LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

minimal free resolutions unique up to chain isomorphism

Statement

Any two minimal degreewise finite free resolutions of a finite module over a nonzero Noetherian local ring are augmentation-preservingly chain-isomorphic, in general noncanonically. In particular their ranks agree in every degree.

Facts & Assumptions

Given: The objects and hypotheses in the statement. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

betti number is rank in minimal resolution: For every minimal degreewise finite free resolution $F_{∙} \to M$ of a finite module over a nonzero Noetherian local ring, $β_{i}^{R} (M) = {rank}_{R} F_{i}$ for all $i \geq 0$ .

[F2]

Projective comparison maps exist: Assume the Axiom of Dependent Choice. Let $u : A \to B$ be a morphism, and let $P_{∙} \to A$ and $Q_{∙} \to B$ be projective resolutions. Then there exists an augmentation-preserving chain map $f_{∙} : P_{∙} \to Q_{∙}$ lifting $u$ .

[F3]

Projective comparison maps are unique up to chain homotopy: Assume the Axiom of Dependent Choice. Any two augmentation-preserving maps between projective resolutions lifting the same object morphism are chain-homotopic.

[F4]

Assuming the Axiom of Choice, Nakayama's lemma: Assume the Axiom of Choice. Let $R$ be a commutative ring, let $I ⊴ R$ satisfy $I \subseteq J (R)$ , and let $M$ be a finitely generated left $R$ -module. If $I M = M$ , then $M = 0$ .

Proof

1.1

Choose comparison maps $f : F \to G$ and $g : G \to F$ lifting the identity on $M$ . Their composites are homotopic to the identities. The cited comparison assertions use DC and supplied resolution data.

F2F3

2.1

After reduction modulo $m$ , all differentials vanish, so the homotopy identities become ${\overset{ˉ}{g}}_{i} {\overset{ˉ}{f}}_{i} = 1$ and ${\overset{ˉ}{f}}_{i} {\overset{ˉ}{g}}_{i} = 1$ . The finite ranks agree, also by the Betti rank theorem. Nakayama makes each $f_{i}$ surjective; equivalently its square matrix has determinant nonzero modulo $m$ , hence unit. Its adjugate gives an inverse over $R$ . These inverses form a chain map since $f$ does. Rank zero causes no difficulty: the unique map between zero modules is invertible.

F1F4step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 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

Remark 12.28, p.121 (standard reference, not scraped)