TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06‡ 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.

Regular Sequences Give Acyclic Koszul Complexes

Statement

Every finite $M$ -regular sequence is $M$ -Koszul-regular: $H_{i} (K (x; M)) = 0$ for $i > 0$ .

Facts & Assumptions

Given: The ring, finite sequence, and module stated in the claim. The declared prerequisites used here are Regular Sequence Koszul Acyclicity Induction, Regular One Element Koszul Acyclicity, Empty Koszul Complex Is The Coefficient Module, and Regular Sequence On A Module.

Proof

technique · direct

1.1

For the empty sequence the Koszul complex is $M$ in degree zero, so the conclusion is immediate. For length one it is the one-element calculation.

givenalgebra

2.1

Suppose the result holds for an initial segment $x$ . Regularity says that the next element $y$ acts injectively on $M / (x) M$ . The induction lemma applied to the already acyclic $K (x; M)$ therefore makes $K (x, y; M)$ acyclic in positive degrees. Induction on the length proves the claim.

step 1.1algebra∎

Depends on

Used by

Dependency tree · two levels

10 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

The Stacks Project, Koszul complexes and regular sequences (standard reference, not scraped)