynomials.",",N_m$ on $U_m$ satisfy the same relation;\n$F_m$ is invertible and $N_m$ is nilpotent. The split maps constructed above\nintertwine both operators. The proof first identifies the independently defined\nkernel through an equivariant linear isomorphism\n\\[\n \\ker N_m\\simeq\\prod_{i\\in I}\\prod_{k<m_i}\\ker N_i.\n\\]\nThe relation, with $q\\ne0$, proves that both $F_i$ and its inverse preserve\n$\\ker N_i$. Thus $F_i|_{\\ker N_i}$ is an actual automorphism, and Lean proves\n\\[\n \\det(XI-F_m|_{\\ker N_m})\n   =\\prod_{i\\in I}\\det(XI-F_i|_{\\ker N_i})^{m_i}.\n\\]\nThe corresponding kernel dimensions and traces are additive with multiplicity.\nCommon constituent kernel polynomials, when supplied through a coefficient-ring\nmap, give the mapped product polynomial for the summand. Whole-space Frobenius\npolynomials are not substituted for these kernel polynomials.","and $N_m$ is nilpotent. The split maps constructed above\nintertwine both operators. The proof first identifies the independently defined\nkernel through an equivariant linear isomorphism\n\\[\n \\ker N_m\\simeq\\prod_{i\\in I}\\prod_{k<m_i}\\ker N_i.\n\\]\nThe relation, with $q\\ne0$, proves that both $F_i$ and its inverse preserve\n$\\ker N_i$. Thus $F_i|_{\\ker N_i}$ is an actual automorphism, and Lean proves\n\\[\n \\det(XI-F_m|_{\\ker N_m})\n   =\\prod_{i\\in I}\\det(XI-F_i|_{\\ker N_i})^{m_i}.\n\\]\nThe corresponding kernel dimensions and traces are additive with multiplicity.\nCommon constituent kernel polynomials, when supplied through a coefficient-ring\nmap, give the mapped product polynomial for the summand. Whole-space Frobenius\npolynomials are not substituted for these kernel polynomials.","The exact special chain has basis $e_0,\\ldots,e_{n-1}$ and operators\n\\[\n Ne_0=0,\\qquad Ne_j=e_{j-1}\\ (j>0),\\qquad Fe_j=aq^{-j}e_j,\n \\qquad a,q\\in K^\\times.\n\\]\nLean proves $N^n=0$, and for $n>0$ also $N^{n-1}\\ne0$ and\n$\\ker N=Ke_0$. Consequently its kernel polynomial is $X-a$, whereas its\nwhole-space polynomial is $\\prod_{j=0}^{n-1}(X-aq^{-j})$.\nThe zero-length chain is included as a zero-dimensional space; the assertion\n$\\ker N=Ke_0$ is made only for positive length.","as basis $e_0,\\ldots,e_{n-1}$ and operators\n\\[\n Ne_0=0,\\qquad Ne_j=e_{j-1}\\ (j>0),\\qquad Fe_j=aq^{-j}e_j,\n \\qquad a,q\\in K^\\times.\n\\]\nLean proves $N^n=0$, and for $n>0$ also $N^{n-1}\\ne0$ and\n$\\ker N=Ke_0$. Consequently its kernel polynomial is $X-a$, whereas its\nwhole-space polynomial is $\\prod_{j=0}^{n-1}(X-aq^{-j})$.\nThe zero-length chain is included as a zero-dimensional space; the assertion\n$\\ker N=Ke_0$ is made only for positive length.","For $n>1$, replacing $N$ by zero retains exactly the same $F$ and the same\nwhole-space characteristic polynomial, while changing the degree of the kernel\npolynomial from one to $n$. Both pairs still satisfy the algebraic monodromy\nrelation. This Lean-checked example shows that whole Frobenius data alone do\nnot determine the monodromy kernel polynomial. It is not asserted to be a\ncounterexample among the intended strictly compatible arithmetic systems.","These statements concern supplied algebraic pairs. No Weil-group or inertia\naction, Hodge--Tate object, arithmetic comparison, or realization of the pair\nfrom a Galois representation is constructed here. The polynomial convention is\n$\\det(XI-F|_{\\ker N})$; no identification with an inertia-dependent local\nEuler factor is asserted. Claim 9.2 remains partial at its full arithmetic scope.","\\paragraph{Checked group, inertia and monodromy finite sums (Block 12).}\nLet $K$ be a field, $G$ a group, and $\\nu:G\\to K^\\times$ a character.\nThe new datum consists of an actual representation $r$ and nilpotent $N$ with\n$r(g)N=\\nu(g)Nr(g)$ for every $g$. For a supplied subgroup $H$, define\n\\[\n E(D,H)=\\ker N\\cap V^H.\n\\]\nThis is the intersection of independently defined kernel and actual invariant\nsubspaces. For normal $H$, the restricted whole-group representation and its\ninverse automorphisms are constructed. A change from $\\varphi$ to $\\varphi h$,\nwith $h\\in H$, leaves the restricted operator itself unchanged.","p, inertia and monodromy finite sums (Block 12).}\nLet $K$ be a field, $G$ a group, and $\\nu:G\\to K^\\times$ a character.\nThe new datum consists of an actual representation $r$ and nilpotent $N$ with\n$r(g)N=\\nu(g)Nr(g)$ for every $g$. For a supplied subgroup $H$, define\n\\[\n E(D,H)=\\ker N\\cap V^H.\n\\]\nThis is the intersection of independently defined kernel and actual invariant\nsubspaces. For normal $H$, the restricted whole-group representation and its\ninverse automorphisms are constructed. A change from $\\varphi$ to $\\varphi h$,\nwith $h\\in H$, leaves the restricted operator itself unchanged.","character.\nThe new datum consists of an actual representation $r$ and nilpotent $N$ with\n$r(g)N=\\nu(g)Nr(g)$ for every $g$. For a supplied subgroup $H$, define\n\\[\n E(D,H)=\\ker N\\cap V^H.\n\\]\nThis is the intersection of independently defined kernel and actual invariant\nsubspaces. For normal $H$, the restricted whole-group representation and its\ninverse automorphisms are constructed. A change from $\\varphi$ to $\\varphi h$,\nwith $h\\in H$, leaves the restricted operator itself unchanged.","he new datum consists of an actual representation $r$ and nilpotent $N$ with\n$r(g)N=\\nu(g)Nr(g)$ for every $g$. For a supplied subgroup $H$, define\n\\[\n E(D,H)=\\ker N\\cap V^H.\n\\]\nThis is the intersection of independently defined kernel and actual invariant\nsubspaces. For normal $H$, the restricted whole-group representation and its\ninverse automorphisms are constructed. A change from $\\varphi$ to $\\varphi h$,\nwith $h\\in H$, leaves the restricted operator itself unchanged.","sts of an actual representation $r$ and nilpotent $N$ with\n$r(g)N=\\nu(g)Nr(g)$ for every $g$. For a supplied subgroup $H$, define\n\\[\n E(D,H)=\\ker N\\cap V^H.\n\\]\nThis is the intersection of independently defined kernel and actual invariant\nsubspaces. For normal $H$, the restricted whole-group representation and its\ninverse automorphisms are constructed. A change from $\\varphi$ to $\\varphi h$,\nwith $h\\in H$, leaves the restricted operator itself unchanged.","For a finite family of finite-dimensional constituents and natural multiplicities,\nthe actual local space of the constructed sum is linearly equivalent to the sum of the constituent local spaces, equivariantly\nfor every group element. Its characteristic polynomials multiply, its dimensions\nand traces add, and maps intertwining both the group and $N$ restrict to the\nlocal spaces. The explicit bounded-copy inclusion and retraction still split\nthere. Finite inertia image and an open inertia kernel are preserved.","Writing $Q_D(T)$ for the reversed local characteristic polynomial, Lean proves\n\\[\n Q_D(T)=\\det(1-T r(\\varphi)\\mid E(D,H)),\\qquad Q_D(0)=1,\n \\qquad\n Q_{\\bigoplus_iD_i^{\\oplus m_i}}(T)=\\prod_iQ_{D_i}(T)^{m_i}.\n\\]\nThe abstract formulas do not need characteristic zero. Normality and finite\ndimension appear explicitly where used. Arithmetic identification of $G,H,\\nu$\nand $\\varphi$ remains separate. In the geometric-Frobenius convention,\n$\\nu(\\varphi)$ is the inverse residue cardinality; it is this unit, rather than\nthe residue cardinality, that enters the earlier special-chain parameter.","l characteristic polynomial, Lean proves\n\\[\n Q_D(T)=\\det(1-T r(\\varphi)\\mid E(D,H)),\\qquad Q_D(0)=1,\n \\qquad\n Q_{\\bigoplus_iD_i^{\\oplus m_i}}(T)=\\prod_iQ_{D_i}(T)^{m_i}.\n\\]\nThe abstract formulas do not need characteristic zero. Normality and finite\ndimension appear explicitly where used. Arithmetic identification of $G,H,\\nu$\nand $\\varphi$ remains separate. In the geometric-Frobenius convention,\n$\\nu(\\varphi)$ is the inverse residue cardinality; it is this unit, rather than\nthe residue cardinality, that enters the earlier special-chain parameter.","An explicit action of $C_2\\times\\mathbb Z$ has $\\nu(\\varphi)=1/3$,\n$F=\\operatorname{diag}(2,6,18)$ and $Ne_1=e_0$, $Ne_2=e_1$, $Ne_0=0$.\nTrivial inertia gives local dimension one and denominator $1-2T$; scalar sign\ninertia gives dimension zero and denominator one. The entire $F$ and $N$ are\nequal in these two examples, and $N^2\\ne0$. Thus inertia cannot be omitted.\nConversely, unramified chains of lengths one and three have equal Euler\ndenominators while only the first has $N=0$. Euler factors therefore do not\nidentify complete monodromy data.","name{diag}(2,6,18)$ and $Ne_1=e_0$, $Ne_2=e_1$, $Ne_0=0$.\nTrivial inertia gives local dimension one and denominator $1-2T$; scalar sign\ninertia gives dimension zero and denominator one. The entire $F$ and $N$ are\nequal in these two examples, and $N^2\\ne0$. Thus inertia cannot be omitted.\nConversely, unramified chains of lengths one and three have equal Euler\ndenominators while only the first has $N=0$. Euler factors therefore do not\nidentify complete monodromy data.","The remaining arithmetic comparison is\n$\\operatorname{WD}(\\bigoplus_i V_i^{\\oplus m_i})\\cong\n\\bigoplus_i\\operatorname{WD}(V_i)^{\\oplus m_i}$ for the actual attached functor,\nintertwining the Weil action and monodromy, together with the intended coefficient\nand Hodge--Tate comparisons. These are not supplied by the abstract datum.\nClaim 9.2 remains partial at its full arithmetic scope.","Recovered multiplicities determine the semisimple isomorphism type relative\nto the specified constituents, not a canonical embedding or projector inside\na repeated isotypic component. For example, the two projectors\n$\\operatorname{diag}(1,0)$ and $\\operatorname{diag}(0,1)$ on a two-dimensional\ntrivial representation are distinct equivariant idempotents with trace one.","\\begin{numericalcertificate}[Multiplicity recovery]\\label{px:9.3}\nThe original companion matrix was not retrieved or rerun. A new exact\nrealization uses the trivial character and two independent sign characters\nof $(\\mathbf Z/2\\mathbf Z)^2$, evaluated at $(0,0),(1,0),(0,1)$:\n\\[\n C=\\begin{pmatrix}1&1&1","\\begin{numericalcertificate}[Multiplicity recovery]\\label{px:9.3}\nThe original companion matrix was not retrieved or rerun. A new exact\nrealization uses the trivial character and two independent sign characters\nof $(\\mathbf Z/2\\mathbf Z)^2$, evaluated at $(0,0),(1,0),(0,1)$:\n\\[\n C=\\begin{pmatrix}1&1&1\\\\1&-1&1\\\\1&1&-1\\end{pmatrix},\n \\qquad \\det C=4,\n \\qquad C\\begin{pmatrix}2\\\\0\\\\1\\end{pmatrix}\n       =\\begin{pmatrix}3\\\\3\\\\1\\end{pmatrix}.\n\\]\nLean checks that these are the traces of actual one-dimensional\nrepresentations, recovers $(2,0,1)$ uniquely, rejects a target whose unique\nsolution is $(1/2,0,0)$, and rejects $(4,0,1)$ under capacities $(3,2,2)$.\nThe separator becomes singular modulo two. These are finite-image\nrepresentation checks, not fabricated Frobenius data for an arithmetic carrier.\n\\end{numericalcertificate}","1\\end{pmatrix}.\n\\]\nLean checks that these are the traces of actual one-dimensional\nrepresentations, recovers $(2,0,1)$ uniquely, rejects a target whose unique\nsolution is $(1/2,0,0)$, and rejects $(4,0,1)$ under capacities $(3,2,2)$.\nThe separator becomes singular modulo two. These are finite-image\nrepresentation checks, not fabricated Frobenius data for an arithmetic carrier.\n\\end{numericalcertificate}","\\paragraph{Finite-exponential reconstruction and a cubic inverse (Block 13).}\nLet $K$ be a characteristic-zero field and let $(r,N)$ be supplied group data\nwith nilpotent $N$ and $r(g)N=\\nu(g)Nr(g)$. For a supplied scalar map satisfying\n\\[\nt(gh)=t(g)+\\nu(g)t(h),\n\\]\nthe formalization constructs\n\\[\n\\rho(g)=\\exp(t(g)N)r(g),\\qquad\n\\exp(aN)=\\sum_{j=0}^{d-1}\\frac{a^j}{j!}N^j\\quad(N^d=0).\n\\]\nLean proves the representation laws, actual finite-copy and split-map\ncompatibility, and joint continuity from continuity of $N$, $r$ and $t$ with\nthe stated continuous vector-space operations. The monodromy fixed kernel\nintersected with subgroup invariants is unchanged by this twist. For normal\ninertia, its restricted action and finite-dimensional Euler denominator agree.\nTwisting again by $-","\\paragraph{Finite-exponential reconstruction and a cubic inverse (Block 13).}\nLet $K$ be a characteristic-zero field and let $(r,N)$ be supplied group data\nwith nilpotent $N$ and $r(g)N=\\nu(g)Nr(g)$. For a supplied scalar map satisfying\n\\[\nt(gh)=t(g)+\\nu(g)t(h),\n\\]\nthe formalization constructs\n\\[\n\\rho(g)=\\exp(t(g)N)r(g),\\qquad\n\\exp(aN)=\\sum_{j=0}^{d-1}\\frac{a^j}{j!}N^j\\quad(N^d=0).\n\\]\nLean proves the representation laws, actual finite-copy and split-map\ncompatibility, and joint continuity from continuity of $N$, $r$ and $t$ with\nthe stated continuous vector-space operations. The monodromy fixed kernel\nintersected with subgroup invariants is unchanged by this twist. For normal\ninertia, its restricted action and finite-dimensional Euler denominator agree.\nTwisting again by $-t$ recovers the datum with $N$ still supplied.","ld and let $(r,N)$ be supplied group data\nwith nilpotent $N$ and $r(g)N=\\nu(g)Nr(g)$. For a supplied scalar map satisfying\n\\[\nt(gh)=t(g)+\\nu(g)t(h),\n\\]\nthe formalization constructs\n\\[\n\\rho(g)=\\exp(t(g)N)r(g),\\qquad\n\\exp(aN)=\\sum_{j=0}^{d-1}\\frac{a^j}{j!}N^j\\quad(N^d=0).\n\\]\nLean proves the representation laws, actual finite-copy and split-map\ncompatibility, and joint continuity from continuity of $N$, $r$ and $t$ with\nthe stated continuous vector-space operations. The monodromy fixed kernel\nintersected with subgroup invariants is unchanged by this twist. For normal\ninertia, its restricted action and finite-dimensional Euler denominator agree.\nTwisting again by $-t$ recovers the datum with $N$ still supplied.","d let $(r,N)$ be supplied group data\nwith nilpotent $N$ and $r(g)N=\\nu(g)Nr(g)$. For a supplied scalar map satisfying\n\\[\nt(gh)=t(g)+\\nu(g)t(h),\n\\]\nthe formalization constructs\n\\[\n\\rho(g)=\\exp(t(g)N)r(g),\\qquad\n\\exp(aN)=\\sum_{j=0}^{d-1}\\frac{a^j}{j!}N^j\\quad(N^d=0).\n\\]\nLean proves the representation laws, actual finite-copy and split-map\ncompatibility, and joint continuity from continuity of $N$, $r$ and $t$ with\nthe stated continuous vector-space operations. The monodromy fixed kernel\nintersected with subgroup invariants is unchanged by this twist. For normal\ninertia, its restricted action and finite-dimensional Euler denominator agree.\nTwisting again by $-t$ recovers the datum with $N$ still supplied.","r,N)$ be supplied group data\nwith nilpotent $N$ and $r(g)N=\\nu(g)Nr(g)$. For a supplied scalar map satisfying\n\\[\nt(gh)=t(g)+\\nu(g)t(h),\n\\]\nthe formalization constructs\n\\[\n\\rho(g)=\\exp(t(g)N)r(g),\\qquad\n\\exp(aN)=\\sum_{j=0}^{d-1}\\frac{a^j}{j!}N^j\\quad(N^d=0).\n\\]\nLean proves the representation laws, actual finite-copy and split-map\ncompatibility, and joint continuity from continuity of $N$, $r$ and $t$ with\nthe stated continuous vector-space operations. The monodromy fixed kernel\nintersected with subgroup invariants is unchanged by this twist. For normal\ninertia, its restricted action and finite-dimensional Euler denominator agree.\nTwisting again by $-t$ recovers the datum with $N$ still supplied.",".\n\\]\nLean proves the representation laws, actual finite-copy and split-map\ncompatibility, and joint continuity from continuity of $N$, $r$ and $t$ with\nthe stated continuous vector-space operations. The monodromy fixed kernel\nintersected with subgroup invariants is unchanged by this twist. For normal\ninertia, its restricted action and finite-dimensional Euler denominator agree.\nTwisting again by $-t$ recovers the datum with $N$ still supplied.","tation laws, actual finite-copy and split-map\ncompatibility, and joint continuity from continuity of $N$, $r$ and $t$ with\nthe stated continuous vector-space operations. The monodromy fixed kernel\nintersected with subgroup invariants is unchanged by this twist. For normal\ninertia, its restricted action and finite-dimensional Euler denominator agree.\nTwisting again by $-t$ recovers the datum with $N$ still supplied.","Block 13 also established an extracted inverse on the explicit class $N^3=0$:\n\\[\n\\log_2 U=(U-I)-\\tfrac12(U-I)^2,\\qquad\n a^{-1}\\log_2(\\exp(aN))=N\\quad(a\\ne0).\n\\]\nBoth exponential/logarithmic inverse identities are checked on the cubic\nnilpotent and unipotent loci. If a supplied common sample $u$ has\n$r_D(u)=r_E(u)=I$ and $t(u)\\ne0$, an intertwiner of the reconstructed cubic\nactions recovers both original group equivariance and monodromy equivariance.\nThe general nilpotence exponential theorem is not restricted to this cubic\nclass, but the extracted inverse in Block 13 was. Block 20 below removes\nthat restriction by using the full finite logarithm.","The rational affine-group example has $u=(1,1)$, $d=(0,1/3)$ and\n\\[\n\\rho(u)=\\begin{pmatrix}1&1&1/2","The rational affine-group example has $u=(1,1)$, $d=(0,1/3)$ and\n\\[\n\\rho(u)=\\begin{pmatrix}1&1&1/2\\\\0&1&1\\\\0&0&1\\end{pmatrix},\n\\qquad \\rho(d)=\\operatorname{diag}(1,3,9).\n\\]\nReversing the factor order or dropping the quadratic term gives checked\nfailures. For the length-four chain, the quadratic logarithm has erroneous\n$e_3$-to-$e_0$ coefficient $-1/3$, so its cubic restriction is substantive.\nA zero-parameter sample also fails to recover $N$.","0&0&1\\end{pmatrix},\n\\qquad \\rho(d)=\\operatorname{diag}(1,3,9).\n\\]\nReversing the factor order or dropping the quadratic term gives checked\nfailures. For the length-four chain, the quadratic logarithm has erroneous\n$e_3$-to-$e_0$ coefficient $-1/3$, so its cubic restriction is substantive.\nA zero-parameter sample also fails to recover $N$.","No arithmetic group, tame cocycle or required sample is constructed by these\nstatements. The attached comparison for every intended representation,\ncoefficient comparisons and Hodge--Tate/period data remain separate. Finite\ninertia image and an open inertia kernel of $r$ are not asserted to persist\nunder this twist to $\\rho$. Claim 9.2 remains partial.","Let \\(E\\) be a field of characteristic zero and let \\(T_1,\\ldots,T_N\\) be commuting operators on a finite-dimensional \\(E\\)-vector space \\(A\\), each diagonalizable over \\(E\\).  Thus the selected spectra split over the coefficient field.  Let \\(a_i\\in E\\) be a selected target scalar for \\(T_i\\), possibly absent from its spectrum.  Define\n\\[\ne_i\n=\n\\prod_{\\mu\\in\\Spec(T_i)\\setminus\\{a_i\\}}\n\\frac{T_i-\\mu I}{a_i-\\mu},\n\\qquad\ne_T=\\prod_{i=1}^{N}e_i,\n\\]\nwith \\(e_i=0\\) if \\(a_i\\) is absent.",",\\ldots,T_N\\) be commuting operators on a finite-dimensional \\(E\\)-vector space \\(A\\), each diagonalizable over \\(E\\).  Thus the selected spectra split over the coefficient field.  Let \\(a_i\\in E\\) be a selected target scalar for \\(T_i\\), possibly absent from its spectrum.  Define\n\\[\ne_i\n=\n\\prod_{\\mu\\in\\Spec(T_i)\\setminus\\{a_i\\}}\n\\frac{T_i-\\mu I}{a_i-\\mu},\n\\qquad\ne_T=\\prod_{i=1}^{N}e_i,\n\\]\nwith \\(e_i=0\\) if \\(a_i\\) is absent.","be commuting operators on a finite-dimensional \\(E\\)-vector space \\(A\\), each diagonalizable over \\(E\\).  Thus the selected spectra split over the coefficient field.  Let \\(a_i\\in E\\) be a selected target scalar for \\(T_i\\), possibly absent from its spectrum.  Define\n\\[\ne_i\n=\n\\prod_{\\mu\\in\\Spec(T_i)\\setminus\\{a_i\\}}\n\\frac{T_i-\\mu I}{a_i-\\mu},\n\\qquad\ne_T=\\prod_{i=1}^{N}e_i,\n\\]\nwith \\(e_i=0\\) if \\(a_i\\) is absent.","\\begin{theorem}[Exact finite occupancy trace]\\label{px:10.1}\nThe operator \\(e_T\\) is the projector onto the joint eigenspace\n\\[\nA_T\n=\n\\{v:T_iv=a_iv\\text{ for every }i\\}.\n\\]\nMoreover,\n\\[\n\\boxed{\n\\Tr(e_T)=\\rank(e_T)=\\dim A_T\\in\\Z_{\\ge0}.\n}\n\\]\nThe trace is identified with its unique nonnegative integer preimage under \\(\\Z\\hookrightarrow E\\).  Thus \\(\\Tr(e_T)>0\\) refers to that integer, not to an ordering of an arbitrary field \\(E\\).  The finite packet is occupied exactly when this integer is positive.  In positive characteristic the field-valued trace can vanish for a nonzero projector; the rank test, not a positive field trace, must then be used.\n\\end{theorem}","occupancy trace]\\label{px:10.1}\nThe operator \\(e_T\\) is the projector onto the joint eigenspace\n\\[\nA_T\n=\n\\{v:T_iv=a_iv\\text{ for every }i\\}.\n\\]\nMoreover,\n\\[\n\\boxed{\n\\Tr(e_T)=\\rank(e_T)=\\dim A_T\\in\\Z_{\\ge0}.\n}\n\\]\nThe trace is identified with its unique nonnegative integer preimage under \\(\\Z\\hookrightarrow E\\).  Thus \\(\\Tr(e_T)>0\\) refers to that integer, not to an ordering of an arbitrary field \\(E\\).  The finite packet is occupied exactly when this integer is positive.  In positive characteristic the field-valued trace can vanish for a nonzero projector; the rank test, not a positive field trace, must then be used.\n\\end{theorem}","\\begin{proof}\nFor each operator, its finite set of actual eigenvalues is used without\nmultiplicity. The polynomial\n\\[\n p_{i,a}(X)=\\prod_{\\mu\\in\\Spec(T_i)\\setminus\\{a\\}}\n (a-\\mu)^{-1}(X-\\mu)\n\\]\nis used when $a$ occurs, and the zero polynomial otherwise. For an eigenvector\n$T_i v=bv$, evaluation gives\n\\[\n p_{i,a}(T_i)v=\\begin{cases}v&b=a,","\\begin{proof}\nFor each operator, its finite set of actual eigenvalues is used without\nmultiplicity. The polynomial\n\\[\n p_{i,a}(X)=\\prod_{\\mu\\in\\Spec(T_i)\\setminus\\{a\\}}\n (a-\\mu)^{-1}(X-\\mu)\n\\]\nis used when $a$ occurs, and the zero polynomial otherwise. For an eigenvector\n$T_i v=bv$, evaluation gives\n\\[\n p_{i,a}(T_i)v=\\begin{cases}v&b=a,\\\\0&b\\ne a.\\end{cases}\n\\]\nBecause the eigenspaces span $A$, this proves that $e_i=p_{i,a_i}(T_i)$ maps\nall of $A$ into the selected eigenspace and fixes that eigenspace pointwise.\nIt therefore proves the individual projector and image assertions, including\nan absent target.","0&b\\ne a.\\end{cases}\n\\]\nBecause the eigenspaces span $A$, this proves that $e_i=p_{i,a_i}(T_i)$ maps\nall of $A$ into the selected eigenspace and fixes that eigenspace pointwise.\nIt therefore proves the individual projector and image assertions, including\nan absent target.","Polynomials in commuting operators commute. If projections $P,Q$ onto\n$U,W$ commute, then $PQ$ maps into both $U$ and $W$, and fixes every vector\nin $U\\cap W$. Induction proves that the actual ordered product $e_T$ is a\nprojection onto $A_T$; commutation also makes the result independent of the\nordering. An empty list of conditions gives the identity.","For a projection onto a subspace, the direct sum of its image and kernel\nidentifies it with $I\\oplus0$. Thus its trace in $E$ is the natural dimension\nof its image cast into $E$. In characteristic zero this cast is injective,\nso there is a unique natural integer represented by the trace. That integer\nis positive exactly when the joint eigenspace contains a nonzero vector.\n\\end{proof}","\\paragraph{Naturality and finite hypotheses.}\nIf $F T_i=U_i F$ and the source operators split, then\n$F e_{T,a}=e_{U,a}F$. This follows first on source eigenvectors and then on\ntheir span. In particular, a change of basis conjugates the constructed\nprojector and preserves its trace. Adding conditions decreases the actual\njoint eigenspace and cannot increase its dimension. These statements do not\nassume an orthonormal basis or a positive definite inner product.","\\paragraph{A specified finite realization.}\nOn $\\Q^4$, take the diagonal operators with diagonal lists\n\\[\n(1,1,1,0),\\quad(2,2,0,0),\\quad(3,0,0,0),\\quad(0,4,4,4)\n\\]\nand target scalars $(1,2,3,4)$. All four targets occur individually. Imposing\nthese conditions successively gives actual projector ranks $3,2,1,0$.\nThe last packet is empty despite individual feasibility. The formal code\nchecks these operators, projector formulas, traces and dimensions. This is\na newly specified realization, not a claim to have recovered the original\ncompanion implementation.","\\paragraph{Why the hypotheses cannot be omitted.}\nFor\n\\[\n P=\\begin{pmatrix}1&0\\\\0&0\\end{pmatrix},\\qquad\n Q=\\tfrac12\\begin{pmatrix}1&1\\\\1&1\\end{pmatrix},\n\\]\neach operator is idempotent, but $PQ$ is not idempotent and has trace $1/2$.\nFor the Jordan block $J=\\left(\\begin{smallmatrix}1&1","1&1\\end{pmatrix},\n\\]\neach operator is idempotent, but $PQ$ is not idempotent and has trace $1/2$.\nFor the Jordan block $J=\\left(\\begin{smallmatrix}1&1","1&1\\end{pmatrix},\n\\]\neach operator is idempotent, but $PQ$ is not idempotent and has trace $1/2$.\nFor the Jordan block $J=\\left(\\begin{smallmatrix}1&1\\\\0&1\\end{smallmatrix}\\right)$,\nthe singleton-spectrum Lagrange polynomial is $1$, so the formula produces\nthe identity although the eigenspace at $1$ is only a line. Both failures\nare exact formal counterexamples to dropping the corresponding hypotheses,\nnot counterexamples to the theorem with those hypotheses retained.","0&1\\end{smallmatrix}\\right)$,\nthe singleton-spectrum Lagrange polynomial is $1$, so the formula produces\nthe identity although the eigenspace at $1$ is only a line. Both failures\nare exact formal counterexamples to dropping the corresponding hypotheses,\nnot counterexamples to the theorem with those hypotheses retained.","\\begin{figure}[H]\n\\centering\n\\includegraphics[width=0.77\\linewidth]{15_finite_occupancy_ranks.pdf}\n\\caption{A consistent sequence of finite conditions narrows the carrier from ranks \\(3\\) to \\(2\\) to \\(1\\).  One inconsistent condition forces the joint projector to zero.}\n\\end{figure}","\\begin{corollary}[Exact location of the remaining existence problem]\\label{px:11.2}\nInverse-limit reconstruction alone is not a nonemptiness theorem.  The existence requirements must retain both residual companions and liftability through every finite depth:\n\\[\n\\boxed{\n\\begin{aligned}\n\\text{companion existence: }&X_{\\lambda,1}(\\Compat)\\ne\\varnothing\n\\text{ for every }\\lambda;\\\\\n\\text{finite-depth liftability: }&X_{\\lambda,m}(\\Compat)\\ne\\varnothing\n\\text{ for every }\\lambda,m\\ge2;\\\\\n\\text{finite occupancy: }&\\Tr(e_T)>0\n\\text{ for every required global packet.}\n\\end{aligned}}\n\\]\nFormal reconstruction does not create a missing residual root, repair an empty higher level, or force global occupancy.  Nonempty finite levels with finite branching supply a branch, after which inverse-limit closure reconstructs an admissible object.\n\\end{corollary}","\\paragraph{An occupied node can have no child.}\nThe actual order-two representation with generator\n\\[\n A=\\begin{pmatrix}1&2\\\\2&3\\end{pmatrix}\\pmod4\n\\]\nsatisfies $A^2=I$ and $\\det(XI-A)=X^2-1$. It has no lift modulo $8$ with\nthe same characteristic polynomial. Indeed, every lift has entries congruent\nto $1,2,2,3$ modulo $4$. Trace zero then forces its determinant to be $3$\nmodulo $8$, whereas $X^2-1$ requires determinant $-1=7$ modulo $8$.\nBoth the occupied node and the absence of any next-level matrix or representation\nwith the prescribed polynomial are Lean-checked. Other branches of the same\nunrestricted involution forest survive, so this example does not contradict\nfinite-branch compactness.","2&3\\end{pmatrix}\\pmod4\n\\]\nsatisfies $A^2=I$ and $\\det(XI-A)=X^2-1$. It has no lift modulo $8$ with\nthe same characteristic polynomial. Indeed, every lift has entries congruent\nto $1,2,2,3$ modulo $4$. Trace zero then forces its determinant to be $3$\nmodulo $8$, whereas $X^2-1$ requires determinant $-1=7$ modulo $8$.\nBoth the occupied node and the absence of any next-level matrix or representation\nwith the prescribed polynomial are Lean-checked. Other branches of the same\nunrestricted involution forest survive, so this example does not contradict\nfinite-branch compactness.","\\paragraph{Local properties need their own limit argument.}\nOver a domain, for a nonzero $\\pi\\in I$, the representations of the additive\ninteger group given by\n\\[\n \\rho_n(g)=\\begin{pmatrix}1&\\pi^n g","\\paragraph{Local properties need their own limit argument.}\nOver a domain, for a nonzero $\\pi\\in I$, the representations of the additive\ninteger group given by\n\\[\n \\rho_n(g)=\\begin{pmatrix}1&\\pi^n g\\\\0&1\\end{pmatrix}\n\\]\nare all nontrivial and converge pointwise in the ideal-adic topology to the\ntrivial representation. The convergence and the failure of limit preservation\nfor nontriviality are formally checked. Thus not every property of approximating\nintegral representations can be asserted for a limit. This example is not\nclaimed to disprove stability for any particular intended Galois local condition.","For each supplied required packet $p$, let $A_p$ be its finite-dimensional carrier\nover a characteristic-zero field $E_p$, and let $e_p$ be the actual joint spectral\nprojector for its finite list of split commuting conditions. No order on $E_p$\nis assumed. The source notation $\\Tr(e_p)>0$ means that the unique natural number\n$n_p$ with $\\Tr(e_p)=(n_p:E_p)$ is positive. The checked projector formula identifies\n$n_p$ with the dimension of the actual joint eigenspace.","ied required packet $p$, let $A_p$ be its finite-dimensional carrier\nover a characteristic-zero field $E_p$, and let $e_p$ be the actual joint spectral\nprojector for its finite list of split commuting conditions. No order on $E_p$\nis assumed. The source notation $\\Tr(e_p)>0$ means that the unique natural number\n$n_p$ with $\\Tr(e_p)=(n_p:E_p)$ is positive. The checked projector formula identifies\n$n_p$ with the dimension of the actual joint eigenspace.","An admissible realization here includes continuous coefficient-field members and\nmembership of every finite reduction in its designated admissible class.\nUnder the completion hypotheses and these supplied carrier hypotheses, Lean proves\n\\[\n\\begin{aligned}\n&\\exists\\,\\rho,\\ (x_p)_p:\\quad\n  \\rho\\text{ is an admissible realization of the full chosen spine},\\quad\n  0\\ne x_p\\in\\operatorname{im}(e_p)\\text{ for every }p\\\\\n&\\quad\\Longleftrightarrow\\quad\n \\bigl(\\forall\\lambda,\\ X_{\\lambda,1}(\\Compat)\\ne\\varnothing\\bigr)\n \\ \\land\\ \\bigl(\\forall\\lambda,\\ \\forall m\\ge2,\n               X_{\\lambda,m}(\\Compat)\\ne\\varnothing\\bigr)\n \\ \\land\\ \\bigl(\\forall p,\\ n_p>0\\bigr).\n\\end{aligned}\n\\]\nThe packet index may be infinite. Each $x_p$ belongs to its own supplied carrier;\nno common vector across distinct carriers or embedding of the reconstructed\nsystem into those carriers is asserted. Neither an unknown carrier nor positive\ntrace is constructed without the corresponding hypotheses.","&\\quad\\Longleftrightarrow\\quad\n \\bigl(\\forall\\lambda,\\ X_{\\lambda,1}(\\Compat)\\ne\\varnothing\\bigr)\n \\ \\land\\ \\bigl(\\forall\\lambda,\\ \\forall m\\ge2,\n               X_{\\lambda,m}(\\Compat)\\ne\\varnothing\\bigr)\n \\ \\land\\ \\bigl(\\forall p,\\ n_p>0\\bigr).\n\\end{aligned}\n\\]\nThe packet index may be infinite. Each $x_p$ belongs to its own supplied carrier;\nno common vector across distinct carriers or embedding of the reconstructed\nsystem into those carriers is asserted. Neither an unknown carrier nor positive\ntrace is constructed without the corresponding hypotheses.","\\paragraph{Every required packet is necessary.}\nFor each omitted index $N\\in\\mathbb N$, consider one-dimensional rational carriers\nwith the identity operator, requiring eigenvalue $0$ at $N$ and eigenvalue $1$\nat every other index. The actual spectral projector has trace $0$ at $N$ and\ntrace $1$ elsewhere. Lean proves that any predetermined finite sample of packet\nindices can miss an empty required packet: choose $N$ outside that sample.\nThe family depends on the omitted index. This does not assert that one fixed\nfamily passes every finite prefix while failing global occupancy.","equired packet is necessary.}\nFor each omitted index $N\\in\\mathbb N$, consider one-dimensional rational carriers\nwith the identity operator, requiring eigenvalue $0$ at $N$ and eigenvalue $1$\nat every other index. The actual spectral projector has trace $0$ at $N$ and\ntrace $1$ elsewhere. Lean proves that any predetermined finite sample of packet\nindices can miss an empty required packet: choose $N$ outside that sample.\nThe family depends on the omitted index. This does not assert that one fixed\nfamily passes every finite prefix while failing global occupancy.","For \\(q=3\\) and formal dimension \\(d=2\\), the finite levels have cardinalities\n\\[\n1,9,81,729,6561,\\ldots.\n\\]\nThe seed zeta function is\n\\[\nZ(T)=\\frac{T}{1-9T}.\n\\]","For weights \\((1,2,4)\\), the counts are\n\\[\n1,2,8,32,256,2048,\\ldots.\n\\]\nThe first coordinate is active immediately, the second one level later, and the third at depth five.  The exact rational function generated by the code is\n\\[\nZ_{(1,2,4)}(T)\n=\n\\frac{T(1-6T-8T^2-32T^3)}{1-8T}.\n\\]\nThe negative signs in this presentation arise from moving the denominator to \\(1-8T\\); every power-series coefficient is positive.","For \\(x^2=0\\) over a residue field of size three, the counts are\n\\[\n1,3,3,9,9,27,27,81,\\ldots,\n\\]\nand\n\\[\nZ(T)=\\frac{T+3T^2}{1-3T^2}.\n\\]\nThe exact branching dimension is \\(1/2\\).","Recovered carrier multiplicities & \\((2,0,1)\\)","Finite occupancy ranks & \\((3,2,1,0)\\)","The complete automated suite contains \\(28\\) tests.  They cover residual partition, no crossing, inverse-limit digits, weighted and singular zeta coefficients, directed source components, cohomology packets, companion forests, common-spine consistency, finite Frobenius separation, multiplicity recovery, and exact occupancy projectors.","Singular power zeta and dimension & proved & scalar equation \\(x^e=0\\)","The residual seed theory becomes precise once vertical and horizontal generation are separated.  Positive-depth perturbations can be complete within one formal component.  They cannot change the residual root.  The universal arithmetic seed is therefore a forest rather than one branch:\n\\[\n\\boxed{\n\\text{residual roots}\n\\longrightarrow\n\\text{finite deformation levels}\n\\longrightarrow\n\\text{inverse-limit branches}.\n}\n\\]","The residual module partitions coordinates modulo \\(p\\), verifies compatibility\n\\[\nx_{m+1}\\equiv x_m\\pmod{p^m},\n\\]\nand reconstructs the digit \\(a_{m-1}\\) by\n\\[\na_{m-1}=\\left\\lfloor\\frac{x_m}{p^{m-1}}\\right\\rfloor\\pmod p.\n\\]","For canonical representatives $0\\le x_m<p^m$, set $x_0=0$. Compatibility also gives\n$a_{m-1}=(x_m-x_{m-1})/p^{m-1}\\in\\{0,\\ldots,p-1\\}$.\nThe floor in the first formula is essential. For example, the compatible\nbase-five residues $4,19$ give the second digit $3$, not the rational number $19/5$.\nThe executable companion already uses integer floor division. This edition\ncorrects the displayed algorithm, not the implementation.","For an integral polynomial \\(f\\), the finite root forest is obtained by exact enumeration of\n\\[\nf(x)\\equiv0\\pmod{p^m}\n\\]\nand by connecting a child \\(x_{m+1}\\) to its reduction modulo \\(p^m\\).  This is intended for small exact examples, not large deformation rings.","Modular Gaussian elimination computes the ranks of the differentials.  For\n\\[\nC^{i-1}\\xrightarrow{d_{i-1}}C^i\\xrightarrow{d_i}C^{i+1},\n\\]\nthe finite dimension is evaluated as\n\\[\n\\dim H^i\n=\n\\dim C^i-\\rank d_i-\\rank d_{i-1}.\n\\]\nPacket subcomplexes are computed separately and checked against the total.","\\paragraph{Finite-group comparison with precision loss (Block 17).}\nLet $R$ be a commutative domain, $\\Gamma$ a finite group, and $\\rho,\\tau$\nactual integral matrix representations. Suppose $a=|\\Gamma|\\ne0$ in $R$ and\n$a\\mid\\pi^e$. An equivariant matrix modulo $(\\pi)^{e+k}$ lifts to an\nintegral equivariant matrix with the prescribed reduction modulo $(\\pi)^k$.\nIndeed, lift its entries to $A$ and form\n\\[\nS=\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nThe finite equations imply $S-aA=\\pi^{e+k}E$ with $E$ integral. Choose $b$\nwith $\\pi^e=ab$ and put $P=A+b\\pi^kE$. Then $aP=S$, so cancellation of\n$a\\ne0$ proves equivariance. Over a local ring, if $\\pi$ is a nonunit,\n$k>0$ and the supplied comparison is invertible, $P$ is an integral unit.\nNeither completeness nor a finit","\\paragraph{Finite-group comparison with precision loss (Block 17).}\nLet $R$ be a commutative domain, $\\Gamma$ a finite group, and $\\rho,\\tau$\nactual integral matrix representations. Suppose $a=|\\Gamma|\\ne0$ in $R$ and\n$a\\mid\\pi^e$. An equivariant matrix modulo $(\\pi)^{e+k}$ lifts to an\nintegral equivariant matrix with the prescribed reduction modulo $(\\pi)^k$.\nIndeed, lift its entries to $A$ and form\n\\[\nS=\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nThe finite equations imply $S-aA=\\pi^{e+k}E$ with $E$ integral. Choose $b$\nwith $\\pi^e=ab$ and put $P=A+b\\pi^kE$. Then $aP=S$, so cancellation of\n$a\\ne0$ proves equivariance. Over a local ring, if $\\pi$ is a nonunit,\n$k>0$ and the supplied comparison is invertible, $P$ is an integral unit.\nNeither completeness nor a finite residue field is needed for this","\\paragraph{Finite-group comparison with precision loss (Block 17).}\nLet $R$ be a commutative domain, $\\Gamma$ a finite group, and $\\rho,\\tau$\nactual integral matrix representations. Suppose $a=|\\Gamma|\\ne0$ in $R$ and\n$a\\mid\\pi^e$. An equivariant matrix modulo $(\\pi)^{e+k}$ lifts to an\nintegral equivariant matrix with the prescribed reduction modulo $(\\pi)^k$.\nIndeed, lift its entries to $A$ and form\n\\[\nS=\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nThe finite equations imply $S-aA=\\pi^{e+k}E$ with $E$ integral. Choose $b$\nwith $\\pi^e=ab$ and put $P=A+b\\pi^kE$. Then $aP=S$, so cancellation of\n$a\\ne0$ proves equivariance. Over a local ring, if $\\pi$ is a nonunit,\n$k>0$ and the supplied comparison is invertible, $P$ is an integral unit.\nNeither completeness nor a finite residue field is needed for this step.","raph{Finite-group comparison with precision loss (Block 17).}\nLet $R$ be a commutative domain, $\\Gamma$ a finite group, and $\\rho,\\tau$\nactual integral matrix representations. Suppose $a=|\\Gamma|\\ne0$ in $R$ and\n$a\\mid\\pi^e$. An equivariant matrix modulo $(\\pi)^{e+k}$ lifts to an\nintegral equivariant matrix with the prescribed reduction modulo $(\\pi)^k$.\nIndeed, lift its entries to $A$ and form\n\\[\nS=\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nThe finite equations imply $S-aA=\\pi^{e+k}E$ with $E$ integral. Choose $b$\nwith $\\pi^e=ab$ and put $P=A+b\\pi^kE$. Then $aP=S$, so cancellation of\n$a\\ne0$ proves equivariance. Over a local ring, if $\\pi$ is a nonunit,\n$k>0$ and the supplied comparison is invertible, $P$ is an integral unit.\nNeither completeness nor a finite residue field is needed for this step.","a commutative domain, $\\Gamma$ a finite group, and $\\rho,\\tau$\nactual integral matrix representations. Suppose $a=|\\Gamma|\\ne0$ in $R$ and\n$a\\mid\\pi^e$. An equivariant matrix modulo $(\\pi)^{e+k}$ lifts to an\nintegral equivariant matrix with the prescribed reduction modulo $(\\pi)^k$.\nIndeed, lift its entries to $A$ and form\n\\[\nS=\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nThe finite equations imply $S-aA=\\pi^{e+k}E$ with $E$ integral. Choose $b$\nwith $\\pi^e=ab$ and put $P=A+b\\pi^kE$. Then $aP=S$, so cancellation of\n$a\\ne0$ proves equivariance. Over a local ring, if $\\pi$ is a nonunit,\n$k>0$ and the supplied comparison is invertible, $P$ is an integral unit.\nNeither completeness nor a finite residue field is needed for this step.","\\Gamma$ a finite group, and $\\rho,\\tau$\nactual integral matrix representations. Suppose $a=|\\Gamma|\\ne0$ in $R$ and\n$a\\mid\\pi^e$. An equivariant matrix modulo $(\\pi)^{e+k}$ lifts to an\nintegral equivariant matrix with the prescribed reduction modulo $(\\pi)^k$.\nIndeed, lift its entries to $A$ and form\n\\[\nS=\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nThe finite equations imply $S-aA=\\pi^{e+k}E$ with $E$ integral. Choose $b$\nwith $\\pi^e=ab$ and put $P=A+b\\pi^kE$. Then $aP=S$, so cancellation of\n$a\\ne0$ proves equivariance. Over a local ring, if $\\pi$ is a nonunit,\n$k>0$ and the supplied comparison is invertible, $P$ is an integral unit.\nNeither completeness nor a finite residue field is needed for this step.","p, and $\\rho,\\tau$\nactual integral matrix representations. Suppose $a=|\\Gamma|\\ne0$ in $R$ and\n$a\\mid\\pi^e$. An equivariant matrix modulo $(\\pi)^{e+k}$ lifts to an\nintegral equivariant matrix with the prescribed reduction modulo $(\\pi)^k$.\nIndeed, lift its entries to $A$ and form\n\\[\nS=\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nThe finite equations imply $S-aA=\\pi^{e+k}E$ with $E$ integral. Choose $b$\nwith $\\pi^e=ab$ and put $P=A+b\\pi^kE$. Then $aP=S$, so cancellation of\n$a\\ne0$ proves equivariance. Over a local ring, if $\\pi$ is a nonunit,\n$k>0$ and the supplied comparison is invertible, $P$ is an integral unit.\nNeither completeness nor a finite residue field is needed for this step.","rho,\\tau$\nactual integral matrix representations. Suppose $a=|\\Gamma|\\ne0$ in $R$ and\n$a\\mid\\pi^e$. An equivariant matrix modulo $(\\pi)^{e+k}$ lifts to an\nintegral equivariant matrix with the prescribed reduction modulo $(\\pi)^k$.\nIndeed, lift its entries to $A$ and form\n\\[\nS=\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nThe finite equations imply $S-aA=\\pi^{e+k}E$ with $E$ integral. Choose $b$\nwith $\\pi^e=ab$ and put $P=A+b\\pi^kE$. Then $aP=S$, so cancellation of\n$a\\ne0$ proves equivariance. Over a local ring, if $\\pi$ is a nonunit,\n$k>0$ and the supplied comparison is invertible, $P$ is an integral unit.\nNeither completeness nor a finite residue field is needed for this step.","tegral matrix representations. Suppose $a=|\\Gamma|\\ne0$ in $R$ and\n$a\\mid\\pi^e$. An equivariant matrix modulo $(\\pi)^{e+k}$ lifts to an\nintegral equivariant matrix with the prescribed reduction modulo $(\\pi)^k$.\nIndeed, lift its entries to $A$ and form\n\\[\nS=\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nThe finite equations imply $S-aA=\\pi^{e+k}E$ with $E$ integral. Choose $b$\nwith $\\pi^e=ab$ and put $P=A+b\\pi^kE$. Then $aP=S$, so cancellation of\n$a\\ne0$ proves equivariance. Over a local ring, if $\\pi$ is a nonunit,\n$k>0$ and the supplied comparison is invertible, $P$ is an integral unit.\nNeither completeness nor a finite residue field is needed for this step.","entations. Suppose $a=|\\Gamma|\\ne0$ in $R$ and\n$a\\mid\\pi^e$. An equivariant matrix modulo $(\\pi)^{e+k}$ lifts to an\nintegral equivariant matrix with the prescribed reduction modulo $(\\pi)^k$.\nIndeed, lift its entries to $A$ and form\n\\[\nS=\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nThe finite equations imply $S-aA=\\pi^{e+k}E$ with $E$ integral. Choose $b$\nwith $\\pi^e=ab$ and put $P=A+b\\pi^kE$. Then $aP=S$, so cancellation of\n$a\\ne0$ proves equivariance. Over a local ring, if $\\pi$ is a nonunit,\n$k>0$ and the supplied comparison is invertible, $P$ is an integral unit.\nNeither completeness nor a finite residue field is needed for this step.","ma|\\ne0$ in $R$ and\n$a\\mid\\pi^e$. An equivariant matrix modulo $(\\pi)^{e+k}$ lifts to an\nintegral equivariant matrix with the prescribed reduction modulo $(\\pi)^k$.\nIndeed, lift its entries to $A$ and form\n\\[\nS=\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nThe finite equations imply $S-aA=\\pi^{e+k}E$ with $E$ integral. Choose $b$\nwith $\\pi^e=ab$ and put $P=A+b\\pi^kE$. Then $aP=S$, so cancellation of\n$a\\ne0$ proves equivariance. Over a local ring, if $\\pi$ is a nonunit,\n$k>0$ and the supplied comparison is invertible, $P$ is an integral unit.\nNeither completeness nor a finite residue field is needed for this step.","trix modulo $(\\pi)^{e+k}$ lifts to an\nintegral equivariant matrix with the prescribed reduction modulo $(\\pi)^k$.\nIndeed, lift its entries to $A$ and form\n\\[\nS=\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nThe finite equations imply $S-aA=\\pi^{e+k}E$ with $E$ integral. Choose $b$\nwith $\\pi^e=ab$ and put $P=A+b\\pi^kE$. Then $aP=S$, so cancellation of\n$a\\ne0$ proves equivariance. Over a local ring, if $\\pi$ is a nonunit,\n$k>0$ and the supplied comparison is invertible, $P$ is an integral unit.\nNeither completeness nor a finite residue field is needed for this step.","e+k}$ lifts to an\nintegral equivariant matrix with the prescribed reduction modulo $(\\pi)^k$.\nIndeed, lift its entries to $A$ and form\n\\[\nS=\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nThe finite equations imply $S-aA=\\pi^{e+k}E$ with $E$ integral. Choose $b$\nwith $\\pi^e=ab$ and put $P=A+b\\pi^kE$. Then $aP=S$, so cancellation of\n$a\\ne0$ proves equivariance. Over a local ring, if $\\pi$ is a nonunit,\n$k>0$ and the supplied comparison is invertible, $P$ is an integral unit.\nNeither completeness nor a finite residue field is needed for this step.","h the prescribed reduction modulo $(\\pi)^k$.\nIndeed, lift its entries to $A$ and form\n\\[\nS=\\sum_{g\\in\\Gamma}\\rho(g)A\\tau(g^{-1}).\n\\]\nThe finite equations imply $S-aA=\\pi^{e+k}E$ with $E$ integral. Choose $b$\nwith $\\pi^e=ab$ and put $P=A+b\\pi^kE$. Then $aP=S$, so cancellation of\n$a\\ne0$ proves equivariance. Over a local ring, if $\\pi$ is a nonunit,\n$k>0$ and the supplied comparison is invertible, $P$ is an integral unit.\nNeither completeness nor a finite residue field is needed for this step.","For a DVR, $e=v_\\pi(|\\Gamma|)$ is derived from the actual group order.\nThe single precision $e+1$, independent of rank, detects integral\nisomorphism. It makes reduction injective on integral isomorphism classes,\nand finite coefficient quotients give finitely many integral classes in\nfixed rank. This is a formalization of classical finite-group rigidity\n(Maranda, 1953, Theorem~2), with an explicit proof of the comparison's\nprecision loss.","For complete local coefficients with finite quotients, downward-compatible\nallowed representations occupied at every precision, and the required\nfinite-group type at depth $e+1$, reconstruction followed by this comparison\nproduces the required integral representation. Continuity is proved for a\nfinite discrete source and for pullback through a fixed continuous map to\nthat source. Occupancy at only one precision is insufficient. The result\nlifts comparisons between actual integral representations; it does not lift\narbitrary quotient representations or supply the missing arithmetic packet\ninterpretation.","For $C_2$ over $\\mathbb Z_2$, take\n\\[\n\\tau(g)=\\begin{pmatrix}1&0","For $C_2$ over $\\mathbb Z_2$, take\n\\[\n\\tau(g)=\\begin{pmatrix}1&0\\\\0&-1\\end{pmatrix},\\qquad\n\\rho(g)=\\begin{pmatrix}1&2\\\\0&-1\\end{pmatrix},\\qquad\nA=\\begin{pmatrix}1&1\\\\0&1\\end{pmatrix}.\n\\]\nThe matrix $A$ intertwines modulo $4$ and its average is\n$P=\\left(\\begin{smallmatrix}1&-1","0&1\\end{pmatrix}.\n\\]\nThe matrix $A$ intertwines modulo $4$ and its average is\n$P=\\left(\\begin{smallmatrix}1&-1","0&1\\end{pmatrix}.\n\\]\nThe matrix $A$ intertwines modulo $4$ and its average is\n$P=\\left(\\begin{smallmatrix}1&-1\\\\0&1\\end{smallmatrix}\\right)$.\nNo exact intertwiner can retain $A$ modulo $4$, although $P$ retains it\nmodulo $2$. The actual trivial and sign representations also agree modulo\n$2$ but are not integrally isomorphic. These checked examples show why\nresidual-only detection and full preservation of the input precision cannot\nbe asserted here.","0&1\\end{smallmatrix}\\right)$.\nNo exact intertwiner can retain $A$ modulo $4$, although $P$ retains it\nmodulo $2$. The actual trivial and sign representations also agree modulo\n$2$ but are not integrally isomorphic. These checked examples show why\nresidual-only detection and full preservation of the input precision cannot\nbe asserted here.","\\paragraph{Lifting a summand inside a supplied finite-group carrier (Block 18).}\nLet $R$ be a commutative integral domain complete and separated for the\n$(\\pi)$-adic filtration. Let $\\Gamma$ be a finite group, and let\n$\\rho:\\Gamma\\to\\mathrm{GL}_n(R)$ be an actual integral representation.\nAssume that $a=|\\Gamma|\\ne0$ in $R$ and $a\\mid\\pi^e$. For $k>0$, suppose\n\\[\n\\bar E\\in M_n\\bigl(R/(\\pi)^{e+k}\\bigr),\\qquad\n\\bar E^2=\\bar E,\\qquad\n\\bar\\rho(g)\\bar E=\\bar E\\bar\\rho(g)\\quad(g\\in\\Gamma).\n\\]\nThere is an actual integral matrix $E$ satisfying\n\\[\nE^2=E,\\qquad \\rho(g)E=E\\rho(g),\\qquad\nE\\bmod(\\pi)^k=\\bar E\\bmod(\\pi)^k.\n\\]\nNo integral summand or coherent family of quotient projectors is assumed.\nNo local-ring assumption or finiteness of the residue field is needed.","ete and separated for the\n$(\\pi)$-adic filtration. Let $\\Gamma$ be a finite group, and let\n$\\rho:\\Gamma\\to\\mathrm{GL}_n(R)$ be an actual integral representation.\nAssume that $a=|\\Gamma|\\ne0$ in $R$ and $a\\mid\\pi^e$. For $k>0$, suppose\n\\[\n\\bar E\\in M_n\\bigl(R/(\\pi)^{e+k}\\bigr),\\qquad\n\\bar E^2=\\bar E,\\qquad\n\\bar\\rho(g)\\bar E=\\bar E\\bar\\rho(g)\\quad(g\\in\\Gamma).\n\\]\nThere is an actual integral matrix $E$ satisfying\n\\[\nE^2=E,\\qquad \\rho(g)E=E\\rho(g),\\qquad\nE\\bmod(\\pi)^k=\\bar E\\bmod(\\pi)^k.\n\\]\nNo integral summand or coherent family of quotient projectors is assumed.\nNo local-ring assumption or finiteness of the residue field is needed.","d for the\n$(\\pi)$-adic filtration. Let $\\Gamma$ be a finite group, and let\n$\\rho:\\Gamma\\to\\mathrm{GL}_n(R)$ be an actual integral representation.\nAssume that $a=|\\Gamma|\\ne0$ in $R$ and $a\\mid\\pi^e$. For $k>0$, suppose\n\\[\n\\bar E\\in M_n\\bigl(R/(\\pi)^{e+k}\\bigr),\\qquad\n\\bar E^2=\\bar E,\\qquad\n\\bar\\rho(g)\\bar E=\\bar E\\bar\\rho(g)\\quad(g\\in\\Gamma).\n\\]\nThere is an actual integral matrix $E$ satisfying\n\\[\nE^2=E,\\qquad \\rho(g)E=E\\rho(g),\\qquad\nE\\bmod(\\pi)^k=\\bar E\\bmod(\\pi)^k.\n\\]\nNo integral summand or coherent family of quotient projectors is assumed.\nNo local-ring assumption or finiteness of the residue field is needed.","To prove this, apply the preceding averaging correction with both\nrepresentations equal to $\\rho$. This produces an exact equivariant matrix\n$A$ with the prescribed reduction at depth $k$, so that\n$A^2-A\\in(\\pi)^kM_n(R)$. Averaging alone need not preserve idempotence.\nUse the integral polynomial\n\\[\nf(A)=3A^2-2A^3.\n\\]\nThe identities\n\\[\nf(A)-A=(1-2A)(A^2-A),\\qquad\nf(A)^2-f(A)=(A^2-A)^2(4A^2-4A-3)\n\\]\nshow that successive corrections preserve the retained reduction and improve\nthe idempotence defect. In these formulas scalar constants denote scalar\nmatrices. The Lean proof establishes at least one additional ideal digit at\neach step, constructs the resulting compatible family of finite-level matrices, and uses\ncompleteness and separatedness to recover $E$. Every matrix commuting with\n$A$ c","To prove this, apply the preceding averaging correction with both\nrepresentations equal to $\\rho$. This produces an exact equivariant matrix\n$A$ with the prescribed reduction at depth $k$, so that\n$A^2-A\\in(\\pi)^kM_n(R)$. Averaging alone need not preserve idempotence.\nUse the integral polynomial\n\\[\nf(A)=3A^2-2A^3.\n\\]\nThe identities\n\\[\nf(A)-A=(1-2A)(A^2-A),\\qquad\nf(A)^2-f(A)=(A^2-A)^2(4A^2-4A-3)\n\\]\nshow that successive corrections preserve the retained reduction and improve\nthe idempotence defect. In these formulas scalar constants denote scalar\nmatrices. The Lean proof establishes at least one additional ideal digit at\neach step, constructs the resulting compatible family of finite-level matrices, and uses\ncompleteness and separatedness to recover $E$. Every matrix commuting with\n$A$ commutes with $E$. There is no division by $2$,","To prove this, apply the preceding averaging correction with both\nrepresentations equal to $\\rho$. This produces an exact equivariant matrix\n$A$ with the prescribed reduction at depth $k$, so that\n$A^2-A\\in(\\pi)^kM_n(R)$. Averaging alone need not preserve idempotence.\nUse the integral polynomial\n\\[\nf(A)=3A^2-2A^3.\n\\]\nThe identities\n\\[\nf(A)-A=(1-2A)(A^2-A),\\qquad\nf(A)^2-f(A)=(A^2-A)^2(4A^2-4A-3)\n\\]\nshow that successive corrections preserve the retained reduction and improve\nthe idempotence defect. In these formulas scalar constants denote scalar\nmatrices. The Lean proof establishes at least one additional ideal digit at\neach step, constructs the resulting compatible family of finite-level matrices, and uses\ncompleteness and separatedness to recover $E$. Every matrix commuting with\n$A$ commutes with $E$. There is no division by $2$, so the correction also\nworks in residue cha","To prove this, apply the preceding averaging correction with both\nrepresentations equal to $\\rho$. This produces an exact equivariant matrix\n$A$ with the prescribed reduction at depth $k$, so that\n$A^2-A\\in(\\pi)^kM_n(R)$. Averaging alone need not preserve idempotence.\nUse the integral polynomial\n\\[\nf(A)=3A^2-2A^3.\n\\]\nThe identities\n\\[\nf(A)-A=(1-2A)(A^2-A),\\qquad\nf(A)^2-f(A)=(A^2-A)^2(4A^2-4A-3)\n\\]\nshow that successive corrections preserve the retained reduction and improve\nthe idempotence defect. In these formulas scalar constants denote scalar\nmatrices. The Lean proof establishes at least one additional ideal digit at\neach step, constructs the resulting compatible family of finite-level matrices, and uses\ncompleteness and separatedness to recover $E$. Every matrix commuting with\n$A$ commutes with $E$. There is no division by $2$, so the correction also\nworks in residue characteristic $2$.","qquad\nf(A)^2-f(A)=(A^2-A)^2(4A^2-4A-3)\n\\]\nshow that successive corrections preserve the retained reduction and improve\nthe idempotence defect. In these formulas scalar constants denote scalar\nmatrices. The Lean proof establishes at least one additional ideal digit at\neach step, constructs the resulting compatible family of finite-level matrices, and uses\ncompleteness and separatedness to recover $E$. Every matrix commuting with\n$A$ commutes with $E$. There is no division by $2$, so the correction also\nworks in residue characteristic $2$.","4A-3)\n\\]\nshow that successive corrections preserve the retained reduction and improve\nthe idempotence defect. In these formulas scalar constants denote scalar\nmatrices. The Lean proof establishes at least one additional ideal digit at\neach step, constructs the resulting compatible family of finite-level matrices, and uses\ncompleteness and separatedness to recover $E$. Every matrix commuting with\n$A$ commutes with $E$. There is no division by $2$, so the correction also\nworks in residue characteristic $2$.","successive corrections preserve the retained reduction and improve\nthe idempotence defect. In these formulas scalar constants denote scalar\nmatrices. The Lean proof establishes at least one additional ideal digit at\neach step, constructs the resulting compatible family of finite-level matrices, and uses\ncompleteness and separatedness to recover $E$. Every matrix commuting with\n$A$ commutes with $E$. There is no division by $2$, so the correction also\nworks in residue characteristic $2$.","rve the retained reduction and improve\nthe idempotence defect. In these formulas scalar constants denote scalar\nmatrices. The Lean proof establishes at least one additional ideal digit at\neach step, constructs the resulting compatible family of finite-level matrices, and uses\ncompleteness and separatedness to recover $E$. Every matrix commuting with\n$A$ commutes with $E$. There is no division by $2$, so the correction also\nworks in residue characteristic $2$.","t. In these formulas scalar constants denote scalar\nmatrices. The Lean proof establishes at least one additional ideal digit at\neach step, constructs the resulting compatible family of finite-level matrices, and uses\ncompleteness and separatedness to recover $E$. Every matrix commuting with\n$A$ commutes with $E$. There is no division by $2$, so the correction also\nworks in residue characteristic $2$.","The actual submodules $U=\\operatorname{im}E$ and $W=\\ker E$ are invariant,\nand addition gives an equivariant isomorphism\n\\[\nU\\oplus W\\simeq R^n.\n\\]\nThe inclusion $i:U\\hookrightarrow R^n$ and retraction $r(v)=Ev$ are\nconstructed as linear maps and satisfy $ri=1_U$ and $ir=E$. Their\nequivariance is proved. Coordinatewise reduction maps $U$ onto\n$\\operatorname{im}(\\bar E\\bmod(\\pi)^k)$ and maps $W$ onto\n$\\ker(\\bar E\\bmod(\\pi)^k)$. Both equalities are surjectivity statements\nonto the actual submodules over the finite-level quotient, not merely containment. The same\nprojector defines a summand after pullback through any group homomorphism\nto $\\Gamma$.","actual submodules $U=\\operatorname{im}E$ and $W=\\ker E$ are invariant,\nand addition gives an equivariant isomorphism\n\\[\nU\\oplus W\\simeq R^n.\n\\]\nThe inclusion $i:U\\hookrightarrow R^n$ and retraction $r(v)=Ev$ are\nconstructed as linear maps and satisfy $ri=1_U$ and $ir=E$. Their\nequivariance is proved. Coordinatewise reduction maps $U$ onto\n$\\operatorname{im}(\\bar E\\bmod(\\pi)^k)$ and maps $W$ onto\n$\\ker(\\bar E\\bmod(\\pi)^k)$. Both equalities are surjectivity statements\nonto the actual submodules over the finite-level quotient, not merely containment. The same\nprojector defines a summand after pullback through any group homomorphism\nto $\\Gamma$.","d satisfy $ri=1_U$ and $ir=E$. Their\nequivariance is proved. Coordinatewise reduction maps $U$ onto\n$\\operatorname{im}(\\bar E\\bmod(\\pi)^k)$ and maps $W$ onto\n$\\ker(\\bar E\\bmod(\\pi)^k)$. Both equalities are surjectivity statements\nonto the actual submodules over the finite-level quotient, not merely containment. The same\nprojector defines a summand after pullback through any group homomorphism\nto $\\Gamma$.","For $R=\\mathbb Z_2$ and the sign carrier\n$\\rho(s)=\\operatorname{diag}(1,-1)$, the projector\n\\[\n\\bar E=\\begin{pmatrix}1&2","For $R=\\mathbb Z_2$ and the sign carrier\n$\\rho(s)=\\operatorname{diag}(1,-1)$, the projector\n\\[\n\\bar E=\\begin{pmatrix}1&2\\\\0&0\\end{pmatrix}\\pmod4\n\\]\nis equivariant and the theorem constructs a nonzero proper equivariant\nprojector retaining its reduction modulo $2$. No exact equivariant\nendomorphism retains this prescribed matrix modulo $4$. At the insufficient\ninput depth $1$, the projector\n$\\left(\\begin{smallmatrix}1&1","0&0\\end{pmatrix}\\pmod4\n\\]\nis equivariant and the theorem constructs a nonzero proper equivariant\nprojector retaining its reduction modulo $2$. No exact equivariant\nendomorphism retains this prescribed matrix modulo $4$. At the insufficient\ninput depth $1$, the projector\n$\\left(\\begin{smallmatrix}1&1","0&0\\end{pmatrix}\\pmod4\n\\]\nis equivariant and the theorem constructs a nonzero proper equivariant\nprojector retaining its reduction modulo $2$. No exact equivariant\nendomorphism retains this prescribed matrix modulo $4$. At the insufficient\ninput depth $1$, the projector\n$\\left(\\begin{smallmatrix}1&1\\\\0&0\\end{smallmatrix}\\right)\\pmod2$\nhas no exact equivariant lift retaining its reduction. These are statements\nabout the supplied carrier and prescribed projectors, not about nonexistence\nof all integral summands.","\\paragraph{Simultaneous projector-family lifting and rigidity (Block 19).}\nKeep the complete domain, supplied finite-group carrier and group-order\nhypotheses of Block 18. Let $J$ be any finite label set and $k>0$. Suppose\n$(\\bar E_i)_{i\\in J}$ is a complete orthogonal family of equivariant\nidempotents modulo $(\\pi)^{e+k}$. There exists an integral equivariant\nfamily $(E_i)_{i\\in J}$ satisfying\n\\[\nE_i^2=E_i,\\qquad E_iE_j=0\\ (i\\ne j),\\qquad\n\\sum_{i\\in J}E_i=1,\\qquad\nE_i\\bmod(\\pi)^k=\\bar E_i\\bmod(\\pi)^k.\n\\]\nThe loss is at most $e$, independent of $|J|$ and the ambient matrix rank.\nThere is no local-ring or finite-residue-field hypothesis for existence.\nThe formal theorem also treats empty label sets whenever their input\nequations can hold.","and group-order\nhypotheses of Block 18. Let $J$ be any finite label set and $k>0$. Suppose\n$(\\bar E_i)_{i\\in J}$ is a complete orthogonal family of equivariant\nidempotents modulo $(\\pi)^{e+k}$. There exists an integral equivariant\nfamily $(E_i)_{i\\in J}$ satisfying\n\\[\nE_i^2=E_i,\\qquad E_iE_j=0\\ (i\\ne j),\\qquad\n\\sum_{i\\in J}E_i=1,\\qquad\nE_i\\bmod(\\pi)^k=\\bar E_i\\bmod(\\pi)^k.\n\\]\nThe loss is at most $e$, independent of $|J|$ and the ambient matrix rank.\nThere is no local-ring or finite-residue-field hypothesis for existence.\nThe formal theorem also treats empty label sets whenever their input\nequations can hold.","theses of Block 18. Let $J$ be any finite label set and $k>0$. Suppose\n$(\\bar E_i)_{i\\in J}$ is a complete orthogonal family of equivariant\nidempotents modulo $(\\pi)^{e+k}$. There exists an integral equivariant\nfamily $(E_i)_{i\\in J}$ satisfying\n\\[\nE_i^2=E_i,\\qquad E_iE_j=0\\ (i\\ne j),\\qquad\n\\sum_{i\\in J}E_i=1,\\qquad\nE_i\\bmod(\\pi)^k=\\bar E_i\\bmod(\\pi)^k.\n\\]\nThe loss is at most $e$, independent of $|J|$ and the ambient matrix rank.\nThere is no local-ring or finite-residue-field hypothesis for existence.\nThe formal theorem also treats empty label sets whenever their input\nequations can hold.","Average every supplied matrix once. Inductively let $C$ be the complement\nof the previously lifted projectors and correct the next averaged matrix\ninside the corner defined by $C$. The integral polynomial correction\npreserves the retained reduction and commutation with $C$ and the carrier.\nMultiplication by $C$ gives the required supported projector. This consumes\nno further precision. Once every label has been lifted, the complementary\nidempotent reduces to zero at positive depth. Adic separatedness forces it\nto vanish, proving the sum-to-identity equation.","The actual invariant images $U_i=\\operatorname{im}E_i$ give an equivariant\nlinear isomorphism\n\\[\n\\bigoplus_{i\\in J}U_i\\simeq R^n,\n\\qquad (v_i)_i\\longmapsto\\sum_i v_i,\n\\qquad v\\longmapsto(E_iv)_i.\n\\]\nLean represents the finite direct sum by the finite product of the image\nsubmodules. The existing coordinatewise reduction theorem maps every\n$U_i$ onto the corresponding reduced projector image. No chosen bases or\nunproved freeness of the images are used.","For rigidity alone, let $R$ be a commutative local ring, let $\\pi$ be a\nnonunit, and let $E,F$ be complete equivariant families on the same carrier\nwith $E_i\\equiv F_i\\pmod{(\\pi)^k}$ for every label and $k>0$. Then\n\\[\nU=\\sum_{i\\in J}F_iE_i,\\qquad\nU\\equiv1\\pmod{(\\pi)^k},\\qquad UE_i=F_iU.\n\\]\nReduction reflects invertibility in this setting, so $U$ is an integral\nunit and $UE_iU^{-1}=F_i$. It commutes with the carrier and induces actual\nequivariant isomorphisms $\\operatorname{im}E_i\\simeq\\operatorname{im}F_i$.\nThis rigidity theorem requires neither a finite acting group, a domain,\nnor completeness. More generally a unit-reflecting coefficient map suffices.\nThe combined result proves existence and uniqueness up to this l","For rigidity alone, let $R$ be a commutative local ring, let $\\pi$ be a\nnonunit, and let $E,F$ be complete equivariant families on the same carrier\nwith $E_i\\equiv F_i\\pmod{(\\pi)^k}$ for every label and $k>0$. Then\n\\[\nU=\\sum_{i\\in J}F_iE_i,\\qquad\nU\\equiv1\\pmod{(\\pi)^k},\\qquad UE_i=F_iU.\n\\]\nReduction reflects invertibility in this setting, so $U$ is an integral\nunit and $UE_iU^{-1}=F_i$. It commutes with the carrier and induces actual\nequivariant isomorphisms $\\operatorname{im}E_i\\simeq\\operatorname{im}F_i$.\nThis rigidity theorem requires neither a finite acting group, a domain,\nnor completeness. More generally a unit-reflecting coefficient map suffices.\nThe combined result proves existence and uniqueness up to this labeled\nconjugacy; exact uniqueness of t","For rigidity alone, let $R$ be a commutative local ring, let $\\pi$ be a\nnonunit, and let $E,F$ be complete equivariant families on the same carrier\nwith $E_i\\equiv F_i\\pmod{(\\pi)^k}$ for every label and $k>0$. Then\n\\[\nU=\\sum_{i\\in J}F_iE_i,\\qquad\nU\\equiv1\\pmod{(\\pi)^k},\\qquad UE_i=F_iU.\n\\]\nReduction reflects invertibility in this setting, so $U$ is an integral\nunit and $UE_iU^{-1}=F_i$. It commutes with the carrier and induces actual\nequivariant isomorphisms $\\operatorname{im}E_i\\simeq\\operatorname{im}F_i$.\nThis rigidity theorem requires neither a finite acting group, a domain,\nnor completeness. More generally a unit-reflecting coefficient map suffices.\nThe combined result proves existence and uniqueness up to this labeled\nconjugacy; exact uniqueness of the matrices is false.","i$ be a\nnonunit, and let $E,F$ be complete equivariant families on the same carrier\nwith $E_i\\equiv F_i\\pmod{(\\pi)^k}$ for every label and $k>0$. Then\n\\[\nU=\\sum_{i\\in J}F_iE_i,\\qquad\nU\\equiv1\\pmod{(\\pi)^k},\\qquad UE_i=F_iU.\n\\]\nReduction reflects invertibility in this setting, so $U$ is an integral\nunit and $UE_iU^{-1}=F_i$. It commutes with the carrier and induces actual\nequivariant isomorphisms $\\operatorname{im}E_i\\simeq\\operatorname{im}F_i$.\nThis rigidity theorem requires neither a finite acting group, a domain,\nnor completeness. More generally a unit-reflecting coefficient map suffices.\nThe combined result proves existence and uniqueness up to this labeled\nconjugacy; exact uniqueness of the matrices is false.","For the continuous $\\mathbb Z_2$ carrier\n$\\rho(s)=\\operatorname{diag}(1,-1,1)$, conjugate the three coordinate\nprojectors by $1+2e_{01}$ and reduce modulo $4$. They form a complete\nequivariant family. The general theorem constructs three nonzero integral\nsummands retaining this family modulo $2$. No exact equivariant family\nretains the whole prescribed family modulo $4$. Separate checked examples\nshow that independent equivariant idempotent lifts can lose orthogonality,\nand that distinct complete families with the same reduction can be related\nby the asserted identity-congruent conjugacy.","\\paragraph{General finite logarithmic recovery (Block 20).}\nLet $A$ be a rational algebra, without any commutativity assumption on $A$.\nFor a unipotent element $U$ and a positive integer $d$ satisfying\n$(U-1)^d=0$, put\n\\[\nL_d(U)=\\sum_{j=1}^{d-1}\\frac{(-1)^{j+1}}{j}(U-1)^j.\n\\]\nThe value is independent of the chosen annihilating exponent. The finite\nlogarithm and the nilpotent exponential satisfy\n\\[\nL(\\exp N)=N,\\qquad \\exp(L(U))=U.\n\\]\nBoth maps preserve every annihilating exponent, and hence preserve the\nexact nilpotency class of $N$ and $\\exp(N)-1$, or of $U-1$ and $L(U)$.\nThe Lean finite sum includes its zero-index term with coefficient zero;\nthe displayed mathematical sum starts at one. The formal statements also\nhandle the zero algebra and the exponent-z","\\paragraph{General finite logarithmic recovery (Block 20).}\nLet $A$ be a rational algebra, without any commutativity assumption on $A$.\nFor a unipotent element $U$ and a positive integer $d$ satisfying\n$(U-1)^d=0$, put\n\\[\nL_d(U)=\\sum_{j=1}^{d-1}\\frac{(-1)^{j+1}}{j}(U-1)^j.\n\\]\nThe value is independent of the chosen annihilating exponent. The finite\nlogarithm and the nilpotent exponential satisfy\n\\[\nL(\\exp N)=N,\\qquad \\exp(L(U))=U.\n\\]\nBoth maps preserve every annihilating exponent, and hence preserve the\nexact nilpotency class of $N$ and $\\exp(N)-1$, or of $U-1$ and $L(U)$.\nThe Lean finite sum includes its zero-index term with coefficient zero;\nthe displayed mathematical sum starts at one. The formal statements also\nhandle the zero algebra and the exponent-zero case.","For a unipotent element $U$ and a positive integer $d$ satisfying\n$(U-1)^d=0$, put\n\\[\nL_d(U)=\\sum_{j=1}^{d-1}\\frac{(-1)^{j+1}}{j}(U-1)^j.\n\\]\nThe value is independent of the chosen annihilating exponent. The finite\nlogarithm and the nilpotent exponential satisfy\n\\[\nL(\\exp N)=N,\\qquad \\exp(L(U))=U.\n\\]\nBoth maps preserve every annihilating exponent, and hence preserve the\nexact nilpotency class of $N$ and $\\exp(N)-1$, or of $U-1$ and $L(U)$.\nThe Lean finite sum includes its zero-index term with coefficient zero;\nthe displayed mathematical sum starts at one. The formal statements also\nhandle the zero algebra and the exponent-zero case.","nipotent element $U$ and a positive integer $d$ satisfying\n$(U-1)^d=0$, put\n\\[\nL_d(U)=\\sum_{j=1}^{d-1}\\frac{(-1)^{j+1}}{j}(U-1)^j.\n\\]\nThe value is independent of the chosen annihilating exponent. The finite\nlogarithm and the nilpotent exponential satisfy\n\\[\nL(\\exp N)=N,\\qquad \\exp(L(U))=U.\n\\]\nBoth maps preserve every annihilating exponent, and hence preserve the\nexact nilpotency class of $N$ and $\\exp(N)-1$, or of $U-1$ and $L(U)$.\nThe Lean finite sum includes its zero-index term with coefficient zero;\nthe displayed mathematical sum starts at one. The formal statements also\nhandle the zero algebra and the exponent-zero case.","$ and a positive integer $d$ satisfying\n$(U-1)^d=0$, put\n\\[\nL_d(U)=\\sum_{j=1}^{d-1}\\frac{(-1)^{j+1}}{j}(U-1)^j.\n\\]\nThe value is independent of the chosen annihilating exponent. The finite\nlogarithm and the nilpotent exponential satisfy\n\\[\nL(\\exp N)=N,\\qquad \\exp(L(U))=U.\n\\]\nBoth maps preserve every annihilating exponent, and hence preserve the\nexact nilpotency class of $N$ and $\\exp(N)-1$, or of $U-1$ and $L(U)$.\nThe Lean finite sum includes its zero-index term with coefficient zero;\nthe displayed mathematical sum starts at one. The formal statements also\nhandle the zero algebra and the exponent-zero case.","sitive integer $d$ satisfying\n$(U-1)^d=0$, put\n\\[\nL_d(U)=\\sum_{j=1}^{d-1}\\frac{(-1)^{j+1}}{j}(U-1)^j.\n\\]\nThe value is independent of the chosen annihilating exponent. The finite\nlogarithm and the nilpotent exponential satisfy\n\\[\nL(\\exp N)=N,\\qquad \\exp(L(U))=U.\n\\]\nBoth maps preserve every annihilating exponent, and hence preserve the\nexact nilpotency class of $N$ and $\\exp(N)-1$, or of $U-1$ and $L(U)$.\nThe Lean finite sum includes its zero-index term with coefficient zero;\nthe displayed mathematical sum starts at one. The formal statements also\nhandle the zero algebra and the exponent-zero case.","Let $K$ be a characteristic-zero field and $N$ an endomorphism with\n$N^d=0$. For $t\\ne0$ the actual recovery formula is\n\\[\nN=\\frac1t\\sum_{j=1}^{d-1}\\frac{(-1)^{j+1}}{j}\n       \\bigl(\\exp(tN)-1\\bigr)^j.\n\\]\nAll coefficients lie in $K$. The vector spaces need not be finite\ndimensional, but nilpotence still means that some finite power of the\nwhole operator is zero. A linear map intertwines two nilpotent operators exactly\nwhen it intertwines their exponential samples at a common nonzero\nparameter. Also\n\\[\n\\ker\\bigl(\\exp(tN)-1\\bigr)=\\ker N.\n\\]","ry formula is\n\\[\nN=\\frac1t\\sum_{j=1}^{d-1}\\frac{(-1)^{j+1}}{j}\n       \\bigl(\\exp(tN)-1\\bigr)^j.\n\\]\nAll coefficients lie in $K$. The vector spaces need not be finite\ndimensional, but nilpotence still means that some finite power of the\nwhole operator is zero. A linear map intertwines two nilpotent operators exactly\nwhen it intertwines their exponential samples at a common nonzero\nparameter. Also\n\\[\n\\ker\\bigl(\\exp(tN)-1\\bigr)=\\ker N.\n\\]","For supplied covariance data and the twisted cocycle of Block 13, let\n$\\rho_D(g)=\\exp(t(g)N_D)r_D(g)$. If a common element $u$ satisfies\n$r_D(u)=r_E(u)=1$ and $t(u)\\ne0$, every intertwiner of $\\rho_D$ and\n$\\rho_E$ recovers both the original $r$-intertwiner and the $N$-intertwiner.\nThere is no cubic nilpotence restriction. This recovers equivariant\nprojectors for both original components. Reconstructed-action isomorphisms\ninduce actual isomorphisms of the inertia-fixed monodromy kernels. Their local characteristic polynomials\nagree under the existing finite-dimensional and normal-subgroup hypotheses.","For actual finite original inertia images the common kernel sample is\nderived. Suppose $u\\in H$ satisfies $\\nu(u)=1$ and $t(u)\\ne0$ and both\n$r_D(H)$ and $r_E(H)$ are finite. Their sampled actions have positive\nfinite orders $a,b$. At $u^{ab}$ both original actions are trivial, and\n\\[\nt(u^{ab})=(ab)t(u)\\ne0\n\\]\nby the cocycle identity and characteristic zero. Thus the same full\nintertwiner equivalence holds using the actual finite-image hypotheses,\nwithout a separately supplied common trivial-action sample. The initial\nnonzero cocycle sample and the arithmetic interpretation of the data\nremain requirements.","Actual affine representations with monodromy chains of every finite\nlength invoke these theorems. In particular a length-six example has\n$N^5\\ne0$ and $N^6=0$ and is recovered from its actual reconstructed action.\nThe length-four example still defeats the old quadratic logarithm, while\nthe full finite logarithm recovers it.","The sample hypotheses retain substantive content. A zero parameter loses\n$N$. For unrestricted supplied algebraic data, one can also absorb\n$\\exp(tN)$ into $r$ and set the new monodromy to zero; the reconstructed\nrepresentation is unchanged. The absorbed original action fails the\nrequired trivial-action sample condition. This is a counterexample in the\nalgebraic datum class, not a failure of uniqueness of arithmetic\nWeil--Deligne attachment with finite original inertia action.","\\paragraph{Interpreted quadratic arithmetic completion (Block 21).}\nFix the continuous quotient\n\\[\nq:G_{\\Q}\\longrightarrow C_2\n\\]\nobtained by the action on a chosen square root $i$ of $-1$ in the algebraic\nclosure. Its surjectivity and the equivalence\n\\[\nq(\\sigma)=1\\quad\\Longleftrightarrow\\quad\\sigma(i)=i\n\\]\nare proved using the existing actual quadratic Galois construction.\nFor a commutative ring $B$, define the arithmetic packet\n\\[\n\\mathcal P_q(B)=\n\\{\\chi:G_{\\Q}\\longrightarrow B^\\times\\mid\\ker q\\subseteq\\ker\\chi\\},\n\\]\nwhere the maps are group homomorphisms. Unique descent through the actual\nsurjective map $q$ is derived, rather than supplied. Every member has finite\nimage and is continuous for the coefficient topology, since its descent\nhas the finite discrete domain $C_2$.","aragraph{Interpreted quadratic arithmetic completion (Block 21).}\nFix the continuous quotient\n\\[\nq:G_{\\Q}\\longrightarrow C_2\n\\]\nobtained by the action on a chosen square root $i$ of $-1$ in the algebraic\nclosure. Its surjectivity and the equivalence\n\\[\nq(\\sigma)=1\\quad\\Longleftrightarrow\\quad\\sigma(i)=i\n\\]\nare proved using the existing actual quadratic Galois construction.\nFor a commutative ring $B$, define the arithmetic packet\n\\[\n\\mathcal P_q(B)=\n\\{\\chi:G_{\\Q}\\longrightarrow B^\\times\\mid\\ker q\\subseteq\\ker\\chi\\},\n\\]\nwhere the maps are group homomorphisms. Unique descent through the actual\nsurjective map $q$ is derived, rather than supplied. Every member has finite\nimage and is continuous for the coefficient topology, since its descent\nhas the finite discrete domain $C_2$.","hmetic completion (Block 21).}\nFix the continuous quotient\n\\[\nq:G_{\\Q}\\longrightarrow C_2\n\\]\nobtained by the action on a chosen square root $i$ of $-1$ in the algebraic\nclosure. Its surjectivity and the equivalence\n\\[\nq(\\sigma)=1\\quad\\Longleftrightarrow\\quad\\sigma(i)=i\n\\]\nare proved using the existing actual quadratic Galois construction.\nFor a commutative ring $B$, define the arithmetic packet\n\\[\n\\mathcal P_q(B)=\n\\{\\chi:G_{\\Q}\\longrightarrow B^\\times\\mid\\ker q\\subseteq\\ker\\chi\\},\n\\]\nwhere the maps are group homomorphisms. Unique descent through the actual\nsurjective map $q$ is derived, rather than supplied. Every member has finite\nimage and is continuous for the coefficient topology, since its descent\nhas the finite discrete domain $C_2$.","ts surjectivity and the equivalence\n\\[\nq(\\sigma)=1\\quad\\Longleftrightarrow\\quad\\sigma(i)=i\n\\]\nare proved using the existing actual quadratic Galois construction.\nFor a commutative ring $B$, define the arithmetic packet\n\\[\n\\mathcal P_q(B)=\n\\{\\chi:G_{\\Q}\\longrightarrow B^\\times\\mid\\ker q\\subseteq\\ker\\chi\\},\n\\]\nwhere the maps are group homomorphisms. Unique descent through the actual\nsurjective map $q$ is derived, rather than supplied. Every member has finite\nimage and is continuous for the coefficient topology, since its descent\nhas the finite discrete domain $C_2$.","a commutative ring $B$, define the arithmetic packet\n\\[\n\\mathcal P_q(B)=\n\\{\\chi:G_{\\Q}\\longrightarrow B^\\times\\mid\\ker q\\subseteq\\ker\\chi\\},\n\\]\nwhere the maps are group homomorphisms. Unique descent through the actual\nsurjective map $q$ is derived, rather than supplied. Every member has finite\nimage and is continuous for the coefficient topology, since its descent\nhas the finite discrete domain $C_2$.","Let $R$ be any commutative ring and $B$ any commutative $R$-algebra. The\ncoordinate ring $A_R=R[X]/(X^2-1)$ represents precisely this specified\npacket:\n\\[\n\\Hom_{R\\text{-alg}}(A_R,B)\n\\simeq\\{x\\in B:x^2=1\\}\n\\simeq\\Hom(C_2,B^\\times)\n\\simeq\\mathcal P_q(B).\n\\]\nThe first map evaluates the coordinate. Its inverse is the quotient\nuniversal property. A root is a unit with itself as inverse and determines\na character by its value on the generator of $C_2$. Finally pull back\nalong $q$. All inverse identities and naturality under coefficient\nhomomorphisms are proved. In particular, the point reductions and actual\narithmetic character reductions coincide. There is no invertibility-of-two,\nreducedness or nontri","Let $R$ be any commutative ring and $B$ any commutative $R$-algebra. The\ncoordinate ring $A_R=R[X]/(X^2-1)$ represents precisely this specified\npacket:\n\\[\n\\Hom_{R\\text{-alg}}(A_R,B)\n\\simeq\\{x\\in B:x^2=1\\}\n\\simeq\\Hom(C_2,B^\\times)\n\\simeq\\mathcal P_q(B).\n\\]\nThe first map evaluates the coordinate. Its inverse is the quotient\nuniversal property. A root is a unit with itself as inverse and determines\na character by its value on the generator of $C_2$. Finally pull back\nalong $q$. All inverse identities and naturality under coefficient\nhomomorphisms are proved. In particular, the point reductions and actual\narithmetic character reductions coincide. There is no invertibility-of-two,\nreducedness or nontriviality hypothesis on $B$. No sm","Let $R$ be any commutative ring and $B$ any commutative $R$-algebra. The\ncoordinate ring $A_R=R[X]/(X^2-1)$ represents precisely this specified\npacket:\n\\[\n\\Hom_{R\\text{-alg}}(A_R,B)\n\\simeq\\{x\\in B:x^2=1\\}\n\\simeq\\Hom(C_2,B^\\times)\n\\simeq\\mathcal P_q(B).\n\\]\nThe first map evaluates the coordinate. Its inverse is the quotient\nuniversal property. A root is a unit with itself as inverse and determines\na character by its value on the generator of $C_2$. Finally pull back\nalong $q$. All inverse identities and naturality under coefficient\nhomomorphisms are proved. In particular, the point reductions and actual\narithmetic character reductions coincide. There is no invertibility-of-two,\nreducedness or nontriviality hypothesis on $B$. No smoothness or\n\\'{e}tal","Let $R$ be any commutative ring and $B$ any commutative $R$-algebra. The\ncoordinate ring $A_R=R[X]/(X^2-1)$ represents precisely this specified\npacket:\n\\[\n\\Hom_{R\\text{-alg}}(A_R,B)\n\\simeq\\{x\\in B:x^2=1\\}\n\\simeq\\Hom(C_2,B^\\times)\n\\simeq\\mathcal P_q(B).\n\\]\nThe first map evaluates the coordinate. Its inverse is the quotient\nuniversal property. A root is a unit with itself as inverse and determines\na character by its value on the generator of $C_2$. Finally pull back\nalong $q$. All inverse identities and naturality under coefficient\nhomomorphisms are proved. In particular, the point reductions and actual\narithmetic character reductions coincide. There is no invertibility-of-two,\nreducedness or nontriviality hypothesis on $B$. No smoothness or\n\\'{e}taleness statement at two is used or asserted.","ommutative $R$-algebra. The\ncoordinate ring $A_R=R[X]/(X^2-1)$ represents precisely this specified\npacket:\n\\[\n\\Hom_{R\\text{-alg}}(A_R,B)\n\\simeq\\{x\\in B:x^2=1\\}\n\\simeq\\Hom(C_2,B^\\times)\n\\simeq\\mathcal P_q(B).\n\\]\nThe first map evaluates the coordinate. Its inverse is the quotient\nuniversal property. A root is a unit with itself as inverse and determines\na character by its value on the generator of $C_2$. Finally pull back\nalong $q$. All inverse identities and naturality under coefficient\nhomomorphisms are proved. In particular, the point reductions and actual\narithmetic character reductions coincide. There is no invertibility-of-two,\nreducedness or nontriviality hypothesis on $B$. No smoothness or\n\\'{e}taleness statement at two is used or asserted.","rdinate ring $A_R=R[X]/(X^2-1)$ represents precisely this specified\npacket:\n\\[\n\\Hom_{R\\text{-alg}}(A_R,B)\n\\simeq\\{x\\in B:x^2=1\\}\n\\simeq\\Hom(C_2,B^\\times)\n\\simeq\\mathcal P_q(B).\n\\]\nThe first map evaluates the coordinate. Its inverse is the quotient\nuniversal property. A root is a unit with itself as inverse and determines\na character by its value on the generator of $C_2$. Finally pull back\nalong $q$. All inverse identities and naturality under coefficient\nhomomorphisms are proved. In particular, the point reductions and actual\narithmetic character reductions coincide. There is no invertibility-of-two,\nreducedness or nontriviality hypothesis on $B$. No smoothness or\n\\'{e}taleness statement at two is used or asserted.","rsal property. A root is a unit with itself as inverse and determines\na character by its value on the generator of $C_2$. Finally pull back\nalong $q$. All inverse identities and naturality under coefficient\nhomomorphisms are proved. In particular, the point reductions and actual\narithmetic character reductions coincide. There is no invertibility-of-two,\nreducedness or nontriviality hypothesis on $B$. No smoothness or\n\\'{e}taleness statement at two is used or asserted.","Let $I\\subseteq R$ and suppose that $R$ is $I$-adically complete. Using\nthe checked ring-map reconstruction on the quadratic coordinate ring,\nLean proves\n\\[\n\\mathcal P_q(R)\\simeq\\varprojlim_{m\\ge0}\\mathcal P_q(R/I^m).\n\\]\nThe zero-precision quotient is included only for the formal indexing.\nThe inverse constructs an actual arithmetic character, with exactly the\nprescribed reductions. It is unique, and its continuity is derived from\nits factorization through the same finite quotient.","There is also a completion theorem without chosen compatible characters.\nSuppose that $R/I^m$ is finite for every $m\\ge1$, and that\n$C_m\\subseteq\\mathcal P_q(R/I^m)$ are allowed subsets such that every\nreduction of an allowed member is allowed. Then\n\\[\n\\begin{split}\n&\\exists\\chi\\in\\mathcal P_q(R),\\quad\n  \\chi\\text{ is continuous and }\\chi\\bmod I^m\\in C_m\n  \\text{ for all }m\\ge1","There is also a completion theorem without chosen compatible characters.\nSuppose that $R/I^m$ is finite for every $m\\ge1$, and that\n$C_m\\subseteq\\mathcal P_q(R/I^m)$ are allowed subsets such that every\nreduction of an allowed member is allowed. Then\n\\[\n\\begin{split}\n&\\exists\\chi\\in\\mathcal P_q(R),\\quad\n  \\chi\\text{ is continuous and }\\chi\\bmod I^m\\in C_m\n  \\text{ for all }m\\ge1\\\\\n&\\hspace{3em}\\Longleftrightarrow\\quad\n  C_m\\ne\\varnothing\\text{ for all }m\\ge1.\n\\end{split}\n\\]\nThe root interpretation makes the allowed finite levels genuinely finite.\nFinite-branch compactness chooses the compatible roots, and the ring\nreconstruction gives their integral root. Naturality turns this into the\nrequired arithmetic character. Surjectivity of transitions and individual\nliftability of every finite node are not assumed. The root-level theorem\nneeds only the allowed root subsets to be finite.","For $R=\\Z_2$, Lean derives finiteness of all positive maximal-ideal\nquotients and applies this completion theorem. Since $\\Z_2$ is a domain,\nits only roots of $X^2-1$ are $1$ and $-1$. Thus the integral packet has\nexactly two distinct characters: the trivial character and the sign\ncharacter describing the action on $i$. They have the same reduction\nmodulo two and distinct reductions modulo four.","The distinction between finite occupancy and continuation is witnessed\nby an actual finite arithmetic character. The roots modulo eight are\n$1,3,5,7$. The root $3$ defines a member of\n$\\mathcal P_q(\\Z/8\\Z)$, but no member of\n$\\mathcal P_q(\\Z/16\\Z)$ reduces to it. Indeed the only scalar lifts of\n$3$ are $3$ and $11$, and both square to $9$ modulo sixteen. The formal\nproof also classifies all roots modulo sixteen and the reductions with\nchildren. Consequently a nonempty finite node need not lie on an\ninfinite branch, even in this explicitly interpreted arithmetic packet.","Block 21 completes this restricted arithmetic interpretation and its\nconditional completion. The packet is defined by one fixed quotient; it\nis not the collection of all quadratic characters of $G_{\\Q}$. No local\nramification criterion, Hodge--Tate weight, Weil--Deligne attachment, or\ncross-coefficient local invariant was established by Block 21. The full\nmanuscript still requires its remaining packet identifications and\narithmetic comparisons. At the Block 21 checkpoint, the ledger recorded\nfifteen checked claims and five partial claims; see the Block 23 conditional-scope\ncorrection above. Universal positive occupancy remains conjectural.","\\paragraph{Actual quadratic local conditions and completion (Block 22).}\nContinue with the fixed packet $\\mathcal P_q$ above. For a\ncharacteristic-zero field $K$, let\n\\[\nj_K:G_K\\longrightarrow G_{\\Q}\n\\]\nbe the actual field-induced homomorphism between absolute Galois groups.\nIt includes a chosen embedding of algebraic closures. Let $i_K$ be the\nimage of the chosen square root $i$ under that embedding. Lean proves\ncompatibility with the Galois actions and hence\n\\[\n(q\\circ j_K)(\\sigma)=1\n\\quad\\Longleftrightarrow\\quad \\sigma(i_K)=i_K.\n\\]\nThe fixed-field theorem for the algebraic closure then gives\n\\[\nq\\circ j_K=1\n\\quad\\Longleftrightarrow\\quad\n\\exists x\\in K,\\ x^2=-1.\n\\]\nIf no such root exists, the local quotient is surjective onto $C_2$.\nThese are statements about the actual local Galois map, not a supplied\na","\\paragraph{Actual quadratic local conditions and completion (Block 22).}\nContinue with the fixed packet $\\mathcal P_q$ above. For a\ncharacteristic-zero field $K$, let\n\\[\nj_K:G_K\\longrightarrow G_{\\Q}\n\\]\nbe the actual field-induced homomorphism between absolute Galois groups.\nIt includes a chosen embedding of algebraic closures. Let $i_K$ be the\nimage of the chosen square root $i$ under that embedding. Lean proves\ncompatibility with the Galois actions and hence\n\\[\n(q\\circ j_K)(\\sigma)=1\n\\quad\\Longleftrightarrow\\quad \\sigma(i_K)=i_K.\n\\]\nThe fixed-field theorem for the algebraic closure then gives\n\\[\nq\\circ j_K=1\n\\quad\\Longleftrightarrow\\quad\n\\exists x\\in K,\\ x^2=-1.\n\\]\nIf no such root exists, the local quotient is surjective onto $C_2$.\nThese are statements about the actual local Galois map, not a supplied\narithmetic interpretation of a scalar label.","22).}\nContinue with the fixed packet $\\mathcal P_q$ above. For a\ncharacteristic-zero field $K$, let\n\\[\nj_K:G_K\\longrightarrow G_{\\Q}\n\\]\nbe the actual field-induced homomorphism between absolute Galois groups.\nIt includes a chosen embedding of algebraic closures. Let $i_K$ be the\nimage of the chosen square root $i$ under that embedding. Lean proves\ncompatibility with the Galois actions and hence\n\\[\n(q\\circ j_K)(\\sigma)=1\n\\quad\\Longleftrightarrow\\quad \\sigma(i_K)=i_K.\n\\]\nThe fixed-field theorem for the algebraic closure then gives\n\\[\nq\\circ j_K=1\n\\quad\\Longleftrightarrow\\quad\n\\exists x\\in K,\\ x^2=-1.\n\\]\nIf no such root exists, the local quotient is surjective onto $C_2$.\nThese are statements about the actual local Galois map, not a supplied\narithmetic interpretation of a scalar label.","e the actual field-induced homomorphism between absolute Galois groups.\nIt includes a chosen embedding of algebraic closures. Let $i_K$ be the\nimage of the chosen square root $i$ under that embedding. Lean proves\ncompatibility with the Galois actions and hence\n\\[\n(q\\circ j_K)(\\sigma)=1\n\\quad\\Longleftrightarrow\\quad \\sigma(i_K)=i_K.\n\\]\nThe fixed-field theorem for the algebraic closure then gives\n\\[\nq\\circ j_K=1\n\\quad\\Longleftrightarrow\\quad\n\\exists x\\in K,\\ x^2=-1.\n\\]\nIf no such root exists, the local quotient is surjective onto $C_2$.\nThese are statements about the actual local Galois map, not a supplied\narithmetic interpretation of a scalar label.","be the\nimage of the chosen square root $i$ under that embedding. Lean proves\ncompatibility with the Galois actions and hence\n\\[\n(q\\circ j_K)(\\sigma)=1\n\\quad\\Longleftrightarrow\\quad \\sigma(i_K)=i_K.\n\\]\nThe fixed-field theorem for the algebraic closure then gives\n\\[\nq\\circ j_K=1\n\\quad\\Longleftrightarrow\\quad\n\\exists x\\in K,\\ x^2=-1.\n\\]\nIf no such root exists, the local quotient is surjective onto $C_2$.\nThese are statements about the actual local Galois map, not a supplied\narithmetic interpretation of a scalar label.","For every commutative coefficient ring $B$ and $\\chi\\in\\mathcal P_q(B)$,\ndefine $\\chi_K=\\chi\\circ j_K$. The local character is continuous and\ncommutes with coefficient change. The precise local admissibility\ncondition is\n\\[\n\\operatorname{LocallyTrivial}_K(\\chi)\n\\quad\\Longleftrightarrow\\quad \\chi_K=1.\n\\]\nDescent through the fixed quadratic quotient proves\n\\[\n\\operatorname{LocallyTrivial}_K(\\chi)\n\\quad\\Longleftrightarrow\\quad\n\\bigl(\\exists x\\in K,\\ x^2=-1\\bigr)\\ \\lor\\ \\chi=1.\n\\]\nNo reducedness, characteristic-zero or domain hypothesis is imposed on\n$B$. In particular, the formula applies to the finite coefficient\nquotients used in completion.","h coefficient change. The precise local admissibility\ncondition is\n\\[\n\\operatorname{LocallyTrivial}_K(\\chi)\n\\quad\\Longleftrightarrow\\quad \\chi_K=1.\n\\]\nDescent through the fixed quadratic quotient proves\n\\[\n\\operatorname{LocallyTrivial}_K(\\chi)\n\\quad\\Longleftrightarrow\\quad\n\\bigl(\\exists x\\in K,\\ x^2=-1\\bigr)\\ \\lor\\ \\chi=1.\n\\]\nNo reducedness, characteristic-zero or domain hypothesis is imposed on\n$B$. In particular, the formula applies to the finite coefficient\nquotients used in completion.","The p-adic square-root criterion is proved for every prime:\n\\[\n\\bigl(\\exists x\\in\\Q_p,\\ x^2=-1\\bigr)\n\\quad\\Longleftrightarrow\\quad p\\equiv1\\pmod4.\n\\]\nA root satisfies $\\lvert x\\rvert_p^2=1$ and is therefore integral.\nReduction modulo three rules out a root in $\\Q_3$, and reduction modulo\nfour rules out a root in $\\Q_2$. At an odd prime, any residue root of\n$X^2+1$ has nonzero derivative and lifts by the pinned Mathlib Hensel\ntheorem. The finite-field square criterion then gives the stated\ncongruence. An explicit Hensel application at the residue root $2$ modulo $5$ supplies\na root in $\\Z_5$. Consequently\n\\[\n\\boxed{\n\\operatorname{LocallyTrivial}_{\\Q_p}(\\chi)\n\\quad\\Longleftrightarrow\\quad p\\equiv1\\pmod4\\ \\lor\\ \\chi=1.\n}\n\\]\nThe arithmetic prime $p$ is independent of the coefficient prime. For\nthe two integral $\\Z_2$-valued characters, the","The p-adic square-root criterion is proved for every prime:\n\\[\n\\bigl(\\exists x\\in\\Q_p,\\ x^2=-1\\bigr)\n\\quad\\Longleftrightarrow\\quad p\\equiv1\\pmod4.\n\\]\nA root satisfies $\\lvert x\\rvert_p^2=1$ and is therefore integral.\nReduction modulo three rules out a root in $\\Q_3$, and reduction modulo\nfour rules out a root in $\\Q_2$. At an odd prime, any residue root of\n$X^2+1$ has nonzero derivative and lifts by the pinned Mathlib Hensel\ntheorem. The finite-field square criterion then gives the stated\ncongruence. An explicit Hensel application at the residue root $2$ modulo $5$ supplies\na root in $\\Z_5$. Consequently\n\\[\n\\boxed{\n\\operatorname{LocallyTrivial}_{\\Q_p}(\\chi)\n\\quad\\Longleftrightarrow\\quad p\\equiv1\\pmod4\\ \\lor\\ \\chi=1.\n}\n\\]\nThe arithmetic prime $p$ is independent of the coefficient prime. For\nthe two integral $\\Z_2$-valued characters, the sign character is locally\ntrivial exactl","The p-adic square-root criterion is proved for every prime:\n\\[\n\\bigl(\\exists x\\in\\Q_p,\\ x^2=-1\\bigr)\n\\quad\\Longleftrightarrow\\quad p\\equiv1\\pmod4.\n\\]\nA root satisfies $\\lvert x\\rvert_p^2=1$ and is therefore integral.\nReduction modulo three rules out a root in $\\Q_3$, and reduction modulo\nfour rules out a root in $\\Q_2$. At an odd prime, any residue root of\n$X^2+1$ has nonzero derivative and lifts by the pinned Mathlib Hensel\ntheorem. The finite-field square criterion then gives the stated\ncongruence. An explicit Hensel application at the residue root $2$ modulo $5$ supplies\na root in $\\Z_5$. Consequently\n\\[\n\\boxed{\n\\operatorname{LocallyTrivial}_{\\Q_p}(\\chi)\n\\quad\\Longleftrightarrow\\quad p\\equiv1\\pmod4\\ \\lor\\ \\chi=1.\n}\n\\]\nThe arithmetic prime $p$ is independent of the coefficient prime. For\nthe two integral $\\Z_2$-valued characters, the sign character is locally\ntrivial exactly at primes congruent to one modulo four. Both distinct\nglobal characters are","The p-adic square-root criterion is proved for every prime:\n\\[\n\\bigl(\\exists x\\in\\Q_p,\\ x^2=-1\\bigr)\n\\quad\\Longleftrightarrow\\quad p\\equiv1\\pmod4.\n\\]\nA root satisfies $\\lvert x\\rvert_p^2=1$ and is therefore integral.\nReduction modulo three rules out a root in $\\Q_3$, and reduction modulo\nfour rules out a root in $\\Q_2$. At an odd prime, any residue root of\n$X^2+1$ has nonzero derivative and lifts by the pinned Mathlib Hensel\ntheorem. The finite-field square criterion then gives the stated\ncongruence. An explicit Hensel application at the residue root $2$ modulo $5$ supplies\na root in $\\Z_5$. Consequently\n\\[\n\\boxed{\n\\operatorname{LocallyTrivial}_{\\Q_p}(\\chi)\n\\quad\\Longleftrightarrow\\quad p\\equiv1\\pmod4\\ \\lor\\ \\chi=1.\n}\n\\]\nThe arithmetic prime $p$ is independent of the coefficient prime. For\nthe two integral $\\Z_2$-valued characters, the sign character is locally\ntrivial exactly at primes congruent to one modulo four. Both distinct\nglobal characters are locally trivial at five, while the sign character\nis","The p-adic square-root criterion is proved for every prime:\n\\[\n\\bigl(\\exists x\\in\\Q_p,\\ x^2=-1\\bigr)\n\\quad\\Longleftrightarrow\\quad p\\equiv1\\pmod4.\n\\]\nA root satisfies $\\lvert x\\rvert_p^2=1$ and is therefore integral.\nReduction modulo three rules out a root in $\\Q_3$, and reduction modulo\nfour rules out a root in $\\Q_2$. At an odd prime, any residue root of\n$X^2+1$ has nonzero derivative and lifts by the pinned Mathlib Hensel\ntheorem. The finite-field square criterion then gives the stated\ncongruence. An explicit Hensel application at the residue root $2$ modulo $5$ supplies\na root in $\\Z_5$. Consequently\n\\[\n\\boxed{\n\\operatorname{LocallyTrivial}_{\\Q_p}(\\chi)\n\\quad\\Longleftrightarrow\\quad p\\equiv1\\pmod4\\ \\lor\\ \\chi=1.\n}\n\\]\nThe arithmetic prime $p$ is independent of the coefficient prime. For\nthe two integral $\\Z_2$-valued characters, the sign character is locally\ntrivial exactly at primes congruent to one modulo four. Both distinct\nglobal characters are locally trivial at five, while the sign character\nis nontrivial on the full local Galois groups at","atisfies $\\lvert x\\rvert_p^2=1$ and is therefore integral.\nReduction modulo three rules out a root in $\\Q_3$, and reduction modulo\nfour rules out a root in $\\Q_2$. At an odd prime, any residue root of\n$X^2+1$ has nonzero derivative and lifts by the pinned Mathlib Hensel\ntheorem. The finite-field square criterion then gives the stated\ncongruence. An explicit Hensel application at the residue root $2$ modulo $5$ supplies\na root in $\\Z_5$. Consequently\n\\[\n\\boxed{\n\\operatorname{LocallyTrivial}_{\\Q_p}(\\chi)\n\\quad\\Longleftrightarrow\\quad p\\equiv1\\pmod4\\ \\lor\\ \\chi=1.\n}\n\\]\nThe arithmetic prime $p$ is independent of the coefficient prime. For\nthe two integral $\\Z_2$-valued characters, the sign character is locally\ntrivial exactly at primes congruent to one modulo four. Both distinct\nglobal characters are locally trivial at five, while the sign character\nis nontrivial on the full local Galois groups at three and two.","lvert x\\rvert_p^2=1$ and is therefore integral.\nReduction modulo three rules out a root in $\\Q_3$, and reduction modulo\nfour rules out a root in $\\Q_2$. At an odd prime, any residue root of\n$X^2+1$ has nonzero derivative and lifts by the pinned Mathlib Hensel\ntheorem. The finite-field square criterion then gives the stated\ncongruence. An explicit Hensel application at the residue root $2$ modulo $5$ supplies\na root in $\\Z_5$. Consequently\n\\[\n\\boxed{\n\\operatorname{LocallyTrivial}_{\\Q_p}(\\chi)\n\\quad\\Longleftrightarrow\\quad p\\equiv1\\pmod4\\ \\lor\\ \\chi=1.\n}\n\\]\nThe arithmetic prime $p$ is independent of the coefficient prime. For\nthe two integral $\\Z_2$-valued characters, the sign character is locally\ntrivial exactly at primes congruent to one modulo four. Both distinct\nglobal characters are locally trivial at five, while the sign character\nis nontrivial on the full local Galois groups at three and two.","d is therefore integral.\nReduction modulo three rules out a root in $\\Q_3$, and reduction modulo\nfour rules out a root in $\\Q_2$. At an odd prime, any residue root of\n$X^2+1$ has nonzero derivative and lifts by the pinned Mathlib Hensel\ntheorem. The finite-field square criterion then gives the stated\ncongruence. An explicit Hensel application at the residue root $2$ modulo $5$ supplies\na root in $\\Z_5$. Consequently\n\\[\n\\boxed{\n\\operatorname{LocallyTrivial}_{\\Q_p}(\\chi)\n\\quad\\Longleftrightarrow\\quad p\\equiv1\\pmod4\\ \\lor\\ \\chi=1.\n}\n\\]\nThe arithmetic prime $p$ is independent of the coefficient prime. For\nthe two integral $\\Z_2$-valued characters, the sign character is locally\ntrivial exactly at primes congruent to one modulo four. Both distinct\nglobal characters are locally trivial at five, while the sign character\nis nontrivial on the full local Galois groups at three and two.","l.\nReduction modulo three rules out a root in $\\Q_3$, and reduction modulo\nfour rules out a root in $\\Q_2$. At an odd prime, any residue root of\n$X^2+1$ has nonzero derivative and lifts by the pinned Mathlib Hensel\ntheorem. The finite-field square criterion then gives the stated\ncongruence. An explicit Hensel application at the residue root $2$ modulo $5$ supplies\na root in $\\Z_5$. Consequently\n\\[\n\\boxed{\n\\operatorname{LocallyTrivial}_{\\Q_p}(\\chi)\n\\quad\\Longleftrightarrow\\quad p\\equiv1\\pmod4\\ \\lor\\ \\chi=1.\n}\n\\]\nThe arithmetic prime $p$ is independent of the coefficient prime. For\nthe two integral $\\Z_2$-valued characters, the sign character is locally\ntrivial exactly at primes congruent to one modulo four. Both distinct\nglobal characters are locally trivial at five, while the sign character\nis nontrivial on the full local Galois groups at three and two.","ot of\n$X^2+1$ has nonzero derivative and lifts by the pinned Mathlib Hensel\ntheorem. The finite-field square criterion then gives the stated\ncongruence. An explicit Hensel application at the residue root $2$ modulo $5$ supplies\na root in $\\Z_5$. Consequently\n\\[\n\\boxed{\n\\operatorname{LocallyTrivial}_{\\Q_p}(\\chi)\n\\quad\\Longleftrightarrow\\quad p\\equiv1\\pmod4\\ \\lor\\ \\chi=1.\n}\n\\]\nThe arithmetic prime $p$ is independent of the coefficient prime. For\nthe two integral $\\Z_2$-valued characters, the sign character is locally\ntrivial exactly at primes congruent to one modulo four. Both distinct\nglobal characters are locally trivial at five, while the sign character\nis nontrivial on the full local Galois groups at three and two.","ib Hensel\ntheorem. The finite-field square criterion then gives the stated\ncongruence. An explicit Hensel application at the residue root $2$ modulo $5$ supplies\na root in $\\Z_5$. Consequently\n\\[\n\\boxed{\n\\operatorname{LocallyTrivial}_{\\Q_p}(\\chi)\n\\quad\\Longleftrightarrow\\quad p\\equiv1\\pmod4\\ \\lor\\ \\chi=1.\n}\n\\]\nThe arithmetic prime $p$ is independent of the coefficient prime. For\nthe two integral $\\Z_2$-valued characters, the sign character is locally\ntrivial exactly at primes congruent to one modulo four. Both distinct\nglobal characters are locally trivial at five, while the sign character\nis nontrivial on the full local Galois groups at three and two.","The inertia theorem uses actual residue actions. For a valuation subring\n$A$ of an extension field $L/K$, let $D_A$ be its decomposition subgroup\nin $\\operatorname{Aut}_K(L)$ and put\n\\[\nI_A=\\ker\\bigl(D_A\\longrightarrow\\operatorname{Aut}(k_A)\\bigr),\n\\]\nwhere $k_A$ is the residue field. If $2$ is a unit in $A$, every member\nof $I_A$ fixes every $y\\in L$ with $y^2=-1$. Indeed $y^{-1}=-y$, so\n$y$ belongs to $A$ and is a unit there. An automorphism sends $y$ to\n$y$ or $-y$. Their residues are distinct because $2$ and $y$ are units,\nwhereas inertia fixes residues. The second possibility is excluded.","tia theorem uses actual residue actions. For a valuation subring\n$A$ of an extension field $L/K$, let $D_A$ be its decomposition subgroup\nin $\\operatorname{Aut}_K(L)$ and put\n\\[\nI_A=\\ker\\bigl(D_A\\longrightarrow\\operatorname{Aut}(k_A)\\bigr),\n\\]\nwhere $k_A$ is the residue field. If $2$ is a unit in $A$, every member\nof $I_A$ fixes every $y\\in L$ with $y^2=-1$. Indeed $y^{-1}=-y$, so\n$y$ belongs to $A$ and is a unit there. An automorphism sends $y$ to\n$y$ or $-y$. Their residues are distinct because $2$ and $y$ are units,\nwhereas inertia fixes residues. The second possibility is excluded.","valuation subring\n$A$ of an extension field $L/K$, let $D_A$ be its decomposition subgroup\nin $\\operatorname{Aut}_K(L)$ and put\n\\[\nI_A=\\ker\\bigl(D_A\\longrightarrow\\operatorname{Aut}(k_A)\\bigr),\n\\]\nwhere $k_A$ is the residue field. If $2$ is a unit in $A$, every member\nof $I_A$ fixes every $y\\in L$ with $y^2=-1$. Indeed $y^{-1}=-y$, so\n$y$ belongs to $A$ and is a unit there. An automorphism sends $y$ to\n$y$ or $-y$. Their residues are distinct because $2$ and $y$ are units,\nwhereas inertia fixes residues. The second possibility is excluded.","on subring\n$A$ of an extension field $L/K$, let $D_A$ be its decomposition subgroup\nin $\\operatorname{Aut}_K(L)$ and put\n\\[\nI_A=\\ker\\bigl(D_A\\longrightarrow\\operatorname{Aut}(k_A)\\bigr),\n\\]\nwhere $k_A$ is the residue field. If $2$ is a unit in $A$, every member\nof $I_A$ fixes every $y\\in L$ with $y^2=-1$. Indeed $y^{-1}=-y$, so\n$y$ belongs to $A$ and is a unit there. An automorphism sends $y$ to\n$y$ or $-y$. Their residues are distinct because $2$ and $y$ are units,\nwhereas inertia fixes residues. The second possibility is excluded.","ing\n$A$ of an extension field $L/K$, let $D_A$ be its decomposition subgroup\nin $\\operatorname{Aut}_K(L)$ and put\n\\[\nI_A=\\ker\\bigl(D_A\\longrightarrow\\operatorname{Aut}(k_A)\\bigr),\n\\]\nwhere $k_A$ is the residue field. If $2$ is a unit in $A$, every member\nof $I_A$ fixes every $y\\in L$ with $y^2=-1$. Indeed $y^{-1}=-y$, so\n$y$ belongs to $A$ and is a unit there. An automorphism sends $y$ to\n$y$ or $-y$. Their residues are distinct because $2$ and $y$ are units,\nwhereas inertia fixes residues. The second possibility is excluded.","D_A$ be its decomposition subgroup\nin $\\operatorname{Aut}_K(L)$ and put\n\\[\nI_A=\\ker\\bigl(D_A\\longrightarrow\\operatorname{Aut}(k_A)\\bigr),\n\\]\nwhere $k_A$ is the residue field. If $2$ is a unit in $A$, every member\nof $I_A$ fixes every $y\\in L$ with $y^2=-1$. Indeed $y^{-1}=-y$, so\n$y$ belongs to $A$ and is a unit there. An automorphism sends $y$ to\n$y$ or $-y$. Their residues are distinct because $2$ and $y$ are units,\nwhereas inertia fixes residues. The second possibility is excluded.","its decomposition subgroup\nin $\\operatorname{Aut}_K(L)$ and put\n\\[\nI_A=\\ker\\bigl(D_A\\longrightarrow\\operatorname{Aut}(k_A)\\bigr),\n\\]\nwhere $k_A$ is the residue field. If $2$ is a unit in $A$, every member\nof $I_A$ fixes every $y\\in L$ with $y^2=-1$. Indeed $y^{-1}=-y$, so\n$y$ belongs to $A$ and is a unit there. An automorphism sends $y$ to\n$y$ or $-y$. Their residues are distinct because $2$ and $y$ are units,\nwhereas inertia fixes residues. The second possibility is excluded.","For every prime $p$, Lean constructs a valuation subring\n$A\\subseteq\\overline{\\Q_p}$ containing the integral image of $\\Z_p$\nsuch that the induced map $\\Z_p\\to A$ is local. For odd $p$, two is\na unit. The preceding argument applies to every valuation subring\ncontaining that integral image, and hence every member of\n$\\mathcal P_q(B)$ is trivial on its inertia subgroup after actual local\nrestriction. The existence theorem supplies a locally dominating choice,\nso this assertion is not vacuous.","This inertia is the actual residue-action kernel inside $D_A$. No equality\n$D_A=G_{\\Q_p}$ is established in this block. Nor is ramification at two\nproved: the nontrivial full local character there is a separate result.\nAt three the $\\Z_2$-valued sign character is nontrivial on the full local\nGalois group but trivial on every inertia subgroup covered by the theorem.\nThus local triviality and the stated inertia triviality are distinguished\nby checked arithmetic examples.","The local conditions survive coefficient completion by a proved theorem.\nFor an $I$-adically separated commutative ring $R$,\n\\[\n\\operatorname{LocallyTrivial}_K(\\chi)\n\\quad\\Longleftrightarrow\\quad\n\\operatorname{LocallyTrivial}_K(\\chi\\bmod I^m)\n\\text{ for every }m\\ge1.\n\\]\nIf $-1$ has a square root in $K$, every packet member satisfies the\ncondition. Otherwise the condition says that the character is globally\ntrivial, and separatedness detects that equality through all reductions.\nThis proves the closure statement for the independently defined arithmetic\nproperty.","Suppose now that $R$ is $I$-adically complete and that every positive\nquotient $R/I^m$ is finite. Let $C_m\\subseteq\\mathcal P_q(R/I^m)$ be\nallowed subsets preserved by reduction, and let $(p_s)_{s\\in S}$ be any\nindexed family of arithmetic primes. The index set $S$ need not be finite.\nLean proves the simultaneous arithmetic completion criterion\n\\[\n\\begin{split}\n&\\exists\\chi\\in\\mathcal P_q(R),\\quad\n \\chi\\text{ continuous},\\quad\n \\chi\\bmod I^m\\in C_m\\ (m\\ge1),\\quad\n \\chi_{\\Q_{p_s}}=1\\ (s\\in S)","Suppose now that $R$ is $I$-adically complete and that every positive\nquotient $R/I^m$ is finite. Let $C_m\\subseteq\\mathcal P_q(R/I^m)$ be\nallowed subsets preserved by reduction, and let $(p_s)_{s\\in S}$ be any\nindexed family of arithmetic primes. The index set $S$ need not be finite.\nLean proves the simultaneous arithmetic completion criterion\n\\[\n\\begin{split}\n&\\exists\\chi\\in\\mathcal P_q(R),\\quad\n \\chi\\text{ continuous},\\quad\n \\chi\\bmod I^m\\in C_m\\ (m\\ge1),\\quad\n \\chi_{\\Q_{p_s}}=1\\ (s\\in S)\\\\\n&\\quad\\Longleftrightarrow\\quad\n \\forall m\\ge1,\\ \\exists\\psi\\in C_m,\\quad\n \\forall s\\in S,\\ p_s\\equiv1\\pmod4\\ \\lor\\ \\psi=1.\n\\end{split}\n\\]\nThe constructed character also satisfies every odd-prime valuation-inertia\ncondition stated above. Finite-branch compactness selects all finite\ncharacters together; no coherent branch or surjective transition maps\nare assumed. The new local-closure implication is proved, not supplied.\nEvery-level occupancy remains a substantive hypothesis. One nonsplit\nselected prime forces the completed global character to be trivial.","Write $\\rho_i$ for the supplied constituents, $c_i$ for the carrier multiplicities and $m_i$ for the selected multiplicities, with $m_i\\leq c_i$. At a distinguished coefficient place $\\lambda$, the input presentations have the form\n\\[\n V_{\\mathrm{car},\\lambda}\\simeq\\bigoplus_i\\rho_{i,\\lambda}^{\\oplus c_i},\n \\qquad U_{0,\\lambda}\\simeq\\bigoplus_i\\rho_{i,\\lambda}^{\\oplus m_i}.\n\\]\nThese are actual equivariant presentations. They are not inferred from an unexplained equality of numerical traces. The selected family\n$\\bigoplus_i\\rho_{i,\\lambda}^{\\oplus m_i}$\nhas a split equivariant inclusion in the carrier. The finite trace sample uniquely determines the multiplicity vector in the declared constituent packet.","icities, with $m_i\\leq c_i$. At a distinguished coefficient place $\\lambda$, the input presentations have the form\n\\[\n V_{\\mathrm{car},\\lambda}\\simeq\\bigoplus_i\\rho_{i,\\lambda}^{\\oplus c_i},\n \\qquad U_{0,\\lambda}\\simeq\\bigoplus_i\\rho_{i,\\lambda}^{\\oplus m_i}.\n\\]\nThese are actual equivariant presentations. They are not inferred from an unexplained equality of numerical traces. The selected family\n$\\bigoplus_i\\rho_{i,\\lambda}^{\\oplus m_i}$\nhas a split equivariant inclusion in the carrier. The finite trace sample uniquely determines the multiplicity vector in the declared constituent packet.","ed by the formal endpoint. A nonzero occupied conclusion additionally requires the manuscript's declared nonzero-multiplicity premise and nonzero constituents. The module-support/Fitting-ideal alternative is logically separate: its checked support formalization is inherited and is not re-proved by this final module. No unrestricted companion existence, automorphy, carrier existence, or universal $p$-adic Hodge comparison follows from this conditional assembly.","lds, derives unramifiedness outside the discriminant, and proves independence from Frobenius choices. Its general arithmetic coverage equivalence retains the prime-existence assertion as a premise. Block 25 proves the system-to-member isomorphism bridge under the required overlap density and proves actual arithmetic Frobenius separation for representations of a fixed cyclotomic Galois group over $\\Q$. General arithmetic Frobenius coverage remains unproved here, so Claim 9.1 remains partial. Claim 11.2 is checked as the full-spine existence boundary with a nonzero joint eigenvector in every supplied required packet. The residual, higher-depth, and positive-trace requirements are retained; universal arithmetic occupancy is not proved. Claim 8.2 is checked as a universally quantified conditional completion theorem, retaining its explicit local-closure and finite-level-existence hypotheses. Earlier accounting included verification of particular arithmetic application premises in this claim; those applications remain separate and unfinished. For the fixed quadratic packet of actual $G_{\\Q…","; universal arithmetic occupancy is not proved. Claim 8.2 is checked as a universally quantified conditional completion theorem, retaining its explicit local-closure and finite-level-existence hypotheses. Earlier accounting included verification of particular arithmetic application premises in this claim; those applications remain separate and unfinished. For the fixed quadratic packet of actual $G_{\\Q}$ characters, local triviality after the field-induced map from $G_{\\Q_p}$ is proved equivalent to $p\\equiv1\\pmod4$ or global triviality. Its preservation under coefficient completion and simultaneous finite-level existence criterion are proved. The packet characters are trivial on actual valuation inertia at every odd prime, with a locally dominating valuation constructed. The equality of the valuation decomposition group with the entire local Galois group is not established, and ramification at two is not claimed. The full packet identifications, Galois-to-Weil--Deligne attachment, and coefficient and Hodge--Tate/period comparisons remain unfinished.\n\\end{minipage}}\n\\end{center}","a universally quantified conditional completion theorem, retaining its explicit local-closure and finite-level-existence hypotheses. Earlier accounting included verification of particular arithmetic application premises in this claim; those applications remain separate and unfinished. For the fixed quadratic packet of actual $G_{\\Q}$ characters, local triviality after the field-induced map from $G_{\\Q_p}$ is proved equivalent to $p\\equiv1\\pmod4$ or global triviality. Its preservation under coefficient completion and simultaneous finite-level existence criterion are proved. The packet characters are trivial on actual valuation inertia at every odd prime, with a locally dominating valuation constructed. The equality of the valuation decomposition group with the entire local Galois group is not established, and ramification at two is not claimed. The full packet identifications, Galois-to-Weil--Deligne attachment, and coefficient and Hodge--Tate/period comparisons remain unfinished.\n\\end{minipage}}\n\\end{center}","al completion theorem, retaining its explicit local-closure and finite-level-existence hypotheses. Earlier accounting included verification of particular arithmetic application premises in this claim; those applications remain separate and unfinished. For the fixed quadratic packet of actual $G_{\\Q}$ characters, local triviality after the field-induced map from $G_{\\Q_p}$ is proved equivalent to $p\\equiv1\\pmod4$ or global triviality. Its preservation under coefficient completion and simultaneous finite-level existence criterion are proved. The packet characters are trivial on actual valuation inertia at every odd prime, with a locally dominating valuation constructed. The equality of the valuation decomposition group with the entire local Galois group is not established, and ramification at two is not claimed. The full packet identifications, Galois-to-Weil--Deligne attachment, and coefficient and Hodge--Tate/period comparisons remain unfinished.\n\\end{minipage}}\n\\end{center}","\\bibitem{Chenevier2014}\nGa{\\\"e}tan Chenevier.\n\\newblock The {$p$}-adic analytic space of pseudocharacters of a profinite\n  group and {Pseudorepresentations} over arbitrary rings.\n\\newblock In Fred Diamond, Payman~L. Kassaei, and Minhyong Kim, editors, {\\em\n  Automorphic Forms and Galois Representations. Volume 1}, volume 414 of {\\em\n  London Mathematical Society Lecture Note Series}, pages 221--285. Cambridge\n  University Press, Cambridge, 2014.","\\bibitem{Denef1984}\nJan Denef.\n\\newblock The rationality of the {Poincar{\\'e}} series associated to the\n  {$p$}-adic points on a variety.\n\\newblock {\\em Inventiones Mathematicae}, 77:1--23, 1984.","\\bibitem{Kisin2010}\nMark Kisin.\n\\newblock Deformations of {$G_{\\mathbb Q_p}$} and {$\\mathrm{GL}_2(\\mathbb\n  Q_p)$} representations.\n\\newblock In Laurent Berger, Christophe Breuil, and Pierre Colmez, editors,\n  {\\em Repr{\\'e}sentations {$p$}-adiques de Groupes {$p$}-adiques II:\n  Repr{\\'e}sentations de {$\\mathrm{GL}_2(\\mathbb Q_p)$} et\n  {$(\\varphi,\\Gamma)$}-modules}, volume 330 of {\\em Ast{\\'e}risque}, pages\n  511--528. Soci{\\'e}t{\\'e} Math{\\'e}matique de France, Paris, 2010.","\\bibitem{LarsenPink1992}\nMichael Larsen and Richard Pink.\n\\newblock On {$\\ell$}-independence of algebraic monodromy groups in compatible\n  systems of representations.\n\\newblock {\\em Inventiones Mathematicae}, 107(3):603--636, 1992.","\\bibitem{Mazur1989}\nBarry Mazur.\n\\newblock Deforming {Galois} representations.\n\\newblock In Yasutaka Ihara, Kenneth Ribet, and Jean-Pierre Serre, editors,\n  {\\em Galois Groups over {$\\mathbb Q$}}, volume~16 of {\\em Mathematical\n  Sciences Research Institute Publications}, pages 385--437. Springer, New\n  York, 1989.","\\bibitem{Serre1998}\nJean-Pierre Serre.\n\\newblock {\\em Abelian {$\\ell$}-adic Representations and Elliptic Curves}.\n\\newblock A K Peters, Wellesley, Massachusetts, 1998.","\\paragraph{Scope and notation.}\nLet $K$ be a number field of degree $d\\geq1$, with ring of integers\n$\\OO=\\OO_K$. Write $\\Nm I=\\#(\\OO/I)$ for the absolute norm of a nonzero\nintegral ideal and $n(x)=\\Nm_{K/\\mathbb Q}(x)\\in\\ZZ$ for $x\\in\\OO$.\nThus $\\Nm(x)=\\absn{n(x)}$. Let $\\iota$ be the mixed embedding in its\nEuclidean coordinates, $L(J)=\\iota(J)$, and $D$ the actual unit-norm\nfundamental cone used in the preceding counting block. Its norm function\nscales by $t^d$ under a positive real dilation $t$.\nThe full-paper ledger remains \\textbf{17 checked claims and 3 partial claims}.","egree $d\\geq1$, with ring of integers\n$\\OO=\\OO_K$. Write $\\Nm I=\\#(\\OO/I)$ for the absolute norm of a nonzero\nintegral ideal and $n(x)=\\Nm_{K/\\mathbb Q}(x)\\in\\ZZ$ for $x\\in\\OO$.\nThus $\\Nm(x)=\\absn{n(x)}$. Let $\\iota$ be the mixed embedding in its\nEuclidean coordinates, $L(J)=\\iota(J)$, and $D$ the actual unit-norm\nfundamental cone used in the preceding counting block. Its norm function\nscales by $t^d$ under a positive real dilation $t$.\nThe full-paper ledger remains \\textbf{17 checked claims and 3 partial claims}.","$\\OO=\\OO_K$. Write $\\Nm I=\\#(\\OO/I)$ for the absolute norm of a nonzero\nintegral ideal and $n(x)=\\Nm_{K/\\mathbb Q}(x)\\in\\ZZ$ for $x\\in\\OO$.\nThus $\\Nm(x)=\\absn{n(x)}$. Let $\\iota$ be the mixed embedding in its\nEuclidean coordinates, $L(J)=\\iota(J)$, and $D$ the actual unit-norm\nfundamental cone used in the preceding counting block. Its norm function\nscales by $t^d$ under a positive real dilation $t$.\nThe full-paper ledger remains \\textbf{17 checked claims and 3 partial claims}.","\\paragraph{Actual additive residues.}\nFix nonzero integral ideals $M\\subseteq J$, and put $q=[J:M]=\\#(J/M)$.\nThe formal quotient of ideal lattices is identified with the additive\ncongruence relation $x-a\\in M$ on actual elements of $J$. In particular,\nthe residue label is not an independent parameter attached to a count.","\\begin{proposition}\\label{prop:residue-count}\nThere is a constant $B\\geq0$, depending on $K,M,J$, such that, for every\n$a\\in J/M$ and every real $t\\geq1$,\n\\[\n \\left|\\#\\{x\\in J:x\\bmod M=a,\\ \\iota(x)\\in tD\\}\n -\\frac{\\vol(D)}{q\\,\\covol(L(J))}t^d\\right|\n \\leq B(t^{d-1}+1).\n\\]\nThe same constant works for all residues $a$.\n\\end{proposition}","The estimate is proved from the actual boundary charts and the fixed-lattice\ndiscrepancy theorem. The index factor comes from the covolume identity\n$\\covol(L(M))=q\\,\\covol(L(J))$.\nFor a complex weight $w:J/M\\to\\CC$, summing the estimates gives the leading\nterm\n\\[\n \\frac{\\vol(D)t^d}{q\\,\\covol(L(J))}\\sum_{a\\in J/M}w(a),\n\\]\nwith error at most $B\\sum_a\\absn{w(a)}(t^{d-1}+1)$.\nIf the finite sum of weights is zero, only that error remains.\nWriting $Y=t^d$ gives an error $O(Y^{1-1/d}+1)$ for the actual norm threshold.\nThese assertions include $d=1$.","charts and the fixed-lattice\ndiscrepancy theorem. The index factor comes from the covolume identity\n$\\covol(L(M))=q\\,\\covol(L(J))$.\nFor a complex weight $w:J/M\\to\\CC$, summing the estimates gives the leading\nterm\n\\[\n \\frac{\\vol(D)t^d}{q\\,\\covol(L(J))}\\sum_{a\\in J/M}w(a),\n\\]\nwith error at most $B\\sum_a\\absn{w(a)}(t^{d-1}+1)$.\nIf the finite sum of weights is zero, only that error remains.\nWriting $Y=t^d$ gives an error $O(Y^{1-1/d}+1)$ for the actual norm threshold.\nThese assertions include $d=1$.","attice\ndiscrepancy theorem. The index factor comes from the covolume identity\n$\\covol(L(M))=q\\,\\covol(L(J))$.\nFor a complex weight $w:J/M\\to\\CC$, summing the estimates gives the leading\nterm\n\\[\n \\frac{\\vol(D)t^d}{q\\,\\covol(L(J))}\\sum_{a\\in J/M}w(a),\n\\]\nwith error at most $B\\sum_a\\absn{w(a)}(t^{d-1}+1)$.\nIf the finite sum of weights is zero, only that error remains.\nWriting $Y=t^d$ gives an error $O(Y^{1-1/d}+1)$ for the actual norm threshold.\nThese assertions include $d=1$.","tity\n$\\covol(L(M))=q\\,\\covol(L(J))$.\nFor a complex weight $w:J/M\\to\\CC$, summing the estimates gives the leading\nterm\n\\[\n \\frac{\\vol(D)t^d}{q\\,\\covol(L(J))}\\sum_{a\\in J/M}w(a),\n\\]\nwith error at most $B\\sum_a\\absn{w(a)}(t^{d-1}+1)$.\nIf the finite sum of weights is zero, only that error remains.\nWriting $Y=t^d$ gives an error $O(Y^{1-1/d}+1)$ for the actual norm threshold.\nThese assertions include $d=1$.","\\newpage\n\\paragraph{The residue norm and its sign.}\nFor $m\\in\\NN$, reducing the multiplication matrices of $\\OO$ modulo $m$\nand taking their determinants constructs a monoid homomorphism\n\\[\n \\overline n:\\OO/m\\OO\\longrightarrow\\ZZ/m\\ZZ,\n \\qquad \\overline n([x])=n(x)\\bmod m.\n\\]\nThe construction is proved independent of the integral basis.\nFor $m\\ne0$, its source is finite of cardinality $m^d$.\nConsequently $x-y\\in m\\OO$ implies $n(x)\\equiv n(y)\\pmod m$.\nNo surjectivity of $\\overline n$ is asserted.","\\paragraph{The residue norm and its sign.}\nFor $m\\in\\NN$, reducing the multiplication matrices of $\\OO$ modulo $m$\nand taking their determinants constructs a monoid homomorphism\n\\[\n \\overline n:\\OO/m\\OO\\longrightarrow\\ZZ/m\\ZZ,\n \\qquad \\overline n([x])=n(x)\\bmod m.\n\\]\nThe construction is proved independent of the integral basis.\nFor $m\\ne0$, its source is finite of cardinality $m^d$.\nConsequently $x-y\\in m\\OO$ implies $n(x)\\equiv n(y)\\pmod m$.\nNo surjectivity of $\\overline n$ is asserted.","norm and its sign.}\nFor $m\\in\\NN$, reducing the multiplication matrices of $\\OO$ modulo $m$\nand taking their determinants constructs a monoid homomorphism\n\\[\n \\overline n:\\OO/m\\OO\\longrightarrow\\ZZ/m\\ZZ,\n \\qquad \\overline n([x])=n(x)\\bmod m.\n\\]\nThe construction is proved independent of the integral basis.\nFor $m\\ne0$, its source is finite of cardinality $m^d$.\nConsequently $x-y\\in m\\OO$ implies $n(x)\\equiv n(y)\\pmod m$.\nNo surjectivity of $\\overline n$ is asserted.","the multiplication matrices of $\\OO$ modulo $m$\nand taking their determinants constructs a monoid homomorphism\n\\[\n \\overline n:\\OO/m\\OO\\longrightarrow\\ZZ/m\\ZZ,\n \\qquad \\overline n([x])=n(x)\\bmod m.\n\\]\nThe construction is proved independent of the integral basis.\nFor $m\\ne0$, its source is finite of cardinality $m^d$.\nConsequently $x-y\\in m\\OO$ implies $n(x)\\equiv n(y)\\pmod m$.\nNo surjectivity of $\\overline n$ is asserted.","The relation to the geometric sign is also proved. If $\\sigma$ ranges over\nreal embeddings and $\\tau$ over one embedding from each complex pair, then\n\\[\n n(x)=\\prod_{\\sigma}\\sigma(x)\\prod_{\\tau}\\absn{\\tau(x)}^2.\n\\]\nFor $x\\ne0$, the second factor is positive. Thus the sign of $n(x)$ is\nexactly the sign of the real-coordinate product. Cutting $D$ into its\npositive and negative portions $D_+,D_-$ adds no boundary outside the\nexisting frontier, and both portions retain the proved boundary atlas.\nThey are disjoint and their union is $D$.","lex pair, then\n\\[\n n(x)=\\prod_{\\sigma}\\sigma(x)\\prod_{\\tau}\\absn{\\tau(x)}^2.\n\\]\nFor $x\\ne0$, the second factor is positive. Thus the sign of $n(x)$ is\nexactly the sign of the real-coordinate product. Cutting $D$ into its\npositive and negative portions $D_+,D_-$ adds no boundary outside the\nexisting frontier, and both portions retain the proved boundary atlas.\nThey are disjoint and their union is $D$.","For $\\varepsilon\\in\\{1,-1\\}$ define\n$\\overline n_{\\varepsilon}(a)=\\varepsilon\\overline n(a)$.\nOn the corresponding sign portion,\n\\[\n \\Nm(x)\\bmod m=\\overline n_{\\varepsilon}([x]).\n\\]\nThe sign cannot be dropped: $-1\\equiv2\\pmod3$, whereas\n$\\Nm(-1)=1$ and $\\Nm(2)=2$ for principal ideals of $\\ZZ$.","\\paragraph{The exact ideal filter.}\nSuppose $IJ=(x)$, $n(x)$ has the indicated sign, and\n$\\gcd(\\Nm J,m)=1$. For $c\\in\\ZZ/m\\ZZ$,\n\\[\n \\Nm I\\equiv c\\pmod m\n \\quad\\Longleftrightarrow\\quad\n \\overline n_{\\varepsilon}([x])=c(\\Nm J\\bmod m).\n\\]\nThis gives an actual finite residue target, which is allowed to be empty.\nThe product identity before cancellation does not require coprimality.","Let $[J]=C^{-1}$ in $\\Cl(\\OO)$, and let $\\mathcal G(J)$ denote the\nactual fundamental-cone generator set. For every predicate $P:\\RR\\to\\mathrm{Prop}$\nand every real $X$, the norm-preserving correspondence gives\n\\begin{align*}\n &\\#\\{x\\in\\mathcal G(J):\\Nm(x)\\leq X\\Nm J,\\ P(\\Nm(x))\\}","Let $[J]=C^{-1}$ in $\\Cl(\\OO)$, and let $\\mathcal G(J)$ denote the\nactual fundamental-cone generator set. For every predicate $P:\\RR\\to\\mathrm{Prop}$\nand every real $X$, the norm-preserving correspondence gives\n\\begin{align*}\n &\\#\\{x\\in\\mathcal G(J):\\Nm(x)\\leq X\\Nm J,\\ P(\\Nm(x))\\}\\\\\n &\\qquad=\\#\\mu(K)\\,\n \\#\\{I\\ne0:[I]=C,\\ \\Nm I\\leq X,\\ P(\\Nm I\\,\\Nm J)\\}.\n\\end{align*}\nThe same norm-preserving equivalence carries arbitrary complex weights.\nIn particular, write $G_{J,\\chi}(Y)$ for the sum of\n$\\chi(\\Nm(x)\\bmod m)$ over these generators with $\\Nm(x)\\leq Y$, and\n$T_{C,\\chi}(X)$ for the corresponding sum over ideals in class $C$.\nFor multiplicative $\\chi$ preserving $0$ and $1$, the exact identity is\n\\begin{equation}\\label{eq:weighted-torsion}\n G_{J,\\chi}(X\\Nm J)\n =\\#\\mu(K)\\,\\chi(\\Nm J\\bmod m)\\,T_{C,\\chi}(X).\n\\end{equation}\nBoth signs are included before applying this full roots-of-unity factor.\nA root of unity can interchange the sign portions, so that factor is not\nautomatically valid for either portion separately.","&\\qquad=\\#\\mu(K)\\,\n \\#\\{I\\ne0:[I]=C,\\ \\Nm I\\leq X,\\ P(\\Nm I\\,\\Nm J)\\}.\n\\end{align*}\nThe same norm-preserving equivalence carries arbitrary complex weights.\nIn particular, write $G_{J,\\chi}(Y)$ for the sum of\n$\\chi(\\Nm(x)\\bmod m)$ over these generators with $\\Nm(x)\\leq Y$, and\n$T_{C,\\chi}(X)$ for the corresponding sum over ideals in class $C$.\nFor multiplicative $\\chi$ preserving $0$ and $1$, the exact identity is\n\\begin{equation}\\label{eq:weighted-torsion}\n G_{J,\\chi}(X\\Nm J)\n =\\#\\mu(K)\\,\\chi(\\Nm J\\bmod m)\\,T_{C,\\chi}(X).\n\\end{equation}\nBoth signs are included before applying this full roots-of-unity factor.\nA root of unity can interchange the sign portions, so that factor is not\nautomatically valid for either portion separately.","=C,\\ \\Nm I\\leq X,\\ P(\\Nm I\\,\\Nm J)\\}.\n\\end{align*}\nThe same norm-preserving equivalence carries arbitrary complex weights.\nIn particular, write $G_{J,\\chi}(Y)$ for the sum of\n$\\chi(\\Nm(x)\\bmod m)$ over these generators with $\\Nm(x)\\leq Y$, and\n$T_{C,\\chi}(X)$ for the corresponding sum over ideals in class $C$.\nFor multiplicative $\\chi$ preserving $0$ and $1$, the exact identity is\n\\begin{equation}\\label{eq:weighted-torsion}\n G_{J,\\chi}(X\\Nm J)\n =\\#\\mu(K)\\,\\chi(\\Nm J\\bmod m)\\,T_{C,\\chi}(X).\n\\end{equation}\nBoth signs are included before applying this full roots-of-unity factor.\nA root of unity can interchange the sign portions, so that factor is not\nautomatically valid for either portion separately.","\\newpage\n\\paragraph{Ordinary class characters: an unconditional power saving.}\nThe preceding block proved, for every ordinary ideal class $C$,\n\\[\n A_C(X)=c_KX+O_{K,C}(X^{1-1/d}),\\qquad\n c_K=\\frac{2^{r_1}(2\\pi)^{r_2}R_K}{\\#\\mu(K)\\sqrt{\\absn{\\Delta_K}}}.\n\\]\nHere $A_C$ counts actual nonzero integral ideals, $r_1,r_2$ count the real\nand complex places, $R_K$ is the regulator, and $\\Delta_K$ is the discriminant. For a weight\n$w:\\Cl(\\OO)\\to\\CC$, define the actual finite ideal sum\n\\[\n S_w(X)=\\sum_{\\substack{I\\ne0","ary class characters: an unconditional power saving.}\nThe preceding block proved, for every ordinary ideal class $C$,\n\\[\n A_C(X)=c_KX+O_{K,C}(X^{1-1/d}),\\qquad\n c_K=\\frac{2^{r_1}(2\\pi)^{r_2}R_K}{\\#\\mu(K)\\sqrt{\\absn{\\Delta_K}}}.\n\\]\nHere $A_C$ counts actual nonzero integral ideals, $r_1,r_2$ count the real\nand complex places, $R_K$ is the regulator, and $\\Delta_K$ is the discriminant. For a weight\n$w:\\Cl(\\OO)\\to\\CC$, define the actual finite ideal sum\n\\[\n S_w(X)=\\sum_{\\substack{I\\ne0","e preceding block proved, for every ordinary ideal class $C$,\n\\[\n A_C(X)=c_KX+O_{K,C}(X^{1-1/d}),\\qquad\n c_K=\\frac{2^{r_1}(2\\pi)^{r_2}R_K}{\\#\\mu(K)\\sqrt{\\absn{\\Delta_K}}}.\n\\]\nHere $A_C$ counts actual nonzero integral ideals, $r_1,r_2$ count the real\nand complex places, $R_K$ is the regulator, and $\\Delta_K$ is the discriminant. For a weight\n$w:\\Cl(\\OO)\\to\\CC$, define the actual finite ideal sum\n\\[\n S_w(X)=\\sum_{\\substack{I\\ne0\\\\\\Nm I\\leq X}}w([I]).\n\\]\nExact grouping by ideal class gives $S_w(X)=\\sum_C w(C)A_C(X)$.\nIf $\\sum_Cw(C)=0$, the common leading coefficient cancels, so there is\n$B_w\\geq0$ such that\n\\[\n \\absn{S_w(X)}\\leq B_wX^{1-1/d}\\qquad(X\\geq1).\n\\]\nFor every nontrivial character $\\chi:\\Cl(\\OO)\\to\\CC^{\\times}$,\nthe finite character sum is zero. Thus the displayed estimate applies\nwithout an additional arithmetic cancellation hypothesis.\nThis conclusion concerns ordinary class characters; it does not identify\nthem with the further norm-residue characters needed for Frobenius separation.","Nm I\\leq X}}w([I]).\n\\]\nExact grouping by ideal class gives $S_w(X)=\\sum_C w(C)A_C(X)$.\nIf $\\sum_Cw(C)=0$, the common leading coefficient cancels, so there is\n$B_w\\geq0$ such that\n\\[\n \\absn{S_w(X)}\\leq B_wX^{1-1/d}\\qquad(X\\geq1).\n\\]\nFor every nontrivial character $\\chi:\\Cl(\\OO)\\to\\CC^{\\times}$,\nthe finite character sum is zero. Thus the displayed estimate applies\nwithout an additional arithmetic cancellation hypothesis.\nThis conclusion concerns ordinary class characters; it does not identify\nthem with the further norm-residue characters needed for Frobenius separation.","\\paragraph{An actual Mellin endpoint.}\nThe counting functions are constant on the integer norm steps. Their\nmeasurability, and hence that of $S_w$, is proved from this fact.\nFor zero-sum $w$, define\n\\[\n F_w(h)=(1+h)\\int_1^{\\infty}x^{-2-h}S_w(x)\\,dx\n \\qquad(h\\geq0).\n\\]\nSince $1-1/d<1$, the logarithmic first moment\n\\[\n M_w=\\int_1^{\\infty}(1+\\log x)\\,\\absn{x^{-2}S_w(x)}\\,dx\n\\]\nis finite. Lean proves the explicit estimate\n$\\absn{F_w(h)-F_w(0)}\\leq hM_w$ for every $h\\geq0$, and therefore\n$F_w(h)=F_w(0)+O(h)$ as $h\\downarrow0$.\nThe function here is the displayed integral of the actual ideal sum.\nIts identification with the intended arithmetic Euler product is a separate step.","llin endpoint.}\nThe counting functions are constant on the integer norm steps. Their\nmeasurability, and hence that of $S_w$, is proved from this fact.\nFor zero-sum $w$, define\n\\[\n F_w(h)=(1+h)\\int_1^{\\infty}x^{-2-h}S_w(x)\\,dx\n \\qquad(h\\geq0).\n\\]\nSince $1-1/d<1$, the logarithmic first moment\n\\[\n M_w=\\int_1^{\\infty}(1+\\log x)\\,\\absn{x^{-2}S_w(x)}\\,dx\n\\]\nis finite. Lean proves the explicit estimate\n$\\absn{F_w(h)-F_w(0)}\\leq hM_w$ for every $h\\geq0$, and therefore\n$F_w(h)=F_w(0)+O(h)$ as $h\\downarrow0$.\nThe function here is the displayed integral of the actual ideal sum.\nIts identification with the intended arithmetic Euler product is a separate step.","constant on the integer norm steps. Their\nmeasurability, and hence that of $S_w$, is proved from this fact.\nFor zero-sum $w$, define\n\\[\n F_w(h)=(1+h)\\int_1^{\\infty}x^{-2-h}S_w(x)\\,dx\n \\qquad(h\\geq0).\n\\]\nSince $1-1/d<1$, the logarithmic first moment\n\\[\n M_w=\\int_1^{\\infty}(1+\\log x)\\,\\absn{x^{-2}S_w(x)}\\,dx\n\\]\nis finite. Lean proves the explicit estimate\n$\\absn{F_w(h)-F_w(0)}\\leq hM_w$ for every $h\\geq0$, and therefore\n$F_w(h)=F_w(0)+O(h)$ as $h\\downarrow0$.\nThe function here is the displayed integral of the actual ideal sum.\nIts identification with the intended arithmetic Euler product is a separate step.","norm steps. Their\nmeasurability, and hence that of $S_w$, is proved from this fact.\nFor zero-sum $w$, define\n\\[\n F_w(h)=(1+h)\\int_1^{\\infty}x^{-2-h}S_w(x)\\,dx\n \\qquad(h\\geq0).\n\\]\nSince $1-1/d<1$, the logarithmic first moment\n\\[\n M_w=\\int_1^{\\infty}(1+\\log x)\\,\\absn{x^{-2}S_w(x)}\\,dx\n\\]\nis finite. Lean proves the explicit estimate\n$\\absn{F_w(h)-F_w(0)}\\leq hM_w$ for every $h\\geq0$, and therefore\n$F_w(h)=F_w(0)+O(h)$ as $h\\downarrow0$.\nThe function here is the displayed integral of the actual ideal sum.\nIts identification with the intended arithmetic Euler product is a separate step.","rability, and hence that of $S_w$, is proved from this fact.\nFor zero-sum $w$, define\n\\[\n F_w(h)=(1+h)\\int_1^{\\infty}x^{-2-h}S_w(x)\\,dx\n \\qquad(h\\geq0).\n\\]\nSince $1-1/d<1$, the logarithmic first moment\n\\[\n M_w=\\int_1^{\\infty}(1+\\log x)\\,\\absn{x^{-2}S_w(x)}\\,dx\n\\]\nis finite. Lean proves the explicit estimate\n$\\absn{F_w(h)-F_w(0)}\\leq hM_w$ for every $h\\geq0$, and therefore\n$F_w(h)=F_w(0)+O(h)$ as $h\\downarrow0$.\nThe function here is the displayed integral of the actual ideal sum.\nIts identification with the intended arithmetic Euler product is a separate step.","For a finite family of nontrivial ordinary class characters, suppose functions\n$p,Z$ satisfy $h\\,p(h)\\to P\\ne0$, $h\\,Z(h)\\to Q\\ne0$, and, for all sufficiently\nsmall $h>0$,\n\\[\n Z(h)=\\left(p(h)\\prod_i F_{\\chi_i}(h)\\right)^e,\n \\qquad e\\in\\NN.\n\\]\nThe checked real-axis criterion then gives $e=1$ and\n$F_{\\chi_i}(0)\\ne0$ for every $i$. The pole limits and product identity\nremain hypotheses of this application.","\\newpage\n\\paragraph{An actual unit action on ideal residues.}\nFix $m>0$ and a nonzero integral ideal $J$.\nMultiplication by $u\\in(\\OO/m\\OO)^{\\times}$ gives a proved permutation of\n$J/mJ$. Its inverse follows from $(ab-1)x\\in mJ$ when\n$ab-1\\in m\\OO$ and $x\\in J$. No surjectivity of\n$J/mJ\\to\\OO/m\\OO$ or coprimality condition on $J$ is used.\nThe well-defined residue norm on this quotient satisfies\n\\[\n \\overline n_J(u\\cdot a)=\\overline n(u)\\,\\overline n_J(a).\n\\]\nFor a multiplicative $\\chi:\\ZZ/m\\ZZ\\to\\CC$ preserving $1$, an exhibited\nunit with\n\\begin{equation}\\label{eq:unit-witness}\n u\\in(\\OO/m\\OO)^{\\times},\\qquad \\chi(\\overline n(u))\\ne1\n\\end{equation}\ntherefore forces\n\\[\n \\sum_{a\\in J/mJ}\\chi\\bigl(s\\,\\overline n_J(a)\\bigr)=0\n \\qquad(s\\in\\ZZ/m\\ZZ).\n\\]\nTaking $s=1,","\\newpage\n\\paragraph{An actual unit action on ideal residues.}\nFix $m>0$ and a nonzero integral ideal $J$.\nMultiplication by $u\\in(\\OO/m\\OO)^{\\times}$ gives a proved permutation of\n$J/mJ$. Its inverse follows from $(ab-1)x\\in mJ$ when\n$ab-1\\in m\\OO$ and $x\\in J$. No surjectivity of\n$J/mJ\\to\\OO/m\\OO$ or coprimality condition on $J$ is used.\nThe well-defined residue norm on this quotient satisfies\n\\[\n \\overline n_J(u\\cdot a)=\\overline n(u)\\,\\overline n_J(a).\n\\]\nFor a multiplicative $\\chi:\\ZZ/m\\ZZ\\to\\CC$ preserving $1$, an exhibited\nunit with\n\\begin{equation}\\label{eq:unit-witness}\n u\\in(\\OO/m\\OO)^{\\times},\\qquad \\chi(\\overline n(u))\\ne1\n\\end{equation}\ntherefore forces\n\\[\n \\sum_{a\\in J/mJ}\\chi\\bigl(s\\,\\overline n_J(a)\\bigr)=0\n \\qquad(s\\in\\ZZ/m\\ZZ).\n\\]\nTaking $s=1,-1$ gives the two finite means ne","\\newpage\n\\paragraph{An actual unit action on ideal residues.}\nFix $m>0$ and a nonzero integral ideal $J$.\nMultiplication by $u\\in(\\OO/m\\OO)^{\\times}$ gives a proved permutation of\n$J/mJ$. Its inverse follows from $(ab-1)x\\in mJ$ when\n$ab-1\\in m\\OO$ and $x\\in J$. No surjectivity of\n$J/mJ\\to\\OO/m\\OO$ or coprimality condition on $J$ is used.\nThe well-defined residue norm on this quotient satisfies\n\\[\n \\overline n_J(u\\cdot a)=\\overline n(u)\\,\\overline n_J(a).\n\\]\nFor a multiplicative $\\chi:\\ZZ/m\\ZZ\\to\\CC$ preserving $1$, an exhibited\nunit with\n\\begin{equation}\\label{eq:unit-witness}\n u\\in(\\OO/m\\OO)^{\\times},\\qquad \\chi(\\overline n(u))\\ne1\n\\end{equation}\ntherefore forces\n\\[\n \\sum_{a\\in J/mJ}\\chi\\bigl(s\\,\\overline n_J(a)\\bigr)=0\n \\qquad(s\\in\\ZZ/m\\ZZ).\n\\]\nTaking $s=1,-1$ gives the two finite means needed by the sign-corrected\nnorm weights. Thei","\\newpage\n\\paragraph{An actual unit action on ideal residues.}\nFix $m>0$ and a nonzero integral ideal $J$.\nMultiplication by $u\\in(\\OO/m\\OO)^{\\times}$ gives a proved permutation of\n$J/mJ$. Its inverse follows from $(ab-1)x\\in mJ$ when\n$ab-1\\in m\\OO$ and $x\\in J$. No surjectivity of\n$J/mJ\\to\\OO/m\\OO$ or coprimality condition on $J$ is used.\nThe well-defined residue norm on this quotient satisfies\n\\[\n \\overline n_J(u\\cdot a)=\\overline n(u)\\,\\overline n_J(a).\n\\]\nFor a multiplicative $\\chi:\\ZZ/m\\ZZ\\to\\CC$ preserving $1$, an exhibited\nunit with\n\\begin{equation}\\label{eq:unit-witness}\n u\\in(\\OO/m\\OO)^{\\times},\\qquad \\chi(\\overline n(u))\\ne1\n\\end{equation}\ntherefore forces\n\\[\n \\sum_{a\\in J/mJ}\\chi\\bigl(s\\,\\overline n_J(a)\\bigr)=0\n \\qquad(s\\in\\ZZ/m\\ZZ).\n\\]\nTaking $s=1,-1$ gives the two finite means needed by the sign-corrected\nnorm weights. Their vanishing follows from the actual p","\\newpage\n\\paragraph{An actual unit action on ideal residues.}\nFix $m>0$ and a nonzero integral ideal $J$.\nMultiplication by $u\\in(\\OO/m\\OO)^{\\times}$ gives a proved permutation of\n$J/mJ$. Its inverse follows from $(ab-1)x\\in mJ$ when\n$ab-1\\in m\\OO$ and $x\\in J$. No surjectivity of\n$J/mJ\\to\\OO/m\\OO$ or coprimality condition on $J$ is used.\nThe well-defined residue norm on this quotient satisfies\n\\[\n \\overline n_J(u\\cdot a)=\\overline n(u)\\,\\overline n_J(a).\n\\]\nFor a multiplicative $\\chi:\\ZZ/m\\ZZ\\to\\CC$ preserving $1$, an exhibited\nunit with\n\\begin{equation}\\label{eq:unit-witness}\n u\\in(\\OO/m\\OO)^{\\times},\\qquad \\chi(\\overline n(u))\\ne1\n\\end{equation}\ntherefore forces\n\\[\n \\sum_{a\\in J/mJ}\\chi\\bigl(s\\,\\overline n_J(a)\\bigr)=0\n \\qquad(s\\in\\ZZ/m\\ZZ).\n\\]\nTaking $s=1,-1$ gives the two finite means needed by the sign-corrected\nnorm weights. Their vanishing follows from the actual permutation.","integral ideal $J$.\nMultiplication by $u\\in(\\OO/m\\OO)^{\\times}$ gives a proved permutation of\n$J/mJ$. Its inverse follows from $(ab-1)x\\in mJ$ when\n$ab-1\\in m\\OO$ and $x\\in J$. No surjectivity of\n$J/mJ\\to\\OO/m\\OO$ or coprimality condition on $J$ is used.\nThe well-defined residue norm on this quotient satisfies\n\\[\n \\overline n_J(u\\cdot a)=\\overline n(u)\\,\\overline n_J(a).\n\\]\nFor a multiplicative $\\chi:\\ZZ/m\\ZZ\\to\\CC$ preserving $1$, an exhibited\nunit with\n\\begin{equation}\\label{eq:unit-witness}\n u\\in(\\OO/m\\OO)^{\\times},\\qquad \\chi(\\overline n(u))\\ne1\n\\end{equation}\ntherefore forces\n\\[\n \\sum_{a\\in J/mJ}\\chi\\bigl(s\\,\\overline n_J(a)\\bigr)=0\n \\qquad(s\\in\\ZZ/m\\ZZ).\n\\]\nTaking $s=1,-1$ gives the two finite means needed by the sign-corrected\nnorm weights. Their vanishing follows from the actual permutation.","on by $u\\in(\\OO/m\\OO)^{\\times}$ gives a proved permutation of\n$J/mJ$. Its inverse follows from $(ab-1)x\\in mJ$ when\n$ab-1\\in m\\OO$ and $x\\in J$. No surjectivity of\n$J/mJ\\to\\OO/m\\OO$ or coprimality condition on $J$ is used.\nThe well-defined residue norm on this quotient satisfies\n\\[\n \\overline n_J(u\\cdot a)=\\overline n(u)\\,\\overline n_J(a).\n\\]\nFor a multiplicative $\\chi:\\ZZ/m\\ZZ\\to\\CC$ preserving $1$, an exhibited\nunit with\n\\begin{equation}\\label{eq:unit-witness}\n u\\in(\\OO/m\\OO)^{\\times},\\qquad \\chi(\\overline n(u))\\ne1\n\\end{equation}\ntherefore forces\n\\[\n \\sum_{a\\in J/mJ}\\chi\\bigl(s\\,\\overline n_J(a)\\bigr)=0\n \\qquad(s\\in\\ZZ/m\\ZZ).\n\\]\nTaking $s=1,-1$ gives the two finite means needed by the sign-corrected\nnorm weights. Their vanishing follows from the actual permutation.","ved permutation of\n$J/mJ$. Its inverse follows from $(ab-1)x\\in mJ$ when\n$ab-1\\in m\\OO$ and $x\\in J$. No surjectivity of\n$J/mJ\\to\\OO/m\\OO$ or coprimality condition on $J$ is used.\nThe well-defined residue norm on this quotient satisfies\n\\[\n \\overline n_J(u\\cdot a)=\\overline n(u)\\,\\overline n_J(a).\n\\]\nFor a multiplicative $\\chi:\\ZZ/m\\ZZ\\to\\CC$ preserving $1$, an exhibited\nunit with\n\\begin{equation}\\label{eq:unit-witness}\n u\\in(\\OO/m\\OO)^{\\times},\\qquad \\chi(\\overline n(u))\\ne1\n\\end{equation}\ntherefore forces\n\\[\n \\sum_{a\\in J/mJ}\\chi\\bigl(s\\,\\overline n_J(a)\\bigr)=0\n \\qquad(s\\in\\ZZ/m\\ZZ).\n\\]\nTaking $s=1,-1$ gives the two finite means needed by the sign-corrected\nnorm weights. Their vanishing follows from the actual permutation.","ty of\n$J/mJ\\to\\OO/m\\OO$ or coprimality condition on $J$ is used.\nThe well-defined residue norm on this quotient satisfies\n\\[\n \\overline n_J(u\\cdot a)=\\overline n(u)\\,\\overline n_J(a).\n\\]\nFor a multiplicative $\\chi:\\ZZ/m\\ZZ\\to\\CC$ preserving $1$, an exhibited\nunit with\n\\begin{equation}\\label{eq:unit-witness}\n u\\in(\\OO/m\\OO)^{\\times},\\qquad \\chi(\\overline n(u))\\ne1\n\\end{equation}\ntherefore forces\n\\[\n \\sum_{a\\in J/mJ}\\chi\\bigl(s\\,\\overline n_J(a)\\bigr)=0\n \\qquad(s\\in\\ZZ/m\\ZZ).\n\\]\nTaking $s=1,-1$ gives the two finite means needed by the sign-corrected\nnorm weights. Their vanishing follows from the actual permutation.","norm on this quotient satisfies\n\\[\n \\overline n_J(u\\cdot a)=\\overline n(u)\\,\\overline n_J(a).\n\\]\nFor a multiplicative $\\chi:\\ZZ/m\\ZZ\\to\\CC$ preserving $1$, an exhibited\nunit with\n\\begin{equation}\\label{eq:unit-witness}\n u\\in(\\OO/m\\OO)^{\\times},\\qquad \\chi(\\overline n(u))\\ne1\n\\end{equation}\ntherefore forces\n\\[\n \\sum_{a\\in J/mJ}\\chi\\bigl(s\\,\\overline n_J(a)\\bigr)=0\n \\qquad(s\\in\\ZZ/m\\ZZ).\n\\]\nTaking $s=1,-1$ gives the two finite means needed by the sign-corrected\nnorm weights. Their vanishing follows from the actual permutation.","\\paragraph{The assembled ideal norm-character estimate.}\n\\begin{proposition}\\label{prop:norm-character}\nLet $C\\in\\Cl(\\OO)$, let $J\\ne0$ be an integral ideal with $[J]=C^{-1}$,\nand let $m>0$. Let $\\chi:\\ZZ/m\\ZZ\\to\\CC$ preserve multiplication,\n$0$, and $1$. Assume a residue-ring unit satisfying\n\\eqref{eq:unit-witness}, and assume\n$\\chi(\\Nm J\\bmod m)\\ne0$. Then\n\\[\n T_{C,\\chi}(X)\n =\\sum_{\\substack{I\\ne0,\\ [I]=C","\\paragraph{The assembled ideal norm-character estimate.}\n\\begin{proposition}\\label{prop:norm-character}\nLet $C\\in\\Cl(\\OO)$, let $J\\ne0$ be an integral ideal with $[J]=C^{-1}$,\nand let $m>0$. Let $\\chi:\\ZZ/m\\ZZ\\to\\CC$ preserve multiplication,\n$0$, and $1$. Assume a residue-ring unit satisfying\n\\eqref{eq:unit-witness}, and assume\n$\\chi(\\Nm J\\bmod m)\\ne0$. Then\n\\[\n T_{C,\\chi}(X)\n =\\sum_{\\substack{I\\ne0,\\ [I]=C\\\\\\Nm I\\leq X}}\n       \\chi(\\Nm I\\bmod m)\n =O_{K,J,m,\\chi}(X^{1-1/d}).\n\\]\nMore precisely, some $B\\geq0$ bounds its absolute value by\n$BX^{1-1/d}$ for every real $X\\geq1$.\n\\end{proposition}","Nm I\\leq X}}\n       \\chi(\\Nm I\\bmod m)\n =O_{K,J,m,\\chi}(X^{1-1/d}).\n\\]\nMore precisely, some $B\\geq0$ bounds its absolute value by\n$BX^{1-1/d}$ for every real $X\\geq1$.\n\\end{proposition}","The assembly identifies the residue sum with the sum over actual generators,\nrecombines the sign portions, and uses $Y=t^d$ to obtain\n$\\absn{G_{J,\\chi}(Y)}\\leq B_0(Y^{1-1/d}+1)$.\nEquation~\\eqref{eq:weighted-torsion} then permits division by the nonzero\nfactor $\\#\\mu(K)\\chi(\\Nm J\\bmod m)$.\nThe final theorem supplies its finite means and counting estimates internally.\nFor $d=1$ the exponent is zero and the conclusion is a bounded sum, so the\nargument does not require $d>1$.","tors,\nrecombines the sign portions, and uses $Y=t^d$ to obtain\n$\\absn{G_{J,\\chi}(Y)}\\leq B_0(Y^{1-1/d}+1)$.\nEquation~\\eqref{eq:weighted-torsion} then permits division by the nonzero\nfactor $\\#\\mu(K)\\chi(\\Nm J\\bmod m)$.\nThe final theorem supplies its finite means and counting estimates internally.\nFor $d=1$ the exponent is zero and the conclusion is a bounded sum, so the\nargument does not require $d>1$.","\\paragraph{What remains open.}\nNontriviality of $\\chi$ on the larger residue ring does not supply\n\\eqref{eq:unit-witness}: its restriction to the image of the norm on units\nmay be trivial. Witness existence for the intended characters, their\nremaining norm-image cases, and suitable inverse-class representatives with\n$\\chi(\\Nm J)\\ne0$ are still arithmetic obligations.\nMeasurability and the Mellin attachment for these norm-character sums remain\nto be checked. The compiled Mellin results above concern ordinary class characters.\nThe resulting estimates must also be attached to the required Euler-product\nidentities and prime-occurrence argument. The conditional pole criterion\ndoes not supply those identities. Claims 9.1, 9.2, and 11.1 remain partia","\\paragraph{What remains open.}\nNontriviality of $\\chi$ on the larger residue ring does not supply\n\\eqref{eq:unit-witness}: its restriction to the image of the norm on units\nmay be trivial. Witness existence for the intended characters, their\nremaining norm-image cases, and suitable inverse-class representatives with\n$\\chi(\\Nm J)\\ne0$ are still arithmetic obligations.\nMeasurability and the Mellin attachment for these norm-character sums remain\nto be checked. The compiled Mellin results above concern ordinary class characters.\nThe resulting estimates must also be attached to the required Euler-product\nidentities and prime-occurrence argument. The conditional pole criterion\ndoes not supply those identities. Claims 9.1, 9.2, and 11.1 remain partial,\nincluding the separate local compatibility obligations. The ledger stays\n\\textbf{17 checked / 3 partial}.","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\draw[black!25] (4.55,0) -- (4.55,4.45);\n\\draw[black!25] (9.25,0) -- (9.25,4.45);\n\\node[font=\\small,align=center] at (2.1,4.35) {A residual root};\n\\node[font=\\small,align=center] at (2.1,3.55) {$x^2+1=0$ over $\\mathbb F_2$};\n\\path[draw,fill=black!12] (2.1,2.55) circle (0.24);\n\\node[font=\\small,align=center] at (2.1,2.55) {$1$};\n\\node[font=\\small,align=center] at (2.1,1.45) {one point mod $2$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (2.1,0.55) {no point mod $4$};\n\\node[font=\\small,align=center] at (6.9,4.35) {A coherent branch};\n\\node[font=\\small,align=center] at (6.9,3.55) {$x^2+1=0$ over $\\mathbb Z_5$};\n\\path[draw,fill=white] (5.3,2.5) circle (0.29);\n\\node[font=\\small,align=cen","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line 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{$u_1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.25,2.3) {$u_2$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.25,1.64) {$u_3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.73,4.18) {$e_0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.39,4.18) {$e_1$};\n\\node[font=\\small,align=center,font=\\small] at (6.48,4) {$d_1:C^1\\to C^2$};\n\\path[draw=black!30,fill=white] (5.1,2.61) rectangle (5.79,3.3);\n\\node[font=\\small,align=center,font=\\small] at (5.445,2.955) {$0$};\n\\path[draw=black!30,fill=black!8] (5.79,2.61) rectangle (6.48,3.3);\n\\node[font=\\small,align=center,font=\\small] at (6.135,2.955) {$1$};\n\\path","[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.25,3.62) {$u_0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.25,2.96) {$u_1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.25,2.3) {$u_2$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.25,1.64) {$u_3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.73,4.18) {$e_0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.39,4.18) {$e_1$};\n\\node[font=\\small,align=center,font=\\small] at (6.48,4) {$d_1:C^1\\to C^2$};\n\\path[draw=black!30,fill=white] (5.1,2.61) rectangle (5.79,3.3);\n\\node[font=\\small,align=center,font=\\small] at (5.445,2.955) {$0$};\n\\path[draw=black!30,fill=black!8] (5.79,2.61) rectangle (6.48,3.3);\n\\node[font=\\small,align=center,font=\\small] at (6.135,2.955) {$1$};\n\\path[draw=black!30,fill=white] (6.48,2.61) rectangle (7.17,3.3);\n\\node[font=\\","u_0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.25,2.96) {$u_1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.25,2.3) {$u_2$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.25,1.64) {$u_3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.73,4.18) {$e_0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.39,4.18) {$e_1$};\n\\node[font=\\small,align=center,font=\\small] at (6.48,4) {$d_1:C^1\\to C^2$};\n\\path[draw=black!30,fill=white] (5.1,2.61) rectangle (5.79,3.3);\n\\node[font=\\small,align=center,font=\\small] at (5.445,2.955) {$0$};\n\\path[draw=black!30,fill=black!8] (5.79,2.61) rectangle (6.48,3.3);\n\\node[font=\\small,align=center,font=\\small] at (6.135,2.955) {$1$};\n\\path[draw=black!30,fill=white] (6.48,2.61) rectangle (7.17,3.3);\n\\node[font=\\small,align=center,font=\\small] at (6.825,2.955) {$0$};\n\\path[draw=black!","e] at (1.25,2.96) {$u_1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.25,2.3) {$u_2$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.25,1.64) {$u_3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.73,4.18) {$e_0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.39,4.18) {$e_1$};\n\\node[font=\\small,align=center,font=\\small] at (6.48,4) {$d_1:C^1\\to C^2$};\n\\path[draw=black!30,fill=white] (5.1,2.61) rectangle (5.79,3.3);\n\\node[font=\\small,align=center,font=\\small] at (5.445,2.955) {$0$};\n\\path[draw=black!30,fill=black!8] (5.79,2.61) rectangle (6.48,3.3);\n\\node[font=\\small,align=center,font=\\small] at (6.135,2.955) {$1$};\n\\path[draw=black!30,fill=white] (6.48,2.61) rectangle (7.17,3.3);\n\\node[font=\\small,align=center,font=\\small] at (6.825,2.955) {$0$};\n\\path[draw=black!30,fill=white] (7.17,2.61) rectangle (7.86,3.3);\n\\node[font=\\small,align=cen","or=east,font=\\scriptsize] at (1.25,1.64) {$u_3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.73,4.18) {$e_0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.39,4.18) {$e_1$};\n\\node[font=\\small,align=center,font=\\small] at (6.48,4) {$d_1:C^1\\to C^2$};\n\\path[draw=black!30,fill=white] (5.1,2.61) rectangle (5.79,3.3);\n\\node[font=\\small,align=center,font=\\small] at (5.445,2.955) {$0$};\n\\path[draw=black!30,fill=black!8] (5.79,2.61) rectangle (6.48,3.3);\n\\node[font=\\small,align=center,font=\\small] at (6.135,2.955) {$1$};\n\\path[draw=black!30,fill=white] (6.48,2.61) rectangle (7.17,3.3);\n\\node[font=\\small,align=center,font=\\small] at (6.825,2.955) {$0$};\n\\path[draw=black!30,fill=white] (7.17,2.61) rectangle (7.86,3.3);\n\\node[font=\\small,align=center,font=\\small] at (7.515,2.955) {$0$};\n\\path[draw=black!30,fill=white] (5.1,1.92) rectangle (5.79,2.61);\n\\node[font=\\small,align=ce","mall,align=center,font=\\scriptsize] at (2.39,4.18) {$e_1$};\n\\node[font=\\small,align=center,font=\\small] at (6.48,4) {$d_1:C^1\\to C^2$};\n\\path[draw=black!30,fill=white] (5.1,2.61) rectangle (5.79,3.3);\n\\node[font=\\small,align=center,font=\\small] at (5.445,2.955) {$0$};\n\\path[draw=black!30,fill=black!8] (5.79,2.61) rectangle (6.48,3.3);\n\\node[font=\\small,align=center,font=\\small] at (6.135,2.955) {$1$};\n\\path[draw=black!30,fill=white] (6.48,2.61) rectangle (7.17,3.3);\n\\node[font=\\small,align=center,font=\\small] at (6.825,2.955) {$0$};\n\\path[draw=black!30,fill=white] (7.17,2.61) rectangle (7.86,3.3);\n\\node[font=\\small,align=center,font=\\small] at (7.515,2.955) {$0$};\n\\path[draw=black!30,fill=white] (5.1,1.92) rectangle (5.79,2.61);\n\\node[font=\\small,align=center,font=\\small] at (5.445,2.265) {$0$};\n\\path[draw=black!30,fill=white] (5.79,1.92) rectangle (6.48,2.61);\n\\node[font=\\small,align=cen",";\n\\path[draw=black!30,fill=white] (5.1,2.61) rectangle (5.79,3.3);\n\\node[font=\\small,align=center,font=\\small] at (5.445,2.955) {$0$};\n\\path[draw=black!30,fill=black!8] (5.79,2.61) rectangle (6.48,3.3);\n\\node[font=\\small,align=center,font=\\small] at (6.135,2.955) {$1$};\n\\path[draw=black!30,fill=white] (6.48,2.61) rectangle (7.17,3.3);\n\\node[font=\\small,align=center,font=\\small] at (6.825,2.955) {$0$};\n\\path[draw=black!30,fill=white] (7.17,2.61) rectangle (7.86,3.3);\n\\node[font=\\small,align=center,font=\\small] at (7.515,2.955) {$0$};\n\\path[draw=black!30,fill=white] (5.1,1.92) rectangle (5.79,2.61);\n\\node[font=\\small,align=center,font=\\small] at (5.445,2.265) {$0$};\n\\path[draw=black!30,fill=white] (5.79,1.92) rectangle (6.48,2.61);\n\\node[font=\\small,align=center,font=\\small] at (6.135,2.265) {$0$};\n\\path[draw=black!30,fill=white] (6.48,1.92) rectangle (7.17,2.61);\n\\node[font=\\small,align=ce","\\path[draw=black!30,fill=black!8] (5.79,2.61) rectangle (6.48,3.3);\n\\node[font=\\small,align=center,font=\\small] at (6.135,2.955) {$1$};\n\\path[draw=black!30,fill=white] (6.48,2.61) rectangle (7.17,3.3);\n\\node[font=\\small,align=center,font=\\small] at (6.825,2.955) {$0$};\n\\path[draw=black!30,fill=white] (7.17,2.61) rectangle (7.86,3.3);\n\\node[font=\\small,align=center,font=\\small] at (7.515,2.955) {$0$};\n\\path[draw=black!30,fill=white] (5.1,1.92) rectangle (5.79,2.61);\n\\node[font=\\small,align=center,font=\\small] at (5.445,2.265) {$0$};\n\\path[draw=black!30,fill=white] (5.79,1.92) rectangle (6.48,2.61);\n\\node[font=\\small,align=center,font=\\small] at (6.135,2.265) {$0$};\n\\path[draw=black!30,fill=white] (6.48,1.92) rectangle (7.17,2.61);\n\\node[font=\\small,align=center,font=\\small] at (6.825,2.265) {$0$};\n\\path[draw=black!30,fill=white] (7.17,1.92) rectangle (7.86,2.61);\n\\node[font=\\small,align=c","};\n\\path[draw=black!30,fill=white] (6.48,2.61) rectangle (7.17,3.3);\n\\node[font=\\small,align=center,font=\\small] at (6.825,2.955) {$0$};\n\\path[draw=black!30,fill=white] (7.17,2.61) rectangle (7.86,3.3);\n\\node[font=\\small,align=center,font=\\small] at (7.515,2.955) {$0$};\n\\path[draw=black!30,fill=white] (5.1,1.92) rectangle (5.79,2.61);\n\\node[font=\\small,align=center,font=\\small] at (5.445,2.265) {$0$};\n\\path[draw=black!30,fill=white] (5.79,1.92) rectangle (6.48,2.61);\n\\node[font=\\small,align=center,font=\\small] at (6.135,2.265) {$0$};\n\\path[draw=black!30,fill=white] (6.48,1.92) rectangle (7.17,2.61);\n\\node[font=\\small,align=center,font=\\small] at (6.825,2.265) {$0$};\n\\path[draw=black!30,fill=white] (7.17,1.92) rectangle (7.86,2.61);\n\\node[font=\\small,align=center,font=\\small] at (7.515,2.265) {$0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.95,2.955) {$f_0$};\n\\node",";\n\\path[draw=black!30,fill=white] (7.17,2.61) rectangle (7.86,3.3);\n\\node[font=\\small,align=center,font=\\small] at (7.515,2.955) {$0$};\n\\path[draw=black!30,fill=white] (5.1,1.92) rectangle (5.79,2.61);\n\\node[font=\\small,align=center,font=\\small] at (5.445,2.265) {$0$};\n\\path[draw=black!30,fill=white] (5.79,1.92) rectangle (6.48,2.61);\n\\node[font=\\small,align=center,font=\\small] at (6.135,2.265) {$0$};\n\\path[draw=black!30,fill=white] (6.48,1.92) rectangle (7.17,2.61);\n\\node[font=\\small,align=center,font=\\small] at (6.825,2.265) {$0$};\n\\path[draw=black!30,fill=white] (7.17,1.92) rectangle (7.86,2.61);\n\\node[font=\\small,align=center,font=\\small] at (7.515,2.265) {$0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.95,2.955) {$f_0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.95,2.265) {$f_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5","\\path[draw=black!30,fill=white] (5.1,1.92) rectangle (5.79,2.61);\n\\node[font=\\small,align=center,font=\\small] at (5.445,2.265) {$0$};\n\\path[draw=black!30,fill=white] (5.79,1.92) rectangle (6.48,2.61);\n\\node[font=\\small,align=center,font=\\small] at (6.135,2.265) {$0$};\n\\path[draw=black!30,fill=white] (6.48,1.92) rectangle (7.17,2.61);\n\\node[font=\\small,align=center,font=\\small] at (6.825,2.265) {$0$};\n\\path[draw=black!30,fill=white] (7.17,1.92) rectangle (7.86,2.61);\n\\node[font=\\small,align=center,font=\\small] at (7.515,2.265) {$0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.95,2.955) {$f_0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.95,2.265) {$f_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.445,3.53) {$u_0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.135,3.53) {$u_1$};\n\\node[font=\\small,align=center,font=\\scri","\\path[draw=black!30,fill=white] (5.79,1.92) rectangle (6.48,2.61);\n\\node[font=\\small,align=center,font=\\small] at (6.135,2.265) {$0$};\n\\path[draw=black!30,fill=white] (6.48,1.92) rectangle (7.17,2.61);\n\\node[font=\\small,align=center,font=\\small] at (6.825,2.265) {$0$};\n\\path[draw=black!30,fill=white] (7.17,1.92) rectangle (7.86,2.61);\n\\node[font=\\small,align=center,font=\\small] at (7.515,2.265) {$0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.95,2.955) {$f_0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.95,2.265) {$f_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.445,3.53) {$u_0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.135,3.53) {$u_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.825,3.53) {$u_2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.515,3.53) {$u_3$};\n\\node[font=\\small,align=cent",",align=center,font=\\small] at (6.135,2.265) {$0$};\n\\path[draw=black!30,fill=white] (6.48,1.92) rectangle (7.17,2.61);\n\\node[font=\\small,align=center,font=\\small] at (6.825,2.265) {$0$};\n\\path[draw=black!30,fill=white] (7.17,1.92) rectangle (7.86,2.61);\n\\node[font=\\small,align=center,font=\\small] at (7.515,2.265) {$0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.95,2.955) {$f_0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.95,2.265) {$f_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.445,3.53) {$u_0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.135,3.53) {$u_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.825,3.53) {$u_2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.515,3.53) {$u_3$};\n\\node[font=\\small,align=center] at (11,3.7) {$d_1d_0=0$};\n\\node[font=\\small,align=center] at (11,2.65) {$H^0=\\lang","48,1.92) rectangle (7.17,2.61);\n\\node[font=\\small,align=center,font=\\small] at (6.825,2.265) {$0$};\n\\path[draw=black!30,fill=white] (7.17,1.92) rectangle (7.86,2.61);\n\\node[font=\\small,align=center,font=\\small] at (7.515,2.265) {$0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.95,2.955) {$f_0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.95,2.265) {$f_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.445,3.53) {$u_0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.135,3.53) {$u_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.825,3.53) {$u_2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.515,3.53) {$u_3$};\n\\node[font=\\small,align=center] at (11,3.7) {$d_1d_0=0$};\n\\node[font=\\small,align=center] at (11,2.65) {$H^0=\\langle[e_1]\\rangle$};\n\\node[font=\\small,align=center] at (11,1.85) {$H^1=\\langle[u_2],[u_3","] at (6.825,2.265) {$0$};\n\\path[draw=black!30,fill=white] (7.17,1.92) rectangle (7.86,2.61);\n\\node[font=\\small,align=center,font=\\small] at (7.515,2.265) {$0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.95,2.955) {$f_0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.95,2.265) {$f_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.445,3.53) {$u_0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.135,3.53) {$u_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.825,3.53) {$u_2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.515,3.53) {$u_3$};\n\\node[font=\\small,align=center] at (11,3.7) {$d_1d_0=0$};\n\\node[font=\\small,align=center] at (11,2.65) {$H^0=\\langle[e_1]\\rangle$};\n\\node[font=\\small,align=center] at (11,1.85) {$H^1=\\langle[u_2],[u_3]\\rangle$};\n\\node[font=\\small,align=center] at (11,1.05) {$H^2=\\langle[f_1","angle (7.86,2.61);\n\\node[font=\\small,align=center,font=\\small] at (7.515,2.265) {$0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.95,2.955) {$f_0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.95,2.265) {$f_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.445,3.53) {$u_0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.135,3.53) {$u_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.825,3.53) {$u_2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.515,3.53) {$u_3$};\n\\node[font=\\small,align=center] at (11,3.7) {$d_1d_0=0$};\n\\node[font=\\small,align=center] at (11,2.65) {$H^0=\\langle[e_1]\\rangle$};\n\\node[font=\\small,align=center] at (11,1.85) {$H^1=\\langle[u_2],[u_3]\\rangle$};\n\\node[font=\\small,align=center] at (11,1.05) {$H^2=\\langle[f_1]\\rangle$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (6.8,0.2",".265) {$0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.95,2.955) {$f_0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.95,2.265) {$f_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.445,3.53) {$u_0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.135,3.53) {$u_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.825,3.53) {$u_2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.515,3.53) {$u_3$};\n\\node[font=\\small,align=center] at (11,3.7) {$d_1d_0=0$};\n\\node[font=\\small,align=center] at (11,2.65) {$H^0=\\langle[e_1]\\rangle$};\n\\node[font=\\small,align=center] at (11,1.85) {$H^1=\\langle[u_2],[u_3]\\rangle$};\n\\node[font=\\small,align=center] at (11,1.05) {$H^2=\\langle[f_1]\\rangle$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (6.8,0.2) {Packet A is exact. 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Packet B has zero differentials and carries all cohomology.};\n\\end{tikzpicture}","er,anchor=east,font=\\scriptsize] at (4.95,2.265) {$f_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.445,3.53) {$u_0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.135,3.53) {$u_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.825,3.53) {$u_2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.515,3.53) {$u_3$};\n\\node[font=\\small,align=center] at (11,3.7) {$d_1d_0=0$};\n\\node[font=\\small,align=center] at (11,2.65) {$H^0=\\langle[e_1]\\rangle$};\n\\node[font=\\small,align=center] at (11,1.85) {$H^1=\\langle[u_2],[u_3]\\rangle$};\n\\node[font=\\small,align=center] at (11,1.05) {$H^2=\\langle[f_1]\\rangle$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (6.8,0.2) {Packet A is exact. Packet B has zero differentials and carries all cohomology.};\n\\end{tikzpicture}","node[font=\\small,align=center,font=\\scriptsize] at (5.445,3.53) {$u_0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.135,3.53) {$u_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.825,3.53) {$u_2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.515,3.53) {$u_3$};\n\\node[font=\\small,align=center] at (11,3.7) {$d_1d_0=0$};\n\\node[font=\\small,align=center] at (11,2.65) {$H^0=\\langle[e_1]\\rangle$};\n\\node[font=\\small,align=center] at (11,1.85) {$H^1=\\langle[u_2],[u_3]\\rangle$};\n\\node[font=\\small,align=center] at (11,1.05) {$H^2=\\langle[f_1]\\rangle$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (6.8,0.2) {Packet A is exact. Packet B has zero differentials and carries all cohomology.};\n\\end{tikzpicture}","node[font=\\small,align=center,font=\\scriptsize] at (6.135,3.53) {$u_1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.825,3.53) {$u_2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.515,3.53) {$u_3$};\n\\node[font=\\small,align=center] at (11,3.7) {$d_1d_0=0$};\n\\node[font=\\small,align=center] at (11,2.65) {$H^0=\\langle[e_1]\\rangle$};\n\\node[font=\\small,align=center] at (11,1.85) {$H^1=\\langle[u_2],[u_3]\\rangle$};\n\\node[font=\\small,align=center] at (11,1.05) {$H^2=\\langle[f_1]\\rangle$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (6.8,0.2) {Packet A is exact. Packet B has zero differentials and carries all cohomology.};\n\\end{tikzpicture}","ont=\\small,align=center,font=\\scriptsize] at (6.825,3.53) {$u_2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.515,3.53) {$u_3$};\n\\node[font=\\small,align=center] at (11,3.7) {$d_1d_0=0$};\n\\node[font=\\small,align=center] at (11,2.65) {$H^0=\\langle[e_1]\\rangle$};\n\\node[font=\\small,align=center] at (11,1.85) {$H^1=\\langle[u_2],[u_3]\\rangle$};\n\\node[font=\\small,align=center] at (11,1.05) {$H^2=\\langle[f_1]\\rangle$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (6.8,0.2) {Packet A is exact. Packet B has zero differentials and carries all cohomology.};\n\\end{tikzpicture}","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center,font=\\small] at (2.47,4.7) {valid $P_A^1$};\n\\path[draw=black!30,fill=black!8] (1.25,3.39) rectangle (1.86,4);\n\\node[font=\\small,align=center,font=\\small] at (1.555,3.695) {$1$};\n\\path[draw=black!30,fill=white] (1.86,3.39) rectangle (2.47,4);\n\\node[font=\\small,align=center,font=\\small] at (2.165,3.695) {$0$};\n\\path[draw=black!30,fill=white] (2.47,3.39) rectangle (3.08,4);\n\\node[font=\\small,align=center,font=\\small] at (2.775,3.695) {$0$};\n\\path[draw=black!30,fill=white] (3.08,3.39) rectangle (3.69,4);\n\\node[font=\\small,align=center,fon","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center,font=\\small] at 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(0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.45,3.3) {$182$};\n\\path[draw,fill=w","er,font=\\scriptsize] at (5.6,2.4) {$-x$};\n\\draw[->] (4.9,3.3) -- (2.25,3.3);\n\\draw[->] (4.9,1.5) -- (2.25,1.5);\n\\node[font=\\small,align=center] at (8.3,4.4) {mod $5^{3}$};\n\\path[draw,fill=white] (8.3,3.3) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.3,3.3) {$57$};\n\\path[draw,fill=white] (8.3,1.5) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.3,1.5) {$68$};\n\\draw[<->] (8.3,3.05) -- (8.3,1.75);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.75,2.4) {$-x$};\n\\draw[->] (8.05,3.3) -- (5.4,3.3);\n\\draw[->] (8.05,1.5) -- (5.4,1.5);\n\\node[font=\\small,align=center] at (11.45,4.4) {mod $5^{4}$};\n\\path[draw,fill=white] (11.45,3.3) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.45,3.3) {$182$};\n\\path[draw,fill=white] (11.45,1.5) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.45,1.5) {$443$};\n\\draw[<->] (","5);\n\\node[font=\\small,align=center] at (8.3,4.4) {mod $5^{3}$};\n\\path[draw,fill=white] (8.3,3.3) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.3,3.3) {$57$};\n\\path[draw,fill=white] (8.3,1.5) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.3,1.5) {$68$};\n\\draw[<->] (8.3,3.05) -- (8.3,1.75);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.75,2.4) {$-x$};\n\\draw[->] (8.05,3.3) -- (5.4,3.3);\n\\draw[->] (8.05,1.5) -- (5.4,1.5);\n\\node[font=\\small,align=center] at (11.45,4.4) {mod $5^{4}$};\n\\path[draw,fill=white] (11.45,3.3) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.45,3.3) {$182$};\n\\path[draw,fill=white] (11.45,1.5) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.45,1.5) {$443$};\n\\draw[<->] (11.45,3.05) -- (11.45,1.75);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.9,2.4) {$-x$};\n\\draw[->","align=center,font=\\scriptsize] at (8.3,3.3) {$57$};\n\\path[draw,fill=white] (8.3,1.5) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.3,1.5) {$68$};\n\\draw[<->] (8.3,3.05) -- (8.3,1.75);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.75,2.4) {$-x$};\n\\draw[->] (8.05,3.3) -- (5.4,3.3);\n\\draw[->] (8.05,1.5) -- (5.4,1.5);\n\\node[font=\\small,align=center] at (11.45,4.4) {mod $5^{4}$};\n\\path[draw,fill=white] (11.45,3.3) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.45,3.3) {$182$};\n\\path[draw,fill=white] (11.45,1.5) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.45,1.5) {$443$};\n\\draw[<->] (11.45,3.05) -- (11.45,1.75);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.9,2.4) {$-x$};\n\\draw[->] (11.2,3.3) -- (8.55,3.3);\n\\draw[->] (11.2,1.5) -- (8.55,1.5);\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,0.3) {Negat","enter,font=\\scriptsize] at (8.3,1.5) {$68$};\n\\draw[<->] (8.3,3.05) -- (8.3,1.75);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.75,2.4) {$-x$};\n\\draw[->] (8.05,3.3) -- (5.4,3.3);\n\\draw[->] (8.05,1.5) -- (5.4,1.5);\n\\node[font=\\small,align=center] at (11.45,4.4) {mod $5^{4}$};\n\\path[draw,fill=white] (11.45,3.3) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.45,3.3) {$182$};\n\\path[draw,fill=white] (11.45,1.5) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.45,1.5) {$443$};\n\\draw[<->] (11.45,3.05) -- (11.45,1.75);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.9,2.4) {$-x$};\n\\draw[->] (11.2,3.3) -- (8.55,3.3);\n\\draw[->] (11.2,1.5) -- (8.55,1.5);\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,0.3) {Negation commutes with reduction. Translation by $1$ fails already at $2\\mapsto3\\mapsto4$ mod $5$.};\n\\end{tikzpicture}","iptsize] at (8.75,2.4) {$-x$};\n\\draw[->] (8.05,3.3) -- (5.4,3.3);\n\\draw[->] (8.05,1.5) -- (5.4,1.5);\n\\node[font=\\small,align=center] at (11.45,4.4) {mod $5^{4}$};\n\\path[draw,fill=white] (11.45,3.3) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.45,3.3) {$182$};\n\\path[draw,fill=white] (11.45,1.5) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.45,1.5) {$443$};\n\\draw[<->] (11.45,3.05) -- (11.45,1.75);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.9,2.4) {$-x$};\n\\draw[->] (11.2,3.3) -- (8.55,3.3);\n\\draw[->] (11.2,1.5) -- (8.55,1.5);\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,0.3) {Negation commutes with reduction. Translation by $1$ fails already at $2\\mapsto3\\mapsto4$ mod $5$.};\n\\end{tikzpicture}","small,align=center] at (11.45,4.4) {mod $5^{4}$};\n\\path[draw,fill=white] (11.45,3.3) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.45,3.3) {$182$};\n\\path[draw,fill=white] (11.45,1.5) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.45,1.5) {$443$};\n\\draw[<->] (11.45,3.05) -- (11.45,1.75);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.9,2.4) {$-x$};\n\\draw[->] (11.2,3.3) -- (8.55,3.3);\n\\draw[->] (11.2,1.5) -- (8.55,1.5);\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,0.3) {Negation commutes with reduction. Translation by $1$ fails already at $2\\mapsto3\\mapsto4$ mod $5$.};\n\\end{tikzpicture}","te] (11.45,1.5) circle (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.45,1.5) {$443$};\n\\draw[<->] (11.45,3.05) -- (11.45,1.75);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.9,2.4) {$-x$};\n\\draw[->] (11.2,3.3) -- (8.55,3.3);\n\\draw[->] (11.2,1.5) -- (8.55,1.5);\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,0.3) {Negation commutes with reduction. Translation by $1$ fails already at $2\\mapsto3\\mapsto4$ mod $5$.};\n\\end{tikzpicture}","e (0.3);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.45,1.5) {$443$};\n\\draw[<->] (11.45,3.05) -- (11.45,1.75);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.9,2.4) {$-x$};\n\\draw[->] (11.2,3.3) -- (8.55,3.3);\n\\draw[->] (11.2,1.5) -- (8.55,1.5);\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,0.3) {Negation commutes with reduction. Translation by $1$ fails already at $2\\mapsto3\\mapsto4$ mod $5$.};\n\\end{tikzpicture}","ll,align=center,font=\\scriptsize] at (11.45,1.5) {$443$};\n\\draw[<->] (11.45,3.05) -- (11.45,1.75);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.9,2.4) {$-x$};\n\\draw[->] (11.2,3.3) -- (8.55,3.3);\n\\draw[->] (11.2,1.5) -- (8.55,1.5);\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,0.3) {Negation commutes with reduction. Translation by $1$ fails already at $2\\mapsto3\\mapsto4$ mod $5$.};\n\\end{tikzpicture}","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center] at (7,4.3) {$P(T)=T^2-5T+6=(T-2)(T-3)$};\n\\node[font=\\small,align=center,font=\\small] at (7.075,3.95) {coefficient images under the declared maps};\n\\path[draw=black!30,fill=black!8] (4.9,2.39) rectangle (6.35,3.25);\n\\node[font=\\small,align=center,font=\\small] at (5.625,2.82) {$1$};\n\\path[draw=black!30,fill=black!8] (6.35,2.39) rectangle (7.8,3.25);\n\\node[font=\\small,align=center,font=\\small] at (7.075,2.82) {$-5$};\n\\path[draw=black!30,fill=black!8] (7.8,2.39) rectangle (9.25,3.25);\n\\node[font=\\small,align=center,font=\\small] at (8.5","lth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center] at (7,4.3) {$P(T)=T^2-5T+6=(T-2)(T-3)$};\n\\node[font=\\small,align=center,font=\\small] at (7.075,3.95) {coefficient images under the declared maps};\n\\path[draw=black!30,fill=black!8] (4.9,2.39) rectangle (6.35,3.25);\n\\node[font=\\small,align=center,font=\\small] at (5.625,2.82) {$1$};\n\\path[draw=black!30,fill=black!8] (6.35,2.39) rectangle (7.8,3.25);\n\\node[font=\\small,align=center,font=\\small] at (7.075,2.82) {$-5$};\n\\path[draw=black!30,fill=black!8] (7.8,2.39) rectangle (9.25,3.25);\n\\node[font=\\small,align=center,font=\\small] at (8.525,2.82) {$6$};\n\\path[draw=black!30,fill=black!8] (4.9,1.53) rectangle (6.35,2.39);\n\\node[font=\\small,align=center,font=\\small] at (5.625,1.96) {$1$};\n\\path[draw=black!30,fill=white] (6.35,1.53) rectangle (7.8,2.39);\n\\node[font=\\small,align=","$P(T)=T^2-5T+6=(T-2)(T-3)$};\n\\node[font=\\small,align=center,font=\\small] at (7.075,3.95) {coefficient images under the declared maps};\n\\path[draw=black!30,fill=black!8] (4.9,2.39) rectangle (6.35,3.25);\n\\node[font=\\small,align=center,font=\\small] at (5.625,2.82) {$1$};\n\\path[draw=black!30,fill=black!8] (6.35,2.39) rectangle (7.8,3.25);\n\\node[font=\\small,align=center,font=\\small] at (7.075,2.82) {$-5$};\n\\path[draw=black!30,fill=black!8] (7.8,2.39) rectangle (9.25,3.25);\n\\node[font=\\small,align=center,font=\\small] at (8.525,2.82) {$6$};\n\\path[draw=black!30,fill=black!8] (4.9,1.53) rectangle (6.35,2.39);\n\\node[font=\\small,align=center,font=\\small] at (5.625,1.96) {$1$};\n\\path[draw=black!30,fill=white] (6.35,1.53) rectangle (7.8,2.39);\n\\node[font=\\small,align=center,font=\\small] at (7.075,1.96) {$0$};\n\\path[draw=black!30,fill=black!8] (7.8,1.53) rectangle (9.25,2.39);\n\\node[font=\\small,align=c","path[draw=black!30,fill=black!8] (4.9,2.39) rectangle (6.35,3.25);\n\\node[font=\\small,align=center,font=\\small] at (5.625,2.82) {$1$};\n\\path[draw=black!30,fill=black!8] (6.35,2.39) rectangle 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(9.25,2.39);\n\\node[font=\\small,align=center,font=\\small] at (8.525,1.96) {$1$};\n\\path[draw=black!30,fill=black!8] (4.9,0.67) rectangle (6.35,1.53);\n\\node[font=\\small,align=center,font=\\small] at (5.625,1.1) {$1$};\n\\path[draw=black!30,fill=black!8] (6.35,0.67) rectangle (7.8,1.53);\n\\node[font=\\small,align=center,font=\\small] at (7.075,1.1) {$2$};\n\\path[draw=black!30,fill=black!8] (7.8,0.67) rectangle (9.25,1.53);\n\\node[font=\\small,align=center,font=\\small] at (8.525,1.1) {$6$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,2.82) {$\\mathbb Z$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,1.96) {$\\mathbb F_5$};\n\\node[font=\\small,align=center,anchor=east,f",";\n\\path[draw=black!30,fill=black!8] (7.8,1.53) rectangle (9.25,2.39);\n\\node[font=\\small,align=center,font=\\small] at (8.525,1.96) {$1$};\n\\path[draw=black!30,fill=black!8] (4.9,0.67) rectangle (6.35,1.53);\n\\node[font=\\small,align=center,font=\\small] at (5.625,1.1) {$1$};\n\\path[draw=black!30,fill=black!8] (6.35,0.67) rectangle (7.8,1.53);\n\\node[font=\\small,align=center,font=\\small] at (7.075,1.1) {$2$};\n\\path[draw=black!30,fill=black!8] (7.8,0.67) rectangle (9.25,1.53);\n\\node[font=\\small,align=center,font=\\small] at (8.525,1.1) {$6$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,2.82) {$\\mathbb Z$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,1.96) {$\\mathbb F_5$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,1.1) {$\\mathbb F_7$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.625,3.48) {$T^2$};\n\\node[font=\\","};\n\\path[draw=black!30,fill=black!8] (4.9,0.67) rectangle (6.35,1.53);\n\\node[font=\\small,align=center,font=\\small] at (5.625,1.1) {$1$};\n\\path[draw=black!30,fill=black!8] (6.35,0.67) rectangle (7.8,1.53);\n\\node[font=\\small,align=center,font=\\small] at (7.075,1.1) {$2$};\n\\path[draw=black!30,fill=black!8] (7.8,0.67) rectangle (9.25,1.53);\n\\node[font=\\small,align=center,font=\\small] at (8.525,1.1) {$6$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,2.82) {$\\mathbb Z$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,1.96) {$\\mathbb F_5$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,1.1) {$\\mathbb F_7$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.625,3.48) {$T^2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.075,3.48) {$T$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.525,3.48) {$1$};\n\\n","small,align=center,font=\\small] at (5.625,1.1) {$1$};\n\\path[draw=black!30,fill=black!8] (6.35,0.67) rectangle (7.8,1.53);\n\\node[font=\\small,align=center,font=\\small] at (7.075,1.1) {$2$};\n\\path[draw=black!30,fill=black!8] (7.8,0.67) rectangle (9.25,1.53);\n\\node[font=\\small,align=center,font=\\small] at (8.525,1.1) {$6$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,2.82) {$\\mathbb Z$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,1.96) {$\\mathbb F_5$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,1.1) {$\\mathbb F_7$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.625,3.48) {$T^2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.075,3.48) {$T$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.525,3.48) {$1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,0.25) {Different residue rows come f","35,0.67) rectangle (7.8,1.53);\n\\node[font=\\small,align=center,font=\\small] at (7.075,1.1) {$2$};\n\\path[draw=black!30,fill=black!8] (7.8,0.67) rectangle (9.25,1.53);\n\\node[font=\\small,align=center,font=\\small] at (8.525,1.1) {$6$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,2.82) {$\\mathbb Z$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,1.96) {$\\mathbb F_5$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,1.1) {$\\mathbb F_7$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.625,3.48) {$T^2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.075,3.48) {$T$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.525,3.48) {$1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,0.25) {Different residue rows come from the same algebraic polynomial.};\n\\end{tikzpicture}","2$};\n\\path[draw=black!30,fill=black!8] (7.8,0.67) rectangle (9.25,1.53);\n\\node[font=\\small,align=center,font=\\small] at (8.525,1.1) {$6$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,2.82) {$\\mathbb Z$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,1.96) {$\\mathbb F_5$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,1.1) {$\\mathbb F_7$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.625,3.48) {$T^2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.075,3.48) {$T$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.525,3.48) {$1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,0.25) {Different residue rows come from the same algebraic polynomial.};\n\\end{tikzpicture}","t=\\small,align=center,font=\\small] at (8.525,1.1) {$6$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,2.82) {$\\mathbb Z$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,1.96) {$\\mathbb F_5$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,1.1) {$\\mathbb F_7$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.625,3.48) {$T^2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.075,3.48) {$T$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.525,3.48) {$1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,0.25) {Different residue rows come from the same algebraic polynomial.};\n\\end{tikzpicture}",",align=center,anchor=east,font=\\scriptsize] at (4.75,2.82) {$\\mathbb Z$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,1.96) {$\\mathbb F_5$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,1.1) {$\\mathbb F_7$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.625,3.48) {$T^2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.075,3.48) {$T$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.525,3.48) {$1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,0.25) {Different residue rows come from the same algebraic polynomial.};\n\\end{tikzpicture}",";\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,1.96) {$\\mathbb F_5$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (4.75,1.1) {$\\mathbb F_7$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.625,3.48) {$T^2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.075,3.48) {$T$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.525,3.48) {$1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,0.25) {Different residue rows come from the same algebraic polynomial.};\n\\end{tikzpicture}","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center,font=\\small] at (3.3,4.55) {$C$: actual $S_3$ character table};\n\\path[draw=black!30,fill=black!8] (2.1,3.05) rectangle (2.9,3.85);\n\\node[font=\\small,align=center,font=\\small] at (2.5,3.45) {$1$};\n\\path[draw=black!30,fill=black!8] (2.9,3.05) rectangle (3.7,3.85);\n\\node[font=\\small,align=center,font=\\small] at (3.3,3.45) 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(7.77,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.19,0.7) {$0$};\n\\path[draw=black!25,fill=white] (7.85,0.45) rectangle (9,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.42,0.7) {$0$};\n\\path[draw=black!25,fill=black!18] (9.08,0.45) rectangle (10.23,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.65,0.7) {$1$};\n\\path[draw=black!25,fill=white] (10.31,0.45) rectangle (11.46,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (10.88,0.7) {$0$};\n\\draw[dashed,line width=.9pt] (2.88,.18) rectangle (6.59,4.2);\n\\node[font=\\small,","th[draw=black!25,fill=white] (5.39,0.45) rectangle (6.54,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.96,0.7) {$0$};\n\\path[draw=black!25,fill=white] (6.62,0.45) rectangle (7.77,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.19,0.7) {$0$};\n\\path[draw=black!25,fill=white] (7.85,0.45) rectangle (9,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.42,0.7) {$0$};\n\\path[draw=black!25,fill=black!18] (9.08,0.45) rectangle (10.23,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.65,0.7) {$1$};\n\\path[draw=black!25,fill=white] (10.31,0.45) rectangle (11.46,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (10.88,0.7) {$0$};\n\\draw[dashed,line width=.9pt] (2.88,.18) rectangle (6.59,4.2);\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.15) {Every $a_k$ is nonzero. Every fixed finite digit window eventually reads only zeros.};","raw=black!25,fill=white] (6.62,0.45) rectangle (7.77,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (7.19,0.7) {$0$};\n\\path[draw=black!25,fill=white] (7.85,0.45) rectangle (9,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.42,0.7) {$0$};\n\\path[draw=black!25,fill=black!18] (9.08,0.45) rectangle (10.23,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.65,0.7) {$1$};\n\\path[draw=black!25,fill=white] (10.31,0.45) rectangle (11.46,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (10.88,0.7) {$0$};\n\\draw[dashed,line width=.9pt] (2.88,.18) rectangle (6.59,4.2);\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.15) {Every $a_k$ is nonzero. Every fixed finite digit window eventually reads only zeros.};\n\\end{tikzpicture}","=black!25,fill=white] (7.85,0.45) rectangle (9,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.42,0.7) {$0$};\n\\path[draw=black!25,fill=black!18] (9.08,0.45) rectangle (10.23,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.65,0.7) {$1$};\n\\path[draw=black!25,fill=white] (10.31,0.45) rectangle (11.46,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (10.88,0.7) {$0$};\n\\draw[dashed,line width=.9pt] (2.88,.18) rectangle (6.59,4.2);\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.15) {Every $a_k$ is nonzero. Every fixed finite digit window eventually reads only zeros.};\n\\end{tikzpicture}","!25,fill=black!18] (9.08,0.45) rectangle (10.23,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.65,0.7) {$1$};\n\\path[draw=black!25,fill=white] (10.31,0.45) rectangle (11.46,0.93);\n\\node[font=\\small,align=center,font=\\scriptsize] at (10.88,0.7) {$0$};\n\\draw[dashed,line width=.9pt] (2.88,.18) rectangle (6.59,4.2);\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.15) {Every $a_k$ is nonzero. Every fixed finite digit window eventually reads only zeros.};\n\\end{tikzpicture}",");\n\\draw[->,line width=.6pt] (1.5,2.75) -- (1.5,2.88);\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.75,2.2) {integer generators};\n\\path[draw,fill=black!15] (3.409091,2.75) circle (0.12);\n\\draw[->] (3.409091,2.55) -- (6.55,1.75);\n\\path[draw,fill=black!15] (8.181818,2.75) circle (0.12);\n\\draw[->] (8.181818,2.55) -- (6.55,1.75);\n\\node[font=\\small,align=center] at (6.55,1.45) {same residue $2\\pmod5$};\n\\node[font=\\small,align=center] at (3,4.35) {$N_{\\mathbb Q/\\mathbb Q}(-3)=-3$};\n\\node[font=\\small,align=center] at (10.5,4.35) {$N_{\\mathbb Q/\\mathbb Q}(2)=2$};\n\\node[font=\\small,align=center] at (3,0.45) {$N((-3))=3$};\n\\node[font=\\small,align=center] at (10.5,0.45) {$N((2))=2$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.15) {The ideal norm is positive. Passing to absolute values must retain the real sign.};\n\\end{tikzpicture}","ode[font=\\small,align=center,font=\\scriptsize] at (6.75,2.2) {integer generators};\n\\path[draw,fill=black!15] (3.409091,2.75) circle (0.12);\n\\draw[->] (3.409091,2.55) -- (6.55,1.75);\n\\path[draw,fill=black!15] (8.181818,2.75) circle (0.12);\n\\draw[->] (8.181818,2.55) -- (6.55,1.75);\n\\node[font=\\small,align=center] at (6.55,1.45) {same residue $2\\pmod5$};\n\\node[font=\\small,align=center] at (3,4.35) {$N_{\\mathbb Q/\\mathbb Q}(-3)=-3$};\n\\node[font=\\small,align=center] at (10.5,4.35) {$N_{\\mathbb Q/\\mathbb Q}(2)=2$};\n\\node[font=\\small,align=center] at (3,0.45) {$N((-3))=3$};\n\\node[font=\\small,align=center] at (10.5,0.45) {$N((2))=2$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.15) {The ideal norm is positive. Passing to absolute values must retain the real sign.};\n\\end{tikzpicture}","\\path[draw,fill=black!15] (3.409091,2.75) circle (0.12);\n\\draw[->] (3.409091,2.55) -- (6.55,1.75);\n\\path[draw,fill=black!15] (8.181818,2.75) circle (0.12);\n\\draw[->] (8.181818,2.55) -- (6.55,1.75);\n\\node[font=\\small,align=center] at (6.55,1.45) {same residue $2\\pmod5$};\n\\node[font=\\small,align=center] at (3,4.35) {$N_{\\mathbb Q/\\mathbb Q}(-3)=-3$};\n\\node[font=\\small,align=center] at (10.5,4.35) {$N_{\\mathbb Q/\\mathbb Q}(2)=2$};\n\\node[font=\\small,align=center] at (3,0.45) {$N((-3))=3$};\n\\node[font=\\small,align=center] at (10.5,0.45) {$N((2))=2$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.15) {The ideal norm is positive. Passing to absolute values must retain the real sign.};\n\\end{tikzpicture}",".55) -- (6.55,1.75);\n\\path[draw,fill=black!15] (8.181818,2.75) circle (0.12);\n\\draw[->] (8.181818,2.55) -- (6.55,1.75);\n\\node[font=\\small,align=center] at (6.55,1.45) {same residue $2\\pmod5$};\n\\node[font=\\small,align=center] at (3,4.35) {$N_{\\mathbb Q/\\mathbb Q}(-3)=-3$};\n\\node[font=\\small,align=center] at (10.5,4.35) {$N_{\\mathbb Q/\\mathbb Q}(2)=2$};\n\\node[font=\\small,align=center] at (3,0.45) {$N((-3))=3$};\n\\node[font=\\small,align=center] at (10.5,0.45) {$N((2))=2$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.15) {The ideal norm is positive. Passing to absolute values must retain the real sign.};\n\\end{tikzpicture}","circle (0.12);\n\\draw[->] (8.181818,2.55) -- (6.55,1.75);\n\\node[font=\\small,align=center] at (6.55,1.45) {same residue $2\\pmod5$};\n\\node[font=\\small,align=center] at (3,4.35) {$N_{\\mathbb Q/\\mathbb Q}(-3)=-3$};\n\\node[font=\\small,align=center] at (10.5,4.35) {$N_{\\mathbb Q/\\mathbb Q}(2)=2$};\n\\node[font=\\small,align=center] at (3,0.45) {$N((-3))=3$};\n\\node[font=\\small,align=center] at (10.5,0.45) {$N((2))=2$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.15) {The ideal norm is positive. Passing to absolute values must retain the real sign.};\n\\end{tikzpicture}","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center,font=\\small] at (3,4.7) {$N(a+bu)=a^2-2b^2\\pmod5$};\n\\path[draw=black!30,fill=white] (1.45,3.38) rectangle (2.07,4);\n\\node[font=\\small,align=center,font=\\small] at (1.76,3.69) {$0$};\n\\path[draw=black!30,fill=black!8] (2.07,3.38) rectangle (2.69,4);\n\\node[font=\\small,align=center,font=\\small] at (2.38,3.69) {$3$};\n\\path[draw=black!30,fill=black!8] (2.69,3.38) rectangle (3.31,4);\n\\node[font=\\small,align=center,font=\\small] at (3,3.69) {$2$};\n\\path[draw=black!30,fill=black!8] (3.31,3.38) rectangle (3.93,4);\n\\node[font=\\small,align=center,font=\\small] at (3.62,3.69) {$2$};\n\\path[draw=black!30,fill=black!8] (3.93,3.38) rectangle (4.55,4);\n\\node[font=\\small,align=center,font=\\small] at (4.24,3.69) 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{$a=3$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.21) {$a=4$};\n\\node[f…","$1$};\n\\path[draw=black!30,fill=black!8] (2.07,0.9) rectangle (2.69,1.52);\n\\node[font=\\small,align=center,font=\\small] at (2.38,1.21) {$4$};\n\\path[draw=black!30,fill=black!8] (2.69,0.9) rectangle (3.31,1.52);\n\\node[font=\\small,align=center,font=\\small] at (3,1.21) {$3$};\n\\path[draw=black!30,fill=black!8] (3.31,0.9) rectangle (3.93,1.52);\n\\node[font=\\small,align=center,font=\\small] at (3.62,1.21) {$3$};\n\\path[draw=black!30,fill=black!8] (3.93,0.9) rectangle (4.55,1.52);\n\\node[font=\\small,align=center,font=\\small] at (4.24,1.21) {$4$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,3.69) {$a=0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,3.07) {$a=1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,2.45) {$a=2$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.83) {$a=3$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.21) {$a=4$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.76,4.23) {$b=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.38,4.23) {$b=…","$4$};\n\\path[draw=black!30,fill=black!8] (2.69,0.9) rectangle (3.31,1.52);\n\\node[font=\\small,align=center,font=\\small] at (3,1.21) {$3$};\n\\path[draw=black!30,fill=black!8] (3.31,0.9) rectangle (3.93,1.52);\n\\node[font=\\small,align=center,font=\\small] at (3.62,1.21) {$3$};\n\\path[draw=black!30,fill=black!8] (3.93,0.9) rectangle (4.55,1.52);\n\\node[font=\\small,align=center,font=\\small] at (4.24,1.21) {$4$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,3.69) {$a=0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,3.07) {$a=1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,2.45) {$a=2$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.83) {$a=3$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.21) {$a=4$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.76,4.23) {$b=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.38,4.23) {$b=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$b=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.62,4…","nt=\\small,align=center,font=\\small] at (3,1.21) {$3$};\n\\path[draw=black!30,fill=black!8] (3.31,0.9) rectangle (3.93,1.52);\n\\node[font=\\small,align=center,font=\\small] at (3.62,1.21) {$3$};\n\\path[draw=black!30,fill=black!8] (3.93,0.9) rectangle (4.55,1.52);\n\\node[font=\\small,align=center,font=\\small] at (4.24,1.21) {$4$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,3.69) {$a=0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,3.07) {$a=1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,2.45) {$a=2$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.83) {$a=3$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.21) {$a=4$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.76,4.23) {$b=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.38,4.23) {$b=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$b=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.62,4.23) {$b=3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (4.24,4.23) {$b=…","k!8] (3.31,0.9) rectangle (3.93,1.52);\n\\node[font=\\small,align=center,font=\\small] at (3.62,1.21) {$3$};\n\\path[draw=black!30,fill=black!8] (3.93,0.9) rectangle (4.55,1.52);\n\\node[font=\\small,align=center,font=\\small] at (4.24,1.21) {$4$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,3.69) {$a=0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,3.07) {$a=1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,2.45) {$a=2$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.83) {$a=3$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.21) {$a=4$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.76,4.23) {$b=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.38,4.23) {$b=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$b=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.62,4.23) {$b=3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (4.24,4.23) {$b=4$};\n\\node[font=\\small,align=center] at (9.6,4.25) {$u^2=2$ in $\\mathbb F_{25}$};\n\\n…","t (3.62,1.21) {$3$};\n\\path[draw=black!30,fill=black!8] (3.93,0.9) rectangle (4.55,1.52);\n\\node[font=\\small,align=center,font=\\small] at (4.24,1.21) {$4$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,3.69) {$a=0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,3.07) {$a=1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,2.45) {$a=2$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.83) {$a=3$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.21) {$a=4$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.76,4.23) {$b=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.38,4.23) {$b=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$b=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.62,4.23) {$b=3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (4.24,4.23) {$b=4$};\n\\node[font=\\small,align=center] at (9.6,4.25) {$u^2=2$ in $\\mathbb F_{25}$};\n\\node[font=\\small,align=center] at (9.6,3.2) {$M_{a+bu}=\\begin{pmatrix}a&2b","52);\n\\node[font=\\small,align=center,font=\\small] at (4.24,1.21) {$4$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,3.69) {$a=0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,3.07) {$a=1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,2.45) {$a=2$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.83) {$a=3$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.21) {$a=4$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.76,4.23) {$b=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.38,4.23) {$b=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$b=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.62,4.23) {$b=3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (4.24,4.23) {$b=4$};\n\\node[font=\\small,align=center] at (9.6,4.25) {$u^2=2$ in $\\mathbb F_{25}$};\n\\node[font=\\small,align=center] at (9.6,3.2) {$M_{a+bu}=\\begin{pmatrix}a&2b","mall,align=center,anchor=east,font=\\scriptsize] at (1.3,3.69) {$a=0$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,3.07) {$a=1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,2.45) {$a=2$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.83) {$a=3$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.21) {$a=4$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.76,4.23) {$b=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.38,4.23) {$b=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$b=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.62,4.23) {$b=3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (4.24,4.23) {$b=4$};\n\\node[font=\\small,align=center] at (9.6,4.25) {$u^2=2$ in $\\mathbb F_{25}$};\n\\node[font=\\small,align=center] at (9.6,3.2) {$M_{a+bu}=\\begin{pmatrix}a&2b","ode[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,3.07) {$a=1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,2.45) {$a=2$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.83) {$a=3$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.21) {$a=4$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.76,4.23) {$b=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.38,4.23) {$b=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$b=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.62,4.23) {$b=3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (4.24,4.23) {$b=4$};\n\\node[font=\\small,align=center] at (9.6,4.25) {$u^2=2$ in $\\mathbb F_{25}$};\n\\node[font=\\small,align=center] at (9.6,3.2) {$M_{a+bu}=\\begin{pmatrix}a&2b","{$a=1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,2.45) {$a=2$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.83) {$a=3$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.21) {$a=4$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.76,4.23) {$b=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.38,4.23) {$b=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$b=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.62,4.23) {$b=3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (4.24,4.23) {$b=4$};\n\\node[font=\\small,align=center] at (9.6,4.25) {$u^2=2$ in $\\mathbb F_{25}$};\n\\node[font=\\small,align=center] at (9.6,3.2) {$M_{a+bu}=\\begin{pmatrix}a&2b","at (1.3,2.45) {$a=2$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.83) {$a=3$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.21) {$a=4$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.76,4.23) {$b=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.38,4.23) {$b=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$b=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.62,4.23) {$b=3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (4.24,4.23) {$b=4$};\n\\node[font=\\small,align=center] at (9.6,4.25) {$u^2=2$ in $\\mathbb F_{25}$};\n\\node[font=\\small,align=center] at (9.6,3.2) {$M_{a+bu}=\\begin{pmatrix}a&2b","criptsize] at (1.3,1.83) {$a=3$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (1.3,1.21) {$a=4$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.76,4.23) {$b=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.38,4.23) {$b=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$b=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.62,4.23) {$b=3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (4.24,4.23) {$b=4$};\n\\node[font=\\small,align=center] at (9.6,4.25) {$u^2=2$ in $\\mathbb F_{25}$};\n\\node[font=\\small,align=center] at (9.6,3.2) {$M_{a+bu}=\\begin{pmatrix}a&2b","ast,font=\\scriptsize] at (1.3,1.21) {$a=4$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.76,4.23) {$b=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.38,4.23) {$b=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$b=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.62,4.23) {$b=3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (4.24,4.23) {$b=4$};\n\\node[font=\\small,align=center] at (9.6,4.25) {$u^2=2$ in $\\mathbb F_{25}$};\n\\node[font=\\small,align=center] at (9.6,3.2) {$M_{a+bu}=\\begin{pmatrix}a&2b","=\\small,align=center,font=\\scriptsize] at (1.76,4.23) {$b=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.38,4.23) {$b=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$b=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.62,4.23) {$b=3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (4.24,4.23) {$b=4$};\n\\node[font=\\small,align=center] at (9.6,4.25) {$u^2=2$ in $\\mathbb F_{25}$};\n\\node[font=\\small,align=center] at (9.6,3.2) {$M_{a+bu}=\\begin{pmatrix}a&2b","gn=center,font=\\scriptsize] at (1.76,4.23) {$b=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.38,4.23) {$b=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$b=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.62,4.23) {$b=3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (4.24,4.23) {$b=4$};\n\\node[font=\\small,align=center] at (9.6,4.25) {$u^2=2$ in $\\mathbb F_{25}$};\n\\node[font=\\small,align=center] at (9.6,3.2) 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{no};\n\\path[draw=black!30,fill=white] (6.3,3.3) rectangle (7.7,3.88);\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,3.6) {no};\n\\path[draw=black!",",align=center,font=\\scriptsize] at (8.5,1.05) {no};\n\\path[draw=black!30,fill=black!16] (9.3,0.75) rectangle (10.7,1.33);\n\\node[font=\\small,align=center,font=\\scriptsize] at (10,1.05) {yes};\n\\path[draw=black!30,fill=black!16] (10.8,0.75) rectangle (12.2,1.33);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.5,1.05) {yes};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,0.05) {For $\\chi_k(2)=i^k$, a witness exists exactly when $kd\\not\\equiv0\\pmod4$.};\n\\end{tikzpicture}","no};\n\\path[draw=black!30,fill=black!16] (9.3,0.75) rectangle (10.7,1.33);\n\\node[font=\\small,align=center,font=\\scriptsize] at (10,1.05) {yes};\n\\path[draw=black!30,fill=black!16] (10.8,0.75) rectangle (12.2,1.33);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.5,1.05) 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(2.5,1.26);\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.14,0.99) {$25$};\n\\path[draw=black!25,fill=white] (2.56,0.68) rectangle (3.28,1.26);\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.92,0.99) {$26$};\n\\path[draw=black!25,fill=white] (3.34,0.68) rectangle (4.06,1.26);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.7,0.99) {$27$};\n\\path[draw=black!25,fill=white] (4.12,0.68) rectangle (4.84,1.26);\n\\node[font=\\small,align=center,font=\\scriptsize] at (4.48,0.99) {$28$};\n\\path[draw=black!25,fill=black!17] (4.9,0.68) rectangle (5.62,1.26);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.26,0.99) {$29$};\n\\path[draw,line width=1pt] (4.87,0.65) rectangle (5.65,1.29);\n\\node[font=\\small,align=center] at (9.8,4.2) {Choose one residue in each quotient};\n\\node[font=\\small,align=center] at (9.8,3.2) {$a\\equiv1\\pmod2$};\n\\node","=black!25,fill=white] (2.56,0.68) rectangle (3.28,1.26);\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.92,0.99) {$26$};\n\\path[draw=black!25,fill=white] (3.34,0.68) rectangle (4.06,1.26);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.7,0.99) {$27$};\n\\path[draw=black!25,fill=white] (4.12,0.68) rectangle (4.84,1.26);\n\\node[font=\\small,align=center,font=\\scriptsize] at (4.48,0.99) {$28$};\n\\path[draw=black!25,fill=black!17] (4.9,0.68) rectangle (5.62,1.26);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.26,0.99) {$29$};\n\\path[draw,line width=1pt] (4.87,0.65) rectangle (5.65,1.29);\n\\node[font=\\small,align=center] at (9.8,4.2) {Choose one residue in each quotient};\n\\node[font=\\small,align=center] at (9.8,3.2) {$a\\equiv1\\pmod2$};\n\\node[font=\\small,align=center] at (9.8,2.45) {$a\\equiv2\\pmod3$};\n\\node[font=\\small,align=center] at (9.8,1.7) {$a\\equiv4\\pmod5$};\n\\node[font=\\s","lack!25,fill=white] (3.34,0.68) rectangle (4.06,1.26);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.7,0.99) {$27$};\n\\path[draw=black!25,fill=white] (4.12,0.68) rectangle (4.84,1.26);\n\\node[font=\\small,align=center,font=\\scriptsize] at (4.48,0.99) {$28$};\n\\path[draw=black!25,fill=black!17] (4.9,0.68) rectangle (5.62,1.26);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.26,0.99) {$29$};\n\\path[draw,line width=1pt] (4.87,0.65) rectangle (5.65,1.29);\n\\node[font=\\small,align=center] at (9.8,4.2) {Choose one residue in each quotient};\n\\node[font=\\small,align=center] at (9.8,3.2) {$a\\equiv1\\pmod2$};\n\\node[font=\\small,align=center] at (9.8,2.45) {$a\\equiv2\\pmod3$};\n\\node[font=\\small,align=center] at (9.8,1.7) {$a\\equiv4\\pmod5$};\n\\node[font=\\small,align=center] at (9.8,0.75) {$a=29,\\quad\\gcd(29,30)=1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.1) {Shaded values a","font=\\small,align=center,font=\\scriptsize] at (4.48,0.99) {$28$};\n\\path[draw=black!25,fill=black!17] (4.9,0.68) rectangle (5.62,1.26);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.26,0.99) {$29$};\n\\path[draw,line width=1pt] (4.87,0.65) rectangle (5.65,1.29);\n\\node[font=\\small,align=center] at (9.8,4.2) {Choose one residue in each quotient};\n\\node[font=\\small,align=center] at (9.8,3.2) {$a\\equiv1\\pmod2$};\n\\node[font=\\small,align=center] at (9.8,2.45) {$a\\equiv2\\pmod3$};\n\\node[font=\\small,align=center] at (9.8,1.7) {$a\\equiv4\\pmod5$};\n\\node[font=\\small,align=center] at (9.8,0.75) {$a=29,\\quad\\gcd(29,30)=1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.1) {Shaded values avoid $2$, $3$, and $5$ simultaneously.};\n\\end{tikzpicture}","\\path[draw=black!25,fill=black!17] (4.9,0.68) rectangle (5.62,1.26);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.26,0.99) {$29$};\n\\path[draw,line width=1pt] (4.87,0.65) rectangle (5.65,1.29);\n\\node[font=\\small,align=center] at (9.8,4.2) {Choose one residue in each quotient};\n\\node[font=\\small,align=center] at (9.8,3.2) {$a\\equiv1\\pmod2$};\n\\node[font=\\small,align=center] at (9.8,2.45) {$a\\equiv2\\pmod3$};\n\\node[font=\\small,align=center] at (9.8,1.7) {$a\\equiv4\\pmod5$};\n\\node[font=\\small,align=center] at (9.8,0.75) {$a=29,\\quad\\gcd(29,30)=1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.1) {Shaded values avoid $2$, $3$, and $5$ simultaneously.};\n\\end{tikzpicture}","6);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.26,0.99) {$29$};\n\\path[draw,line width=1pt] (4.87,0.65) rectangle (5.65,1.29);\n\\node[font=\\small,align=center] at (9.8,4.2) {Choose one residue in each quotient};\n\\node[font=\\small,align=center] at (9.8,3.2) {$a\\equiv1\\pmod2$};\n\\node[font=\\small,align=center] at (9.8,2.45) {$a\\equiv2\\pmod3$};\n\\node[font=\\small,align=center] at (9.8,1.7) {$a\\equiv4\\pmod5$};\n\\node[font=\\small,align=center] at (9.8,0.75) {$a=29,\\quad\\gcd(29,30)=1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.1) {Shaded values avoid $2$, $3$, and $5$ simultaneously.};\n\\end{tikzpicture}",") {$29$};\n\\path[draw,line width=1pt] (4.87,0.65) rectangle (5.65,1.29);\n\\node[font=\\small,align=center] at (9.8,4.2) {Choose one residue in each quotient};\n\\node[font=\\small,align=center] at (9.8,3.2) {$a\\equiv1\\pmod2$};\n\\node[font=\\small,align=center] at (9.8,2.45) {$a\\equiv2\\pmod3$};\n\\node[font=\\small,align=center] at (9.8,1.7) {$a\\equiv4\\pmod5$};\n\\node[font=\\small,align=center] at (9.8,0.75) {$a=29,\\quad\\gcd(29,30)=1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.1) {Shaded values avoid $2$, $3$, and $5$ simultaneously.};\n\\end{tikzpicture}","4.2) {Choose one residue in each quotient};\n\\node[font=\\small,align=center] at (9.8,3.2) {$a\\equiv1\\pmod2$};\n\\node[font=\\small,align=center] at (9.8,2.45) {$a\\equiv2\\pmod3$};\n\\node[font=\\small,align=center] at (9.8,1.7) {$a\\equiv4\\pmod5$};\n\\node[font=\\small,align=center] at (9.8,0.75) {$a=29,\\quad\\gcd(29,30)=1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.1) {Shaded values avoid $2$, $3$, and $5$ simultaneously.};\n\\end{tikzpicture}","{Choose one residue in each quotient};\n\\node[font=\\small,align=center] at (9.8,3.2) {$a\\equiv1\\pmod2$};\n\\node[font=\\small,align=center] at (9.8,2.45) {$a\\equiv2\\pmod3$};\n\\node[font=\\small,align=center] at (9.8,1.7) {$a\\equiv4\\pmod5$};\n\\node[font=\\small,align=center] at (9.8,0.75) {$a=29,\\quad\\gcd(29,30)=1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.1) {Shaded values avoid $2$, $3$, and $5$ simultaneously.};\n\\end{tikzpicture}","ne residue in each quotient};\n\\node[font=\\small,align=center] at (9.8,3.2) {$a\\equiv1\\pmod2$};\n\\node[font=\\small,align=center] at (9.8,2.45) {$a\\equiv2\\pmod3$};\n\\node[font=\\small,align=center] at (9.8,1.7) {$a\\equiv4\\pmod5$};\n\\node[font=\\small,align=center] at (9.8,0.75) {$a=29,\\quad\\gcd(29,30)=1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.1) {Shaded values avoid $2$, $3$, and $5$ simultaneously.};\n\\end{tikzpicture}","raw=black!22,fill=black!13] (4.5,3.65) rectangle (4.8,3.95);\n\\path[draw,fill=black] (4.65,3.8) circle (0.035);\n\\path[draw,line width=.8pt] (1.2,0.35) rectangle (2.7,1.85);\n\\path[draw,line width=.8pt] (1.2,1.85) rectangle (2.7,3.35);\n\\path[draw,line width=.8pt] (2.7,0.35) rectangle (4.2,1.85);\n\\path[draw,line width=.8pt] (2.7,1.85) rectangle (4.2,3.35);\n\\node[font=\\small,align=center] at (3,4.4) {$[0,12)^2\\cap\\mathbb Z^2$};\n\\node[font=\\small,align=center] at (9.8,3.75) {four full $5\\times5$ cells};\n\\node[font=\\small,align=center] at (9.8,2.85) {$100$ bulk points: sum $0$};\n\\node[font=\\small,align=center] at (9.8,1.95) {$44$ boundary points};\n\\node[font=\\small,align=center] at (9.8,1.05) {actual total $S_{12}=1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.1) {$|S_L|\\le L^2-(L-(L\\bmod5))^2$ for this square-lattice model.};\n\\end{tikzpicture}","4.65,3.8) circle (0.035);\n\\path[draw,line width=.8pt] (1.2,0.35) rectangle (2.7,1.85);\n\\path[draw,line width=.8pt] (1.2,1.85) rectangle (2.7,3.35);\n\\path[draw,line width=.8pt] (2.7,0.35) rectangle (4.2,1.85);\n\\path[draw,line width=.8pt] (2.7,1.85) rectangle (4.2,3.35);\n\\node[font=\\small,align=center] at (3,4.4) {$[0,12)^2\\cap\\mathbb Z^2$};\n\\node[font=\\small,align=center] at (9.8,3.75) {four full $5\\times5$ cells};\n\\node[font=\\small,align=center] at (9.8,2.85) {$100$ bulk points: sum $0$};\n\\node[font=\\small,align=center] at (9.8,1.95) {$44$ boundary points};\n\\node[font=\\small,align=center] at (9.8,1.05) {actual total $S_{12}=1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.1) {$|S_L|\\le L^2-(L-(L\\bmod5))^2$ for this square-lattice model.};\n\\end{tikzpicture}","ectangle (2.7,1.85);\n\\path[draw,line width=.8pt] (1.2,1.85) rectangle (2.7,3.35);\n\\path[draw,line width=.8pt] (2.7,0.35) rectangle (4.2,1.85);\n\\path[draw,line width=.8pt] (2.7,1.85) rectangle (4.2,3.35);\n\\node[font=\\small,align=center] at (3,4.4) {$[0,12)^2\\cap\\mathbb Z^2$};\n\\node[font=\\small,align=center] at (9.8,3.75) {four full $5\\times5$ cells};\n\\node[font=\\small,align=center] at (9.8,2.85) {$100$ bulk points: sum $0$};\n\\node[font=\\small,align=center] at (9.8,1.95) {$44$ boundary points};\n\\node[font=\\small,align=center] at (9.8,1.05) {actual total $S_{12}=1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.1) {$|S_L|\\le L^2-(L-(L\\bmod5))^2$ for this square-lattice model.};\n\\end{tikzpicture}","ath[draw,line width=.8pt] (1.2,1.85) rectangle (2.7,3.35);\n\\path[draw,line width=.8pt] (2.7,0.35) rectangle (4.2,1.85);\n\\path[draw,line width=.8pt] (2.7,1.85) rectangle (4.2,3.35);\n\\node[font=\\small,align=center] at (3,4.4) {$[0,12)^2\\cap\\mathbb Z^2$};\n\\node[font=\\small,align=center] at (9.8,3.75) {four full $5\\times5$ cells};\n\\node[font=\\small,align=center] at (9.8,2.85) {$100$ bulk points: sum $0$};\n\\node[font=\\small,align=center] at (9.8,1.95) {$44$ boundary points};\n\\node[font=\\small,align=center] at (9.8,1.05) {actual total $S_{12}=1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.1) {$|S_L|\\le L^2-(L-(L\\bmod5))^2$ for this square-lattice model.};\n\\end{tikzpicture}",".35);\n\\path[draw,line width=.8pt] (2.7,0.35) rectangle (4.2,1.85);\n\\path[draw,line width=.8pt] (2.7,1.85) rectangle (4.2,3.35);\n\\node[font=\\small,align=center] at (3,4.4) {$[0,12)^2\\cap\\mathbb Z^2$};\n\\node[font=\\small,align=center] at (9.8,3.75) {four full $5\\times5$ cells};\n\\node[font=\\small,align=center] at (9.8,2.85) {$100$ bulk points: sum $0$};\n\\node[font=\\small,align=center] at (9.8,1.95) {$44$ boundary points};\n\\node[font=\\small,align=center] at (9.8,1.05) {actual total $S_{12}=1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.1) {$|S_L|\\le L^2-(L-(L\\bmod5))^2$ for this square-lattice model.};\n\\end{tikzpicture}","width=.8pt] (2.7,1.85) rectangle (4.2,3.35);\n\\node[font=\\small,align=center] at (3,4.4) {$[0,12)^2\\cap\\mathbb Z^2$};\n\\node[font=\\small,align=center] at (9.8,3.75) {four full $5\\times5$ cells};\n\\node[font=\\small,align=center] at (9.8,2.85) {$100$ bulk points: sum $0$};\n\\node[font=\\small,align=center] at (9.8,1.95) {$44$ boundary points};\n\\node[font=\\small,align=center] at (9.8,1.05) {actual total $S_{12}=1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.1) {$|S_L|\\le L^2-(L-(L\\bmod5))^2$ for this square-lattice model.};\n\\end{tikzpicture}","at (3,4.4) {$[0,12)^2\\cap\\mathbb Z^2$};\n\\node[font=\\small,align=center] at (9.8,3.75) {four full $5\\times5$ cells};\n\\node[font=\\small,align=center] at (9.8,2.85) {$100$ bulk points: sum $0$};\n\\node[font=\\small,align=center] at (9.8,1.95) {$44$ boundary points};\n\\node[font=\\small,align=center] at (9.8,1.05) {actual total $S_{12}=1$};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.1) {$|S_L|\\le L^2-(L-(L\\bmod5))^2$ for this square-lattice model.};\n\\end{tikzpicture}","-- (6.1,3.05);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.94,3.05) {3};\n\\draw[black!13,line width=.25pt] (1.1,3.8) -- (6.1,3.8);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.94,3.8) {4};\n\\draw[->,line width=.6pt] (1.1,0.8) -- (6.22,0.8);\n\\draw[->,line width=.6pt] (1.1,0.8) -- (1.1,3.92);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.6,0.25) {$s$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (1.1,4.2) {$I(s)=s/(s-1/2)$};\n\\draw[line width=1.1pt] plot coordinates {(1.1,3.425) (1.15,3.352885) (1.2,3.286111) (1.25,3.224107) (1.3,3.166379) (1.35,3.1125) (1.4,3.062097) (1.45,3.014844) (1.5,2.970455) (1.55,2.928676) (1.6,2.889286) (1.65,2.852083) (1.7,2.816892) (1.75,2.783553) (1.8,2.751923) (1.85,2.721875) (1.9,2.693293) (1.95,2.666071) (2,2.640116) (2.05,2.615341) (2.1,2.591667) (2.15,2.569022) (2.2,2.54734) (2.25,2.5265",".94,3.05) {3};\n\\draw[black!13,line width=.25pt] (1.1,3.8) -- (6.1,3.8);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.94,3.8) {4};\n\\draw[->,line width=.6pt] (1.1,0.8) -- (6.22,0.8);\n\\draw[->,line width=.6pt] (1.1,0.8) -- (1.1,3.92);\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.6,0.25) {$s$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (1.1,4.2) {$I(s)=s/(s-1/2)$};\n\\draw[line width=1.1pt] plot coordinates {(1.1,3.425) (1.15,3.352885) (1.2,3.286111) (1.25,3.224107) (1.3,3.166379) (1.35,3.1125) (1.4,3.062097) (1.45,3.014844) (1.5,2.970455) (1.55,2.928676) (1.6,2.889286) (1.65,2.852083) (1.7,2.816892) (1.75,2.783553) (1.8,2.751923) (1.85,2.721875) (1.9,2.693293) (1.95,2.666071) (2,2.640116) (2.05,2.615341) (2.1,2.591667) (2.15,2.569022) (2.2,2.54734) (2.25,2.526562) (2.3,2.506633) (2.35,2.4875) (2.4,2.469118) (2.45,2.451442) (2.5,2.434434) (2.55,2.418056)","t=\\small,align=center,anchor=east,font=\\scriptsize] at (8.04,2.228571) {0.5};\n\\draw[black!13,line width=.25pt] (8.2,3.657143) -- (12.8,3.657143);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.04,3.657143) {1};\n\\draw[->,line width=.6pt] (8.2,0.8) -- (12.92,0.8);\n\\draw[->,line width=.6pt] (8.2,0.8) -- (8.2,3.92);\n\\node[font=\\small,align=center,font=\\scriptsize] at (10.5,0.25) {$h$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (8.2,4.2) {endpoint difference};\n\\draw[line width=1.1pt] plot coordinates {(8.2,3.657143) (8.246,3.545098) (8.292,3.437363) (8.338,3.333693) (8.384,3.233862) (8.43,3.137662) (8.476,3.044898) (8.522,2.955388) (8.568,2.868966) (8.614,2.785472) (8.66,2.704762) (8.706,2.626698) (8.752,2.551152) (8.798,2.478005) (8.844,2.407143) (8.89,2.338462) (8.936,2.271861) (8.982,2.207249) (9.028,2.144538) (9.074,2.083644) (9.12,2.02449) (9.166,","(12.018,2.685714) (12.064,2.742857) (12.11,2.8) (12.156,2.857143) (12.202,2.914286) (12.248,2.971429) (12.294,3.028571) (12.34,3.085714) (12.386,3.142857) (12.432,3.2) (12.478,3.257143) (12.524,3.314286) (12.57,3.371429) (12.616,3.428571) (12.662,3.485714) (12.708,3.542857) (12.754,3.6) (12.8,3.657143)};\n\\node[font=\\small,align=center,font=\\scriptsize] at (10.5,-0.28) {solid: exact\\quad dashed: $4|h|$};\n\\end{tikzpicture}","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.25,4) {$3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.25,3.3) {$0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.85,4) {$5$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.85,3.3) {$2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.45,4) 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(7,2.65);\n\\node[font=\\small,align=center,font=\\small] at (6.6,2.25) {$1$};\n\\path[draw=black!30,fill=white] (7,1.85) rectangle (7.8,2.65);\n\\node[font=\\small,align=center,font=\\small] at (7.4,2.25) {$0$};\n\\node[font=\\small,align=center] at (7,0.85) {$A^2+I=0$};\n\\draw[black!25] (9.3,0.3) -- (9.3,4.45);\n\\node[font=\\small,align=center] at (11.7,4.3) {Scalar over $\\mathbb F_9$};\n\\node[font=\\small,align=center] at (11.7,3.25) {$u^2=-1$};\n\\node[font=\\small,align=center] at (11.7,2.2) {roots $u,-u$};\n\\node[font=\\small,align=center] at (11.7,1.15) {coefficient extension};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.05) {Fix the coefficient field, dimension, and point category before d","7,2.65) rectangle (7.8,3.45);\n\\node[font=\\small,align=center,font=\\small] at (7.4,3.05) {$-1$};\n\\path[draw=black!30,fill=black!8] (6.2,1.85) rectangle (7,2.65);\n\\node[font=\\small,align=center,font=\\small] at (6.6,2.25) {$1$};\n\\path[draw=black!30,fill=white] (7,1.85) rectangle (7.8,2.65);\n\\node[font=\\small,align=center,font=\\small] at (7.4,2.25) {$0$};\n\\node[font=\\small,align=center] at (7,0.85) {$A^2+I=0$};\n\\draw[black!25] (9.3,0.3) -- (9.3,4.45);\n\\node[font=\\small,align=center] at (11.7,4.3) {Scalar over $\\mathbb F_9$};\n\\node[font=\\small,align=center] at (11.7,3.25) {$u^2=-1$};\n\\node[font=\\small,align=center] at (11.7,2.2) {roots $u,-u$};\n\\node[font=\\small,align=center] at (11.7,1.15) {coefficient extension};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.05) {Fix the coefficient field, dimension, and point category before declaring an occupancy failure.};\n\\end{tikzpicture}","!30,fill=black!8] (6.2,1.85) rectangle (7,2.65);\n\\node[font=\\small,align=center,font=\\small] at (6.6,2.25) {$1$};\n\\path[draw=black!30,fill=white] (7,1.85) rectangle (7.8,2.65);\n\\node[font=\\small,align=center,font=\\small] at (7.4,2.25) {$0$};\n\\node[font=\\small,align=center] at (7,0.85) {$A^2+I=0$};\n\\draw[black!25] (9.3,0.3) -- (9.3,4.45);\n\\node[font=\\small,align=center] at (11.7,4.3) {Scalar over $\\mathbb F_9$};\n\\node[font=\\small,align=center] at (11.7,3.25) {$u^2=-1$};\n\\node[font=\\small,align=center] at (11.7,2.2) {roots $u,-u$};\n\\node[font=\\small,align=center] at (11.7,1.15) {coefficient extension};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.05) {Fix the coefficient field, dimension, and point category before declaring an occupancy failure.};\n\\end{tikzpicture}","ll,align=center,font=\\small] at (6.6,2.25) {$1$};\n\\path[draw=black!30,fill=white] (7,1.85) rectangle (7.8,2.65);\n\\node[font=\\small,align=center,font=\\small] at (7.4,2.25) {$0$};\n\\node[font=\\small,align=center] at (7,0.85) {$A^2+I=0$};\n\\draw[black!25] (9.3,0.3) -- (9.3,4.45);\n\\node[font=\\small,align=center] at (11.7,4.3) {Scalar over $\\mathbb F_9$};\n\\node[font=\\small,align=center] at (11.7,3.25) {$u^2=-1$};\n\\node[font=\\small,align=center] at (11.7,2.2) {roots $u,-u$};\n\\node[font=\\small,align=center] at (11.7,1.15) {coefficient extension};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.05) {Fix the coefficient field, dimension, and point category before declaring an occupancy failure.};\n\\end{tikzpicture}","ck!30,fill=white] (7,1.85) rectangle (7.8,2.65);\n\\node[font=\\small,align=center,font=\\small] at (7.4,2.25) {$0$};\n\\node[font=\\small,align=center] at (7,0.85) {$A^2+I=0$};\n\\draw[black!25] (9.3,0.3) -- (9.3,4.45);\n\\node[font=\\small,align=center] at (11.7,4.3) {Scalar over $\\mathbb F_9$};\n\\node[font=\\small,align=center] at (11.7,3.25) {$u^2=-1$};\n\\node[font=\\small,align=center] at (11.7,2.2) {roots $u,-u$};\n\\node[font=\\small,align=center] at (11.7,1.15) {coefficient extension};\n\\node[font=\\small,align=center,font=\\footnotesize] at (7,-0.05) {Fix the coefficient field, dimension, and point category before declaring an occupancy failure.};\n\\end{tikzpicture}","lier scalar/Mellin checkpoints are historical.}\nPaper X separates three questions that can look deceptively similar. Does a residual object exist? Does it extend coherently through every precision? Does the resulting object occur in the required carrier? A successful answer to one question does not supply the next.\n\\vfig{01_three_questions}\n\\head{Three models to keep in view}\nThe scalar equation $x^2+1=0$ has the root $1$ modulo two, but no root modulo four. Over powers of five, the residues $2,7,57,182$ form the beginning of a compatible branch. Neither fact constructs an automorphic carrier. In a supplied finite carrier, an idempotent such as $\\diag(1,0)$ detects an occupied subspace; it does not explain where that carrier came from.","kpoints are historical.}\nPaper X separates three questions that can look deceptively similar. Does a residual object exist? Does it extend coherently through every precision? Does the resulting object occur in the required carrier? A successful answer to one question does not supply the next.\n\\vfig{01_three_questions}\n\\head{Three models to keep in view}\nThe scalar equation $x^2+1=0$ has the root $1$ modulo two, but no root modulo four. Over powers of five, the residues $2,7,57,182$ form the beginning of a compatible branch. Neither fact constructs an automorphic carrier. In a supplied finite carrier, an idempotent such as $\\diag(1,0)$ detects an occupied subspace; it does not explain where that carrier came from.","ceptively similar. Does a residual object exist? Does it extend coherently through every precision? Does the resulting object occur in the required carrier? A successful answer to one question does not supply the next.\n\\vfig{01_three_questions}\n\\head{Three models to keep in view}\nThe scalar equation $x^2+1=0$ has the root $1$ modulo two, but no root modulo four. Over powers of five, the residues $2,7,57,182$ form the beginning of a compatible branch. Neither fact constructs an automorphic carrier. In a supplied finite carrier, an idempotent such as $\\diag(1,0)$ detects an occupied subspace; it does not explain where that carrier came from.","A successful answer to one question does not supply the next.\n\\vfig{01_three_questions}\n\\head{Three models to keep in view}\nThe scalar equation $x^2+1=0$ has the root $1$ modulo two, but no root modulo four. Over powers of five, the residues $2,7,57,182$ form the beginning of a compatible branch. Neither fact constructs an automorphic carrier. In a supplied finite carrier, an idempotent such as $\\diag(1,0)$ detects an occupied subspace; it does not explain where that carrier came from.","midrule\n$\\OO,\\varpi,k$ & Complete valuation ring, uniformizer, and residue field $k$ of size $q$. & A represented deformation functor.","$\\bar x$ & One residual object in the declared moduli problem. & Other residual roots or their transports.","$X_m(\\bar x)$ & Admissible objects modulo $\\varpi^m$ reducing to $\\bar x$. & A lift of every individual member.","$(x_m)_m$ & A branch satisfying $x_{m+1}\\bmod\\varpi^m=x_m$. & A global object without the stated reconstruction hypotheses.","$w_i$ & A coordinate constrained to begin in $\\varpi^{w_i}\\OO$. & A classification of all singular deformation rings.","$Z(T)$ & The formal series $\\sum_{m\\ge1}\\#X_mT^m$. & The cardinality of the inverse limit.","$C^\\bullet,d$ & Cochains and a differential with $d^2=0$. & Its arithmetic identification with a Galois problem.","$G$ & The graph of supplied admissible residual transports. & A proof that a proposed arithmetic edge exists.","$\\mathscr C$ & Common algebraic polynomials and declared local data. & A member realizing those data at every prime.","$C\\bm m=\\bm b$ & A finite trace system for carrier multiplicities. & Integrality, positivity, or available capacity automatically.","$e_TV$ & The joint eigenspace inside an existing finite carrier $V$. & Nonzero occupancy for every desired joint label.","bottomrule\n\\end{tabularx}\n\\head{Three indices that must not be conflated}\nThe depth $m$ counts retained digits. A coordinate index $i$ selects a parameter or a packet. A coefficient place $\\lambda$ selects a different local field. Increasing $m$ at one $\\lambda$ neither changes $i$ nor constructs the member at another $\\lambda$.\n\\takeaway{Specify the ambient object before applying a projector, a reduction, a rank, or a counting formula. The same-looking symbol can refer to quite different spaces.}","\\vpage{3}{Higher digits cannot change the root}{RESIDUAL PARTITION}{Main text: Theorems 2.1--2.2 and Corollary 2.3.}\nWork in $\\ZZ/27\\ZZ$. Every element has a unique form $x=3q+r$ with $r\\in\\{0,1,2\\}$ and $q\\in\\{0,\\ldots,8\\}$. The row below is the residual root. The column contains all the remaining information at this precision.\n\\vfig{03_residual_grid}\n\\head{Read the operation, not just the arrow}\nAn operation $x\\mapsto x+3h$ preserves $r$, because its correction vanishes modulo three. Repeated corrections of this form can visit every column in a row, but can never enter another row. By contrast, adding one changes the residual class. It is therefore not a positive-depth perturbation in this model.","r digits cannot change the root}{RESIDUAL PARTITION}{Main text: Theorems 2.1--2.2 and Corollary 2.3.}\nWork in $\\ZZ/27\\ZZ$. Every element has a unique form $x=3q+r$ with $r\\in\\{0,1,2\\}$ and $q\\in\\{0,\\ldots,8\\}$. The row below is the residual root. The column contains all the remaining information at this precision.\n\\vfig{03_residual_grid}\n\\head{Read the operation, not just the arrow}\nAn operation $x\\mapsto x+3h$ preserves $r$, because its correction vanishes modulo three. Repeated corrections of this form can visit every column in a row, but can never enter another row. By contrast, adding one changes the residual class. It is therefore not a positive-depth perturbation in this model.","root}{RESIDUAL PARTITION}{Main text: Theorems 2.1--2.2 and Corollary 2.3.}\nWork in $\\ZZ/27\\ZZ$. Every element has a unique form $x=3q+r$ with $r\\in\\{0,1,2\\}$ and $q\\in\\{0,\\ldots,8\\}$. The row below is the residual root. The column contains all the remaining information at this precision.\n\\vfig{03_residual_grid}\n\\head{Read the operation, not just the arrow}\nAn operation $x\\mapsto x+3h$ preserves $r$, because its correction vanishes modulo three. Repeated corrections of this form can visit every column in a row, but can never enter another row. By contrast, adding one changes the residual class. It is therefore not a positive-depth perturbation in this model.","For example, the addresses $1,4,7,10,\\ldots,25$ may be deeply connected by higher-digit changes. They remain disjoint from the addresses reducing to zero. In a genuine deformation problem, the same conclusion applies to operations whose reduction is the identity, not to every operation that happens to be called a deformation.\n\\[\n\\operatorname{red}(x+3h)=\\operatorname{red}(x),\n\\qquad\n\\mathcal M(A)=\\bigsqcup_{\\bar x}\\mathcal M_{\\bar x}(A).\n\\]\n\\takeaway{Vertical completeness can fill one entire residual fiber without reaching a single point in a different fiber.}\n\\captiontext{All 27 addresses and their finite corrections are checked by integer reduction. Passing to an infinite perturbation stream additionally requires the convergence condi","For example, the addresses $1,4,7,10,\\ldots,25$ may be deeply connected by higher-digit changes. They remain disjoint from the addresses reducing to zero. In a genuine deformation problem, the same conclusion applies to operations whose reduction is the identity, not to every operation that happens to be called a deformation.\n\\[\n\\operatorname{red}(x+3h)=\\operatorname{red}(x),\n\\qquad\n\\mathcal M(A)=\\bigsqcup_{\\bar x}\\mathcal M_{\\bar x}(A).\n\\]\n\\takeaway{Vertical completeness can fill one entire residual fiber without reaching a single point in a different fiber.}\n\\captiontext{All 27 addresses and their finite corrections are checked by integer reduction. Passing to an infinite perturbation stream additionally requires the convergence condition recorded in the formal no-crossing statement.}","\\vpage{4}{A branch fixes digits permanently}{INVERSE-LIMIT RECONSTRUCTION}{Main text: Theorem 3.2 and the compatible-digit example in §3.}\nThe rational number $1/4$ is a five-adic integer. Its normalized residues are obtained by solving $4x_m\\equiv1\\pmod{5^m}$, not by taking decimal approximations. At the first six depths they are $4,19,94,469,2344,11719$.\n\\vfig{04_digits}\n\\head{Compatibility is stronger than independent solvability}\nThe residues satisfy both\n\\[\nx_{m+1}\\equiv x_m\\pmod{5^m},\n\\qquad 4x_m-1=3\\cdot5^m.\n\\]\nThe latter identity shows that the residual error tends to zero in the five-adic norm. The stabilized digit sequence is\n\\[\n\\frac14=4+3\\cdot5+3\\cdot5^2+3\\cdot5^3+\\cdots\\quad\\text{in }\\ZZ_5.\n\\]\nThis series is not convergent in the ordinary real norm. The topology is part of the statement. To view the branch as a positive-depth coordinate a","\\vpage{4}{A branch fixes digits permanently}{INVERSE-LIMIT RECONSTRUCTION}{Main text: Theorem 3.2 and the compatible-digit example in §3.}\nThe rational number $1/4$ is a five-adic integer. Its normalized residues are obtained by solving $4x_m\\equiv1\\pmod{5^m}$, not by taking decimal approximations. At the first six depths they are $4,19,94,469,2344,11719$.\n\\vfig{04_digits}\n\\head{Compatibility is stronger than independent solvability}\nThe residues satisfy both\n\\[\nx_{m+1}\\equiv x_m\\pmod{5^m},\n\\qquad 4x_m-1=3\\cdot5^m.\n\\]\nThe latter identity shows that the residual error tends to zero in the five-adic norm. The stabilized digit sequence is\n\\[\n\\frac14=4+3\\cdot5+3\\cdot5^2+3\\cdot5^3+\\cdots\\quad\\text{in }\\ZZ_5.\n\\]\nThis series is not convergent in the ordinary real norm. The topology is part of the statement. To view the branch as a positive-depth coordinate above the residual root $4$, use $t=x-4\\in5\\ZZ_5$.","RUCTION}{Main text: Theorem 3.2 and the compatible-digit example in §3.}\nThe rational number $1/4$ is a five-adic integer. Its normalized residues are obtained by solving $4x_m\\equiv1\\pmod{5^m}$, not by taking decimal approximations. At the first six depths they are $4,19,94,469,2344,11719$.\n\\vfig{04_digits}\n\\head{Compatibility is stronger than independent solvability}\nThe residues satisfy both\n\\[\nx_{m+1}\\equiv x_m\\pmod{5^m},\n\\qquad 4x_m-1=3\\cdot5^m.\n\\]\nThe latter identity shows that the residual error tends to zero in the five-adic norm. The stabilized digit sequence is\n\\[\n\\frac14=4+3\\cdot5+3\\cdot5^2+3\\cdot5^3+\\cdots\\quad\\text{in }\\ZZ_5.\n\\]\nThis series is not convergent in the ordinary real norm. The topology is part of the statement. To view the branch as a positive-depth coordinate above the residual root $4$, use $t=x-4\\in5\\ZZ_5$.","iv1\\pmod{5^m}$, not by taking decimal approximations. At the first six depths they are $4,19,94,469,2344,11719$.\n\\vfig{04_digits}\n\\head{Compatibility is stronger than independent solvability}\nThe residues satisfy both\n\\[\nx_{m+1}\\equiv x_m\\pmod{5^m},\n\\qquad 4x_m-1=3\\cdot5^m.\n\\]\nThe latter identity shows that the residual error tends to zero in the five-adic norm. The stabilized digit sequence is\n\\[\n\\frac14=4+3\\cdot5+3\\cdot5^2+3\\cdot5^3+\\cdots\\quad\\text{in }\\ZZ_5.\n\\]\nThis series is not convergent in the ordinary real norm. The topology is part of the statement. To view the branch as a positive-depth coordinate above the residual root $4$, use $t=x-4\\in5\\ZZ_5$.","od{5^m},\n\\qquad 4x_m-1=3\\cdot5^m.\n\\]\nThe latter identity shows that the residual error tends to zero in the five-adic norm. The stabilized digit sequence is\n\\[\n\\frac14=4+3\\cdot5+3\\cdot5^2+3\\cdot5^3+\\cdots\\quad\\text{in }\\ZZ_5.\n\\]\nThis series is not convergent in the ordinary real norm. The topology is part of the statement. To view the branch as a positive-depth coordinate above the residual root $4$, use $t=x-4\\in5\\ZZ_5$.","\\qquad 4x_m-1=3\\cdot5^m.\n\\]\nThe latter identity shows that the residual error tends to zero in the five-adic norm. The stabilized digit sequence is\n\\[\n\\frac14=4+3\\cdot5+3\\cdot5^2+3\\cdot5^3+\\cdots\\quad\\text{in }\\ZZ_5.\n\\]\nThis series is not convergent in the ordinary real norm. The topology is part of the statement. To view the branch as a positive-depth coordinate above the residual root $4$, use $t=x-4\\in5\\ZZ_5$.","\\vpage{5}{A smooth chart has independent choices}{ACTUAL FINITE POINTS}{Main text: Proposition 3.4 and Theorem 6.2.}\nTake $R=\\ZZ_3[[t_1,t_2]]$ and coefficient-preserving local maps to $\\ZZ/27\\ZZ$. Locality forces both parameters into the maximal ideal. Each therefore has nine possible values, $0,3,6,\\ldots,24$.\n\\vfig{05_smooth_grid}\n\\head{Why the exponent is $d(m-1)$}\nAt depth $m$, the image of the maximal ideal contains $3^{m-1}$ elements. Two independent parameters give $3^{2(m-1)}$ points. The grid displays all $81$ points at depth three, not a random sample. Reduction to depth two maps each coordinate to one of $0,3,6$, and every parent has nine children.\n\\[\n\\#X_m=3^{2(m-1)},\\qquad\nZ(T)=\\sum_{m\\ge1}\\#X_mT^m=\\frac{T}{1-9T}.\n\\]\nCounting all $27^2$ parameter assignments at depth three would ignore locality. Counting only tangent","\\vpage{5}{A smooth chart has independent choices}{ACTUAL FINITE POINTS}{Main text: Proposition 3.4 and Theorem 6.2.}\nTake $R=\\ZZ_3[[t_1,t_2]]$ and coefficient-preserving local maps to $\\ZZ/27\\ZZ$. Locality forces both parameters into the maximal ideal. Each therefore has nine possible values, $0,3,6,\\ldots,24$.\n\\vfig{05_smooth_grid}\n\\head{Why the exponent is $d(m-1)$}\nAt depth $m$, the image of the maximal ideal contains $3^{m-1}$ elements. Two independent parameters give $3^{2(m-1)}$ points. The grid displays all $81$ points at depth three, not a random sample. Reduction to depth two maps each coordinate to one of $0,3,6$, and every parent has nine children.\n\\[\n\\#X_m=3^{2(m-1)},\\qquad\nZ(T)=\\sum_{m\\ge1}\\#X_mT^m=\\frac{T}{1-9T}.\n\\]\nCounting all $27^2$ parameter assignments at depth three would ignore locality. Counting only tangent directions would describe the first layer but not al","\\vpage{5}{A smooth chart has independent choices}{ACTUAL FINITE POINTS}{Main text: Proposition 3.4 and Theorem 6.2.}\nTake $R=\\ZZ_3[[t_1,t_2]]$ and coefficient-preserving local maps to $\\ZZ/27\\ZZ$. Locality forces both parameters into the maximal ideal. Each therefore has nine possible values, $0,3,6,\\ldots,24$.\n\\vfig{05_smooth_grid}\n\\head{Why the exponent is $d(m-1)$}\nAt depth $m$, the image of the maximal ideal contains $3^{m-1}$ elements. Two independent parameters give $3^{2(m-1)}$ points. The grid displays all $81$ points at depth three, not a random sample. Reduction to depth two maps each coordinate to one of $0,3,6$, and every parent has nine children.\n\\[\n\\#X_m=3^{2(m-1)},\\qquad\nZ(T)=\\sum_{m\\ge1}\\#X_mT^m=\\frac{T}{1-9T}.\n\\]\nCounting all $27^2$ parameter assignments at depth three would ignore locality. Counting only tangent directions would describe the first layer but not all later layers. The power-series presentation explains why these choices stay independent through every precision.","as independent choices}{ACTUAL FINITE POINTS}{Main text: Proposition 3.4 and Theorem 6.2.}\nTake $R=\\ZZ_3[[t_1,t_2]]$ and coefficient-preserving local maps to $\\ZZ/27\\ZZ$. Locality forces both parameters into the maximal ideal. Each therefore has nine possible values, $0,3,6,\\ldots,24$.\n\\vfig{05_smooth_grid}\n\\head{Why the exponent is $d(m-1)$}\nAt depth $m$, the image of the maximal ideal contains $3^{m-1}$ elements. Two independent parameters give $3^{2(m-1)}$ points. The grid displays all $81$ points at depth three, not a random sample. Reduction to depth two maps each coordinate to one of $0,3,6$, and every parent has nine children.\n\\[\n\\#X_m=3^{2(m-1)},\\qquad\nZ(T)=\\sum_{m\\ge1}\\#X_mT^m=\\frac{T}{1-9T}.\n\\]\nCounting all $27^2$ parameter assignments at depth three would ignore locality. Counting only tangent directions would describe the first layer but not all later layers. The power-series presentation explains why these choices stay independent through every precision.","ext: Proposition 3.4 and Theorem 6.2.}\nTake $R=\\ZZ_3[[t_1,t_2]]$ and coefficient-preserving local maps to $\\ZZ/27\\ZZ$. Locality forces both parameters into the maximal ideal. Each therefore has nine possible values, $0,3,6,\\ldots,24$.\n\\vfig{05_smooth_grid}\n\\head{Why the exponent is $d(m-1)$}\nAt depth $m$, the image of the maximal ideal contains $3^{m-1}$ elements. Two independent parameters give $3^{2(m-1)}$ points. The grid displays all $81$ points at depth three, not a random sample. Reduction to depth two maps each coordinate to one of $0,3,6$, and every parent has nine children.\n\\[\n\\#X_m=3^{2(m-1)},\\qquad\nZ(T)=\\sum_{m\\ge1}\\#X_mT^m=\\frac{T}{1-9T}.\n\\]\nCounting all $27^2$ parameter assignments at depth three would ignore locality. Counting only tangent directions would describe the first layer but not all later layers. The power-series presentation explains why these choices stay independent through every precision.","$R=\\ZZ_3[[t_1,t_2]]$ and coefficient-preserving local maps to $\\ZZ/27\\ZZ$. Locality forces both parameters into the maximal ideal. Each therefore has nine possible values, $0,3,6,\\ldots,24$.\n\\vfig{05_smooth_grid}\n\\head{Why the exponent is $d(m-1)$}\nAt depth $m$, the image of the maximal ideal contains $3^{m-1}$ elements. Two independent parameters give $3^{2(m-1)}$ points. The grid displays all $81$ points at depth three, not a random sample. Reduction to depth two maps each coordinate to one of $0,3,6$, and every parent has nine children.\n\\[\n\\#X_m=3^{2(m-1)},\\qquad\nZ(T)=\\sum_{m\\ge1}\\#X_mT^m=\\frac{T}{1-9T}.\n\\]\nCounting all $27^2$ parameter assignments at depth three would ignore locality. Counting only tangent directions would describe the first layer but not all later layers. The power-series presentation explains why these choices stay independent through every precision.","ters into the maximal ideal. Each therefore has nine possible values, $0,3,6,\\ldots,24$.\n\\vfig{05_smooth_grid}\n\\head{Why the exponent is $d(m-1)$}\nAt depth $m$, the image of the maximal ideal contains $3^{m-1}$ elements. Two independent parameters give $3^{2(m-1)}$ points. The grid displays all $81$ points at depth three, not a random sample. Reduction to depth two maps each coordinate to one of $0,3,6$, and every parent has nine children.\n\\[\n\\#X_m=3^{2(m-1)},\\qquad\nZ(T)=\\sum_{m\\ge1}\\#X_mT^m=\\frac{T}{1-9T}.\n\\]\nCounting all $27^2$ parameter assignments at depth three would ignore locality. Counting only tangent directions would describe the first layer but not all later layers. The power-series presentation explains why these choices stay independent through every precision.","s nine possible values, $0,3,6,\\ldots,24$.\n\\vfig{05_smooth_grid}\n\\head{Why the exponent is $d(m-1)$}\nAt depth $m$, the image of the maximal ideal contains $3^{m-1}$ elements. Two independent parameters give $3^{2(m-1)}$ points. The grid displays all $81$ points at depth three, not a random sample. Reduction to depth two maps each coordinate to one of $0,3,6$, and every parent has nine children.\n\\[\n\\#X_m=3^{2(m-1)},\\qquad\nZ(T)=\\sum_{m\\ge1}\\#X_mT^m=\\frac{T}{1-9T}.\n\\]\nCounting all $27^2$ parameter assignments at depth three would ignore locality. Counting only tangent directions would describe the first layer but not all later layers. The power-series presentation explains why these choices stay independent through every precision.","ent is $d(m-1)$}\nAt depth $m$, the image of the maximal ideal contains $3^{m-1}$ elements. Two independent parameters give $3^{2(m-1)}$ points. The grid displays all $81$ points at depth three, not a random sample. Reduction to depth two maps each coordinate to one of $0,3,6$, and every parent has nine children.\n\\[\n\\#X_m=3^{2(m-1)},\\qquad\nZ(T)=\\sum_{m\\ge1}\\#X_mT^m=\\frac{T}{1-9T}.\n\\]\nCounting all $27^2$ parameter assignments at depth three would ignore locality. Counting only tangent directions would describe the first layer but not all later layers. The power-series presentation explains why these choices stay independent through every precision.","\\vpage{6}{Weights delay visibility}{WEIGHTED DYADIC CHARTS}{Main text: Proposition 5.3, Theorem 6.3, and §12.2.}\nOver $\\ZZ_2$, require three coordinates to lie in $2\\ZZ_2$, $2^2\\ZZ_2$, and $2^4\\ZZ_2$. At depth $m$, coordinate $i$ has $\\max(m-w_i,0)$ free binary digits. A zero in the table means only one allowed residue: zero.\n\\vfig{06_weights}\n\\head{The two different thresholds}\nA coordinate of weight $w_i$ first becomes nontrivial when $m=w_i+1$. By contrast, the first separation depth of two distinct fixed values $a,b$ is $v_2(a-b)+1$. Thus $1$ and $-1$ separate at depth two, because their difference has valuation one. Neither threshold is simply the printed coordinate label.","{6}{Weights delay visibility}{WEIGHTED DYADIC CHARTS}{Main text: Proposition 5.3, Theorem 6.3, and §12.2.}\nOver $\\ZZ_2$, require three coordinates to lie in $2\\ZZ_2$, $2^2\\ZZ_2$, and $2^4\\ZZ_2$. At depth $m$, coordinate $i$ has $\\max(m-w_i,0)$ free binary digits. A zero in the table means only one allowed residue: zero.\n\\vfig{06_weights}\n\\head{The two different thresholds}\nA coordinate of weight $w_i$ first becomes nontrivial when $m=w_i+1$. By contrast, the first separation depth of two distinct fixed values $a,b$ is $v_2(a-b)+1$. Thus $1$ and $-1$ separate at depth two, because their difference has valuation one. Neither threshold is simply the printed coordinate label.","ED DYADIC CHARTS}{Main text: Proposition 5.3, Theorem 6.3, and §12.2.}\nOver $\\ZZ_2$, require three coordinates to lie in $2\\ZZ_2$, $2^2\\ZZ_2$, and $2^4\\ZZ_2$. At depth $m$, coordinate $i$ has $\\max(m-w_i,0)$ free binary digits. A zero in the table means only one allowed residue: zero.\n\\vfig{06_weights}\n\\head{The two different thresholds}\nA coordinate of weight $w_i$ first becomes nontrivial when $m=w_i+1$. By contrast, the first separation depth of two distinct fixed values $a,b$ is $v_2(a-b)+1$. Thus $1$ and $-1$ separate at depth two, because their difference has valuation one. Neither threshold is simply the printed coordinate label.","_2$, require three coordinates to lie in $2\\ZZ_2$, $2^2\\ZZ_2$, and $2^4\\ZZ_2$. At depth $m$, coordinate $i$ has $\\max(m-w_i,0)$ free binary digits. A zero in the table means only one allowed residue: zero.\n\\vfig{06_weights}\n\\head{The two different thresholds}\nA coordinate of weight $w_i$ first becomes nontrivial when $m=w_i+1$. By contrast, the first separation depth of two distinct fixed values $a,b$ is $v_2(a-b)+1$. Thus $1$ and $-1$ separate at depth two, because their difference has valuation one. Neither threshold is simply the printed coordinate label.","ire three coordinates to lie in $2\\ZZ_2$, $2^2\\ZZ_2$, and $2^4\\ZZ_2$. At depth $m$, coordinate $i$ has $\\max(m-w_i,0)$ free binary digits. A zero in the table means only one allowed residue: zero.\n\\vfig{06_weights}\n\\head{The two different thresholds}\nA coordinate of weight $w_i$ first becomes nontrivial when $m=w_i+1$. By contrast, the first separation depth of two distinct fixed values $a,b$ is $v_2(a-b)+1$. Thus $1$ and $-1$ separate at depth two, because their difference has valuation one. Neither threshold is simply the printed coordinate label.","es to lie in $2\\ZZ_2$, $2^2\\ZZ_2$, and $2^4\\ZZ_2$. At depth $m$, coordinate $i$ has $\\max(m-w_i,0)$ free binary digits. A zero in the table means only one allowed residue: zero.\n\\vfig{06_weights}\n\\head{The two different thresholds}\nA coordinate of weight $w_i$ first becomes nontrivial when $m=w_i+1$. By contrast, the first separation depth of two distinct fixed values $a,b$ is $v_2(a-b)+1$. Thus $1$ and $-1$ separate at depth two, because their difference has valuation one. Neither threshold is simply the printed coordinate label.","e in $2\\ZZ_2$, $2^2\\ZZ_2$, and $2^4\\ZZ_2$. At depth $m$, coordinate $i$ has $\\max(m-w_i,0)$ free binary digits. A zero in the table means only one allowed residue: zero.\n\\vfig{06_weights}\n\\head{The two different thresholds}\nA coordinate of weight $w_i$ first becomes nontrivial when $m=w_i+1$. By contrast, the first separation depth of two distinct fixed values $a,b$ is $v_2(a-b)+1$. Thus $1$ and $-1$ separate at depth two, because their difference has valuation one. Neither threshold is simply the printed coordinate label.","For weights $(1,2,4)$,\n\\[\nN(m)=2^{\\max(m-1,0)+\\max(m-2,0)+\\max(m-4,0)}.\n\\]\nThe first counts are $1,2,8,32,256,2048$. Once all three coordinates are active, every additional precision digit multiplies the count by eight. Separating the initial levels from that geometric tail gives\n\\[\nZ(T)=\\frac{T(1-6T-8T^2-32T^3)}{1-8T}.\n\\]\nNegative coefficients in the numerator do not mean negative numbers of points; expanding the whole rational expression gives positive coefficients.\n\\takeaway{Weights change the finite-depth delay. After activation, this chart still has three independent digit streams.}","\\vpage{7}{Many approximate roots, one limit}{SINGULAR POWER SEEDS}{Main text: Theorem 6.4 and Corollary 6.6.}\nNow impose $x^2=0$ modulo $3^m$. The condition is exactly $2v_3(x)\\ge m$, so the roots are the multiples of $3^{\\lceil m/2\\rceil}$ in that quotient. Their number is $3^{m-\\lceil m/2\\rceil}$.\n\\vfig{07_singular_forest}\n\\head{Count vertices and count branches separately}\nThere are three roots modulo nine: $0,3,6$. But the only roots modulo $27$ are $0,9,18$, and all three reduce to zero modulo nine. The other two vertices at the previous level have no children. Counting them as permanent branches would be incorrect.\n\\[\n\\#X_m=1,3,3,9,9,27,\\ldots,\n\\qquad Z(T)=\\frac{T+3T^2}{1-3T^2}.\n\\]\nThe count-growth exponent is $\\lim_m\\log_3(\\#X_m)/m=1/2$. Nevertheless, a compatible branch reconstructs $x\\in\\ZZ_3$ with $x^2=0$.","\\vpage{7}{Many approximate roots, one limit}{SINGULAR POWER SEEDS}{Main text: Theorem 6.4 and Corollary 6.6.}\nNow impose $x^2=0$ modulo $3^m$. The condition is exactly $2v_3(x)\\ge m$, so the roots are the multiples of $3^{\\lceil m/2\\rceil}$ in that quotient. Their number is $3^{m-\\lceil m/2\\rceil}$.\n\\vfig{07_singular_forest}\n\\head{Count vertices and count branches separately}\nThere are three roots modulo nine: $0,3,6$. But the only roots modulo $27$ are $0,9,18$, and all three reduce to zero modulo nine. The other two vertices at the previous level have no children. Counting them as permanent branches would be incorrect.\n\\[\n\\#X_m=1,3,3,9,9,27,\\ldots,\n\\qquad Z(T)=\\frac{T+3T^2}{1-3T^2}.\n\\]\nThe count-growth exponent is $\\lim_m\\log_3(\\#X_m)/m=1/2$. Nevertheless, a compatible branch reconstructs $x\\in\\ZZ_3$ with $x^2=0$. Since $\\ZZ_3","\\vpage{7}{Many approximate roots, one limit}{SINGULAR POWER SEEDS}{Main text: Theorem 6.4 and Corollary 6.6.}\nNow impose $x^2=0$ modulo $3^m$. The condition is exactly $2v_3(x)\\ge m$, so the roots are the multiples of $3^{\\lceil m/2\\rceil}$ in that quotient. Their number is $3^{m-\\lceil m/2\\rceil}$.\n\\vfig{07_singular_forest}\n\\head{Count vertices and count branches separately}\nThere are three roots modulo nine: $0,3,6$. But the only roots modulo $27$ are $0,9,18$, and all three reduce to zero modulo nine. The other two vertices at the previous level have no children. Counting them as permanent branches would be incorrect.\n\\[\n\\#X_m=1,3,3,9,9,27,\\ldots,\n\\qquad Z(T)=\\frac{T+3T^2}{1-3T^2}.\n\\]\nThe count-growth exponent is $\\lim_m\\log_3(\\#X_m)/m=1/2$. Nevertheless, a compatible branch reconstructs $x\\in\\ZZ_3$ with $x^2=0$. Since $\\ZZ_3$ has no nonzero nilpotent, the only such","\\vpage{7}{Many approximate roots, one limit}{SINGULAR POWER SEEDS}{Main text: Theorem 6.4 and Corollary 6.6.}\nNow impose $x^2=0$ modulo $3^m$. The condition is exactly $2v_3(x)\\ge m$, so the roots are the multiples of $3^{\\lceil m/2\\rceil}$ in that quotient. Their number is $3^{m-\\lceil m/2\\rceil}$.\n\\vfig{07_singular_forest}\n\\head{Count vertices and count branches separately}\nThere are three roots modulo nine: $0,3,6$. But the only roots modulo $27$ are $0,9,18$, and all three reduce to zero modulo nine. The other two vertices at the previous level have no children. Counting them as permanent branches would be incorrect.\n\\[\n\\#X_m=1,3,3,9,9,27,\\ldots,\n\\qquad Z(T)=\\frac{T+3T^2}{1-3T^2}.\n\\]\nThe count-growth exponent is $\\lim_m\\log_3(\\#X_m)/m=1/2$. Nevertheless, a compatible branch reconstructs $x\\in\\ZZ_3$ with $x^2=0$. Since $\\ZZ_3$ has no nonzero nilpotent, the only such branch is the zero branch.","approximate roots, one limit}{SINGULAR POWER SEEDS}{Main text: Theorem 6.4 and Corollary 6.6.}\nNow impose $x^2=0$ modulo $3^m$. The condition is exactly $2v_3(x)\\ge m$, so the roots are the multiples of $3^{\\lceil m/2\\rceil}$ in that quotient. Their number is $3^{m-\\lceil m/2\\rceil}$.\n\\vfig{07_singular_forest}\n\\head{Count vertices and count branches separately}\nThere are three roots modulo nine: $0,3,6$. But the only roots modulo $27$ are $0,9,18$, and all three reduce to zero modulo nine. The other two vertices at the previous level have no children. Counting them as permanent branches would be incorrect.\n\\[\n\\#X_m=1,3,3,9,9,27,\\ldots,\n\\qquad Z(T)=\\frac{T+3T^2}{1-3T^2}.\n\\]\nThe count-growth exponent is $\\lim_m\\log_3(\\#X_m)/m=1/2$. Nevertheless, a compatible branch reconstructs $x\\in\\ZZ_3$ with $x^2=0$. Since $\\ZZ_3$ has no nonzero nilpotent, the only such branch is the zero branch.","LAR POWER SEEDS}{Main text: Theorem 6.4 and Corollary 6.6.}\nNow impose $x^2=0$ modulo $3^m$. The condition is exactly $2v_3(x)\\ge m$, so the roots are the multiples of $3^{\\lceil m/2\\rceil}$ in that quotient. Their number is $3^{m-\\lceil m/2\\rceil}$.\n\\vfig{07_singular_forest}\n\\head{Count vertices and count branches separately}\nThere are three roots modulo nine: $0,3,6$. But the only roots modulo $27$ are $0,9,18$, and all three reduce to zero modulo nine. The other two vertices at the previous level have no children. Counting them as permanent branches would be incorrect.\n\\[\n\\#X_m=1,3,3,9,9,27,\\ldots,\n\\qquad Z(T)=\\frac{T+3T^2}{1-3T^2}.\n\\]\nThe count-growth exponent is $\\lim_m\\log_3(\\#X_m)/m=1/2$. Nevertheless, a compatible branch reconstructs $x\\in\\ZZ_3$ with $x^2=0$. Since $\\ZZ_3$ has no nonzero nilpotent, the only such branch is the zero branch.","SEEDS}{Main text: Theorem 6.4 and Corollary 6.6.}\nNow impose $x^2=0$ modulo $3^m$. The condition is exactly $2v_3(x)\\ge m$, so the roots are the multiples of $3^{\\lceil m/2\\rceil}$ in that quotient. Their number is $3^{m-\\lceil m/2\\rceil}$.\n\\vfig{07_singular_forest}\n\\head{Count vertices and count branches separately}\nThere are three roots modulo nine: $0,3,6$. But the only roots modulo $27$ are $0,9,18$, and all three reduce to zero modulo nine. The other two vertices at the previous level have no children. Counting them as permanent branches would be incorrect.\n\\[\n\\#X_m=1,3,3,9,9,27,\\ldots,\n\\qquad Z(T)=\\frac{T+3T^2}{1-3T^2}.\n\\]\nThe count-growth exponent is $\\lim_m\\log_3(\\#X_m)/m=1/2$. Nevertheless, a compatible branch reconstructs $x\\in\\ZZ_3$ with $x^2=0$. Since $\\ZZ_3$ has no nonzero nilpotent, the only such branch is the zero branch.","2\\rceil}$ in that quotient. Their number is $3^{m-\\lceil m/2\\rceil}$.\n\\vfig{07_singular_forest}\n\\head{Count vertices and count branches separately}\nThere are three roots modulo nine: $0,3,6$. But the only roots modulo $27$ are $0,9,18$, and all three reduce to zero modulo nine. The other two vertices at the previous level have no children. Counting them as permanent branches would be incorrect.\n\\[\n\\#X_m=1,3,3,9,9,27,\\ldots,\n\\qquad Z(T)=\\frac{T+3T^2}{1-3T^2}.\n\\]\nThe count-growth exponent is $\\lim_m\\log_3(\\#X_m)/m=1/2$. Nevertheless, a compatible branch reconstructs $x\\in\\ZZ_3$ with $x^2=0$. Since $\\ZZ_3$ has no nonzero nilpotent, the only such branch is the zero branch.","\\head{Count vertices and count branches separately}\nThere are three roots modulo nine: $0,3,6$. But the only roots modulo $27$ are $0,9,18$, and all three reduce to zero modulo nine. The other two vertices at the previous level have no children. Counting them as permanent branches would be incorrect.\n\\[\n\\#X_m=1,3,3,9,9,27,\\ldots,\n\\qquad Z(T)=\\frac{T+3T^2}{1-3T^2}.\n\\]\nThe count-growth exponent is $\\lim_m\\log_3(\\#X_m)/m=1/2$. Nevertheless, a compatible branch reconstructs $x\\in\\ZZ_3$ with $x^2=0$. Since $\\ZZ_3$ has no nonzero nilpotent, the only such branch is the zero branch.","dulo nine: $0,3,6$. But the only roots modulo $27$ are $0,9,18$, and all three reduce to zero modulo nine. The other two vertices at the previous level have no children. Counting them as permanent branches would be incorrect.\n\\[\n\\#X_m=1,3,3,9,9,27,\\ldots,\n\\qquad Z(T)=\\frac{T+3T^2}{1-3T^2}.\n\\]\nThe count-growth exponent is $\\lim_m\\log_3(\\#X_m)/m=1/2$. Nevertheless, a compatible branch reconstructs $x\\in\\ZZ_3$ with $x^2=0$. Since $\\ZZ_3$ has no nonzero nilpotent, the only such branch is the zero branch.","$. But the only roots modulo $27$ are $0,9,18$, and all three reduce to zero modulo nine. The other two vertices at the previous level have no children. Counting them as permanent branches would be incorrect.\n\\[\n\\#X_m=1,3,3,9,9,27,\\ldots,\n\\qquad Z(T)=\\frac{T+3T^2}{1-3T^2}.\n\\]\nThe count-growth exponent is $\\lim_m\\log_3(\\#X_m)/m=1/2$. Nevertheless, a compatible branch reconstructs $x\\in\\ZZ_3$ with $x^2=0$. Since $\\ZZ_3$ has no nonzero nilpotent, the only such branch is the zero branch.","roots modulo $27$ are $0,9,18$, and all three reduce to zero modulo nine. The other two vertices at the previous level have no children. Counting them as permanent branches would be incorrect.\n\\[\n\\#X_m=1,3,3,9,9,27,\\ldots,\n\\qquad Z(T)=\\frac{T+3T^2}{1-3T^2}.\n\\]\nThe count-growth exponent is $\\lim_m\\log_3(\\#X_m)/m=1/2$. Nevertheless, a compatible branch reconstructs $x\\in\\ZZ_3$ with $x^2=0$. Since $\\ZZ_3$ has no nonzero nilpotent, the only such branch is the zero branch.","More generally, $x^e=0$ has count $q^{m-\\lceil m/e\\rceil}$ and count-growth exponent $1-1/e$. These exponents describe finite thickenings. They do not assert that the set of complete roots has a fractional geometric dimension or more than one point.\n\\takeaway{The seed zeta function counts finite-level solutions. It is not a generating function for distinct infinite branches.}","\\vpage{8}{One equation, several local outcomes}{A SCALAR COMPANION MODEL}{Main text: Theorem 8.2, Corollary 8.3, and Example 8.4.}\nConsider the fixed scalar equation $x^2+1=0$ in $\\ZZ/p^m\\ZZ$. The following are exact counts for five primes and three depths. The coefficient rings and the scalar nature of the unknown are fixed throughout this page.\n\\vfig{08_prime_forests}\n\\head{Where failure occurs matters}\nAt $p=3$ and $p=7$, no residual root exists. At $p=2$, the residual root exists but dies immediately: every odd square is one modulo four. At $p=5$ and $p=13$, the two residual roots are simple, and the usual one-digit equation gives one child above each root at every depth.","equation, several local outcomes}{A SCALAR COMPANION MODEL}{Main text: Theorem 8.2, Corollary 8.3, and Example 8.4.}\nConsider the fixed scalar equation $x^2+1=0$ in $\\ZZ/p^m\\ZZ$. The following are exact counts for five primes and three depths. The coefficient rings and the scalar nature of the unknown are fixed throughout this page.\n\\vfig{08_prime_forests}\n\\head{Where failure occurs matters}\nAt $p=3$ and $p=7$, no residual root exists. At $p=2$, the residual root exists but dies immediately: every odd square is one modulo four. At $p=5$ and $p=13$, the two residual roots are simple, and the usual one-digit equation gives one child above each root at every depth.","several local outcomes}{A SCALAR COMPANION MODEL}{Main text: Theorem 8.2, Corollary 8.3, and Example 8.4.}\nConsider the fixed scalar equation $x^2+1=0$ in $\\ZZ/p^m\\ZZ$. The following are exact counts for five primes and three depths. The coefficient rings and the scalar nature of the unknown are fixed throughout this page.\n\\vfig{08_prime_forests}\n\\head{Where failure occurs matters}\nAt $p=3$ and $p=7$, no residual root exists. At $p=2$, the residual root exists but dies immediately: every odd square is one modulo four. At $p=5$ and $p=13$, the two residual roots are simple, and the usual one-digit equation gives one child above each root at every depth.","OMPANION MODEL}{Main text: Theorem 8.2, Corollary 8.3, and Example 8.4.}\nConsider the fixed scalar equation $x^2+1=0$ in $\\ZZ/p^m\\ZZ$. The following are exact counts for five primes and three depths. The coefficient rings and the scalar nature of the unknown are fixed throughout this page.\n\\vfig{08_prime_forests}\n\\head{Where failure occurs matters}\nAt $p=3$ and $p=7$, no residual root exists. At $p=2$, the residual root exists but dies immediately: every odd square is one modulo four. At $p=5$ and $p=13$, the two residual roots are simple, and the usual one-digit equation gives one child above each root at every depth.","alar equation $x^2+1=0$ in $\\ZZ/p^m\\ZZ$. The following are exact counts for five primes and three depths. The coefficient rings and the scalar nature of the unknown are fixed throughout this page.\n\\vfig{08_prime_forests}\n\\head{Where failure occurs matters}\nAt $p=3$ and $p=7$, no residual root exists. At $p=2$, the residual root exists but dies immediately: every odd square is one modulo four. At $p=5$ and $p=13$, the two residual roots are simple, and the usual one-digit equation gives one child above each root at every depth.","ion $x^2+1=0$ in $\\ZZ/p^m\\ZZ$. The following are exact counts for five primes and three depths. The coefficient rings and the scalar nature of the unknown are fixed throughout this page.\n\\vfig{08_prime_forests}\n\\head{Where failure occurs matters}\nAt $p=3$ and $p=7$, no residual root exists. At $p=2$, the residual root exists but dies immediately: every odd square is one modulo four. At $p=5$ and $p=13$, the two residual roots are simple, and the usual one-digit equation gives one child above each root at every depth.","At five, one branch begins $2,7,57,182$; the other is its negative modulo each power of five. Arbitrarily many digits in either branch cannot supply the absent scalar root over $\\ZZ_3$. Nor can one root modulo two satisfy the hypothesis that every finite level is nonempty.","\\vpage{9}{Why finiteness enters compactness}{AN INFINITE-LEVEL COUNTERMODEL}{Main text: Theorem 3.3 and its finite-branching hypotheses.}\nLet $X_m=\\{n\\in\\ZZ:n\\ge m\\}$, with the inclusion $X_{m+1}\\hookrightarrow X_m$ as transition map. Every level is nonempty. Every vertex has at most one child. Yet there is no compatible infinite branch.\n\\vfig{09_infinite_tail}\n\\head{The missing hypothesis is not surjectivity}\nCompatibility would force all coordinates of a branch to be one fixed integer $n$. Membership at every level would then require $n\\ge m$ for every $m$, which is impossible. Each vertical column in the drawing eventually ends. A finite window always contains columns extending beyond it, but no single column survives all windows.","em 3.3 and its finite-branching hypotheses.}\nLet $X_m=\\{n\\in\\ZZ:n\\ge m\\}$, with the inclusion $X_{m+1}\\hookrightarrow X_m$ as transition map. Every level is nonempty. Every vertex has at most one child. Yet there is no compatible infinite branch.\n\\vfig{09_infinite_tail}\n\\head{The missing hypothesis is not surjectivity}\nCompatibility would force all coordinates of a branch to be one fixed integer $n$. Membership at every level would then require $n\\ge m$ for every $m$, which is impossible. Each vertical column in the drawing eventually ends. A finite window always contains columns extending beyond it, but no single column survives all windows.","X_m=\\{n\\in\\ZZ:n\\ge m\\}$, with the inclusion $X_{m+1}\\hookrightarrow X_m$ as transition map. Every level is nonempty. Every vertex has at most one child. Yet there is no compatible infinite branch.\n\\vfig{09_infinite_tail}\n\\head{The missing hypothesis is not surjectivity}\nCompatibility would force all coordinates of a branch to be one fixed integer $n$. Membership at every level would then require $n\\ge m$ for every $m$, which is impossible. Each vertical column in the drawing eventually ends. A finite window always contains columns extending beyond it, but no single column survives all windows.","m\\}$, with the inclusion $X_{m+1}\\hookrightarrow X_m$ as transition map. Every level is nonempty. Every vertex has at most one child. Yet there is no compatible infinite branch.\n\\vfig{09_infinite_tail}\n\\head{The missing hypothesis is not surjectivity}\nCompatibility would force all coordinates of a branch to be one fixed integer $n$. Membership at every level would then require $n\\ge m$ for every $m$, which is impossible. Each vertical column in the drawing eventually ends. A finite window always contains columns extending beyond it, but no single column survives all windows.","In the paper, finite nonempty levels and actual total reduction maps provide compactness even when some vertices die and some transition maps are not onto. One can retain only vertices with descendants at arbitrarily large depths. Finiteness makes that surviving part nonempty and permits compatible choices.\n\\[\nX_m\\ne\\varnothing\\ \\text{for all }m\n\\quad\\not\\Longrightarrow\\quad\n\\varprojlim_m X_m\\ne\\varnothing\n\\quad\\text{without the compactness hypotheses.}\n\\]\n\\takeaway{Arbitrarily long finite paths are not automatically one infinite path. The finite-level hypothesis prevents escape to new roots.}","\\vpage{10}{See cycles, boundaries, and classes}{A TWO-PACKET COCHAIN COMPLEX}{Main text: Theorem 4.2 and the finite-complex example.}\nOver $\\FF_5$, use bases $(e_0,e_1)$, $(u_0,u_1,u_2,u_3)$, and $(f_0,f_1)$ in degrees zero, one, and two. The only nonzero differential values are $d_0e_0=u_0$ and $d_1u_1=f_0$. Consequently $d_1d_0=0$.\n\\vfig{10_cohomology}\n\\head{The quotient is the important step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. A","\\vpage{10}{See cycles, boundaries, and classes}{A TWO-PACKET COCHAIN COMPLEX}{Main text: Theorem 4.2 and the finite-complex example.}\nOver $\\FF_5$, use bases $(e_0,e_1)$, $(u_0,u_1,u_2,u_3)$, and $(f_0,f_1)$ in degrees zero, one, and two. The only nonzero differential values are $d_0e_0=u_0$ and $d_1u_1=f_0$. Consequently $d_1d_0=0$.\n\\vfig{10_cohomology}\n\\head{The quotient is the important step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomo","\\vpage{10}{See cycles, boundaries, and classes}{A TWO-PACKET COCHAIN COMPLEX}{Main text: Theorem 4.2 and the finite-complex example.}\nOver $\\FF_5$, use bases $(e_0,e_1)$, $(u_0,u_1,u_2,u_3)$, and $(f_0,f_1)$ in degrees zero, one, and two. The only nonzero differential values are $d_0e_0=u_0$ and $d_1u_1=f_0$. Consequently $d_1d_0=0$.\n\\vfig{10_cohomology}\n\\head{The quotient is the important step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomologies reconstructs t","\\vpage{10}{See cycles, boundaries, and classes}{A TWO-PACKET COCHAIN COMPLEX}{Main text: Theorem 4.2 and the finite-complex example.}\nOver $\\FF_5$, use bases $(e_0,e_1)$, $(u_0,u_1,u_2,u_3)$, and $(f_0,f_1)$ in degrees zero, one, and two. The only nonzero differential values are $d_0e_0=u_0$ and $d_1u_1=f_0$. Consequently $d_1d_0=0$.\n\\vfig{10_cohomology}\n\\head{The quotient is the important step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomologies reconstructs the total.","ies, and classes}{A TWO-PACKET COCHAIN COMPLEX}{Main text: Theorem 4.2 and the finite-complex example.}\nOver $\\FF_5$, use bases $(e_0,e_1)$, $(u_0,u_1,u_2,u_3)$, and $(f_0,f_1)$ in degrees zero, one, and two. The only nonzero differential values are $d_0e_0=u_0$ and $d_1u_1=f_0$. Consequently $d_1d_0=0$.\n\\vfig{10_cohomology}\n\\head{The quotient is the important step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomologies reconstructs the total.","CKET COCHAIN COMPLEX}{Main text: Theorem 4.2 and the finite-complex example.}\nOver $\\FF_5$, use bases $(e_0,e_1)$, $(u_0,u_1,u_2,u_3)$, and $(f_0,f_1)$ in degrees zero, one, and two. The only nonzero differential values are $d_0e_0=u_0$ and $d_1u_1=f_0$. Consequently $d_1d_0=0$.\n\\vfig{10_cohomology}\n\\head{The quotient is the important step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomologies reconstructs the total.","-complex example.}\nOver $\\FF_5$, use bases $(e_0,e_1)$, $(u_0,u_1,u_2,u_3)$, and $(f_0,f_1)$ in degrees zero, one, and two. The only nonzero differential values are $d_0e_0=u_0$ and $d_1u_1=f_0$. Consequently $d_1d_0=0$.\n\\vfig{10_cohomology}\n\\head{The quotient is the important step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomologies reconstructs the total.","s $(e_0,e_1)$, $(u_0,u_1,u_2,u_3)$, and $(f_0,f_1)$ in degrees zero, one, and two. The only nonzero differential values are $d_0e_0=u_0$ and $d_1u_1=f_0$. Consequently $d_1d_0=0$.\n\\vfig{10_cohomology}\n\\head{The quotient is the important step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomologies reconstructs the total.","degrees zero, one, and two. The only nonzero differential values are $d_0e_0=u_0$ and $d_1u_1=f_0$. Consequently $d_1d_0=0$.\n\\vfig{10_cohomology}\n\\head{The quotient is the important step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomologies reconstructs the total.","ero, one, and two. The only nonzero differential values are $d_0e_0=u_0$ and $d_1u_1=f_0$. Consequently $d_1d_0=0$.\n\\vfig{10_cohomology}\n\\head{The quotient is the important step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomologies reconstructs the total.",", and two. The only nonzero differential values are $d_0e_0=u_0$ and $d_1u_1=f_0$. Consequently $d_1d_0=0$.\n\\vfig{10_cohomology}\n\\head{The quotient is the important step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomologies reconstructs the total.","e $d_0e_0=u_0$ and $d_1u_1=f_0$. Consequently $d_1d_0=0$.\n\\vfig{10_cohomology}\n\\head{The quotient is the important step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomologies reconstructs the total.","_0=u_0$ and $d_1u_1=f_0$. Consequently $d_1d_0=0$.\n\\vfig{10_cohomology}\n\\head{The quotient is the important step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomologies reconstructs the total.","1u_1=f_0$. Consequently $d_1d_0=0$.\n\\vfig{10_cohomology}\n\\head{The quotient is the important step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomologies reconstructs the total.","\\vfig{10_cohomology}\n\\head{The quotient is the important step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomologies reconstructs the total.","he quotient is the important step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomologies reconstructs the total.","ient is the important step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomologies reconstructs the total.","ortant step}\nIn degree one, the cycles are $\\Span(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomologies reconstructs the total.","n(u_0,u_2,u_3)$, but $u_0$ is a boundary. Thus the cohomology is represented by $u_2,u_3$, not by every vector killed by $d_1$. In the other degrees the surviving classes are $e_1$ and $f_1$.\n\\[\n\\dim(H^0,H^1,H^2)=(1,2,1).\n\\]\nPacket A contains $e_0$, $u_0,u_1$, and $f_0$. It is exact and contributes $(0,0,0)$. Packet B contains $e_1$, $u_2,u_3$, and $f_1$, with zero differentials, and contributes $(1,2,1)$. Adding the packet cohomologies reconstructs the total.","\\vpage{11}{An arbitrary cut need not descend}{THE CHAIN-MAP CONDITION}{Main text: Theorem 4.2 and the scope of the cohomology splitting.}\nRetain the previous complex. The valid A-packet projectors are $P^0=\\diag(1,0)$, $P^1=\\diag(1,1,0,0)$, and $P^2=\\diag(1,0)$. They satisfy $d_rP^r=P^{r+1}d_r$.\n\\vfig{11_chain_defect}\n\\head{A boundary can cross an incompatible cut}\nReplace the degree-one projector by $Q^1=\\diag(0,1,0,0)$ while keeping $P^0$. Then\n\\[\nd_0P^0e_0=u_0,\n\\qquad Q^1d_0e_0=0,\n\\qquad(d_0P^0-Q^1d_0)e_0=u_0\\ne0.\n\\]\nThe selected degree-zero component is sent outside the selected degree-one component. There is no subcomplex with these cuts, so the cohomology packet theorem cannot be applied to them. Both matrices remain perfectly valid idempotents; the failure is compatibility with the differential.","e{11}{An arbitrary cut need not descend}{THE CHAIN-MAP CONDITION}{Main text: Theorem 4.2 and the scope of the cohomology splitting.}\nRetain the previous complex. The valid A-packet projectors are $P^0=\\diag(1,0)$, $P^1=\\diag(1,1,0,0)$, and $P^2=\\diag(1,0)$. They satisfy $d_rP^r=P^{r+1}d_r$.\n\\vfig{11_chain_defect}\n\\head{A boundary can cross an incompatible cut}\nReplace the degree-one projector by $Q^1=\\diag(0,1,0,0)$ while keeping $P^0$. Then\n\\[\nd_0P^0e_0=u_0,\n\\qquad Q^1d_0e_0=0,\n\\qquad(d_0P^0-Q^1d_0)e_0=u_0\\ne0.\n\\]\nThe selected degree-zero component is sent outside the selected degree-one component. There is no subcomplex with these cuts, so the cohomology packet theorem cannot be applied to them. Both matrices remain perfectly valid idempotents; the failure is compatibility with the differential.","cend}{THE CHAIN-MAP CONDITION}{Main text: Theorem 4.2 and the scope of the cohomology splitting.}\nRetain the previous complex. The valid A-packet projectors are $P^0=\\diag(1,0)$, $P^1=\\diag(1,1,0,0)$, and $P^2=\\diag(1,0)$. They satisfy $d_rP^r=P^{r+1}d_r$.\n\\vfig{11_chain_defect}\n\\head{A boundary can cross an incompatible cut}\nReplace the degree-one projector by $Q^1=\\diag(0,1,0,0)$ while keeping $P^0$. Then\n\\[\nd_0P^0e_0=u_0,\n\\qquad Q^1d_0e_0=0,\n\\qquad(d_0P^0-Q^1d_0)e_0=u_0\\ne0.\n\\]\nThe selected degree-zero component is sent outside the selected degree-one component. There is no subcomplex with these cuts, so the cohomology packet theorem cannot be applied to them. Both matrices remain perfectly valid idempotents; the failure is compatibility with the differential.","IN-MAP CONDITION}{Main text: Theorem 4.2 and the scope of the cohomology splitting.}\nRetain the previous complex. The valid A-packet projectors are $P^0=\\diag(1,0)$, $P^1=\\diag(1,1,0,0)$, and $P^2=\\diag(1,0)$. They satisfy $d_rP^r=P^{r+1}d_r$.\n\\vfig{11_chain_defect}\n\\head{A boundary can cross an incompatible cut}\nReplace the degree-one projector by $Q^1=\\diag(0,1,0,0)$ while keeping $P^0$. Then\n\\[\nd_0P^0e_0=u_0,\n\\qquad Q^1d_0e_0=0,\n\\qquad(d_0P^0-Q^1d_0)e_0=u_0\\ne0.\n\\]\nThe selected degree-zero component is sent outside the selected degree-one component. There is no subcomplex with these cuts, so the cohomology packet theorem cannot be applied to them. Both matrices remain perfectly valid idempotents; the failure is compatibility with the differential.","This example separates two kinds of obstruction. The nonzero matrix above obstructs this proposed decomposition. A nonzero class in $H^2$ may obstruct a lift only after the deformation problem and its obstruction map have been identified. One should not equate every failed projector commutator with an arithmetic obstruction class.","\\vpage{12}{How many starting seeds are necessary?}{DIRECTED TRANSPORT}{Main text: Theorem 7.3 and the seven-root example.}\nThe arrows below are the entire supplied transport graph. Vertices joined by paths in both directions form strongly connected components. The components $\\{A,B\\}$, $\\{D,E\\}$, and $\\{G\\}$ have no incoming edge from outside themselves.\n\\vfig{12_source_scc}\n\\head{A lower bound and a construction}\nAny seed set reaching every vertex must meet all three source components. Otherwise there is no path into the omitted source. Conversely, one representative from each source reaches $C$ and $F$ as well. The directed seed rank is therefore three.","aths in both directions form strongly connected components. The components $\\{A,B\\}$, $\\{D,E\\}$, and $\\{G\\}$ have no incoming edge from outside themselves.\n\\vfig{12_source_scc}\n\\head{A lower bound and a construction}\nAny seed set reaching every vertex must meet all three source components. Otherwise there is no path into the omitted source. Conversely, one representative from each source reaches $C$ and $F$ as well. The directed seed rank is therefore three.","both directions form strongly connected components. The components $\\{A,B\\}$, $\\{D,E\\}$, and $\\{G\\}$ have no incoming edge from outside themselves.\n\\vfig{12_source_scc}\n\\head{A lower bound and a construction}\nAny seed set reaching every vertex must meet all three source components. Otherwise there is no path into the omitted source. Conversely, one representative from each source reaches $C$ and $F$ as well. The directed seed rank is therefore three.","There are precisely four minimum seed sets:\n\\[\n\\{A,D,G\\},\\quad\\{A,E,G\\},\\quad\n\\{B,D,G\\},\\quad\\{B,E,G\\}.\n\\]\nThe example checks this by exhaustive enumeration of all $2^7$ subsets, independently of the strongly connected component calculation. Adding an admissible edge cannot increase the required seed count. In particular, adding $F\\to G$ makes $G$ reachable from the other sources, reducing the count to two.","\\vpage{13}{A transport must work at every depth}{ONE REVERSIBLE EXAMPLE}{Main text: Definition 7.1 and the reduction-compatibility requirement.}\nFor the two roots of $x^2+1=0$ modulo powers of five, negation is an actual reversible transport. It preserves the equation, is its own inverse, and commutes with reduction.\n\\vfig{13_transport}\n\\head{One successful input is not a natural transformation}\nAt every depth $m$,\n\\[\nN_m(x)=-x\\pmod{5^m},\\qquad\nr_{m+1,m}N_{m+1}=N_mr_{m+1,m}.\n\\]\nThe two compatible branches beginning at $2$ and $3$ are exchanged. Because $2\\not\\equiv3\\pmod5$, this transport changes the residual root. It is horizontal rather than a positive-depth correction.","ansport must work at every depth}{ONE REVERSIBLE EXAMPLE}{Main text: Definition 7.1 and the reduction-compatibility requirement.}\nFor the two roots of $x^2+1=0$ modulo powers of five, negation is an actual reversible transport. It preserves the equation, is its own inverse, and commutes with reduction.\n\\vfig{13_transport}\n\\head{One successful input is not a natural transformation}\nAt every depth $m$,\n\\[\nN_m(x)=-x\\pmod{5^m},\\qquad\nr_{m+1,m}N_{m+1}=N_mr_{m+1,m}.\n\\]\nThe two compatible branches beginning at $2$ and $3$ are exchanged. Because $2\\not\\equiv3\\pmod5$, this transport changes the residual root. It is horizontal rather than a positive-depth correction.","t must work at every depth}{ONE REVERSIBLE EXAMPLE}{Main text: Definition 7.1 and the reduction-compatibility requirement.}\nFor the two roots of $x^2+1=0$ modulo powers of five, negation is an actual reversible transport. It preserves the equation, is its own inverse, and commutes with reduction.\n\\vfig{13_transport}\n\\head{One successful input is not a natural transformation}\nAt every depth $m$,\n\\[\nN_m(x)=-x\\pmod{5^m},\\qquad\nr_{m+1,m}N_{m+1}=N_mr_{m+1,m}.\n\\]\nThe two compatible branches beginning at $2$ and $3$ are exchanged. Because $2\\not\\equiv3\\pmod5$, this transport changes the residual root. It is horizontal rather than a positive-depth correction.","ility requirement.}\nFor the two roots of $x^2+1=0$ modulo powers of five, negation is an actual reversible transport. It preserves the equation, is its own inverse, and commutes with reduction.\n\\vfig{13_transport}\n\\head{One successful input is not a natural transformation}\nAt every depth $m$,\n\\[\nN_m(x)=-x\\pmod{5^m},\\qquad\nr_{m+1,m}N_{m+1}=N_mr_{m+1,m}.\n\\]\nThe two compatible branches beginning at $2$ and $3$ are exchanged. Because $2\\not\\equiv3\\pmod5$, this transport changes the residual root. It is horizontal rather than a positive-depth correction.","quirement.}\nFor the two roots of $x^2+1=0$ modulo powers of five, negation is an actual reversible transport. It preserves the equation, is its own inverse, and commutes with reduction.\n\\vfig{13_transport}\n\\head{One successful input is not a natural transformation}\nAt every depth $m$,\n\\[\nN_m(x)=-x\\pmod{5^m},\\qquad\nr_{m+1,m}N_{m+1}=N_mr_{m+1,m}.\n\\]\nThe two compatible branches beginning at $2$ and $3$ are exchanged. Because $2\\not\\equiv3\\pmod5$, this transport changes the residual root. It is horizontal rather than a positive-depth correction.","ts of $x^2+1=0$ modulo powers of five, negation is an actual reversible transport. It preserves the equation, is its own inverse, and commutes with reduction.\n\\vfig{13_transport}\n\\head{One successful input is not a natural transformation}\nAt every depth $m$,\n\\[\nN_m(x)=-x\\pmod{5^m},\\qquad\nr_{m+1,m}N_{m+1}=N_mr_{m+1,m}.\n\\]\nThe two compatible branches beginning at $2$ and $3$ are exchanged. Because $2\\not\\equiv3\\pmod5$, this transport changes the residual root. It is horizontal rather than a positive-depth correction.","Translation by one is a useful false control. Modulo five it takes the root $2$ to the root $3$, but takes the root $3$ to $4$, which is not a root. A single successful arrow between selected points does not define an operation on the whole deformation problem.","Even negation is admissible only for the equation and conditions stated here. An extra condition that fixes the residual root at $2$ would not be preserved. Likewise, a representation-theoretic dual or twist needs its own compatibility checks for the chosen local data and carrier.\n\\takeaway{A transport is a family of maps respecting equations and reductions. A convenient pair of endpoints is not enough.}\n\\captiontext{Both branches are computed by exact one-digit lifting. Negation is checked on all roots at the displayed depths; its universal reduction identity follows directly from ring homomorphism laws.}","\\vpage{14}{Common coefficients, different residues}{THE ARITHMETIC SPINE}{Main text: Definition 8.1 and Theorem 8.2.}\nA common spine records algebraic data before reducing at a coefficient prime. For a small model, take the one polynomial\n\\[\nP(T)=T^2-5T+6=(T-2)(T-3)\\in\\ZZ[T].\n\\]\nIts reductions at five and seven look different, but they come from the same coefficients.\n\\vfig{14_spine}\n\\head{Agreement means descent from supplied data}\nModulo five the polynomial is $T^2+1$; modulo seven it is $T^2+2T+6$. The roots $2$ and $3$ are the reductions of the same integer roots in both places. The coefficient lists are not required to be equal as raw residues in different rings.","PINE}{Main text: Definition 8.1 and Theorem 8.2.}\nA common spine records algebraic data before reducing at a coefficient prime. For a small model, take the one polynomial\n\\[\nP(T)=T^2-5T+6=(T-2)(T-3)\\in\\ZZ[T].\n\\]\nIts reductions at five and seven look different, but they come from the same coefficients.\n\\vfig{14_spine}\n\\head{Agreement means descent from supplied data}\nModulo five the polynomial is $T^2+1$; modulo seven it is $T^2+2T+6$. The roots $2$ and $3$ are the reductions of the same integer roots in both places. The coefficient lists are not required to be equal as raw residues in different rings.","8.1 and Theorem 8.2.}\nA common spine records algebraic data before reducing at a coefficient prime. For a small model, take the one polynomial\n\\[\nP(T)=T^2-5T+6=(T-2)(T-3)\\in\\ZZ[T].\n\\]\nIts reductions at five and seven look different, but they come from the same coefficients.\n\\vfig{14_spine}\n\\head{Agreement means descent from supplied data}\nModulo five the polynomial is $T^2+1$; modulo seven it is $T^2+2T+6$. The roots $2$ and $3$ are the reductions of the same integer roots in both places. The coefficient lists are not required to be equal as raw residues in different rings.","A common spine records algebraic data before reducing at a coefficient prime. For a small model, take the one polynomial\n\\[\nP(T)=T^2-5T+6=(T-2)(T-3)\\in\\ZZ[T].\n\\]\nIts reductions at five and seven look different, but they come from the same coefficients.\n\\vfig{14_spine}\n\\head{Agreement means descent from supplied data}\nModulo five the polynomial is $T^2+1$; modulo seven it is $T^2+2T+6$. The roots $2$ and $3$ are the reductions of the same integer roots in both places. The coefficient lists are not required to be equal as raw residues in different rings.","spine records algebraic data before reducing at a coefficient prime. For a small model, take the one polynomial\n\\[\nP(T)=T^2-5T+6=(T-2)(T-3)\\in\\ZZ[T].\n\\]\nIts reductions at five and seven look different, but they come from the same coefficients.\n\\vfig{14_spine}\n\\head{Agreement means descent from supplied data}\nModulo five the polynomial is $T^2+1$; modulo seven it is $T^2+2T+6$. The roots $2$ and $3$ are the reductions of the same integer roots in both places. The coefficient lists are not required to be equal as raw residues in different rings.","Finite agreement also leaves ambiguity. Adding $35$ to an integer coefficient changes neither its residue modulo five nor its residue modulo seven. Thus two finite coefficient observations do not determine a unique global polynomial without a height bound or further exact information. Conversely, one arbitrarily chosen family of local objects need not descend to an algebraic spine.\n\\takeaway{The common spine is input shared across coefficient places. It is not created by comparing unrelated residue labels after the fact.}","\\vpage{15}{A finite trace matrix can separate types}{CHARACTERS BEFORE ARITHMETIC PRIMES}{Main text: Theorem 9.1; retained scalar/Mellin continuation: arithmetic-system separation.}\nUse the three irreducible characters of $S_3$: trivial, sign, and standard. On the identity, a transposition, and a three-cycle, their values form the matrix below. Rows are samples; columns are constituent types.\n\\vfig{15_separator}\n\\head{What full rank buys}\nSince $\\det C=6$, these samples distinguish every rational multiplicity vector. A target with two trivial constituents, no sign constituent, and one standard constituent has\n\\[\n\\bm m=\\mat{2","s}{CHARACTERS BEFORE ARITHMETIC PRIMES}{Main text: Theorem 9.1; retained scalar/Mellin continuation: arithmetic-system separation.}\nUse the three irreducible characters of $S_3$: trivial, sign, and standard. On the identity, a transposition, and a three-cycle, their values form the matrix below. Rows are samples; columns are constituent types.\n\\vfig{15_separator}\n\\head{What full rank buys}\nSince $\\det C=6$, these samples distinguish every rational multiplicity vector. A target with two trivial constituents, no sign constituent, and one standard constituent has\n\\[\n\\bm m=\\mat{2","of $S_3$: trivial, sign, and standard. On the identity, a transposition, and a three-cycle, their values form the matrix below. Rows are samples; columns are constituent types.\n\\vfig{15_separator}\n\\head{What full rank buys}\nSince $\\det C=6$, these samples distinguish every rational multiplicity vector. A target with two trivial constituents, no sign constituent, and one standard constituent has\n\\[\n\\bm m=\\mat{2\\\\0\\\\1},\\qquad\n\\bm b=C\\bm m=\\mat{4\\\\2\\\\1}.\n\\]\nSolving $C\\bm m=\\bm b$ recovers the unique rational vector. The standard character is not introduced as an arbitrary list: the checks construct matrices for a transposition and a three-cycle, verify their group relations, and obtain their traces.","1}.\n\\]\nSolving $C\\bm m=\\bm b$ recovers the unique rational vector. The standard character is not introduced as an arbitrary list: the checks construct matrices for a transposition and a three-cycle, verify their group relations, and obtain their traces.","\\vpage{16}{A solution is not yet a summand}{INTEGRAL, NONNEGATIVE, WITHIN CAPACITY}{Main text: Theorem 9.2 and its multiplicity-recovery example.}\nKeep the same separating matrix $C$ and suppose the available carrier has capacities $\\bm c=(2,1,1)$. A recovered vector describes a genuine sub-sum of that carrier only if each entry is an integer and $0\\le m_i\\le c_i$.\n\\vfig{16_multiplicities}\n\\head{Three independent rejection tests}\nThe trace vector $(1,0,0)$ gives $(1/6,1/6,1/3)$, so it fails integrality. The vector $(2,-2,-1)$ gives $(-1,1,1)$, which is integral but not nonnegative. The vector $(5,3,2)$ gives $(3,0,1)$, which exceeds the available capacities. In each case the linear equation itself has a unique rational solution.",", NONNEGATIVE, WITHIN CAPACITY}{Main text: Theorem 9.2 and its multiplicity-recovery example.}\nKeep the same separating matrix $C$ and suppose the available carrier has capacities $\\bm c=(2,1,1)$. A recovered vector describes a genuine sub-sum of that carrier only if each entry is an integer and $0\\le m_i\\le c_i$.\n\\vfig{16_multiplicities}\n\\head{Three independent rejection tests}\nThe trace vector $(1,0,0)$ gives $(1/6,1/6,1/3)$, so it fails integrality. The vector $(2,-2,-1)$ gives $(-1,1,1)$, which is integral but not nonnegative. The vector $(5,3,2)$ gives $(3,0,1)$, which exceeds the available capacities. In each case the linear equation itself has a unique rational solution.","ITHIN CAPACITY}{Main text: Theorem 9.2 and its multiplicity-recovery example.}\nKeep the same separating matrix $C$ and suppose the available carrier has capacities $\\bm c=(2,1,1)$. A recovered vector describes a genuine sub-sum of that carrier only if each entry is an integer and $0\\le m_i\\le c_i$.\n\\vfig{16_multiplicities}\n\\head{Three independent rejection tests}\nThe trace vector $(1,0,0)$ gives $(1/6,1/6,1/3)$, so it fails integrality. The vector $(2,-2,-1)$ gives $(-1,1,1)$, which is integral but not nonnegative. The vector $(5,3,2)$ gives $(3,0,1)$, which exceeds the available capacities. In each case the linear equation itself has a unique rational solution.","licity-recovery example.}\nKeep the same separating matrix $C$ and suppose the available carrier has capacities $\\bm c=(2,1,1)$. A recovered vector describes a genuine sub-sum of that carrier only if each entry is an integer and $0\\le m_i\\le c_i$.\n\\vfig{16_multiplicities}\n\\head{Three independent rejection tests}\nThe trace vector $(1,0,0)$ gives $(1/6,1/6,1/3)$, so it fails integrality. The vector $(2,-2,-1)$ gives $(-1,1,1)$, which is integral but not nonnegative. The vector $(5,3,2)$ gives $(3,0,1)$, which exceeds the available capacities. In each case the linear equation itself has a unique rational solution.","ample.}\nKeep the same separating matrix $C$ and suppose the available carrier has capacities $\\bm c=(2,1,1)$. A recovered vector describes a genuine sub-sum of that carrier only if each entry is an integer and $0\\le m_i\\le c_i$.\n\\vfig{16_multiplicities}\n\\head{Three independent rejection tests}\nThe trace vector $(1,0,0)$ gives $(1/6,1/6,1/3)$, so it fails integrality. The vector $(2,-2,-1)$ gives $(-1,1,1)$, which is integral but not nonnegative. The vector $(5,3,2)$ gives $(3,0,1)$, which exceeds the available capacities. In each case the linear equation itself has a unique rational solution.","ailable carrier has capacities $\\bm c=(2,1,1)$. A recovered vector describes a genuine sub-sum of that carrier only if each entry is an integer and $0\\le m_i\\le c_i$.\n\\vfig{16_multiplicities}\n\\head{Three independent rejection tests}\nThe trace vector $(1,0,0)$ gives $(1/6,1/6,1/3)$, so it fails integrality. The vector $(2,-2,-1)$ gives $(-1,1,1)$, which is integral but not nonnegative. The vector $(5,3,2)$ gives $(3,0,1)$, which exceeds the available capacities. In each case the linear equation itself has a unique rational solution.","has capacities $\\bm c=(2,1,1)$. A recovered vector describes a genuine sub-sum of that carrier only if each entry is an integer and $0\\le m_i\\le c_i$.\n\\vfig{16_multiplicities}\n\\head{Three independent rejection tests}\nThe trace vector $(1,0,0)$ gives $(1/6,1/6,1/3)$, so it fails integrality. The vector $(2,-2,-1)$ gives $(-1,1,1)$, which is integral but not nonnegative. The vector $(5,3,2)$ gives $(3,0,1)$, which exceeds the available capacities. In each case the linear equation itself has a unique rational solution.","The accepted example $\\bm b=(4,2,1)$ recovers $(2,0,1)$. When compatible constituent families are already supplied, the same admissible multiplicities define the corresponding direct sum at every coefficient place. This is the finite algebraic core of summand transport.","\\vpage{17}{Separation can consume a digit}{EXACT PRECISION LOSS}{Main text: The finite trace-recovery discussion in §9.}\nThe character matrix has determinant six, so its inverse contains a factor of two in the denominator. That arithmetic fact is visible as a loss of one binary digit in trace reconstruction.\n\\vfig{17_precision}\n\\head{A concrete ambiguity modulo eight}\nThe multiplicity difference $\\delta\\bm m=(4,4,0)$ is nonzero modulo eight, but\n\\[\nC\\,\\delta\\bm m=(8,0,8)\\equiv0\\pmod8.\n\\]\nThus traces known modulo eight do not distinguish those two residue vectors. The same trace difference is nonzero modulo sixteen, so one more trace digit resolves this ambiguity.","CISION LOSS}{Main text: The finite trace-recovery discussion in §9.}\nThe character matrix has determinant six, so its inverse contains a factor of two in the denominator. That arithmetic fact is visible as a loss of one binary digit in trace reconstruction.\n\\vfig{17_precision}\n\\head{A concrete ambiguity modulo eight}\nThe multiplicity difference $\\delta\\bm m=(4,4,0)$ is nonzero modulo eight, but\n\\[\nC\\,\\delta\\bm m=(8,0,8)\\equiv0\\pmod8.\n\\]\nThus traces known modulo eight do not distinguish those two residue vectors. The same trace difference is nonzero modulo sixteen, so one more trace digit resolves this ambiguity.","More generally, the kernel of $C$ modulo $2^n$ has exactly two elements:\n\\[\n(0,0,0),\\qquad(2^{n-1},2^{n-1},0).\n\\]\nTo see this, the second equation gives $m_1=m_2$; the third gives $m_3=2m_1$; the first then gives $6m_1=0$. Three is a unit modulo $2^n$, leaving precisely the displayed possibilities.","Conversely, $C\\delta\\bm m\\equiv0\\pmod{2^{n+1}}$ implies $6\\delta\\bm m\\equiv0\\pmod{2^{n+1}}$ by the adjugate identity, hence $\\delta\\bm m\\equiv0\\pmod{2^n}$. Capacity bounds can remove ambiguity earlier in a particular carrier, but do not change this universal linear precision calculation.\n\\takeaway{Full rank over characteristic zero is not the same as invertibility over every residue ring. Denominator valuations determine the needed extra precision.}","\\vpage{18}{Pairwise occupancy can still fail jointly}{EXACT JOINT PROJECTORS}{Main text: Theorem 10.1 and the finite occupancy models.}\nLet a six-dimensional rational carrier have the joint eigenvalue labels shown below for three commuting diagonal operators $T_1,T_2,T_3$. Each coordinate is zero or one. Selecting a zero uses $I-T_i$; selecting a one uses $T_i$.\n\\vfig{18_occupancy}\n\\head{The full conjunction is a different test}\nFor target $(0,0,0)$, successive conditions reduce the ranks from $6$ to $3$, then $2$, then $1$. The joint projector is nonzero, so that label is occupied in this carrier.","jointly}{EXACT JOINT PROJECTORS}{Main text: Theorem 10.1 and the finite occupancy models.}\nLet a six-dimensional rational carrier have the joint eigenvalue labels shown below for three commuting diagonal operators $T_1,T_2,T_3$. Each coordinate is zero or one. Selecting a zero uses $I-T_i$; selecting a one uses $T_i$.\n\\vfig{18_occupancy}\n\\head{The full conjunction is a different test}\nFor target $(0,0,0)$, successive conditions reduce the ranks from $6$ to $3$, then $2$, then $1$. The joint projector is nonzero, so that label is occupied in this carrier.","1 and the finite occupancy models.}\nLet a six-dimensional rational carrier have the joint eigenvalue labels shown below for three commuting diagonal operators $T_1,T_2,T_3$. Each coordinate is zero or one. Selecting a zero uses $I-T_i$; selecting a one uses $T_i$.\n\\vfig{18_occupancy}\n\\head{The full conjunction is a different test}\nFor target $(0,0,0)$, successive conditions reduce the ranks from $6$ to $3$, then $2$, then $1$. The joint projector is nonzero, so that label is occupied in this carrier.","he finite occupancy models.}\nLet a six-dimensional rational carrier have the joint eigenvalue labels shown below for three commuting diagonal operators $T_1,T_2,T_3$. Each coordinate is zero or one. Selecting a zero uses $I-T_i$; selecting a one uses $T_i$.\n\\vfig{18_occupancy}\n\\head{The full conjunction is a different test}\nFor target $(0,0,0)$, successive conditions reduce the ranks from $6$ to $3$, then $2$, then $1$. The joint projector is nonzero, so that label is occupied in this carrier.","occupancy models.}\nLet a six-dimensional rational carrier have the joint eigenvalue labels shown below for three commuting diagonal operators $T_1,T_2,T_3$. Each coordinate is zero or one. Selecting a zero uses $I-T_i$; selecting a one uses $T_i$.\n\\vfig{18_occupancy}\n\\head{The full conjunction is a different test}\nFor target $(0,0,0)$, successive conditions reduce the ranks from $6$ to $3$, then $2$, then $1$. The joint projector is nonzero, so that label is occupied in this carrier.","models.}\nLet a six-dimensional rational carrier have the joint eigenvalue labels shown below for three commuting diagonal operators $T_1,T_2,T_3$. Each coordinate is zero or one. Selecting a zero uses $I-T_i$; selecting a one uses $T_i$.\n\\vfig{18_occupancy}\n\\head{The full conjunction is a different test}\nFor target $(0,0,0)$, successive conditions reduce the ranks from $6$ to $3$, then $2$, then $1$. The joint projector is nonzero, so that label is occupied in this carrier.","For target $(0,1,1)$, every pair of conditions is occupied: $(0,1,0)$ witnesses the first two, $(0,0,1)$ the first and third, and $(1,1,1)$ the second and third. But the triple itself is absent. Multiplying all three selectors gives the zero matrix.\n\\[\nE_{\\bm a}=\\prod_{i=1}^{3}\n\\begin{cases}I-T_i,&a_i=0,","For target $(0,1,1)$, every pair of conditions is occupied: $(0,1,0)$ witnesses the first two, $(0,0,1)$ the first and third, and $(1,1,1)$ the second and third. But the triple itself is absent. Multiplying all three selectors gives the zero matrix.\n\\[\nE_{\\bm a}=\\prod_{i=1}^{3}\n\\begin{cases}I-T_i,&a_i=0,\\\\T_i,&a_i=1.\\end{cases}\n\\qquad \\Tr(E_{\\bm a})=\\rank(E_{\\bm a}).\n\\]\nThe checks compute all eight joint projectors, their pairwise orthogonality, and their sum. The sum is the identity even though two labels are absent: those two projectors are simply zero.\n\\takeaway{A formula for a projector does not force its image to be nonzero. Occupancy must be tested on the complete set of required conditions.}","theses do real work}{TRACE AND SEMISIMPLICITY}{Main text: Theorem 10.1 and the checked false controls following it.}\nThe finite occupancy theorem uses a split semisimple commuting family in characteristic zero. Two tiny matrices show why neither the coefficient condition nor semisimplicity should disappear from the statement.\n\\vfig{19_hypotheses}\n\\head{Trace zero need not mean an empty image}\nOn $\\FF_5^5$, the identity is an idempotent of rank five, but its field-valued trace is $5=0$ in $\\FF_5$. Rank and localization still detect its nonzero image. A characteristic-zero integer trace cannot be silently transferred to this setting.\n\\head{An eigenvalue label need not select ordinary eigenvectors}\nFor the nilpotent matrix $N=\\left(\\begin{smallmatrix}0&1","d false controls following it.}\nThe finite occupancy theorem uses a split semisimple commuting family in characteristic zero. Two tiny matrices show why neither the coefficient condition nor semisimplicity should disappear from the statement.\n\\vfig{19_hypotheses}\n\\head{Trace zero need not mean an empty image}\nOn $\\FF_5^5$, the identity is an idempotent of rank five, but its field-valued trace is $5=0$ in $\\FF_5$. Rank and localization still detect its nonzero image. A characteristic-zero integer trace cannot be silently transferred to this setting.\n\\head{An eigenvalue label need not select ordinary eigenvectors}\nFor the nilpotent matrix $N=\\left(\\begin{smallmatrix}0&1","ontrols following it.}\nThe finite occupancy theorem uses a split semisimple commuting family in characteristic zero. Two tiny matrices show why neither the coefficient condition nor semisimplicity should disappear from the statement.\n\\vfig{19_hypotheses}\n\\head{Trace zero need not mean an empty image}\nOn $\\FF_5^5$, the identity is an idempotent of rank five, but its field-valued trace is $5=0$ in $\\FF_5$. Rank and localization still detect its nonzero image. A characteristic-zero integer trace cannot be silently transferred to this setting.\n\\head{An eigenvalue label need not select ordinary eigenvectors}\nFor the nilpotent matrix $N=\\left(\\begin{smallmatrix}0&1","ig{19_hypotheses}\n\\head{Trace zero need not mean an empty image}\nOn $\\FF_5^5$, the identity is an idempotent of rank five, but its field-valued trace is $5=0$ in $\\FF_5$. Rank and localization still detect its nonzero image. A characteristic-zero integer trace cannot be silently transferred to this setting.\n\\head{An eigenvalue label need not select ordinary eigenvectors}\nFor the nilpotent matrix $N=\\left(\\begin{smallmatrix}0&1\\\\0&0\\end{smallmatrix}\\right)$, the only eigenvalue is zero. A Lagrange selector that multiplies over all other eigenvalues is an empty product, hence $I$. Its image has dimension two, while $\\ker N$ has dimension one. The selector does not project onto the ordinary eigenspace.","0&0\\end{smallmatrix}\\right)$, the only eigenvalue is zero. A Lagrange selector that multiplies over all other eigenvalues is an empty product, hence $I$. Its image has dimension two, while $\\ker N$ has dimension one. The selector does not project onto the ordinary eigenspace.","\\vpage{20}{A condition must survive the limit}{CLOSEDNESS IS NOT DECORATION}{Main text: Theorems 8.2 and 11.1; the inverse-limit closure hypotheses.}\nIn $\\ZZ_3$, every term of the sequence $x_k=3^k$ is nonzero. Yet $x_k\\to0$ in the three-adic topology. The property ``nonzero'' is not closed, so a limit of objects satisfying it can lose that property.\n\\vfig{20_closedness}\n\\head{Polynomial conditions behave differently}\nFor a continuous polynomial $f$, if a convergent sequence satisfies $f(x_k)=0$ for every $k$, then continuity gives $f(x)=0$ at its limit. More generally, a compatible residue branch satisfying $f(x_m)\\equiv0\\pmod{3^m}$ reconstructs an exact root once the complete ring and its reduction maps have been identified.","EDNESS IS NOT DECORATION}{Main text: Theorems 8.2 and 11.1; the inverse-limit closure hypotheses.}\nIn $\\ZZ_3$, every term of the sequence $x_k=3^k$ is nonzero. Yet $x_k\\to0$ in the three-adic topology. The property ``nonzero'' is not closed, so a limit of objects satisfying it can lose that property.\n\\vfig{20_closedness}\n\\head{Polynomial conditions behave differently}\nFor a continuous polynomial $f$, if a convergent sequence satisfies $f(x_k)=0$ for every $k$, then continuity gives $f(x)=0$ at its limit. More generally, a compatible residue branch satisfying $f(x_m)\\equiv0\\pmod{3^m}$ reconstructs an exact root once the complete ring and its reduction maps have been identified.","orems 8.2 and 11.1; the inverse-limit closure hypotheses.}\nIn $\\ZZ_3$, every term of the sequence $x_k=3^k$ is nonzero. Yet $x_k\\to0$ in the three-adic topology. The property ``nonzero'' is not closed, so a limit of objects satisfying it can lose that property.\n\\vfig{20_closedness}\n\\head{Polynomial conditions behave differently}\nFor a continuous polynomial $f$, if a convergent sequence satisfies $f(x_k)=0$ for every $k$, then continuity gives $f(x)=0$ at its limit. More generally, a compatible residue branch satisfying $f(x_m)\\equiv0\\pmod{3^m}$ reconstructs an exact root once the complete ring and its reduction maps have been identified.","he inverse-limit closure hypotheses.}\nIn $\\ZZ_3$, every term of the sequence $x_k=3^k$ is nonzero. Yet $x_k\\to0$ in the three-adic topology. The property ``nonzero'' is not closed, so a limit of objects satisfying it can lose that property.\n\\vfig{20_closedness}\n\\head{Polynomial conditions behave differently}\nFor a continuous polynomial $f$, if a convergent sequence satisfies $f(x_k)=0$ for every $k$, then continuity gives $f(x)=0$ at its limit. More generally, a compatible residue branch satisfying $f(x_m)\\equiv0\\pmod{3^m}$ reconstructs an exact root once the complete ring and its reduction maps have been identified.","potheses.}\nIn $\\ZZ_3$, every term of the sequence $x_k=3^k$ is nonzero. Yet $x_k\\to0$ in the three-adic topology. The property ``nonzero'' is not closed, so a limit of objects satisfying it can lose that property.\n\\vfig{20_closedness}\n\\head{Polynomial conditions behave differently}\nFor a continuous polynomial $f$, if a convergent sequence satisfies $f(x_k)=0$ for every $k$, then continuity gives $f(x)=0$ at its limit. More generally, a compatible residue branch satisfying $f(x_m)\\equiv0\\pmod{3^m}$ reconstructs an exact root once the complete ring and its reduction maps have been identified.","_k\\to0$ in the three-adic topology. The property ``nonzero'' is not closed, so a limit of objects satisfying it can lose that property.\n\\vfig{20_closedness}\n\\head{Polynomial conditions behave differently}\nFor a continuous polynomial $f$, if a convergent sequence satisfies $f(x_k)=0$ for every $k$, then continuity gives $f(x)=0$ at its limit. More generally, a compatible residue branch satisfying $f(x_m)\\equiv0\\pmod{3^m}$ reconstructs an exact root once the complete ring and its reduction maps have been identified.","The sequence $3^k$ here is \\emph{not} presented as a compatible branch of nonzero residues modulo $3^k$: those particular residues are zero. It is a convergent sequence of nonzero elements of the complete ring. The distinction prevents an incorrect finite-level interpretation of the picture.","requires a proof that the chosen condition is preserved by the limit under construction. The word ``admissible'' does not itself provide that proof.\n\\takeaway{Compatibility reconstructs an object. Closedness is what carries the declared equations or conditions to that object.}\n\\captiontext{The finite diagram displays the retreat of the first nonzero digit. The convergence assertion follows from $|3^k|_3=3^{-k}$, not from extrapolating a finite numerical plot.}","\\vpage{21}{The sign of a norm is arithmetic data}{ELEMENTS VERSUS IDEALS}{Main text: §9, arithmetic residue-counting continuation; signed-norm comparison.}\nEven in the number field $\\QQ$, congruent generators need not have congruent absolute norms. The integers $2$ and $-3$ are congruent modulo five. Their signed field norms are congruent, but their absolute values are not.\n\\vfig{21_signed_norm}\n\\head{The smallest counterexample is already decisive}\nFor $K=\\QQ$,\n\\[\nN_{K/\\QQ}(2)=2,\\qquad N_{K/\\QQ}(-3)=-3\\equiv2\\pmod5,\n\\]\nwhereas\n\\[\nN((2))=2,\\qquad N((-3))=3\\not\\equiv2\\pmod5.\n\\]\nHere $N((a))$ is the positive index of the ideal $(a)$, equal to $|a|$. The determinant norm of an element is signed. Taking absolute values is not a ring operation and does not commute with reduction in this example.","VERSUS IDEALS}{Main text: §9, arithmetic residue-counting continuation; signed-norm comparison.}\nEven in the number field $\\QQ$, congruent generators need not have congruent absolute norms. The integers $2$ and $-3$ are congruent modulo five. Their signed field norms are congruent, but their absolute values are not.\n\\vfig{21_signed_norm}\n\\head{The smallest counterexample is already decisive}\nFor $K=\\QQ$,\n\\[\nN_{K/\\QQ}(2)=2,\\qquad N_{K/\\QQ}(-3)=-3\\equiv2\\pmod5,\n\\]\nwhereas\n\\[\nN((2))=2,\\qquad N((-3))=3\\not\\equiv2\\pmod5.\n\\]\nHere $N((a))$ is the positive index of the ideal $(a)$, equal to $|a|$. The determinant norm of an element is signed. Taking absolute values is not a ring operation and does not commute with reduction in this example.","ALS}{Main text: §9, arithmetic residue-counting continuation; signed-norm comparison.}\nEven in the number field $\\QQ$, congruent generators need not have congruent absolute norms. The integers $2$ and $-3$ are congruent modulo five. Their signed field norms are congruent, but their absolute values are not.\n\\vfig{21_signed_norm}\n\\head{The smallest counterexample is already decisive}\nFor $K=\\QQ$,\n\\[\nN_{K/\\QQ}(2)=2,\\qquad N_{K/\\QQ}(-3)=-3\\equiv2\\pmod5,\n\\]\nwhereas\n\\[\nN((2))=2,\\qquad N((-3))=3\\not\\equiv2\\pmod5.\n\\]\nHere $N((a))$ is the positive index of the ideal $(a)$, equal to $|a|$. The determinant norm of an element is signed. Taking absolute values is not a ring operation and does not commute with reduction in this example.","d-norm comparison.}\nEven in the number field $\\QQ$, congruent generators need not have congruent absolute norms. The integers $2$ and $-3$ are congruent modulo five. Their signed field norms are congruent, but their absolute values are not.\n\\vfig{21_signed_norm}\n\\head{The smallest counterexample is already decisive}\nFor $K=\\QQ$,\n\\[\nN_{K/\\QQ}(2)=2,\\qquad N_{K/\\QQ}(-3)=-3\\equiv2\\pmod5,\n\\]\nwhereas\n\\[\nN((2))=2,\\qquad N((-3))=3\\not\\equiv2\\pmod5.\n\\]\nHere $N((a))$ is the positive index of the ideal $(a)$, equal to $|a|$. The determinant norm of an element is signed. Taking absolute values is not a ring operation and does not commute with reduction in this example.","ngruent generators need not have congruent absolute norms. The integers $2$ and $-3$ are congruent modulo five. Their signed field norms are congruent, but their absolute values are not.\n\\vfig{21_signed_norm}\n\\head{The smallest counterexample is already decisive}\nFor $K=\\QQ$,\n\\[\nN_{K/\\QQ}(2)=2,\\qquad N_{K/\\QQ}(-3)=-3\\equiv2\\pmod5,\n\\]\nwhereas\n\\[\nN((2))=2,\\qquad N((-3))=3\\not\\equiv2\\pmod5.\n\\]\nHere $N((a))$ is the positive index of the ideal $(a)$, equal to $|a|$. The determinant norm of an element is signed. Taking absolute values is not a ring operation and does not commute with reduction in this example.","bsolute norms. The integers $2$ and $-3$ are congruent modulo five. Their signed field norms are congruent, but their absolute values are not.\n\\vfig{21_signed_norm}\n\\head{The smallest counterexample is already decisive}\nFor $K=\\QQ$,\n\\[\nN_{K/\\QQ}(2)=2,\\qquad N_{K/\\QQ}(-3)=-3\\equiv2\\pmod5,\n\\]\nwhereas\n\\[\nN((2))=2,\\qquad N((-3))=3\\not\\equiv2\\pmod5.\n\\]\nHere $N((a))$ is the positive index of the ideal $(a)$, equal to $|a|$. The determinant norm of an element is signed. Taking absolute values is not a ring operation and does not commute with reduction in this example.","he integers $2$ and $-3$ are congruent modulo five. Their signed field norms are congruent, but their absolute values are not.\n\\vfig{21_signed_norm}\n\\head{The smallest counterexample is already decisive}\nFor $K=\\QQ$,\n\\[\nN_{K/\\QQ}(2)=2,\\qquad N_{K/\\QQ}(-3)=-3\\equiv2\\pmod5,\n\\]\nwhereas\n\\[\nN((2))=2,\\qquad N((-3))=3\\not\\equiv2\\pmod5.\n\\]\nHere $N((a))$ is the positive index of the ideal $(a)$, equal to $|a|$. The determinant norm of an element is signed. Taking absolute values is not a ring operation and does not commute with reduction in this example.","\\vpage{22}{A norm is a determinant you can compute}{THE QUOTIENT ALGEBRA $\\FF_{25}$}{Main text: §9; retained scalar/Mellin continuation: actual quotient determinant norms.}\nUse $A=\\FF_5[u]/(u^2-2)$. Since two is not a square modulo five, this quotient is a field with 25 elements. Every element is uniquely $a+bu$, with $a,b\\in\\FF_5$.\n\\vfig{22_norm_table}\n\\head{Coordinates, multiplication, and norm agree}\nIn the basis $(1,u)$, multiplication by $a+bu$ has matrix\n\\[\nM_{a,b}=\\mat{a&2b","a determinant you can compute}{THE QUOTIENT ALGEBRA $\\FF_{25}$}{Main text: §9; retained scalar/Mellin continuation: actual quotient determinant norms.}\nUse $A=\\FF_5[u]/(u^2-2)$. Since two is not a square modulo five, this quotient is a field with 25 elements. Every element is uniquely $a+bu$, with $a,b\\in\\FF_5$.\n\\vfig{22_norm_table}\n\\head{Coordinates, multiplication, and norm agree}\nIn the basis $(1,u)$, multiplication by $a+bu$ has matrix\n\\[\nM_{a,b}=\\mat{a&2b","te}{THE QUOTIENT ALGEBRA $\\FF_{25}$}{Main text: §9; retained scalar/Mellin continuation: actual quotient determinant norms.}\nUse $A=\\FF_5[u]/(u^2-2)$. Since two is not a square modulo five, this quotient is a field with 25 elements. Every element is uniquely $a+bu$, with $a,b\\in\\FF_5$.\n\\vfig{22_norm_table}\n\\head{Coordinates, multiplication, and norm agree}\nIn the basis $(1,u)$, multiplication by $a+bu$ has matrix\n\\[\nM_{a,b}=\\mat{a&2b","GEBRA $\\FF_{25}$}{Main text: §9; retained scalar/Mellin continuation: actual quotient determinant norms.}\nUse $A=\\FF_5[u]/(u^2-2)$. Since two is not a square modulo five, this quotient is a field with 25 elements. Every element is uniquely $a+bu$, with $a,b\\in\\FF_5$.\n\\vfig{22_norm_table}\n\\head{Coordinates, multiplication, and norm agree}\nIn the basis $(1,u)$, multiplication by $a+bu$ has matrix\n\\[\nM_{a,b}=\\mat{a&2b\\\\b&a},\n\\qquad N(a+bu)=\\det M_{a,b}=a^2-2b^2\\pmod5.\n\\]\nThe table shows every norm value. Only the zero element has norm zero. Each nonzero norm value has exactly six preimages. This makes the later character cancellation visible without estimating any sums.","This model is the reduction of the quadratic integer algebra $\\ZZ[\\sqrt2]$ modulo five. Its finite arithmetic illustrates the determinant construction used in the continuation; it is not a computation of all ideal classes or a general ideal-counting theorem.\n\\takeaway{A quotient norm is built from the actual multiplication operator. Its basis independence is a determinant identity, not a choice of convenient coordinates.}","\\vpage{23}{Permutation forces a zero mean}{A NONSCALAR UNIT WITNESS}{Main text: §9; retained scalar/Mellin continuation: norm-residue lattice mean cancellation.}\nLet $\\chi$ be the quadratic character modulo five, extended by $\\chi(0)=0$. In the field on the preceding page, choose $v=2u$. Its norm is $N(v)=2$, so $\\chi(N(v))=-1$.\n\\vfig{23_mean_cancellation}\n\\head{The whole sum is paired with its negative}\nMultiplication by $v$ permutes the 25 elements. Norm multiplicativity gives\n\\[\n\\sum_{z\\in A}\\chi(N(z))\n=\\sum_{z\\in A}\\chi(N(vz))\n=-\\sum_{z\\in A}\\chi(N(z)).\n\\]\nThe sum is an integer, so it is zero. The origin contributes zero. The other 24 elements fall into three cycles of length eight under multiplication by $v$; the character signs alternate around each cycle.","es a zero mean}{A NONSCALAR UNIT WITNESS}{Main text: §9; retained scalar/Mellin continuation: norm-residue lattice mean cancellation.}\nLet $\\chi$ be the quadratic character modulo five, extended by $\\chi(0)=0$. In the field on the preceding page, choose $v=2u$. Its norm is $N(v)=2$, so $\\chi(N(v))=-1$.\n\\vfig{23_mean_cancellation}\n\\head{The whole sum is paired with its negative}\nMultiplication by $v$ permutes the 25 elements. Norm multiplicativity gives\n\\[\n\\sum_{z\\in A}\\chi(N(z))\n=\\sum_{z\\in A}\\chi(N(vz))\n=-\\sum_{z\\in A}\\chi(N(z)).\n\\]\nThe sum is an integer, so it is zero. The origin contributes zero. The other 24 elements fall into three cycles of length eight under multiplication by $v$; the character signs alternate around each cycle.","tained scalar/Mellin continuation: norm-residue lattice mean cancellation.}\nLet $\\chi$ be the quadratic character modulo five, extended by $\\chi(0)=0$. In the field on the preceding page, choose $v=2u$. Its norm is $N(v)=2$, so $\\chi(N(v))=-1$.\n\\vfig{23_mean_cancellation}\n\\head{The whole sum is paired with its negative}\nMultiplication by $v$ permutes the 25 elements. Norm multiplicativity gives\n\\[\n\\sum_{z\\in A}\\chi(N(z))\n=\\sum_{z\\in A}\\chi(N(vz))\n=-\\sum_{z\\in A}\\chi(N(z)).\n\\]\nThe sum is an integer, so it is zero. The origin contributes zero. The other 24 elements fall into three cycles of length eight under multiplication by $v$; the character signs alternate around each cycle.","N(v))=-1$.\n\\vfig{23_mean_cancellation}\n\\head{The whole sum is paired with its negative}\nMultiplication by $v$ permutes the 25 elements. Norm multiplicativity gives\n\\[\n\\sum_{z\\in A}\\chi(N(z))\n=\\sum_{z\\in A}\\chi(N(vz))\n=-\\sum_{z\\in A}\\chi(N(z)).\n\\]\nThe sum is an integer, so it is zero. The origin contributes zero. The other 24 elements fall into three cycles of length eight under multiplication by $v$; the character signs alternate around each cycle.","The same reasoning works with a complex root of unity $\\chi(N(v))\\ne1$: subtracting the two equal sums gives $(1-\\chi(N(v)))S=0$. The essential inputs are an actual permutation, a multiplicative norm, and a nontrivial character value. They are not consequences of drawing an alternating cycle.","In the arithmetic continuation the quotient is the actual additive module $J/mJ$ associated with an ideal representative. This finite-field example isolates the cancellation mechanism while keeping that broader module and lattice comparison separate.\n\\takeaway{The witness is one unit that changes the character value. Once it exists, the permutation cancels the entire finite mean.}","\\vpage{24}{A nontrivial character may still be exceptional}{THE EXACT SCALAR TEST}{Main text: §9; scalar/Mellin checkpoint: scalar-power and nonscalar-witness boundaries.}\nA scalar $a\\in\\FF_5^\\times$ acts on a degree-$d$ algebra with determinant $a^d$. Therefore the scalar cancellation route tests $\\chi(a^d)$, not merely $\\chi(a)$.\n\\vfig{24_scalar_exception}\n\\head{A failed sufficient test is not an impossibility theorem}\nLet $\\chi_4(2)=i$ generate the character group of $\\FF_5^\\times$, and write $\\chi_k=\\chi_4^k$. Since $2$ generates the unit group,\n\\[\n\\exists a:\\chi_k(a^d)\\ne1\n\\quad\\Longleftrightarrow\\quad kd\\not\\equiv0\\pmod4.\n\\]\nFor the quadratic character $k=2$ and degree $d=2$, every scalar test is trivial. Yet page \\pagerefV{23} supplies the nonscalar witness $2u$, proving cancellation in that specific algebra. Scalar exceptional does not mean exceptional for every possible quotie","\\vpage{24}{A nontrivial character may still be exceptional}{THE EXACT SCALAR TEST}{Main text: §9; scalar/Mellin checkpoint: scalar-power and nonscalar-witness boundaries.}\nA scalar $a\\in\\FF_5^\\times$ acts on a degree-$d$ algebra with determinant $a^d$. Therefore the scalar cancellation route tests $\\chi(a^d)$, not merely $\\chi(a)$.\n\\vfig{24_scalar_exception}\n\\head{A failed sufficient test is not an impossibility theorem}\nLet $\\chi_4(2)=i$ generate the character group of $\\FF_5^\\times$, and write $\\chi_k=\\chi_4^k$. Since $2$ generates the unit group,\n\\[\n\\exists a:\\chi_k(a^d)\\ne1\n\\quad\\Longleftrightarrow\\quad kd\\not\\equiv0\\pmod4.\n\\]\nFor the quadratic character $k=2$ and degree $d=2$, every scalar test is trivial. Yet page \\pagerefV{23} supplies the nonscalar witness $2u$, proving cancellation in that specific algebra. Scalar exceptional does not mean exceptional for every possible quotient unit.","ter may still be exceptional}{THE EXACT SCALAR TEST}{Main text: §9; scalar/Mellin checkpoint: scalar-power and nonscalar-witness boundaries.}\nA scalar $a\\in\\FF_5^\\times$ acts on a degree-$d$ algebra with determinant $a^d$. Therefore the scalar cancellation route tests $\\chi(a^d)$, not merely $\\chi(a)$.\n\\vfig{24_scalar_exception}\n\\head{A failed sufficient test is not an impossibility theorem}\nLet $\\chi_4(2)=i$ generate the character group of $\\FF_5^\\times$, and write $\\chi_k=\\chi_4^k$. Since $2$ generates the unit group,\n\\[\n\\exists a:\\chi_k(a^d)\\ne1\n\\quad\\Longleftrightarrow\\quad kd\\not\\equiv0\\pmod4.\n\\]\nFor the quadratic character $k=2$ and degree $d=2$, every scalar test is trivial. Yet page \\pagerefV{23} supplies the nonscalar witness $2u$, proving cancellation in that specific algebra. Scalar exceptional does not mean exceptional for every possible quotient unit.","TEST}{Main text: §9; scalar/Mellin checkpoint: scalar-power and nonscalar-witness boundaries.}\nA scalar $a\\in\\FF_5^\\times$ acts on a degree-$d$ algebra with determinant $a^d$. Therefore the scalar cancellation route tests $\\chi(a^d)$, not merely $\\chi(a)$.\n\\vfig{24_scalar_exception}\n\\head{A failed sufficient test is not an impossibility theorem}\nLet $\\chi_4(2)=i$ generate the character group of $\\FF_5^\\times$, and write $\\chi_k=\\chi_4^k$. Since $2$ generates the unit group,\n\\[\n\\exists a:\\chi_k(a^d)\\ne1\n\\quad\\Longleftrightarrow\\quad kd\\not\\equiv0\\pmod4.\n\\]\nFor the quadratic character $k=2$ and degree $d=2$, every scalar test is trivial. Yet page \\pagerefV{23} supplies the nonscalar witness $2u$, proving cancellation in that specific algebra. Scalar exceptional does not mean exceptional for every possible quotient unit.","ar/Mellin checkpoint: scalar-power and nonscalar-witness boundaries.}\nA scalar $a\\in\\FF_5^\\times$ acts on a degree-$d$ algebra with determinant $a^d$. Therefore the scalar cancellation route tests $\\chi(a^d)$, not merely $\\chi(a)$.\n\\vfig{24_scalar_exception}\n\\head{A failed sufficient test is not an impossibility theorem}\nLet $\\chi_4(2)=i$ generate the character group of $\\FF_5^\\times$, and write $\\chi_k=\\chi_4^k$. Since $2$ generates the unit group,\n\\[\n\\exists a:\\chi_k(a^d)\\ne1\n\\quad\\Longleftrightarrow\\quad kd\\not\\equiv0\\pmod4.\n\\]\nFor the quadratic character $k=2$ and degree $d=2$, every scalar test is trivial. Yet page \\pagerefV{23} supplies the nonscalar witness $2u$, proving cancellation in that specific algebra. Scalar exceptional does not mean exceptional for every possible quotient unit.","lar-power and nonscalar-witness boundaries.}\nA scalar $a\\in\\FF_5^\\times$ acts on a degree-$d$ algebra with determinant $a^d$. Therefore the scalar cancellation route tests $\\chi(a^d)$, not merely $\\chi(a)$.\n\\vfig{24_scalar_exception}\n\\head{A failed sufficient test is not an impossibility theorem}\nLet $\\chi_4(2)=i$ generate the character group of $\\FF_5^\\times$, and write $\\chi_k=\\chi_4^k$. Since $2$ generates the unit group,\n\\[\n\\exists a:\\chi_k(a^d)\\ne1\n\\quad\\Longleftrightarrow\\quad kd\\not\\equiv0\\pmod4.\n\\]\nFor the quadratic character $k=2$ and degree $d=2$, every scalar test is trivial. Yet page \\pagerefV{23} supplies the nonscalar witness $2u$, proving cancellation in that specific algebra. Scalar exceptional does not mean exceptional for every possible quotient unit.","boundaries.}\nA scalar $a\\in\\FF_5^\\times$ acts on a degree-$d$ algebra with determinant $a^d$. Therefore the scalar cancellation route tests $\\chi(a^d)$, not merely $\\chi(a)$.\n\\vfig{24_scalar_exception}\n\\head{A failed sufficient test is not an impossibility theorem}\nLet $\\chi_4(2)=i$ generate the character group of $\\FF_5^\\times$, and write $\\chi_k=\\chi_4^k$. Since $2$ generates the unit group,\n\\[\n\\exists a:\\chi_k(a^d)\\ne1\n\\quad\\Longleftrightarrow\\quad kd\\not\\equiv0\\pmod4.\n\\]\nFor the quadratic character $k=2$ and degree $d=2$, every scalar test is trivial. Yet page \\pagerefV{23} supplies the nonscalar witness $2u$, proving cancellation in that specific algebra. Scalar exceptional does not mean exceptional for every possible quotient unit.","calar cancellation route tests $\\chi(a^d)$, not merely $\\chi(a)$.\n\\vfig{24_scalar_exception}\n\\head{A failed sufficient test is not an impossibility theorem}\nLet $\\chi_4(2)=i$ generate the character group of $\\FF_5^\\times$, and write $\\chi_k=\\chi_4^k$. Since $2$ generates the unit group,\n\\[\n\\exists a:\\chi_k(a^d)\\ne1\n\\quad\\Longleftrightarrow\\quad kd\\not\\equiv0\\pmod4.\n\\]\nFor the quadratic character $k=2$ and degree $d=2$, every scalar test is trivial. Yet page \\pagerefV{23} supplies the nonscalar witness $2u$, proving cancellation in that specific algebra. Scalar exceptional does not mean exceptional for every possible quotient unit.","n route tests $\\chi(a^d)$, not merely $\\chi(a)$.\n\\vfig{24_scalar_exception}\n\\head{A failed sufficient test is not an impossibility theorem}\nLet $\\chi_4(2)=i$ generate the character group of $\\FF_5^\\times$, and write $\\chi_k=\\chi_4^k$. Since $2$ generates the unit group,\n\\[\n\\exists a:\\chi_k(a^d)\\ne1\n\\quad\\Longleftrightarrow\\quad kd\\not\\equiv0\\pmod4.\n\\]\nFor the quadratic character $k=2$ and degree $d=2$, every scalar test is trivial. Yet page \\pagerefV{23} supplies the nonscalar witness $2u$, proving cancellation in that specific algebra. Scalar exceptional does not mean exceptional for every possible quotient unit.","sufficient test is not an impossibility theorem}\nLet $\\chi_4(2)=i$ generate the character group of $\\FF_5^\\times$, and write $\\chi_k=\\chi_4^k$. Since $2$ generates the unit group,\n\\[\n\\exists a:\\chi_k(a^d)\\ne1\n\\quad\\Longleftrightarrow\\quad kd\\not\\equiv0\\pmod4.\n\\]\nFor the quadratic character $k=2$ and degree $d=2$, every scalar test is trivial. Yet page \\pagerefV{23} supplies the nonscalar witness $2u$, proving cancellation in that specific algebra. Scalar exceptional does not mean exceptional for every possible quotient unit.","Conversely, the quartic character $k=1$ works in degree two even though $\\gcd(2,4)\\ne1$. Thus the simpler degree--totient coprimality criterion is sufficient, not necessary. With degree three, every nontrivial character passes because exponentiation by three permutes the units.","\\vpage{25}{Avoid finitely many primes constructively}{A CRT MODEL FOR REPRESENTATIVE CHOICE}{Main text: §9; retained scalar/Mellin continuation: finite-prime avoidance and automatic representatives.}\nA representative needed by an arithmetic argument often has to avoid a prescribed finite collection of primes. A small integer version asks for\n\\[\nx\\equiv1\\pmod2,\\qquad x\\equiv2\\pmod3,\\qquad x\\equiv4\\pmod5.\n\\]\nThere is exactly one solution modulo 30: $x=29$.\n\\vfig{25_avoidance}\n\\head{Avoidance and ideal class are different constraints}\nThe selected residues are nonzero modulo each forbidden prime, so $\\gcd(29,30)=1$. The eight units modulo 30 are displayed separately. They all avoid the primes, but only one also satisfies the three specific residue prescriptions.","}{A CRT MODEL FOR REPRESENTATIVE CHOICE}{Main text: §9; retained scalar/Mellin continuation: finite-prime avoidance and automatic representatives.}\nA representative needed by an arithmetic argument often has to avoid a prescribed finite collection of primes. A small integer version asks for\n\\[\nx\\equiv1\\pmod2,\\qquad x\\equiv2\\pmod3,\\qquad x\\equiv4\\pmod5.\n\\]\nThere is exactly one solution modulo 30: $x=29$.\n\\vfig{25_avoidance}\n\\head{Avoidance and ideal class are different constraints}\nThe selected residues are nonzero modulo each forbidden prime, so $\\gcd(29,30)=1$. The eight units modulo 30 are displayed separately. They all avoid the primes, but only one also satisfies the three specific residue prescriptions.","resentative needed by an arithmetic argument often has to avoid a prescribed finite collection of primes. A small integer version asks for\n\\[\nx\\equiv1\\pmod2,\\qquad x\\equiv2\\pmod3,\\qquad x\\equiv4\\pmod5.\n\\]\nThere is exactly one solution modulo 30: $x=29$.\n\\vfig{25_avoidance}\n\\head{Avoidance and ideal class are different constraints}\nThe selected residues are nonzero modulo each forbidden prime, so $\\gcd(29,30)=1$. The eight units modulo 30 are displayed separately. They all avoid the primes, but only one also satisfies the three specific residue prescriptions.","\\vpage{26}{Cancellation leaves a boundary problem}{A FULL LATTICE-BOX CALCULATION}{Main text: §9; retained scalar/Mellin continuation: additive residue counts and norm-character bounds.}\nExtend the function $f(a,b)=\\chi(a^2-2b^2\\bmod5)$ periodically to $\\ZZ^2$. Every complete $5\\times5$ residue tile has sum zero by the finite-field calculation. Only incomplete tiles can contribute to a box sum.\n\\vfig{26_boundary_count}\n\\head{Separate bulk cancellation from boundary size}\nFor $0\\le a,b<12$, four complete tiles fill a $10\\times10$ square and contribute zero. The remaining 44 points lie in the two boundary strips. Their exact sum is one. Since $|f|\\le1$, a bound of 44 follows without needing that final cancellation.","}{Main text: §9; retained scalar/Mellin continuation: additive residue counts and norm-character bounds.}\nExtend the function $f(a,b)=\\chi(a^2-2b^2\\bmod5)$ periodically to $\\ZZ^2$. Every complete $5\\times5$ residue tile has sum zero by the finite-field calculation. Only incomplete tiles can contribute to a box sum.\n\\vfig{26_boundary_count}\n\\head{Separate bulk cancellation from boundary size}\nFor $0\\le a,b<12$, four complete tiles fill a $10\\times10$ square and contribute zero. The remaining 44 points lie in the two boundary strips. Their exact sum is one. Since $|f|\\le1$, a bound of 44 follows without needing that final cancellation.","ntinuation: additive residue counts and norm-character bounds.}\nExtend the function $f(a,b)=\\chi(a^2-2b^2\\bmod5)$ periodically to $\\ZZ^2$. Every complete $5\\times5$ residue tile has sum zero by the finite-field calculation. Only incomplete tiles can contribute to a box sum.\n\\vfig{26_boundary_count}\n\\head{Separate bulk cancellation from boundary size}\nFor $0\\le a,b<12$, four complete tiles fill a $10\\times10$ square and contribute zero. The remaining 44 points lie in the two boundary strips. Their exact sum is one. Since $|f|\\le1$, a bound of 44 follows without needing that final cancellation.","to $\\ZZ^2$. Every complete $5\\times5$ residue tile has sum zero by the finite-field calculation. Only incomplete tiles can contribute to a box sum.\n\\vfig{26_boundary_count}\n\\head{Separate bulk cancellation from boundary size}\nFor $0\\le a,b<12$, four complete tiles fill a $10\\times10$ square and contribute zero. The remaining 44 points lie in the two boundary strips. Their exact sum is one. Since $|f|\\le1$, a bound of 44 follows without needing that final cancellation.","For a general positive integer $L$, write $L=5q+r$, $0\\le r<5$. The complete tiled square has side $L-r$, so\n\\[\n|S_L|\\le L^2-(L-r)^2=2Lr-r^2\\le8L.\n\\]\nThe area is $X=L^2$, making this a square-root-in-area bound for this square-box model. The proof is the exact tile decomposition; checking many $L$ values is only a regression check of the implementation.","Number-field ideal sums involve different regions, lattice indices, signs, and generator multiplicities. Their $X^{1-1/d}$ exponent cannot be justified simply by relabeling this square picture as an ideal-norm ball.\n\\takeaway{A zero finite mean removes the bulk. The remaining estimate depends on the actual geometry of the region and its boundary.}","\\vpage{27}{What a power-saving bound buys analytically}{A MELLIN ENDPOINT MODEL}{Main text: §9; retained scalar/Mellin continuation: summatory measurability and first-order Mellin control.}\nUse the explicit envelope $A(x)=x^{1/2}$ for $x\\ge1$. It is not an ideal-counting function. It is a model showing why a power saving can make a Mellin expression regular at $s=1$.\n\\vfig{27_mellin}\n\\head{Compute the endpoint rather than just naming it}\nFor real $s>1/2$,\n\\[\nI(s)=s\\int_1^\\infty A(x)x^{-s-1}\\,\\dd x\n=\\frac{s}{s-1/2}.\n\\]\nAt $s=1+h$ this gives\n\\[\nI(1+h)-I(1)=-\\frac{h}{h+1/2},\n\\qquad |I(1+h)-I(1)|\\le4|h|\n\\quad (|h|\\le1/4).\n\\]\nThe derivative at one is $-2$. The logarithmic moment is finite as well: $\\int_1^\\infty x^{-3/2}\\log x\\,\\dd x=4$. These are explicit calculations; the drawn curve does not establish differentiability or convergence.","ly}{A MELLIN ENDPOINT MODEL}{Main text: §9; retained scalar/Mellin continuation: summatory measurability and first-order Mellin control.}\nUse the explicit envelope $A(x)=x^{1/2}$ for $x\\ge1$. It is not an ideal-counting function. It is a model showing why a power saving can make a Mellin expression regular at $s=1$.\n\\vfig{27_mellin}\n\\head{Compute the endpoint rather than just naming it}\nFor real $s>1/2$,\n\\[\nI(s)=s\\int_1^\\infty A(x)x^{-s-1}\\,\\dd x\n=\\frac{s}{s-1/2}.\n\\]\nAt $s=1+h$ this gives\n\\[\nI(1+h)-I(1)=-\\frac{h}{h+1/2},\n\\qquad |I(1+h)-I(1)|\\le4|h|\n\\quad (|h|\\le1/4).\n\\]\nThe derivative at one is $-2$. The logarithmic moment is finite as well: $\\int_1^\\infty x^{-3/2}\\log x\\,\\dd x=4$. These are explicit calculations; the drawn curve does not establish differentiability or convergence.","IN ENDPOINT MODEL}{Main text: §9; retained scalar/Mellin continuation: summatory measurability and first-order Mellin control.}\nUse the explicit envelope $A(x)=x^{1/2}$ for $x\\ge1$. It is not an ideal-counting function. It is a model showing why a power saving can make a Mellin expression regular at $s=1$.\n\\vfig{27_mellin}\n\\head{Compute the endpoint rather than just naming it}\nFor real $s>1/2$,\n\\[\nI(s)=s\\int_1^\\infty A(x)x^{-s-1}\\,\\dd x\n=\\frac{s}{s-1/2}.\n\\]\nAt $s=1+h$ this gives\n\\[\nI(1+h)-I(1)=-\\frac{h}{h+1/2},\n\\qquad |I(1+h)-I(1)|\\le4|h|\n\\quad (|h|\\le1/4).\n\\]\nThe derivative at one is $-2$. The logarithmic moment is finite as well: $\\int_1^\\infty x^{-3/2}\\log x\\,\\dd x=4$. These are explicit calculations; the drawn curve does not establish differentiability or convergence.","ion: summatory measurability and first-order Mellin control.}\nUse the explicit envelope $A(x)=x^{1/2}$ for $x\\ge1$. It is not an ideal-counting function. It is a model showing why a power saving can make a Mellin expression regular at $s=1$.\n\\vfig{27_mellin}\n\\head{Compute the endpoint rather than just naming it}\nFor real $s>1/2$,\n\\[\nI(s)=s\\int_1^\\infty A(x)x^{-s-1}\\,\\dd x\n=\\frac{s}{s-1/2}.\n\\]\nAt $s=1+h$ this gives\n\\[\nI(1+h)-I(1)=-\\frac{h}{h+1/2},\n\\qquad |I(1+h)-I(1)|\\le4|h|\n\\quad (|h|\\le1/4).\n\\]\nThe derivative at one is $-2$. The logarithmic moment is finite as well: $\\int_1^\\infty x^{-3/2}\\log x\\,\\dd x=4$. These are explicit calculations; the drawn curve does not establish differentiability or convergence.","rability and first-order Mellin control.}\nUse the explicit envelope $A(x)=x^{1/2}$ for $x\\ge1$. It is not an ideal-counting function. It is a model showing why a power saving can make a Mellin expression regular at $s=1$.\n\\vfig{27_mellin}\n\\head{Compute the endpoint rather than just naming it}\nFor real $s>1/2$,\n\\[\nI(s)=s\\int_1^\\infty A(x)x^{-s-1}\\,\\dd x\n=\\frac{s}{s-1/2}.\n\\]\nAt $s=1+h$ this gives\n\\[\nI(1+h)-I(1)=-\\frac{h}{h+1/2},\n\\qquad |I(1+h)-I(1)|\\le4|h|\n\\quad (|h|\\le1/4).\n\\]\nThe derivative at one is $-2$. The logarithmic moment is finite as well: $\\int_1^\\infty x^{-3/2}\\log x\\,\\dd x=4$. These are explicit calculations; the drawn curve does not establish differentiability or convergence.","n ideal-counting function. It is a model showing why a power saving can make a Mellin expression regular at $s=1$.\n\\vfig{27_mellin}\n\\head{Compute the endpoint rather than just naming it}\nFor real $s>1/2$,\n\\[\nI(s)=s\\int_1^\\infty A(x)x^{-s-1}\\,\\dd x\n=\\frac{s}{s-1/2}.\n\\]\nAt $s=1+h$ this gives\n\\[\nI(1+h)-I(1)=-\\frac{h}{h+1/2},\n\\qquad |I(1+h)-I(1)|\\le4|h|\n\\quad (|h|\\le1/4).\n\\]\nThe derivative at one is $-2$. The logarithmic moment is finite as well: $\\int_1^\\infty x^{-3/2}\\log x\\,\\dd x=4$. These are explicit calculations; the drawn curve does not establish differentiability or convergence.","why a power saving can make a Mellin expression regular at $s=1$.\n\\vfig{27_mellin}\n\\head{Compute the endpoint rather than just naming it}\nFor real $s>1/2$,\n\\[\nI(s)=s\\int_1^\\infty A(x)x^{-s-1}\\,\\dd x\n=\\frac{s}{s-1/2}.\n\\]\nAt $s=1+h$ this gives\n\\[\nI(1+h)-I(1)=-\\frac{h}{h+1/2},\n\\qquad |I(1+h)-I(1)|\\le4|h|\n\\quad (|h|\\le1/4).\n\\]\nThe derivative at one is $-2$. The logarithmic moment is finite as well: $\\int_1^\\infty x^{-3/2}\\log x\\,\\dd x=4$. These are explicit calculations; the drawn curve does not establish differentiability or convergence.","\\vpage{28}{Finite splitting data are not a density theorem}{THE REMAINING FROBENIUS STEP}{Main text: Theorem 9.1. Latest continuation: GeneralArithmeticCoverage and ArithmeticSystemSeparation.}\nFor the quadratic polynomial $X^2+1$, reduction modulo an odd prime either gives two roots or no roots. This finite scan displays both behaviors. It is a useful model for the distinction between splitting and nonsplitting classes.\n\\vfig{28_frobenius_boundary}\n\\head{Existence must survive arbitrary exclusions}\nThe continuation's missing general coverage proposition asks for an actual unramified rational prime in a prescribed conjugacy class, outside any prescribed finite forbidden set, together with the corresponding absolute Frobenius witness. A list of small primes does not establish that quantifier.","The finite character argument on page \\pagerefV{15} also needs samples that jointly separate the constituents. Repeating the identity-class row $(1,1,2)$ of the $S_3$ table gives rank one no matter how often it is repeated. Many samples are not necessarily informative samples.","\\vpage{29}{Changing coefficients changes the problem}{SCALAR ABSENCE VERSUS REPRESENTATION EXISTENCE}{Main text: Example 8.4 and the declared coefficient categories in §§8--9.}\nOver $\\FF_3$, no scalar solves $x^2+1=0$. But the matrix\n\\[\nA=\\mat{0&-1","\\vpage{29}{Changing coefficients changes the problem}{SCALAR ABSENCE VERSUS REPRESENTATION EXISTENCE}{Main text: Example 8.4 and the declared coefficient categories in §§8--9.}\nOver $\\FF_3$, no scalar solves $x^2+1=0$. But the matrix\n\\[\nA=\\mat{0&-1\\\\1&0}\\in M_2(\\FF_3)\n\\quad\\text{satisfies}\\quad A^2+I=0.\n\\]\nAnd over $\\FF_9=\\FF_3[u]/(u^2+1)$, the scalars $u$ and $-u$ solve the same polynomial.\n\\vfig{29_coefficients}\n\\head{Three outputs with different types}\nThe failed scalar problem over $\\FF_3$, the successful matrix problem over $\\FF_3$, and the successful scalar problem over $\\FF_9$ are not contradictory answers. They have different allowed objects. The matrix is multiplication by $u$ in the $\\FF_3$-basis $(1,u)$ of the extension field.","1&0}\\in M_2(\\FF_3)\n\\quad\\text{satisfies}\\quad A^2+I=0.\n\\]\nAnd over $\\FF_9=\\FF_3[u]/(u^2+1)$, the scalars $u$ and $-u$ solve the same polynomial.\n\\vfig{29_coefficients}\n\\head{Three outputs with different types}\nThe failed scalar problem over $\\FF_3$, the successful matrix problem over $\\FF_3$, and the successful scalar problem over $\\FF_9$ are not contradictory answers. They have different allowed objects. The matrix is multiplication by $u$ in the $\\FF_3$-basis $(1,u)$ of the extension field.","uad A^2+I=0.\n\\]\nAnd over $\\FF_9=\\FF_3[u]/(u^2+1)$, the scalars $u$ and $-u$ solve the same polynomial.\n\\vfig{29_coefficients}\n\\head{Three outputs with different types}\nThe failed scalar problem over $\\FF_3$, the successful matrix problem over $\\FF_3$, and the successful scalar problem over $\\FF_9$ are not contradictory answers. They have different allowed objects. The matrix is multiplication by $u$ in the $\\FF_3$-basis $(1,u)$ of the extension field.",".\n\\]\nAnd over $\\FF_9=\\FF_3[u]/(u^2+1)$, the scalars $u$ and $-u$ solve the same polynomial.\n\\vfig{29_coefficients}\n\\head{Three outputs with different types}\nThe failed scalar problem over $\\FF_3$, the successful matrix problem over $\\FF_3$, and the successful scalar problem over $\\FF_9$ are not contradictory answers. They have different allowed objects. The matrix is multiplication by $u$ in the $\\FF_3$-basis $(1,u)$ of the extension field.","$\\FF_9=\\FF_3[u]/(u^2+1)$, the scalars $u$ and $-u$ solve the same polynomial.\n\\vfig{29_coefficients}\n\\head{Three outputs with different types}\nThe failed scalar problem over $\\FF_3$, the successful matrix problem over $\\FF_3$, and the successful scalar problem over $\\FF_9$ are not contradictory answers. They have different allowed objects. The matrix is multiplication by $u$ in the $\\FF_3$-basis $(1,u)$ of the extension field.","This also has a characteristic-zero analogue. The matrix with the same entries gives a rational two-dimensional representation of a cyclic group of order four. Its scalar eigenvalues require a field containing a square root of $-1$. No one-dimensional rational eigencharacter is created by writing the two-dimensional matrix."]}