LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedverified 2026-09-24 (gpt-6-sol)

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.

Noetherian domains are atomic

Statement

Assume Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain). Every nonzero nonunit in a Noetherian integral domain is a finite product of irreducible elements.

Facts & Assumptions

Given: Dependent Choice, a Noetherian integral domain $R$ , and a nonzero nonunit $a \in R$ .

[L1]

Under Dependent Choice, every nonempty family of ideals in a Noetherian ring has an inclusion-maximal member (Noetherian commutative rings and modules, A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member). The use in step 1.1 is the only choice-dependent step.

[L2]

Divisibility and associates are those of Divisibility and associates in an integral domain, and irreducible elements are those of Irreducible and prime elements of an integral domain.

Proof

technique · direct

1.1L1L2choose

Suppose the statement were false, and let $S$ be the set of nonzero nonunits that are not finite products of irreducibles. By [L1], the family of principal ideals $(x)$ with $x \in S$ has a maximal member; choose $a \in S$ with $(a)$ maximal. The element $a$ is not irreducible, so write $a = b c$ with $b$ and $c$ nonunits.

2.1step 1.1L2algebra

Since $a$ is not a finite product of irreducibles, at least one of $b$ or $c$ lies in $S$ ; choose $b$ if possible, otherwise choose $c$ . Also $a = b c$ makes $a \in (b)$ and $a \in (c)$ , while neither $b$ nor $c$ is associate to $a$ because both are nonunits. Hence $(a) ⊊ (b)$ and $(a) ⊊ (c)$ , contradicting the maximal choice of $(a)$ .

3.1step 2.1∎

The contradiction in step 2.1 shows $S$ is empty. Therefore every nonzero nonunit of $R$ factors into irreducibles.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 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, Lemma 10.120.3 (standard reference, not scraped)