TheoremStatement: AI-adaptedProof: AI-adaptedverified 2026-09-24 (gpt-6-sol)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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A finite flat module over a local ring is free

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $(R, m)$ be a Noetherian local ring and let $M$ be a finite flat $R$ -module. Then $M$ is free.

Facts & Assumptions

Given: The Axiom of Choice, a Noetherian local ring $(R, m)$ and a finite flat $R$ -module $M$ .

[L1]

Under the assumed Axiom of Choice, if lifts of residue classes generate $M / m M$ , they generate $M$ ; likewise a finite module $K$ with $K / m K = 0$ vanishes by the same lifting corollary applied to the empty list (Assuming the Axiom of Choice, generators modulo an ideal in the Jacobson radical lift to generators).

[L2]

Over a Noetherian ring, finite modules are finitely presented, so the kernel of a map from a finite free module to a finite module is finite (Over a Noetherian ring a module is Noetherian exactly when it is finitely generated, exactly when it is finitely presented).

[L3]

A short exact sequence with flat quotient remains short exact after tensoring with any module (A short exact sequence with flat quotient remains short exact after tensoring).

Proof

technique · direct

1.1L1givenchoose

Choose elements $x_{1}, \dots, x_{r} \in M$ whose residue classes form a basis of the vector space $M / m M$ . By [L1], they generate $M$ . Thus there is a surjection $φ : R^{r} ↠ M$ sending the $i$ th standard basis vector to $x_{i}$ .

1.2L2L3algebra

Let $K = \ker φ$ . Because $R$ is Noetherian and $R^{r}$ is finite, [L2] makes $K$ finitely generated. Tensoring $0 \to K \to R^{r} \to M \to 0$ with the residue field $k = R / m$ remains exact by [L3], so $0 \to K / m K \to k^{r} \to M / m M \to 0.$ The last map is an isomorphism by the choice of the $x_{i}$ , hence $K / m K = 0$ .

2.1L1step 1.2algebra

Applying [L1] to the finite module $K$ and the empty list of generators gives $K = 0$ . Thus $φ$ is an isomorphism and $M ≅ R^{r}$ is free.

A finite flat module over a local ring is free

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources