CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Simple roots lift uniquely in Z_p

Statement

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

$f (a_{0}) \equiv 0 (m o d p), f^{'} (a_{0}) \equiv̸ 0 (m o d p),$

then there is a unique $a \in Z_{p}$ such that $a \equiv a_{0} (m o d p)$ and $f (a) = 0$ .

Facts & Assumptions

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

[L1]

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

[L2]

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

Proof

technique · constructive

1.1L1giveninductionconstruct

Construct $a_{n} \in Z_{p}$ inductively so that $a_{n} \equiv a_{0} (m o d p), f (a_{n}) \in p^{n + 1} Z_{p} .$ Given $a_{n}$ , write $a_{n + 1} = a_{n} + t_{n} p^{n + 1}$ with $t_{n} \in {0, \dots, p - 1}$ . Taylor expansion modulo $p^{n + 2}$ gives $f (a_{n + 1}) \equiv f (a_{n}) + t_{n} p^{n + 1} f^{'} (a_{n}) (m o d p^{n + 2}) .$ Since $f^{'} (a_{n}) \equiv f^{'} (a_{0}) \equiv̸ 0 (m o d p)$ , there is a unique choice of $t_{n}$ modulo $p$ making the right-hand side $0$ modulo $p^{n + 2}$ .

2.1L2step 1.1algebra

The differences satisfy $a_{n + 1} - a_{n} \in p^{n + 1} Z_{p}$ , so $(a_{n})$ is a $p$ -adic Cauchy sequence. By [L2] it converges to some $a \in Z_{p}$ , and continuity of polynomial evaluation gives $f (a) = 0$ . Also $a - a_{0} \in p Z_{p}$ , so $a \equiv a_{0} (m o d p)$ .

3.1L1step 2.1algebradischarge-construct∎

If $b$ is another root with $b \equiv a_{0} (m o d p)$ , then $0 = f (a) - f (b) = (a - b) g (a, b)$ for the usual divided-difference polynomial $g$ . Modulo $p$ , one has $g (a, b) \equiv f^{'} (a_{0}) \equiv̸ 0$ , so $g (a, b)$ is a unit of $Z_{p}$ by [L1]. Therefore $a - b = 0$ .

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
Almost all local completions are isotropic in dimension at least three Theorem
Square criterion in Qₚ for odd p Theorem
The odd-prime Hilbert symbol formula 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 2.1 (standard reference, not scraped)
Andrew V. Sutherland, 18.782 Lecture 8, Theorem 8.8 (standard reference, not scraped)