Paper IX: Arithmetic Packets, Dyadic Towers, and Local Polynomial Deformation

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1 \hypertarget{cont6:paper-ix-weighted-nonlinear-lifting}{%
2 \subsection{Paper IX: weighted nonlinear
3 lifting}\label{cont6:paper-ix-weighted-nonlinear-lifting}}
4 
5 \textbf{Komorebi P. --- 8 September 2026.}
6 
7 \hypertarget{cont6:the-theorem}{%
8 \subsubsection{The theorem}\label{cont6:the-theorem}}
9 
10 Let \(p\) be prime, \(R=\mathbb Z_p\), \(K=\mathbb Q_p\), and
11 \(d,s,r\in\mathbb N\). Let
12 \(F=(F_1,\ldots,F_d)\in R[X_1,\ldots,X_d]^d\), \(a\in R^d\), and let \[
13 J=DF(a)=\bigl(\partial_jF_i(a)\bigr)_{i,j}.
14 \] Assume \[
15 J\in\operatorname{GL}_d(K),\qquad p^sJ^{-1}\in M_d(R),\qquad
16 F(a)\in p^rR^d,\qquad r>2s.
17 \] Then \[
18 \boxed{\exists!\,b\in a+p^{s+1}R^d:\ F(b)=0,}
19 \qquad
20 \boxed{b-a\in p^{r-s}R^d.}
21 \] All vector divisibility assertions are coordinatewise. The uniqueness
22 predicate is the entire ball of depth \(s+1\): competitors need not be
23 assumed to have depth \(r-s\). The theorem permits arbitrary integral
24 p-adic coefficients and centers, every prime including two, and
25 dimension zero. It retains the supplied inverse-denominator exponent
26 \(s\); there is no replacement by a determinant valuation. Integrality
27 of \(p^sJ^{-1}\) is expressed in Lean by existence of an integral matrix
28 with exactly that image under the coefficient inclusion
29 \(R\hookrightarrow K\).
30 
31 \hypertarget{cont6:proof}{%
32 \subsubsection{Proof}\label{cont6:proof}}
33 
34 Write \(A=p^sJ^{-1}\in M_d(R)\). Then \[
35 AJ=JA=p^sI.
36 \] The formal matrix bridge proves the equivalence with the stated
37 fraction-field hypotheses, including nonsingularity. In particular,
38 multiplication by \(A\) is injective, since multiplication by \(J\)
39 sends \(Av=0\) to \(p^sv=0\).
40 
41 Put \(G=AF\), \(t=p^{s+1}\), and \(k=r-(2s+1)\). Taylor expansion of the
42 actual preconditioned polynomials gives one polynomial system \(Q\) such
43 that \[
44 G(a+tu)=G(a)+p^{2s+1}u+p^{2s+2}Q(u),\qquad Q(0)=0.
45 \] This expansion is proved first over every commutative ring. Its
46 remainder is a polynomial valid for every \(u\), not a pointwise
47 quotient depending on \(u\). Since \(G(a)\in p^rR^d\), it can be written
48 \(p^{2s+1}C\) with \(C\in p^kR^d\). Thus \[
49 G(a+tu)=p^{2s+1}\bigl(C+u+pQ(u)\bigr).
50 \]
51 
52 Integral polynomial evaluation is 1-Lipschitz in the supremum p-adic
53 metric. Consequently \(T(u)=-C-pQ(u)\) has contraction constant
54 \(|p|_p<1\) on the complete space \(R^d\), and has exactly one fixed
55 point \(u\).
56 
57 For completeness, the sharp precision is additional to this fixed-point
58 existence. Inductively, if \(u\in p^nR^d\), then polynomial congruence
59 and \(Q(0)=0\) imply \(Q(u)\in p^nR^d\). For \(n<k\), the equation
60 \(u=-C-pQ(u)\) therefore gives \(u\in p^{n+1}R^d\). Starting at \(n=0\),
61 this proves \(u\in p^kR^d\). Hence \(b=a+tu\) satisfies \[
62 b-a\in p^{s+1+k}R^d=p^{r-s}R^d.
63 \] The normalization and injectivity of \(A\) give \(F(b)=0\).
64 
65 Every other \(z\in a+p^{s+1}R^d\) has the form \(a+tw\). If \(F(z)=0\),
66 the same normalization, with cancellation of the nonzero scalar
67 \(p^{2s+1}\), gives \(w=T(w)\). Fixed-point uniqueness implies \(w=u\),
68 hence \(z=b\).
69 
70 \hypertarget{cont6:boundary-and-sharpness}{%
71 \subsubsection{Boundary and sharpness}\label{cont6:boundary-and-sharpness}}
72 
73 The strict inequality cannot in general be weakened to \(r\ge2s\): for
74 \(p=2\), \(F(X)=X^2+3\), and \(a=1\), we have \(F(a)=4\), \(DF(a)=2\),
75 and may take \(s=1,r=2\). Yet no root exists in \(\mathbb Z_2\), since
76 there is none modulo eight. The auxiliary Lean controls prove this
77 obstruction and the exact residual depth.
78 
79 The displacement exponent cannot uniformly be increased by one. For
80 \(F(X)=p^sX+p^r\) at \(a=0\), with \(r>2s\), its root is \(b=-p^{r-s}\),
81 which is not divisible by \(p^{r-s+1}\). The controls prove the linear
82 root and the absence of that extra digit for all primes.

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