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.

Local Koszul Acyclicity Inductive Converse

Statement

Assume the Axiom of Choice. Let $(R, m)$ be Noetherian local, $M$ finite, and $x = (x_{1}, \dots, x_{n}) \subseteq m$ be nonempty, with $M / (x) M \neq 0$ . If $K (x; M)$ has no positive homology, then the shorter complex is acyclic and the last element is injective on its preceding quotient.

Facts & Assumptions

Given: The Axiom of Choice and the rings, finite sequences, modules, and local hypotheses stated in the claim. The declared prerequisites used here are Koszul Mapping Cone Homology Exact Sequence, Regular Sequence On A Module, A local ring is a nonzero commutative ring with a unique maximal ideal, Left and right Noetherian rings, Noetherian modules: every submodule is finitely generated, Generated submodule, cyclic and finitely generated modules, module basis and free module, Assuming the Axiom of Choice, Nakayama's lemma, and The Axiom of Choice.

Proof

technique · direct

1.1givenalgebra

Put $C = K (x_{1}, \dots, x_{n - 1}; M)$ and $y = x_{n}$ . For every $j > 0$ , the cone exact sequence and $H_{j + 1} (K (x; M)) = 0$ show that multiplication by $y$ on $H_{j} (C)$ is surjective. Each $H_{j} (C)$ is finite because $C$ is a bounded complex of finite modules over the Noetherian ring $R$ .

2.1step 1.1algebra∎

Since $y \in m$ , apply the AC-qualified Nakayama lemma to the surjections in step 1.1; this is the precise use of Choice and gives $H_{j} (C) = 0$ for every $j > 0$ . The segment of the same exact sequence ending in $H_{1} (K (x; M)) = 0$ then shows that multiplication by $y$ on $H_{0} (C) = M / (x_{1}, \dots, x_{n - 1}) M$ is injective. These are the two asserted conclusions.

Depends on

Dependency tree · 2 levels

Sources

The Stacks Project, Koszul complexes and regular sequences