Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-09-23 (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.

residue field splits off reduced maximal ideal

Statement

Assume the Axiom of Choice. Let (R,m,k) be nonzero Noetherian local and x∈m∖m2 a nonzerodivisor. Over S=R/(x) the sequence 0→(x)/(xm)→m/xm→m/(x)→0 splits, and (x)/(xm)≅k. Consequently finite pd⁡Rk implies finite pd⁡Sk.

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]

regular element reduction preserves minimal resolution: 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.

[F2]

auslander buchsbaum syzygy projective dimension: Let 0→K→F0→M→0 be the initial minimal presentation of a nonzero finite module over a nonzero Noetherian local ring. If 0<n=pd⁡M<∞, then K≠0 and pd⁡K=n−1.

[F3]

Projective dimension at most n iff higher Ext vanishes: Assume the Axiom of Dependent Choice. In an abelian category with enough projectives and enough injectives, fix supplied projective and injective resolution data on all objects. Let M be an object and n≥0. The following are equivalent: 1. pd⁡(M)≤n; 2. Ext⁡k(M,N)=0 for every object N and every k>n; 3. Ext⁡n+1(M,N)=0 for every object N.

[F4]

The canonical iterated free cover gives projective resolution data under AC (The iterated free-module resolution is canonical in ZF). Module categories are Grothendieck (Module categories are Grothendieck categories), and under AC their functorial injective embeddings can be iterated to give injective resolutions (Grothendieck abelian categories have functorial injective embeddings, Every module admits an injective resolution).

Proof

1.1givenalgebra

The sequence is the quotient sequence for xm⊆(x)⊆m; x kills every term. Multiplication by x identifies R/m with (x)/(xm) because cancellation is valid. Choose a k-linear functional on m/m2 taking the class of x to 1. Composing with m/xm→m/m2 gives an S-linear retraction onto k. Thus the sequence splits.

2.1givenF1F2F3F4step 1.1∎

If pd⁡Rk is finite, it is positive: projectivity of k would split R→k, giving a nontrivial idempotent unless m=0, impossible here. Its first minimal syzygy m therefore has finite projective dimension. The element x acts injectively on this ideal, so reduction gives finite pd⁡S(m/xm). The stated AC meets [F1]'s premise and implies the DC premise of [F3]. For S-modules, [F4] supplies the projective and injective resolution data required by [F3], and the corresponding enough-projectives and enough-injectives conditions. Apply [F3] to the finite projective dimension of m/xm. By step 1.1, it is k⊕m/(x); finite direct-sum additivity of Ext makes Ext⁡Sj(k,N) a summand of Ext⁡Sj(m/xm,N) for every j,N. It therefore vanishes above the same finite degree. The reverse implication of [F3] gives finite pd⁡Sk.

Depends on

Used by

Dependency tree · two levels

40 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