CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 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.

Factor lifting implies simple-root lifting

Statement

Let $(A, m)$ be a Henselian local ring with residue field $k = A / m$ . Let $f \in A [T]$ be monic, and let $\overline{a} \in k$ be a simple root of $\overline{f}$ . Then there exists a unique $a \in A$ lifting $\overline{a}$ such that $f (a) = 0$ .

Facts & Assumptions

Given: A Henselian local ring $(A, m)$ , a monic polynomial $f \in A [T]$ , and a simple residue root $\overline{a}$ of $\overline{f}$ .

[L1]

A simple residue root gives a coprime factorization $\overline{f} = (T - \overline{a}) \overline{h}$ in $k [T]$ (A simple residue root determines a coprime residue factorisation).

[L2]

A Henselian pair lifts coprime monic factorizations uniquely (Henselian pairs and Henselian local rings, Lifted coprime factorisations are unique).

Proof

technique · lift the linear factor

1.1L1L2given

By [L1], write $\overline{f} = (T - \overline{a}) \overline{h}$ with $T - \overline{a}$ and $\overline{h}$ coprime. Since $(A, m)$ is Henselian, [L2] gives a lifted factorization $f = (T - a) h$ with $a \in A$ reducing to $\overline{a}$ . Evaluating at $T = a$ gives $f (a) = 0$ .

2.1L2step 1.1given

If $a^{'}$ is another lift of $\overline{a}$ with $f (a^{'}) = 0$ , then $f = (T - a^{'}) h^{'}$ for some monic $h^{'}$ , and this is another lift of the same residue factorization. By [L2], the lifted factorization is unique, so $T - a = T - a^{'}$ and hence $a = a^{'}$ .

3.1step 1.1step 2.1∎

Therefore factor lifting implies unique lifting of every simple residue root.

Depends on

Used by

Dependency tree · two levels

8 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

The Stacks Project, Section 10.153: Henselian local rings (standard reference, not scraped)