B.7 Paper IX: weighted first-order root motion
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Komorebi P. — 8 September 2026.
B.7.1 The theorem
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Let \(p\) be prime, \(R=\mathbb Z_p\), \(K=\mathbb Q_p\), and \(d,s,M\in\mathbb N\). Let \(F,H\in R[X_1,\ldots,X_d]^d\), let \(a\in R^d\) satisfy \(F(a)=0\), and set
Assume
Then there is exactly one root \(b\) of \(G\) in \(a+p^{s+1}R^d\). It satisfies
and there is \(\varepsilon\in p^{2M-3s}R^d\) such that, in \(K^d\),
The inverse in this formula belongs to the original Jacobian \(DF(a)\). The theorem constructs the perturbed root from these hypotheses and proves uniqueness throughout the full ball of depth \(s+1\). It does not require an additional Jacobian hypothesis on \(G\). Coefficients and centers are arbitrary integral p-adic elements. Every prime and finite square dimension are included, including \(p=2\) and dimension zero.
The error depth exceeds the displacement depth by
Thus the first-order term determines strictly more digits than the displacement bound alone. These are guaranteed divisibility bounds; equality of valuations is not asserted for every system.
B.7.2 Root construction with the original inverse
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Write \(A=p^sJ^{-1}\in M_d(R)\), so \(AJ=JA=p^sI\). Let \(t=p^{s+1}\) and \(k=M-(2s+1)\ge0\). The actual polynomial Taylor and difference identities give polynomial systems \(Q_F,Q_H\), both vanishing at zero, such that
Consequently
where
All exponents are nonnegative under \(M>2s\). Actual integral polynomial evaluation is 1-Lipschitz, so \(u\mapsto-C-pQ(u)\) is a strict contraction on the complete space \(R^d\). Its unique fixed point inherits the depth \(k\) of \(C\), because \(Q(0)=0\). Thus \(b=a+tu\) has displacement depth \(s+1+k=M-s\).
The certificate \(JA=p^sI\) makes multiplication by \(A\) injective, so the normalized zero is an actual root of \(G\). Every root in \(a+tR^d\) has the form \(a+tw\) and satisfies the same fixed-point equation. This proves uniqueness on the entire stated ball using the original matrix certificate.
B.7.3 The first-order error
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Put \(\delta=b-a\) and \(L=M-s\). The displacement form of Taylor's theorem, proved over any commutative ring, gives
Polynomial congruence gives
Since \(F(a)=0\) and \(G(b)=0\),
We have \(M+L\ge2L\), so the right side lies in \(p^{2L}R^d\). Cancelling the nonzero scalar \(p^s\) proves the integral estimate
Define this integral vector to be \(\varepsilon\). Under \(R\hookrightarrow K\), the exact identity \(p^{M-s}A=p^MJ^{-1}\) gives the displayed first-order formula. The Lean bridge checks matrix evaluation and all coefficient inclusions explicitly.
B.7.4 Sharpness and the strict threshold
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For \(p=2\), take \(F(X)=X+X^2\), \(H(X)=1\), \(a=0\), \(s=0\), \(M=1\). There is a unique root \(b\in2\mathbb Z_2\) of \(X+X^2+2\). Its equation gives \(b+2=-b^2\). It also forces \(b\equiv2\pmod4\). Therefore
The accompanying Lean controls show that the error is divisible by four and is not divisible by eight. This refutes a uniform replacement of \(2M-3s\) by \(2M-3s+1\), even for a one-variable unit Jacobian.
The prior strict-threshold example also applies to perturbations: take \(F(X)=X^2-1\), \(H(X)=1\), \(a=1\), \(p=2\), \(s=1\), \(M=2\). Then \(F(a)=0\), \(DF(a)=2\), and \(F+2^2H=X^2+3\) has no root modulo eight, hence none in \(\mathbb Z_2\). The new explanatory identification uses the previously checked no-root polynomial. The strict inequality \(M>2s\) cannot uniformly be replaced by \(M\ge2s\).