LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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 global dimension equals residue field projective dimension

Statement

For a nonzero Noetherian local ring $(R, m, k)$ , $gldim R = {pd}_{R} k$ , allowing infinity. If this common value is $n < \infty$ , every $R$ -module has projective dimension at most $n$ .

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]

global dimension is detected on cyclic modules: For a unital ring $R$ , its left global dimension equals $\sup_{I} {pd}_{R} (R / I)$ over all left ideals $I$ , and equals the supremum of the injective dimensions of all left modules. The equalities allow infinity; in the commutative Noetherian case the cyclic modules are finite.

[F2]

projective dimension from last nonzero betti number: For a nonzero finite module $M$ over a nonzero Noetherian local ring, ${pd}_{R} M = \sup {i \geq 0 : β_{i}^{R} (M) \neq 0}$ , allowing infinity. For each integer $q \geq 0$ , ${pd}_{R} M \leq q$ if and only if ${Tor}_{q + 1}^{R} (k, M) = 0$ .

[F3]

Tor is symmetric over a commutative ring: If $R$ is commutative and $M, N$ are $R$ -modules, then ${Tor}_{i}^{R} (M, N) ≅ {Tor}_{i}^{R} (N, M)$ naturally.

Proof

1.1

The lower bound is immediate because $k$ is an $R$ -module. If its projective dimension is infinite this already proves the equality. Otherwise let $n = pd k$ . Compute Tor using a length- $n$ resolution of $k$ and use symmetry to get ${Tor}_{n + 1}^{R} (k, M) = 0$ for every module $M$ .

F3given

2.1

For each nonzero finite $M$ , the minimal-resolution criterion gives $pd M \leq n$ ; the zero module is projective as well. In particular every cyclic module has that bound. Cyclic detection extends it to all modules and hence bounds global dimension by $n$ . This includes $n = 0$ .

F2F1step 1.1∎

Depends on

Used by

auslander buchsbaum serre regularity criterion Theorem

Dependency tree · two levels

21 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

Corollary 12.30, p.121 (standard reference, not scraped)