TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-31

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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.
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Finitely generated submodules of projective Dedekind modules are projective

Statement

Assume the Axiom of Choice. Let $R$ be a Dedekind domain, let $P$ be a projective $R$ -module, and let $N \subseteq P$ be a finitely generated submodule. Then $N$ is projective. Consequently every finitely generated torsion-free module over a Dedekind domain is projective.

Facts & Assumptions

Given: A Dedekind domain $R$ , a projective $R$ -module $P$ , and a finitely generated submodule $N \subseteq P$ .

[L1]

Every projective module is flat (Every projective module over a commutative ring is flat).

[L2]

Every finite torsion-free module over a Dedekind domain is a direct sum of invertible ideal summands, hence projective (Finite torsion-free Dedekind modules split into invertible ideal summands).

Proof

technique · direct

1.1

By [L1], the module $P$ is flat. Over a domain, every flat module is torsion-free, so the submodule $N \subseteq P$ is torsion-free. Because $N$ is finitely generated by hypothesis, [L2] applies and makes $N$ projective.

L1L2givenalgebra

2.1

The final sentence is exactly the projectivity conclusion already recorded in [L2].

L2∎

Remarks

The stronger arbitrary-rank hereditary statement and the flatness upgrade for all torsion-free modules are left outside this draft. The written proof here establishes exactly the finitely generated submodule form.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

10 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

J. P. May, Notes on Dedekind Rings (standard reference, not scraped)
The Stacks Project, Lemma 15.22.11 (standard reference, not scraped)