Strong Artin: source and evidence

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1 \section{Formal mechanisms and their exact hypotheses}
2 \label{sec:formal-mechanisms}
3 
4 The formal work is most useful when its hypotheses remain visible. A theorem
5 that identifies an obstruction is different from a theorem that makes that
6 obstruction vanish. A theorem that reconstructs a matrix from compatible
7 residues is different from a theorem that attaches a Galois representation to
8 an automorphic object. This section develops the algebra behind the checked
9 interfaces, including examples where an attractive stronger statement fails.
10 The purpose is to make the output of each mechanism clear enough to use in a
11 future arithmetic argument.
12 
13 The coefficient rings, groups, and local restrictions below are part of the
14 input. Their connection to a particular number field or automorphic packet
15 requires additional theorems. Nothing here constructs a global automorphic
16 carrier. The matrix results do give actual integral matrices, rather than a
17 formal symbol for a proposed limit. Their scope is therefore substantial, but
18 local and explicit.
19 
20 \subsection{A cyclic cocycle is constrained by the norm}
21 
22 Let $C_e=\langle\gamma\rangle$ be a finite cyclic group of order $e$, acting
23 linearly on a module $M$. Write $T$ for the action of $\gamma$ and set
24 \[
25  N=1+T+\cdots+T^{e-1}.
26 \]
27 The relation $T^e=1$ implies $N(T-1)=(T-1)N=0$. These identities explain why
28 the image of $T-1$ lies in the kernel of $N$, but they do not identify the two
29 spaces.
30 
31 A one-cocycle is a function $a:C_e\to M$ satisfying
32 \[
33  a(gh)=a(g)+g\cdot a(h).
34 \]
35 Putting $g=h=1$ gives $a(1)=0$. If $x=a(\gamma)$, repeated application of the
36 cocycle equation gives
37 \[
38  a(\gamma^r)=\sum_{j=0}^{r-1}T^j x\qquad(1\le r\le e).
39 \]
40 The group relation $\gamma^e=1$ therefore forces $Nx=0$. Conversely, if
41 $Nx=0$, this formula defines a cocycle. Whenever exponents wrap around $e$,
42 the discrepancy is a translate of $Nx$, which is zero. Thus evaluating at
43 the chosen generator identifies cocycles with $\ker N$.
44 
45 For a zero-cochain $y\in M$, its coboundary at $\gamma$ is $(T-1)y$. Taking
46 the quotient by these changes gives the precise presentation
47 \begin{equation}
48  H^1(C_e,M)\simeq \ker N/\operatorname{im}(T-1).
49  \label{eq:appendix-cyclic-H1}
50 \end{equation}
51 If the norm operator is identically zero, this reduces to
52 $M/\operatorname{im}(T-1)$. That reduction has a hypothesis. Dropping the
53 kernel from the general formula introduces classes that cannot satisfy the
54 cyclic relation.
55 
56 Suppose now that $e$ is invertible in the coefficient ring. Define
57 $S=\sum_{j=0}^{e-1}jT^j$. A direct coefficient calculation gives
58 \[
59  (T-1)S=e-N.
60 \]
61 For $x\in\ker N$, take $y=e^{-1}Sx$. Then $(T-1)y=x$, so every cocycle is
62 a coboundary and $H^1(C_e,M)=0$. The averaging argument uses division by
63 $e$ at its decisive step. In residue characteristic dividing $e$, the same
64 display remains an identity but no longer constructs $y$.
65 
66 \begin{example}[The same generator action gives different cyclic cohomology]
67 Over $\mathbb F_3$, let $M=\operatorname{Mat}_2(\mathbb F_3)$ and let $T$
68 be conjugation by $F=\operatorname{diag}(1,-1)$. The diagonal entries of a
69 matrix form the $+1$ eigenspace of $T$. The off-diagonal entries form its
70 $-1$ eigenspace. Both spaces have dimension two.
71 
72 For $C_2$, the norm $1+T$ is multiplication by $2$ on the diagonal space and
73 zero on the off-diagonal space. Hence its kernel is the off-diagonal space.
74 The operator $T-1$ is invertible on that space and zero on the diagonal
75 space. Formula~\eqref{eq:appendix-cyclic-H1} gives $H^1(C_2,M)=0$.
76 
77 Let instead the generator of $C_6$ act by the same $T$. On the diagonal
78 space, its norm is multiplication by $6=0$ in $\mathbb F_3$. On the
79 off-diagonal space, the alternating six-term sum is also zero. Now $N=0$
80 on all of $M$, while the image of $T-1$ is still only the off-diagonal
81 space. Therefore $H^1(C_6,M)$ has dimension two. The generator matrix
82 alone does not determine this cohomology without the order and coefficient
83 characteristic.
84 \end{example}
85 
86 In Frobenius coordinates the relevant action is often $T(X)=FXF^{-1}$.
87 The images of $T-1$ and $1-T$ coincide, since multiplication of a source
88 vector by $-1$ exchanges the signs. This is the bridge to the concrete
89 quotient by Frobenius coboundaries. The recovered v0.18 source proves that
90 bridge under the explicit zero-norm hypothesis. It does not claim that every
91 Frobenius quotient is automatically an arithmetic local cohomology group.
92 One must first identify the acting group and the coefficient module.
93 
94 \subsection{A square-zero defect gives a genuine obstruction class}
95 
96 Let $q:R'\to R$ be a ring map with kernel $I$, and assume $I^2=0$. Let
97 $\rho:G\to R^\times$ be a group representation. Choose unit-valued lifts
98 $r_g\in (R')^\times$ with $q(r_g)=\rho(g)$ and $r_1=1$. Such a choice is
99 an input to this calculation. It is not already a multiplicative lift.
100 Define its normalized defect by
101 \begin{equation}
102  c(g,h)=(r_gr_h-r_{gh})r_{gh}^{-1}\in I.
103  \label{eq:appendix-normalized-defect}
104 \end{equation}
105 Equivalently, $r_gr_h=(1+c(g,h))r_{gh}$. The coefficients are elements of
106 the actual ring kernel, with its additive group, rather than an unrelated
107 vector space of the same dimension.
108 
109 The action of $G$ on $I$ is $g\cdot x=r_gxr_g^{-1}$. It is independent of
110 the chosen unit lift. Indeed, any other unit lift has the form $(1+b)r_g$
111 with $b\in I$. Since $(1+b)^{-1}=1-b$ and products of two elements of $I$
112 vanish, conjugating $x\in I$ by $1+b$ leaves $x$ unchanged. The same
113 observation shows that the defect in the multiplication of the $r_g$ does
114 not prevent these conjugations from forming a genuine group action on $I$.
115 
116 Compute $(r_gr_h)r_\ell$ and $r_g(r_hr_\ell)$. After expressing both in
117 terms of $r_{gh\ell}$, the first is
118 \[
119  (1+c(g,h)+c(gh,\ell))r_{gh\ell},
120 \]
121 while the second is
122 \[
123  (1+g\cdot c(h,\ell)+c(g,h\ell))r_{gh\ell}.
124 \]
125 All omitted cross products belong to $I^2$. Associativity therefore yields
126 \begin{equation}
127  g\cdot c(h,\ell)-c(gh,\ell)+c(g,h\ell)-c(g,h)=0.
128  \label{eq:appendix-cocycle}
129 \end{equation}
130 This is the usual inhomogeneous two-cocycle identity. Normalization at the
131 identity follows from $r_1=1$.
132 
133 Change the chosen lifts to $r'_g=(1+b(g))r_g$, where $b:G\to I$ and
134 $b(1)=0$. Expanding without commuting the factors gives
135 \[
136  r'_gr'_h
137  =(1+b(g)+g\cdot b(h)+c(g,h))r_{gh}.
138 \]
139 To compare with $r'_{gh}$, multiply by $1-b(gh)$. The new defect is
140 \begin{equation}
141  c'(g,h)=c(g,h)+g\cdot b(h)-b(gh)+b(g).
142  \label{eq:appendix-defect-change}
143 \end{equation}
144 Thus changing the section adds a two-coboundary. The class $[c]\in H^2(G,I)$
145 does not depend on the section.
146 
147 \begin{proposition}[Exact lifting criterion]
148 With the hypotheses above, a unit-valued representation lifting $\rho$
149 exists if and only if $[c]=0$.
150 \end{proposition}
151 
152 \begin{proof}
153 If $c=\delta b$, replace $r_g$ by $(1-b(g))r_g$. Equation
154 \eqref{eq:appendix-defect-change} shows that the corrected defect is zero.
155 The corrected units are therefore multiplicative and reduce to $\rho$.
156 The equality $c(1,1)=0$ forces $b(1)=0$ for a primitive in the displayed
157 normalization. Conversely, a multiplicative unit lift has zero defect.
158 Its difference from the original section is a kernel-valued one-cochain,
159 so the same change formula proves that the original class is zero.
160 \end{proof}
161 
162 The cocycle identity proves compatibility of the defect. It does not prove
163 that its cohomology class vanishes. This distinction is indispensable when
164 an obstruction is used as the target of a resolution argument.
165 
166 \begin{example}[An obstruction that really does not vanish]
167 Reduce $\mathbb Z/27\mathbb Z$ to $\mathbb Z/9\mathbb Z$. Its kernel is
168 $9\mathbb Z/27\mathbb Z$, whose square is zero. Map a generator of $C_3$
169 to the unit $4$ modulo $9$. This defines a representation because
170 $4^3=1$ modulo $9$.
171 
172 A lift of the generator must be $4+9t$ modulo $27$, for $t=0,1,2$.
173 Direct expansion gives
174 \[
175  (4+9t)^3\equiv 4^3\equiv 10\pmod {27}.
176 \]
177 None has cube $1$. Hence no representation lift exists and the obstruction
178 class is nonzero. Conjugation is trivial in this scalar example. The
179 kernel is additively $\mathbb F_3$, and the standard cyclic resolution gives
180 $H^2(C_3,\mathbb F_3)=\mathbb F_3$ for the trivial action. This is a
181 failure of lifting, rather than a failure to find a convenient section.
182 
183 The defect can also be seen without an existence argument. Choose the
184 section $r_1=1$, $r_\gamma=4$, and $r_{\gamma^2}=16$ modulo $27$. These
185 reduce to the required units modulo $9$. The products of consecutive
186 powers have no defect until the exponent passes through three. At that
187 point $4^3=10=1+9$ modulo $27$. Consequently the normalized defect is
188 $9$ when the exponents wrap around and zero otherwise. Under the
189 identification $I\simeq\mathbb F_3$, this is the ordinary carry cocycle.
190 
191 For a proposed correction $b$, sum the equations
192 $c(\gamma^i,\gamma)=b(\gamma)-b(\gamma^{i+1})+b(\gamma^i)$ for
193 $i=0,1,2$. The right side telescopes to $3b(\gamma)=0$, using
194 $b(1)=0$ and the trivial action. The left side is $9\ne0$ in $I$.
195 There is therefore no primitive. This explicit calculation recovers the
196 nonzero obstruction class directly from its normalized representative.
197 \end{example}
198 
199 For topological groups the cochains and correcting primitives must belong
200 to the required continuous complex. An abstract primitive need not be
201 continuous. The recovered continuous modules impose continuous sections and
202 local restrictions, with a discrete coefficient ring. They preserve those
203 hypotheses through their lifting criteria. Replacing continuous cohomology
204 by abstract group cohomology would require a separate comparison.
205 
206 \subsection{Local conjugacy leaves a second gluing obstruction}
207 
208 Suppose an absolute two-cocycle has been cancelled by a continuous
209 one-cochain $b_0$. Prescribed local lifts need not agree with the resulting
210 global lift. Their remaining differences are local one-cocycles. Changing
211 $b_0$ to another absolute primitive changes these differences by the
212 restriction of a global one-cocycle. Conjugating the local lift at index
213 $i$ by a unit $1+x_i$, with $x_i\in I$, changes the local cocycle by a
214 local coboundary. The conjugators $x_i$ are independently chosen at each
215 local index. Requiring one common conjugator is a stronger problem.
216 
217 After taking the local coboundary quotients, the remaining class belongs to
218 \begin{equation}
219  \operatorname{coker}\left(
220  H^1_{\mathrm{cont}}(G,I)\longrightarrow
221  \prod_i H^1_{\mathrm{cont}}(G_i,I)\right).
222  \label{eq:appendix-gauge-cokernel}
223 \end{equation}
224 Here the local groups, their maps into $G$, and the precise continuous
225 coefficient action must be supplied. The recovered source constructs the
226 corresponding quotients directly in its square-zero data.
227 
228 To see the criterion, let $a_i$ be the local mismatch after $b_0$. A match
229 up to the allowed local conjugations exists exactly when
230 \[
231  a_i=\operatorname{res}_i(z)+\delta x_i
232 \]
233 for a global continuous one-cocycle $z$ and kernel elements $x_i$. This is
234 exactly the assertion that the image of $(a_i)_i$ in
235 \eqref{eq:appendix-gauge-cokernel} is zero. If $b_1$ is another absolute
236 primitive, then $b_1-b_0$ is a global cocycle. Its restriction disappears in
237 the cokernel, so the remaining class does not depend on which primitive
238 was used.
239 
240 The v0.29 exact sequence identifies this secondary cokernel with the kernel
241 of the map that forgets local gauge data from the relative obstruction
242 space. Exactness at that position does not make the forgetting map
243 surjective. It also does not make the localization map in
244 \eqref{eq:appendix-gauge-cokernel} surjective. These are separate arithmetic
245 questions. In particular, solving the absolute $H^2$ obstruction does not
246 by itself solve a family of prescribed local conditions.
247 
248 \subsection{The reachable equation target is determined by syzygies}
249 
250 Let $J:V\to W$ be a linear map over a field $k$. The correction equation
251 $Ju+e=0$ has a solution precisely when $e\in\operatorname{im}J$. A useful
252 description of this image uses all the dual relations among the equations.
253 Let
254 \[
255  S=\ker(J^*:W^*\to V^*),\qquad
256  C=\{w\in W:\phi(w)=0\text{ for every }\phi\in S\}.
257 \]
258 Every $Ju$ belongs to $C$. Conversely, if $w$ does not belong to
259 $\operatorname{im}J$, its nonzero class in $W/\operatorname{im}J$ is
260 detected by an algebraic linear functional. Pulling that functional back
261 to $W$ gives an element of $S$ that does not annihilate $w$. Hence
262 \begin{equation}
263  \operatorname{im}J=C.
264  \label{eq:appendix-compatible-target}
265 \end{equation}
266 This algebraic dual argument works without a finite-dimensional hypothesis.
267 It makes no assertion about continuous duals in a different topology.
268 
269 Corestricting $J$ to $C$ makes it surjective onto its actual reachable
270 target. The equations have not changed. By contrast, adjoining variables
271 that absorb arbitrary errors changes the system unless their removal is
272 proved. A solution of an augmented equation is not automatically a solution
273 of the original equation. The original target and the quotient that records
274 unreachable errors must be kept separate.
275 
276 \begin{example}[A reachable target with a visible failed correction]
277 Let $V=k^2$, $W=k^3$, and
278 $J(u,v)=(u,v,u+v)$. Its image is the plane $w_3=w_1+w_2$. The dual
279 kernel is generated by the functional
280 $\phi(w_1,w_2,w_3)=w_3-w_1-w_2$. The error $(1,2,3)$ is compatible
281 in every characteristic and is cancelled by $(u,v)=(-1,-2)$. The error
282 $(0,0,1)$ is incompatible in every field, since its value under $\phi$
283 is $1$. No choice of the two correction variables can cancel it.
284 
285 If a third variable $t$ is added so that the derivative becomes
286 $J'(u,v,t)=(u,v,u+v+t)$, the enlarged map is surjective. It cancels the
287 second error with $(u,v,t)=(0,0,-1)$. But the nonzero value of $t$ records
288 exactly the failed original relation. This enlarged solution cannot be
289 declared a solution of the original equations by forgetting $t$. One must
290 show that the additional variable is removable, or that the target error
291 already lies in the original compatible plane.
292 \end{example}
293 
294 For a nonlinear system $F(x)=0$, a relation among the residual Jacobian
295 rows is only a first-order identity. Consider
296 \[
297  F(x)=(x,x^2),\qquad J_x(h)=(h,2xh).
298 \]
299 At the origin the derivative has horizontal image. The functional selecting
300 the second coordinate is a residual syzygy, but it does not annihilate
301 $F(x)$ for general $x$. The exact relation is the moving relation
302 \[
303  (-x,1)F(x)=0.
304 \]
305 Furthermore,
306 \[
307  F(x+h)-F(x)-J_x(h)=(0,h^2).
308 \]
309 Modulo $8$, taking $x=0$ and $h=2$ gives remainder $(0,4)$, outside the
310 horizontal target. Keeping only the syzygy at the origin would therefore
311 lose a genuine nonlinear error even within the same residual fibre.
312 
313 The graded correction argument retains an exact moving relation
314 $B_xF(x)=0$ and a residual exactness statement
315 \begin{equation}
316  \operatorname{im}\overline J=\ker\overline B.
317  \label{eq:appendix-residual-exactness}
318 \end{equation}
319 If $F(x)=\pi^r e$, the moving identity gives $\pi^r B_xe=0$. A faithful
320 leading layer allows this equality to imply
321 $\overline B\,\overline e=0$. Equation
322 \eqref{eq:appendix-residual-exactness} then supplies a residual correction
323 $\overline u$ with $\overline J\,\overline u=-\overline e$. Lift
324 $\overline u$ to $u$ and put $x'=x+\pi^r u$. The linear part of $F(x')$
325 is divisible by $\pi^{r+1}$. For polynomial systems the remainder contains
326 at least two copies of $\pi^r$, so it is divisible by $\pi^{2r}$ and hence
327 by $\pi^{r+1}$ when $r\ge1$. This produces the next layer.
328 
329 Over a suitable truncated principal coefficient ring, iteration terminates
330 at $\pi^N=0$ and produces a root of the original system. The recovered
331 v0.25 modules derive the necessary faithful-layer property under their
332 stated principal residue and nilpotence hypotheses. They still require
333 the moving identity and exactness. Equality of selected dimensions alone
334 does not prove either one. A correct dimension count can help prove
335 exactness after an inclusion is established, but cannot replace that
336 inclusion.
337 
338 The normalization by $\pi^r$ deserves care. Writing
339 $F(x)=\pi^r e$ need not determine $e$ uniquely in a ring with torsion.
340 If $e'$ is another choice, then $\pi^r(e-e')=0$. The faithful leading
341 layer used by the theorem ensures that $e-e'$ has zero residue below the
342 terminal cutoff. Thus the residue correction problem is independent of
343 the chosen normalized error. The same hypothesis converts
344 $\pi^r B_xe=0$ into the residue relation needed for exactness. These are
345 two concrete reasons why the coefficient-ring condition is part of the
346 theorem. Formal cancellation of $\pi^r$ in a ring where it is a zero
347 divisor would be invalid.
348 
349 The moving relation is evaluated at the current approximate root, rather
350 than at the initial point alone. The residual class of that root is kept
351 fixed during every correction, so its residual derivative and residual
352 relation operator remain the same. The exact relation changes over the
353 coefficient ring as the root changes. This is how the iteration retains
354 the full nonlinear equations while continuing to solve one fixed
355 residual linear problem. The distinction between a fixed residual
356 operator and a fixed integral operator is the mechanism that permits
357 redundant equations without an assumed integral right inverse.
358 
359 There is also an integral warning. In $R=\mathbb Z/4\mathbb Z$, the
360 submodule $C=2R$ contains $2$, and $2$ is divisible by $2$ in $R$. Yet
361 $2$ is not divisible by $2$ inside $C$, since $2C=0$. Thus an ambient
362 divisibility statement need not be an internal divisibility statement in a
363 compatible target. An integral splitting cannot be inferred from the
364 residual vector-space picture. The graded argument avoids silently
365 assuming such a splitting by working with the normalized residue of each
366 actual error.
367 
368 \subsection{Ordered powers have a uniform ultrametric remainder}
369 
370 Let $E$ be a complete normed ring with $\norm{1}=1$, a submultiplicative
371 norm, and the ultrametric inequality
372 \[
373  \norm{x+y}\le\max(\norm{x},\norm{y}).
374 \]
375 Assume $\norm A\le1$. For a nonnegative integer $d$, define
376 \begin{align}
377  D_{A,d}(h)&=\sum_{i=0}^{d-1}A^ihA^{d-1-i},
378  \label{eq:appendix-ordered-derivative}\\
379  R_{A,d}(h)&=(A+h)^d-A^d-D_{A,d}(h).
380  \label{eq:appendix-ordered-remainder}
381 \end{align}
382 For $d=0$ the sum is empty and both maps are zero. The order of the
383 factors in \eqref{eq:appendix-ordered-derivative} matters. Replacing this
384 operator by $dA^{d-1}h$ requires $A$ and $h$ to commute. The replacement
385 is invalid for general matrices.
386 
387 The recursive identities are
388 \begin{align*}
389  D_{A,d+1}(h)&=D_{A,d}(h)A+A^dh,\\
390  R_{A,d+1}(h)&=R_{A,d}(h)(A+h)+D_{A,d}(h)h.
391 \end{align*}
392 They follow by multiplying $(A+h)^d$ on the right by $A+h$ and collecting
393 the terms with zero, one, and at least two occurrences of $h$. This
394 derivation does not rearrange any noncommuting factors.
395 
396 For degree three, the linear term is
397 \[
398  D_{A,3}(h)=A^2h+AhA+hA^2,
399 \]
400 and the nonlinear remainder is
401 \[
402  R_{A,3}(h)=Ah^2+hAh+h^2A+h^3.
403 \]
404 The middle word $hAh$ cannot be absorbed into $Ah^2$ in a matrix ring.
405 The remainder estimate keeps all four words and bounds each by its
406 norm, so it respects the ordered expansion. This also explains why the
407 estimate is not an assertion of commutative binomial calculus in
408 disguise. The scalar bound remains uniform because the norm of a finite
409 ultrametric sum is controlled by its largest summand, rather than by the
410 number of summands.
411 
412 \begin{lemma}[Uniform remainder estimate]
413 For $0\le r\le1$ and $\norm h,\norm k\le r$,
414 \[
415  \norm{D_{A,d}(h)}\le\norm h,
416  \qquad
417  \norm{R_{A,d}(h)-R_{A,d}(k)}\le r\norm{h-k}.
418 \]
419 In particular $\norm{R_{A,d}(h)}\le r\norm h$.
420 \end{lemma}
421 
422 \begin{proof}
423 The recursive formula for $D$ shows its additivity. Each term of its
424 ordered sum has norm at most $\norm h$, because every power of $A$ has
425 norm at most one. A finite ultrametric sum has norm at most the largest
426 term, proving the first estimate without a factor $d$.
427 
428 Induct on $d$ for the second estimate. The case $d=0$ is immediate. Write
429 $D_d=D_{A,d}$ and $R_d=R_{A,d}$. Subtracting the remainder recursions gives
430 \begin{align*}
431  R_{d+1}(h)-R_{d+1}(k)
432  ={}&(R_d(h)-R_d(k))(A+h)\\
433  &+R_d(k)(h-k)\\
434  &+(D_d(h)-D_d(k))h+D_d(k)(h-k).
435 \end{align*}
436 The first term has norm at most $r\norm{h-k}$ by induction and
437 $\norm{A+h}\le1$. Taking $h=k$ and the other argument zero in the
438 inductive estimate gives $\norm{R_d(k)}\le r\norm k\le r$, so the
439 second term has the same bound. Additivity gives
440 $\norm{D_d(h)-D_d(k)}\le\norm{h-k}$, which bounds the third term.
441 Finally $\norm{D_d(k)}\le\norm k\le r$ bounds the fourth term.
442 Ultrametricity bounds their sum by the same maximum. Since $R_{A,d}(0)=0$,
443 taking $k=0$ gives the final claim.
444 \end{proof}
445 
446 For $E=\operatorname{Mat}_n(\mathbb Q_p)$ use the entrywise supremum norm
447 and normalize $\abs p_p=p^{-1}$. Matrix multiplication satisfies
448 \[
449  \abs{(XY)_{ij}}_p
450  =\left|\sum_tX_{it}Y_{tj}\right|_p
451  \le\max_t\abs{X_{it}}_p\abs{Y_{tj}}_p
452  \le\norm X\norm Y.
453 \]
454 Taking the maximum over the finite set of entries proves
455 submultiplicativity. The same ultrametric calculation removes a matrix
456 dimension factor. Completeness is inherited from the finite product of
457 copies of $\mathbb Q_p$. These are the hypotheses required by the actual
458 matrix application, rather than properties borrowed from a complex matrix
459 norm.
460 
461 \subsection{An exact inverse turns the local image into one finite test}
462 
463 Assume that $D=D_{A,d}$ has an additive two-sided inverse $Q$, and that
464 $\norm{Qz}\le K\norm z$, with $K\ge0$. Choose $0\le r\le1$ with $Kr<1$. For a
465 target $U$, set $\Delta=U-A^d$. The proposed correction map is
466 \begin{equation}
467  \Phi(h)=Q(\Delta-R_{A,d}(h)).
468  \label{eq:appendix-contraction-map}
469 \end{equation}
470 Both inverse identities will be used. $DQ=1$ turns a fixed point into a
471 root. $QD=1$ proves the necessary condition from an existing root.
472 
473 \begin{theorem}[Exact image of a small power ball]
474 Under these hypotheses, there exists a unique $h$ with $\norm h\le r$
475 and $(A+h)^d=U$ if and only if $\norm{Q\Delta}\le r$.
476 \label{thm:appendix-local-image}
477 \end{theorem}
478 
479 \begin{proof}
480 Suppose $\norm{Q\Delta}\le r$. On the closed $r$-ball, the remainder
481 estimate and ultrametricity give
482 \[
483  \norm{\Phi(h)}
484  \le\max\bigl(\norm{Q\Delta},K\norm{R_{A,d}(h)}\bigr)
485  \le\max(r,Kr\norm h)\le r.
486 \]
487 For $h,k$ in that ball,
488 \[
489  \norm{\Phi(h)-\Phi(k)}\le Kr\norm{h-k}.
490 \]
491 The ball is complete and $Kr<1$. The contraction theorem gives a unique
492 fixed point $h$. Apply $D$ to its equation to obtain
493 $Dh=\Delta-R_{A,d}(h)$, which is exactly $(A+h)^d=U$.
494 
495 Conversely, a root in the ball satisfies $\Delta=Dh+R_{A,d}(h)$. Applying
496 $Q$ gives
497 \[
498  Q\Delta=h+QR_{A,d}(h).
499 \]
500 Its norm is at most $r$, since the second summand has norm at most
501 $Kr\norm h\le r$. Any such root is a fixed point of
502 \eqref{eq:appendix-contraction-map}, so the contraction also proves
503 uniqueness among all roots in this ball.
504 \end{proof}
505 
506 The proof gives a controlled approximation process. Start at $h_0=0$ and
507 iterate $h_{j+1}=\Phi(h_j)$. Then $h_1=Q\Delta$, and
508 \[
509  \norm{h_{j+1}-h_j}\le (Kr)^j\norm{h_1}.
510 \]
511 For $t>j$, ultrametricity bounds $\norm{h_t-h_j}$ by the largest of
512 the intervening increments, which is at most $(Kr)^j\norm{h_1}$. The
513 limit therefore satisfies the same bound for its distance from $h_j$.
514 With $r=p^{-m}$ and $K=p^s$, every iteration gains at least $m-s$
515 digits in this estimate. This is a quantitative consequence after the
516 inverse and target test have been certified. It is not a search over
517 unrelated roots at successively deeper levels.
518 
519 Uniqueness is restricted to the chosen ball. Two different balls can each
520 contain a root of the same target. A failed target congruence rules out
521 the specified residue prefix, while an accepted congruence determines
522 one root with that prefix. Covering all possible prefixes requires a
523 separate argument, especially near centers where the derivative is
524 singular or where the available precision is too shallow.
525 
526 For integral matrices suppose the inverse has the certified form
527 \[
528  Q=p^{-s}B,\qquad s\in\mathbb N,
529 \]
530 where $B$ is an integral linear operator on the matrix entries. In a fixed
531 ordering of the $n^2$ entries it is an $n^2$ by $n^2$ matrix over
532 $\mathbb Z_p$. This is an operator on the matrix space, not ordinary
533 left multiplication by an $n$ by $n$ matrix. Integral coefficients imply
534 $\norm{Bz}\le\norm z$, so $K=p^s$ is a valid bound.
535 
536 There is a finite exact certificate for the inverse. For every matrix unit
537 $E_{ij}$ check
538 \[
539  QD(E_{ij})=E_{ij},\qquad DQ(E_{ij})=E_{ij}.
540 \]
541 Linearity and the matrix-unit basis extend both identities to every matrix
542 over $\mathbb Q_p$. These equalities are exact in the coefficient field.
543 They are not merely congruences modulo $p$. In particular, an operator
544 singular over $\mathbb Q_p$ cannot satisfy them.
545 
546 Take $r=p^{-m}$ with $m>s$. Then $Kr=p^{s-m}<1$, and
547 \begin{align}
548  \norm{Q(U-A^d)}\le p^{-m}
549  &\Longleftrightarrow \norm{B(U-A^d)}\le p^{-(m+s)}\notag\\
550  &\Longleftrightarrow
551  B\operatorname{vec}(U-A^d)=0\pmod {p^{m+s}}.
552  \label{eq:appendix-finite-criterion}
553 \end{align}
554 The last congruence is entrywise. It depends only on the reductions of the
555 target, center, and integral operator modulo $p^{m+s}$. The theorem
556 therefore decides the existence and uniqueness of a root in
557 $A+p^m\operatorname{Mat}_n(\mathbb Z_p)$ by one finite target test, once
558 the exact inverse certificate is supplied.
559 
560 The root constructed over $\mathbb Q_p$ is integral. Its center is
561 integral, and every correction entry has norm at most $p^{-m}\le1$.
562 If $d>0$ and $U$ is a unit, the root $F$ is also a unit. Indeed, $F$
563 commutes with $U=F^d$ and hence with $U^{-1}$, and
564 $F^{d-1}U^{-1}$ is a two-sided inverse of $F$. Thus the unit version has
565 an actual output in $\operatorname{GL}_n(\mathbb Z_p)$.
566 
567 The value of $s$ measures denominators of the whole inverse derivative.
568 It is not automatically $v_p(d)$. The proof makes no statement about
569 shallow balls with $m\le s$, where its contraction estimate is insufficient.
570 It also does not provide automatic certificate discovery for arbitrary
571 $p$-adic input. Exact rational examples can supply certificates by ordinary
572 linear algebra, but a general certified implementation is an additional
573 task.
574 
575 \subsection{A noncommuting unipotent square example}
576 
577 Work over $\mathbb Z_2$ and take
578 \[
579  A=\begin{pmatrix}1&1\\0&1\end{pmatrix},\qquad d=2.
580 \]
581 For $X=\begin{pmatrix}a&b\\c&e\end{pmatrix}$, the ordered derivative is
582 \[
583  D_{A,2}(X)=AX+XA
584  =\begin{pmatrix}2a+c&a+2b+e\\2c&c+2e\end{pmatrix}.
585 \]
586 Using the entry order $(a,b,c,e)$, its matrix and a scaled inverse are
587 \[
588  J=\begin{pmatrix}
589  2&0&1&0\\1&2&0&1\\0&0&2&0\\0&0&1&2
590  \end{pmatrix},\qquad
591  B=\begin{pmatrix}
592  2&0&-1&0\\-1&2&1&-1\\0&0&2&0\\0&0&-1&2
593  \end{pmatrix}.
594 \]
595 Direct multiplication gives $BJ=JB=4I_4$. Therefore $Q=\tfrac14 B$
596 is the exact inverse and $s=2$. The entry $\tfrac14$ in this inverse
597 prevents clearing its denominators with only one factor of $2$.
598 
599 The residual derivative is singular. In characteristic two,
600 $D_{A,2}(I_2)=2A=0$, while $I_2\ne0$. Nevertheless $J$ is invertible
601 over $\mathbb Q_2$, with determinant $16$. Residual singularity and
602 singularity over the fraction field are different tests.
603 
604 Choose $m=3$, so the local root ball is $A+8\operatorname{Mat}_2(\mathbb Z_2)$
605 and the finite target test is modulo $2^{m+s}=32$. Consider
606 \[
607  F=\begin{pmatrix}1&1\\8&1\end{pmatrix},\qquad
608  U=F^2=\begin{pmatrix}9&2\\16&9\end{pmatrix}.
609 \]
610 Here
611 \[
612  AF=\begin{pmatrix}9&2\\8&1\end{pmatrix},\qquad
613  FA=\begin{pmatrix}1&2\\8&9\end{pmatrix},
614 \]
615 so the root does not commute with the center. Since
616 $A^2=\begin{pmatrix}1&2\\0&1\end{pmatrix}$, the target difference has
617 entry vector $(8,0,16,8)$. Applying $B$ gives
618 \[
619  B\operatorname{vec}(U-A^2)=(0,0,32,0).
620 \]
621 The congruence modulo $32$ passes. The theorem supplies uniqueness in the
622 specified ball, beyond the direct multiplication that exhibits this root.
623 
624 For a negative control, take
625 \[
626  U_{\mathrm{bad}}=\begin{pmatrix}1&10\\0&1\end{pmatrix}.
627 \]
628 The difference from $A^2$ is $(0,8,0,0)$ in entry coordinates, and $B$
629 sends it to $(0,16,0,0)$. This is nonzero modulo $32$, so no square root
630 of $U_{\mathrm{bad}}$ lies in that ball. This is a failure for the
631 specified prefix. It is not a claim that the target has no square root
632 anywhere in a larger coefficient field.
633 
634 The recovered exact checker compares the finite image with the finite
635 criterion. Its unipotent square case tests $256$ corrections at the
636 stated horizon and $4096$ corrections one level deeper. Both give the same
637 $16$ target residues as the criterion. These are bounded corroborating
638 calculations. Infinite existence and uniqueness follow from the general
639 contraction proof and its exact hypotheses.
640 
641 There is a genuinely singular contrast. Let
642 $A_0=\operatorname{diag}(1,-1)$ over $\mathbb Q_p$. Then
643 $D_{A_0,2}(E_{12})=0$, although $E_{12}\ne0$. Thus this derivative has
644 no inverse over the fraction field, including when $p$ is odd and $2$ is
645 a unit. The certificate theorem cannot cover this center. A universal
646 claim that invertible centers always have invertible power derivatives
647 would be false.
648 
649 \subsection{Finite levels give a coherent branch only for a joint problem}
650 
651 Let $X_r$ be finite sets with genuine reduction maps
652 $\pi_{r,t}:X_t\to X_r$ for $r\le t$. Assume identity and composition
653 laws for these maps. Let $P_r\subseteq X_r$ be solution sets, and assume
654 reduction preserves solutions. Then
655 \begin{equation}
656  \left(\forall r,\ P_r\ne\varnothing\right)
657  \Longleftrightarrow
658  \left(\exists (x_r)_r,\ x_r\in P_r\text{ and }
659  \pi_{r,t}(x_t)=x_r\text{ for }r\le t\right).
660  \label{eq:appendix-branch-criterion}
661 \end{equation}
662 The nontrivial implication is finite branch compactness. For each depth,
663 a solution at that depth gives a finite path of solutions at all earlier
664 depths. There are arbitrarily long paths. The resulting tree has finite
665 branching because every level is finite. An infinite branch follows from
666 the finite-branch form of K\"onig's lemma. The reduction maps on the
667 solution sets need not be surjective. Compactness selects persistent
668 choices instead of extending a previously chosen node at every step.
669 
670 This matters for simultaneous matrix prescriptions. At level $r$, a node
671 can contain a shared matrix $F_r$ together with one local conjugator
672 $z_{i,r}$ for each member of a finite family. The equations are
673 \[
674  z_{i,r}F_r^{d_i}z_{i,r}^{-1}=U_i\pmod {p^r},
675 \]
676 with the prescribed prefix of $F_r$ and prescribed residual identity for
677 each conjugator. These are a single joint predicate on a finite tuple.
678 The arithmetic equations and prefixes are preserved by reduction.
679 Compactness therefore produces compatible towers for the shared matrix
680 and for all local conjugators, provided this joint problem is nonempty at
681 every depth.
682 
683 An individual solution for each prescription is weaker. Even the scalar
684 requirements $F=1$ and $F=-1$ modulo an odd prime have solutions
685 separately, but have no common solution. Likewise, solving every member
686 of a family at each depth with a different shared matrix does not establish
687 the premise of \eqref{eq:appendix-branch-criterion}. The compatible branch
688 must come from one family of jointly satisfiable nodes.
689 
690 Nor does a successful shallow node necessarily extend. For example, the
691 scalar prefix $F=4$ modulo $9$ satisfies $F^3=1$ there. Its lifts modulo
692 $27$ are $4,13,22$, all with cube $10$. That branch stops at the next
693 level. Merely checking a finite number of successful levels gives no
694 all-depth conclusion without another theorem.
695 
696 Conversely, for a downward-closed tower of finite solution sets indexed
697 by all natural-number levels, absence of an infinite branch implies that
698 some finite level is empty. This gives a finite
699 obstruction witness in principle. It supplies no uniform computable bound
700 on the depth of that witness. The matrix inverse theorem above provides
701 an explicit horizon only in its certified nonsingular sector with $m>s$.
702 It removes the all-depth premise for one local power problem in that sector.
703 It does not remove the joint all-depth premise for arbitrary simultaneous
704 prescriptions.
705 
706 Finally, compatible integral matrix residues determine an actual matrix
707 over $\mathbb Z_p$ entry by entry. Compatible inverse residues determine
708 its inverse when the nodes are units. Equality of all reductions then
709 implies equality of the integral matrices. These reconstruction steps turn
710 the selected branch into a genuine integral solution. The remaining
711 arithmetic attachment is not part of that reconstruction.
712 
713 \subsection{What the formal receipts establish}
714 
715 The recovered archives preserve the source statements and historical
716 verification receipts. The v0.38 receipt records a successful build of its
717 seven new modules against the pinned Mathlib environment. The earlier
718 interfaces described here are drawn from their recovered source files and
719 their recorded histories. Archive authentication verifies the recovered
720 bytes. It is not a new compilation of the mathematical statements.
721 
722 A fresh Lean replay was attempted during preparation of this edition, but
723 the portable kit could not be recovered because its download returned
724 HTTP 403. No fresh compiler invocation was therefore completed for this
725 edition. The accompanying coverage report records that failure and
726 distinguishes it from the historical successful receipt. The new exact
727 finite controls also have their own tested domains. They do not substitute
728 for a kernel replay or extend a conditional theorem to a missing
729 arithmetic hypothesis.
730 
731 The formal program supplies explicit cocycle quotients, exact obstruction
732 criteria, careful nonlinear correction interfaces, compatible tower
733 reconstruction, and an effective finite criterion for a substantial local
734 matrix sector. To reach Strong Artin, these mechanisms still need the
735 arithmetic identifications and global existence arguments stated in the
736 research boundary. Their mathematical value comes from making those
737 remaining premises visible, while proving the local implications in full.

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