Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-09-26 (gpt-6-sol)
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auslander buchsbaum projective dimension one

Statement

Assume the Axiom of Choice. If M is a nonzero finite module of projective dimension one over a nonzero Noetherian local ring R, then depth⁡R≥1 and depth⁡M=depth⁡R−1.

Facts & Assumptions

Given: The objects and hypotheses in the statement, including the Axiom of Choice (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:Rs→Rt be a map between finite free modules all of whose matrix entries lie in m. Then Ext⁡Ri(k,α)=0 for every i≥0.

[F2]

projective dimension from last nonzero betti number: For a nonzero finite module M over a nonzero Noetherian local ring, pd⁡RM=sup⁡{i≥0:βiR(M)≠0}, allowing infinity. For each integer q≥0, pd⁡RM≤q if and only if Tor⁡q+1R(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≥0:Ext⁡Ri(R/I,M)≠0}, where the infimum of the empty set is ∞.

[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→N′→N→N′′→0 and every M, there is a natural exact sequence 0→Hom⁡(M,N′)→Hom⁡(M,N)→Hom⁡(M,N′′)→δ0Ext⁡1(M,N′)→Ext⁡1(M,N)→⋯ , where δq:Ext⁡q(M,N′′)→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

1.1F1F2F3F4F5

Choose the minimal resolution supplied by [F5]. Since pd⁡RM=1, [F2] says that its last nonzero term is F1≅Rs with s>0 and Fi=0 for i≥2. Thus it gives a minimal exact sequence 0→Rs→αRt→M→0. Write r=depth⁡R. The map on every Ext⁡Ri(k,−) induced by α is zero. If r=0, the injection Hom⁡(k,Rs)→Hom⁡(k,Rt) would be zero with nonzero source, impossible. Hence r≥1.

2.1F4F3F1step 1.1∎

For i<r−1, the adjacent Ext terms for the free modules vanish, so Ext⁡Ri(k,M)=0. At i=r−1, exactness and the zero map in degree r identify Ext⁡Rr−1(k,M) with Ext⁡Rr(k,Rs)≠0. Thus the first nonzero Ext degree is r−1, proving the depth formula, also for r=1.

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