LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-09-23 (gpt-6-sol)

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finite residue field projective dimension forces depth equals dimension

Statement

Assume the Axiom of Choice. If the residue field of a nonzero Noetherian local ring $R$ has finite projective dimension, then $R$ is regular and $depth R = \dim R = edim R$ .

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]

auslander buchsbaum formula: For a nonzero finite module $M$ of finite projective dimension over a nonzero Noetherian local ring $R$ , ${pd}_{R} M + {depth}_{R} M = depth R$ . Consequently such an $M$ with $depth M = depth R$ is free.

[F2]

positive depth ring has regular minimal generator: If a nonzero Noetherian local ring $(R, m, k)$ has positive depth, then some $x \in m ∖ m^{2}$ is a nonzerodivisor. The residue field need not be infinite.

[F3]

residue field splits off reduced maximal ideal: Let $(R, m, k)$ be nonzero Noetherian local and $x \in m ∖ m^{2}$ a nonzerodivisor. Over $S = R / (x)$ the sequence $0 \to (x) / (x m) \to m / x m \to m / (x) \to 0$ splits, and $(x) / (x m) ≅ k$ . Consequently finite ${pd}_{R} k$ implies finite ${pd}_{S} k$ .

[F4]

quotient and lifting regularity across a regular element: Let $(R, m)$ be nonzero Noetherian local. If $x \in m$ is a nonzerodivisor and $R / (x)$ is regular, then $R$ is regular and $x \notin m^{2}$ . For every nonzerodivisor $x \in m$ , $\dim (R / (x)) = \dim R - 1$ . If $R$ is regular and $0 \neq x \in m$ , then $R / (x)$ is regular if and only if $x \notin m^{2}$ .

[F5]

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 $0 \leq c \leq d$ .

[F6]

Depth drops by one after quotienting by a regular element: Let $R$ be Noetherian, let $M$ be finite, let $I$ lie in the Jacobson radical, and let $x \in I$ be $M$ -regular. Then ${depth}_{I} (M / x M) = {depth}_{I} (M) - 1.$

Proof

1.1F1

Induct on the finite integer $r = depth R$ . The residue field has depth zero, since every member of $m$ kills it. If $r = 0$ , Auslander–Buchsbaum gives $pd k = 0$ and its freeness consequence makes $k$ nonzero free. A nonzero free module has zero annihilator, so $m = 0$ and $R = k$ is a field.

2.1givenF2F3F6step 1.1

For $r > 0$ , the stated AC premise permits [F2] and [F3]; choose a nonzerodivisor $x \in m ∖ m^{2}$ . The splitting lemma makes ${pd}_{R / (x)} k$ finite, and the regular-element depth formula gives depth $r - 1$ for the quotient. Depth of this annihilated module over $R$ equals its depth over $R / (x)$ : lift sequences from the quotient or project sequences from $R$ ; multiplication and all successive quotients are identical.

3.1F4F5step 2.1∎

The inductive assertion makes $R / (x)$ regular. Lifting across the nonzerodivisor makes $R$ regular; its regular parameters make it Cohen–Macaulay, so depth equals dimension, and regularity equates that dimension with embedding dimension. This completes the induction.

Depends on

Dependency tree · 2 levels

Sources

Theorem 12.33, p.123