LemmaStatement: 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.

Two lifted factorisations agree modulo every ideal power

Statement

Let $A$ be a commutative ring, let $I \subseteq A$ be an ideal, let $f \in A [T]$ , and let $f = g h = g^{'} h^{'}$ with $g, h, g^{'}, h^{'}$ monic, $\overline{g} = \overline{g^{'}}$ , $\overline{h} = \overline{h^{'}}$ , with $g, g^{'}$ of the same degree $m$ and $h, h^{'}$ of the same degree $n$ , and with $\overline{g}, \overline{h}$ coprime in $(A / I) [T]$ . Then $g \equiv g^{'} (m o d I^{r} [T]) and h \equiv h^{'} (m o d I^{r} [T])$ for every $r \geq 1$ .

Facts & Assumptions

Given: Two monic lifts $f = g h = g^{'} h^{'}$ of the same coprime residue factorization.

[L1]

Coprime residue factors admit a lifted Bezout identity modulo $I$ (Lift a Bezout identity for coprime residue factors).

Proof

technique · improve congruence one power at a time

1.1given

The congruences modulo $I$ hold by hypothesis, so the claim is true for $r = 1$ .

2.1step 1.1givenassume-hypalgebra

Assume $g \equiv g^{'} (m o d I^{r} [T])$ and $h \equiv h^{'} (m o d I^{r} [T])$ for some $r \geq 1$ . Write $g^{'} = g + u$ and $h^{'} = h + v$ with $u, v \in I^{r} [T]$ . Since each pair consists of monic polynomials of the same degree, $\deg u < m$ and $\deg v < n$ . From $g h = (g + u) (h + v)$ we get $u h + v g + u v = 0.$ Modulo $I^{r + 1}$ the term $u v$ vanishes, so $u h + v g \equiv 0 (m o d I^{r + 1}) .$

3.1L1step 2.1algebradischarge-induction

Put $M = I^{r} / I^{r + 1}$ , and let $\overset{ˉ}{u}, \overset{ˉ}{v} \in M [T]$ be the classes of $u, v$ . Step 2.1 gives $\overset{ˉ}{u} h + \overset{ˉ}{v} g = 0$ . Choose $a, b$ with $a g + b h \equiv 1 (m o d I)$ by [L1]. Modulo the monic polynomial $g$ , multiplication by $h$ is invertible with inverse $b$ , so $\overset{ˉ}{u} h = 0$ implies $\overset{ˉ}{u} = 0$ modulo $g$ . Since $\deg \overset{ˉ}{u} < m = \deg g$ , this gives $\overset{ˉ}{u} = 0$ . The equation then becomes $\overset{ˉ}{v} g = 0$ , and multiplication by the monic polynomial $g$ is injective on $M [T]$ , so $\overset{ˉ}{v} = 0$ . Hence $u, v \in I^{r + 1} [T]$ , proving the two congruences modulo $I^{r + 1}$ .

4.1step 1.1step 3.1∎

By induction, the two lifted factorisations agree modulo every power $I^{r}$ .

Depends on

Lift a Bezout identity for coprime residue factors

Used by

Complete separated adic pairs are Henselian Corollary
Lifted coprime factorisations are unique Proposition

Dependency tree · two levels

3 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

Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, 13th ed., Chapter 22 (standard reference, not scraped)