LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-09-26 (gpt-6-sol)

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regular local residue field koszul resolution

Statement

Assume the Axiom of Choice. For a regular local ring $(R, m, k)$ of dimension $d$ , the Koszul complex on any regular system of parameters is a minimal free resolution of $k$ of length $d$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement, including the Axiom of Choice (The Axiom of Choice).

[F1]

regular local rings are domains and cohen macaulay: A regular local ring $R$ of dimension $d$ is a domain and Cohen–Macaulay. For every regular system $(x_{1}, \dots, x_{d})$ , the tuple is $R$ -regular and $R / (x_{1}, \dots, x_{c})$ is regular local of dimension $d - c$ for all $0 \leq c \leq d$ .

[F2]

Koszul Complex Resolves A Regular Quotient: If $M$ is finite free and $x$ is $M$ -regular, then $K (x; M)$ is a finite free resolution of $M / (x) M$ .

[F3]

minimal free resolution differentials land in maximal ideal: For an augmented degreewise finite free resolution over a nonzero Noetherian local ring $(R, m)$ , minimality means that every positive differential matrix has entries in $m$ . Equivalently no positive differential admits a unit pivot, or a nonzero two-term identity direct summand. A unit pivot can be cancelled without changing the resolved module.

Proof

1.1F1F2

The parameters form an $R$ -regular sequence and generate $m$ . Koszul acyclicity for a finite free coefficient module gives a free resolution of $R / m = k$ .

2.1F3step 1.1algebra∎

Every differential entry is a parameter up to sign and hence lies in $m$ , so the resolution is minimal. Its degree- $i$ module is $⋀^{i} R^{d}$ , zero for $i > d$ and rank one in degree $d$ . When $d = 0$ , it is just $R = k$ in degree zero, the Koszul complex on the empty tuple.

Depends on

Used by

Dependency tree · two levels

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Sources

Theorem 12.33 forward proof, p.123 (standard reference, not scraped)