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

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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.

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auslander buchsbaum first syzygy depth

Statement

Assume the Axiom of Choice (The Axiom of Choice). In a minimal presentation $0 \to K \to F \to M \to 0$ of a nonzero finite module over a nonzero Noetherian local ring, let $n = pd M \geq 2$ be finite. If $depth K = depth R - (n - 1)$ , then $depth M = depth K - 1$ .

Facts & Assumptions

Given: The Axiom of Choice; The objects and hypotheses in the statement.

[F1]

A free module has projective dimension zero (Projective dimension of an object).

[F2]

The three Depth Lemma inequalities: Let $(R, m)$ be a Noetherian local ring and $0 \to A \to B \to C \to 0$ a short exact sequence of finite $R$ -modules. With $a = {depth}_{R} (A)$ , $b = {depth}_{R} (B)$ , and $c = {depth}_{R} (C)$ , $b \geq \min {a, c}, a \geq \min {b, c + 1}, c \geq \min {a - 1, b} .$ The last inequality is vacuous when $a = 0$ .

Proof

1.1F1givenalgebra

The presentation has $F \neq 0$ because $M \neq 0$ . If $K = 0$ , then $M ≅ F$ is free and has projective dimension zero by [F1], contrary to $n \geq 2$ . Thus $K \neq 0$ ; it is finite as a submodule of the finite module $F$ over the Noetherian ring $R$ . Write $a = depth K$ , $b = depth F = depth R$ , and $c = depth M$ . The equality for $F$ follows since an element is injective on a nonzero finite direct sum of $R$ exactly when it is injective on $R$ , also after successive quotients. The hypothesis gives $a = b - (n - 1) < b$ .

2.1F2step 1.1algebra∎

The depth inequality $a \geq \min (b, c + 1)$ forces $c + 1 \leq a$ , since $b > a$ . In particular $a \geq 1$ . The other inequality $c \geq \min (a - 1, b) = a - 1$ now applies and yields $c = a - 1$ . This uses only the stated depth hypothesis on $K$ , not the formula being proved by induction later.

Depends on

Used by

auslander buchsbaum formula Theorem

Dependency tree · two levels

9 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

Theorem 1.53, final induction step, p.25 (standard reference, not scraped)