ExampleConstruction: Literature-sourcedVerification: Literature-sourcedverified 2026-09-24 (gpt-6-sol)

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.

A residue-field basis lifts to a basis of a finite flat module over a local ring

Example

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. If ${\overset{ˉ}{x}}_{1}, \dots, {\overset{ˉ}{x}}_{r}$ is a basis of the residue vector space $M / m M$ , then any lifts $x_{1}, \dots, x_{r} \in M$ form an $R$ -basis of $M$ .

Facts & Assumptions

Given: The Axiom of Choice, a Noetherian local ring $(R, m)$ , a finite flat $R$ -module $M$ , a basis ${\overset{ˉ}{x}}_{1}, \dots, {\overset{ˉ}{x}}_{r}$ of $M / m M$ , and lifts $x_{1}, \dots, x_{r} \in M$ .

[L1]

Under the assumed Axiom of Choice, a finite flat module over a Noetherian local ring is free (A finite flat module over a local ring is free).

[L2]

Under the same assumption, every element outside the unique maximal ideal of a local ring is a unit (Assuming the Axiom of Choice, a nonzero commutative ring is local exactly when its nonunits form an ideal, exactly when one of $x$ and $1 - x$ is a unit for every $x$ ).

Verification

technique · direct

1.1L1givenchoose

By [L1], the module $M$ is free of rank $r$ , because reducing a free basis modulo $m$ gives a basis of $M / m M$ . Fix a free basis $e_{1}, \dots, e_{r}$ of $M$ .

2.1L2step 1.1algebra

Let $B \in M_{r} (R)$ be the coordinate matrix whose columns are the $x_{i}$ in the basis $e_{j}$ . Reducing modulo $m$ gives an invertible matrix because the ${\overset{ˉ}{x}}_{i}$ form a residue basis, so $\det B \notin m$ . By [L2], $\det B$ is a unit. The adjugate identity $B adj (B) = (\det B) I_{r}$ therefore makes $B$ invertible over $R$ . Thus its columns $x_{1}, \dots, x_{r}$ are a basis. For $r = 0$ , both bases are empty and the conclusion is immediate.

3.1step 2.1∎

So residue-field bases lift to actual bases in the finite flat local case under the stated Choice assumption.

Depends on

Used by

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Dependency tree · two levels

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Sources

Stacks Project, Section 10.78: Finite projective modules (standard reference, not scraped)