TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Newton's criterion in Q_p

Statement

Let $f \in Z_{p} [X]$ and $a_{0} \in Z_{p}$ . If

$∣ f (a_{0}) ∣_{p} < ∣ f^{'} (a_{0}) ∣_{p}^{2},$

then the Newton iterates

$a_{n + 1} : = a_{n} - \frac{f (a_{n})}{f^{'} (a_{n})}$

are defined, converge in $Z_{p}$ to a root $a$ of $f$ , satisfy $∣ a - a_{0} ∣_{p} \leq ∣ f (a_{0}) / f^{'} (a_{0}) ∣_{p}$ , and that root is unique in the closed ball of that radius around $a_{0}$ .

Facts & Assumptions

Given: A polynomial $f \in Z_{p} [X]$ and $a_{0} \in Z_{p}$ with $∣ f (a_{0}) ∣_{p} < ∣ f^{'} (a_{0}) ∣_{p}^{2}$ .

[L1]

$Z_{p}$ is the valuation ring of $Q_{p}$ (Z_p is the valuation ring of Q_p).

[L2]

$Q_{p}$ is a complete nonarchimedean valued field (The p-adic completion is a complete valued field).

Proof

technique · constructive

1.1L1L2givenconstruct

Put $h_{n} : = f (a_{n}) / f^{'} (a_{n})$ whenever $a_{n}$ is defined. Since $∣ h_{0} ∣_{p} < ∣ f^{'} (a_{0}) ∣_{p} \leq 1$ , we have $h_{0} \in p Z_{p}$ , so $a_{1} = a_{0} - h_{0}$ lies in $Z_{p}$ . For any $x \in Z_{p}$ , Taylor expansion gives $f (x - h) = f (x) - h f^{'} (x) + h^{2} g_{x} (h)$ with $g_{x} (h) \in Z_{p}$ . Hence $f (a_{n + 1}) = h_{n}^{2} g_{a_{n}} (h_{n}),$ so $∣ f (a_{n + 1}) ∣_{p} \leq ∣ h_{n} ∣_{p}^{2}$ .

2.1L2step 1.1induction

The derivative also satisfies $f^{'} (a_{n + 1}) = f^{'} (a_{n}) + h_{n} u_{n}$ for some $u_{n} \in Z_{p}$ . Because $∣ h_{n} ∣_{p} < ∣ f^{'} (a_{n}) ∣_{p}$ , the ultrametric inequality gives $∣ f^{'} (a_{n + 1}) ∣_{p} = ∣ f^{'} (a_{n}) ∣_{p} .$ By induction, every iterate is defined, each $f^{'} (a_{n})$ has the same nonzero absolute value, and $∣ a_{n + 1} - a_{n} ∣_{p} = ∣ h_{n} ∣_{p} \leq \frac{∣ f (a_{n}) ∣_{p}}{∣ f^{'} (a_{0}) ∣_{p}} \leq {(\frac{∣ f (a_{0}) ∣_{p}}{∣ f^{'} (a_{0}) ∣_{p}^{2}})}^{2^{n} - 1} ∣ h_{0} ∣_{p} .$ So the differences tend to $0$ quadratically.

3.1L2step 2.1algebra

The series of successive differences is therefore Cauchy, so $(a_{n})$ converges in $Z_{p}$ by [L2]; call its limit $a$ . Continuity of polynomial evaluation gives $f (a) = 0$ , and the ultrametric inequality applied to $a - a_{0} = \sum_{n \geq 0} (a_{n + 1} - a_{n})$ shows $∣ a - a_{0} ∣_{p} \leq ∣ h_{0} ∣_{p} = ∣ f (a_{0}) / f^{'} (a_{0}) ∣_{p}$ .

4.1L1step 3.1algebradischarge-construct∎

If $b$ is another root in the closed ball of radius $∣ h_{0} ∣_{p}$ about $a_{0}$ , then $0 = f (b) - f (a) = (b - a) v$ for the usual divided-difference element $v \in Z_{p}$ . Because $a - a_{0}$ and $b - a_{0}$ both have absolute value at most $∣ h_{0} ∣_{p}$ , each term of $v - f^{'} (a_{0})$ contains one factor from $(a - a_{0}) Z_{p}$ or $(b - a_{0}) Z_{p}$ . Thus $∣ v - f^{'} (a_{0}) ∣_{p} \leq ∣ h_{0} ∣_{p} < ∣ f^{'} (a_{0}) ∣_{p} .$ The ultrametric inequality therefore gives $∣ v ∣_{p} = ∣ f^{'} (a_{0}) ∣_{p} \neq 0$ , so $v$ is nonzero and hence $b - a = 0$ . Therefore $b = a$ .

Depends on

Used by

Selmer's cubic is locally soluble but globally insoluble Counterexample
A square root of minus one in Q₅ Example
Hensel lifting a simple root of X squared minus 2 in Z₇ Example
Square criterion in Q₂ Theorem

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Keith Conrad, Hensel's Lemma, Theorem 4.1 (standard reference, not scraped)
J. S. Milne, Algebraic Number Theory, Theorem 7.32 (standard reference, not scraped)