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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-09-23 (gpt-6-sol)
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regular element reduction preserves minimal resolution

Statement

Assume the Axiom of Choice. Let (R,m,k) be nonzero Noetherian local, let M be a nonzero finite module, and let x∈m be a nonzerodivisor on both R and M. Reducing a minimal free resolution of M modulo x gives a minimal free resolution of M/xM over S=R/(x). Moreover pd⁡S(M/xM)=pd⁡RM, including infinity. For M=0 the zero-complex assertion also holds, with both projective dimensions zero.

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]

finite local modules admit minimal free resolutions: Every finite module M over a nonzero Noetherian local ring (R,m,k) has an augmented resolution ⋯→F1→F0→M→0 by finite-rank free modules, with di(Fi)⊆mFi−1 for i>0. Such a resolution is called minimal; it need not be bounded. This extends the bounded terminology without changing it.

[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]

The balanced Tor bifunctor: For a right R-module N, a left R-module M, and i≥0, define Tor⁡iR(N,M) to be either Hi(N⊗RP∙) for a projective resolution of M or Hi(Q∙⊗RM) for a projective resolution of N, identified by the preceding natural balance isomorphism. On maps it uses the homology maps induced by comparison maps; coherence makes this a well-defined covariant bifunctor.

[F4]

Tor is symmetric over a commutative ring: If R is commutative and M,N are R-modules, then Tor⁡iR(M,N)≅Tor⁡iR(N,M) naturally.

[F5]

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

Proof

1.1F3F4

The complex 0→R→xR→S→0 resolves S. Tensoring it with M has no positive homology since multiplication by x is injective on M. Balance and symmetry of Tor show that reducing any free resolution of M modulo x has no positive homology and degree-zero homology M/xM.

2.1givenF1F2F5step 1.1∎

AC licenses the existence in [F1], the use of Nakayama [F5], and the projective-dimension criterion [F2]. A minimal degreewise finite resolution exists, and its matrices reduce to entries in m/(x). Its finite ranks do not change on reduction to the nonzero local ring S. Nakayama gives M/xM≠0, so the last-nonzero-Betti criterion identifies both projective dimensions with the same last nonzero rank, or infinity if ranks persist arbitrarily far. For M=0 choose the zero complex on both sides.

Depends on

Used by

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Sources