LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07‡ rests on unproved material (inherited)

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.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

auslander buchsbaum projective dimension one

Statement

If $M$ is a nonzero finite module of projective dimension one over a nonzero Noetherian local ring $R$ , then $depth R \geq 1$ and $depth M = depth R - 1$ .

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]

minimal free matrix induces zero on residue ext: For a nonzero commutative Noetherian local ring $(R, m, k)$ , let $α : R^{s} \to R^{t}$ be a map between finite free modules all of whose matrix entries lie in $m$ . Then ${Ext}_{R}^{i} (k, α) = 0$ for every $i \geq 0$ .

[F2]

projective dimension from last nonzero betti number: For a nonzero finite module $M$ over a nonzero Noetherian local ring, ${pd}_{R} M = \sup {i \geq 0 : β_{i}^{R} (M) \neq 0}$ , allowing infinity. For each integer $q \geq 0$ , ${pd}_{R} M \leq q$ if and only if ${Tor}_{q + 1}^{R} (k, M) = 0$ .

[F3]

Depth as the first nonzero Ext degree: Let $R$ be Noetherian, let $M$ be finite, and let $I$ lie in the Jacobson radical. Then ${depth}_{I} (M) = \inf {i \geq 0 : {Ext}_{R}^{i} (R / I, M) \neq 0},$ where the infimum of the empty set is $\infty$ .

[F4]

The long exact Ext sequence in the second variable: Assume the Axiom of Dependent Choice. Let $A$ be abelian with enough projectives and enough injectives, and fix supplied projective and injective resolution data on all its objects. For $0 \to N^{'} \to N \to N^{''} \to 0$ and every $M$ , there is a natural exact sequence $0 \to Hom (M, N^{'}) \to Hom (M, N) \to Hom (M, N^{''}) \overset{δ^{0}}{\to} {Ext}^{1} (M, N^{'}) \to {Ext}^{1} (M, N) \to \dots,$ where $δ^{q} : {Ext}^{q} (M, N^{''}) \to {Ext}^{q + 1} (M, N^{'})$ ; it is natural in the short exact sequence and contravariantly natural in $M$ .

[F5]

finite local modules admit minimal free resolutions: Every finite module over a nonzero Noetherian local ring has a degreewise finite minimal free resolution.

Proof

1.1

Choose the minimal resolution supplied by [F5]. Since ${pd}_{R} M = 1$ , [F2] says that its last nonzero term is $F_{1} ≅ R^{s}$ with $s > 0$ and $F_{i} = 0$ for $i \geq 2$ . Thus it gives a minimal exact sequence $0 \to R^{s} \overset{α}{\to} R^{t} \to M \to 0$ . Write $r = depth R$ . The map on every ${Ext}_{R}^{i} (k, -)$ induced by $α$ is zero. If $r = 0$ , the injection $Hom (k, R^{s}) \to Hom (k, R^{t})$ would be zero with nonzero source, impossible. Hence $r \geq 1$ .

F1F2F3F4F5

2.1

For $i < r - 1$ , the adjacent Ext terms for the free modules vanish, so ${Ext}_{R}^{i} (k, M) = 0$ . At $i = r - 1$ , exactness and the zero map in degree $r$ identify ${Ext}_{R}^{r - 1} (k, M)$ with ${Ext}_{R}^{r} (k, R^{s}) \neq 0$ . Thus the first nonzero Ext degree is $r - 1$ , proving the depth formula, also for $r = 1$ .

F4F3F1step 1.1∎

Depends on

Used by

auslander buchsbaum formula Theorem

Dependency tree · two levels

21 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

Theorem 1.53, pp.24–25 (standard reference, not scraped)