LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30

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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.
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AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A finite module with free fibre and flat base is free over a Noetherian target

Statement

Assume the Axiom of Choice. Let $R \to S$ be a local homomorphism of Noetherian local rings, with maximal ideal $m \subset R$ , and let $M$ be a nonzero finite $S$ -module. If $M$ is flat over $R$ and $M / m M$ is free over $S / m S$ , then $M$ is a finite free $S$ -module and $S$ is flat over $R$ .

Facts & Assumptions

Given: The local Noetherian map, nonzero finite module, base-flatness, and free fibre.

[F1]

A map from a finite $S$ -module to an $R$ -flat $S$ -module whose reduction modulo $m$ is injective is itself injective (Fibrewise injectivity lifts and leaves a flat cokernel over a Noetherian target).

[F2]

If $C$ is finite over local $S$ and $C / m S C = 0$ , then $C = 0$ by Nakayama, since $m S$ lies in the maximal ideal of $S$ (Assuming the Axiom of Choice, Nakayama's lemma).

[F3]

Direct summands of flat modules are flat, by the ideal-tensor criterion (Flatness is equivalent to preserving injections and to the ideal and finitely generated ideal tests).

Proof

technique · lift a fibre basis, use fibrewise injection and Nakayama, and recover flatness of the middle ring as a summand

1.1F1

Since $M$ is finite over $S$ , its free $S / m S$ -fibre has finite rank $r$ . Choose a basis ${\overline{x}}_{1}, \dots, {\overline{x}}_{r}$ and lifts $x_{1}, \dots, x_{r} \in M$ . They give an $S$ -linear map $u : S^{r} \to M$ whose reduction modulo $m$ is an isomorphism. By [F1], $u$ is injective.

2.1F2step 1.1

Its cokernel $C$ is a finite $S$ -module with $C / m C = 0$ because the fibre map is surjective. By [F2], $C = 0$ , hence $M ≅ S^{r}$ . Since $M \neq 0$ , the rank $r$ is positive.

3.1F1F3step 2.1∎

The $R$ -module $M ≅ S^{r}$ is flat by hypothesis; as $r \geq 1$ , $S$ is a direct summand of it. By [F3], $S$ is flat over $R$ . AC covers the basis selection and the cited fibre-injection boundary.

Depends on

Used by

The flat locus of a finitely presented algebra is open Lemma

Dependency tree · two levels

18 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, Algebra, Lemma 10.99.4 (tag 00MH), free fibre and flatness (standard reference, not scraped)