PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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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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Extension of scalars carries flat modules to flat modules

Statement

Let $R \to S$ be a homomorphism of commutative rings. If $M$ is a flat $R$ -module, then its extension of scalars $S \otimes_{R} M$ is a flat $S$ -module.

Facts & Assumptions

Given: A ring map $R \to S$ and a flat $R$ -module $M$ .

[L1]

For every right $S$ -module $N$ , there is a natural isomorphism $N \otimes_{S} (S \otimes_{R} M) ≅ N \otimes_{R} M$ after restriction of scalars (Change of rings: $N \otimes_{R} M ≅ N \otimes_{S} (S \otimes_{R} M)$ ).

[L2]

A module is flat exactly when tensoring with it preserves injections (Flatness is equivalent to preserving injections and to the ideal and finitely generated ideal tests).

[L3]

Restriction of scalars leaves the underlying abelian group and function of a module map unchanged (Restriction of scalars and extension of scalars $S \otimes_{R} M$ along a ring homomorphism $R \to S$ ).

Proof

technique · direct

1.1

Let $u : N \to N^{'}$ be an injection of right $S$ -modules. By [L3], it is still injective after restriction to $R$ -modules.

givenL3

2.1

Since $M$ is flat, [L2] makes $u \otimes_{R} 1_{M} : N \otimes_{R} M \to N^{'} \otimes_{R} M$ injective.

step 1.1L2

3.1

Under the natural isomorphisms [L1], the map in step 2.1 is precisely $u \otimes_{S} 1_{S \otimes_{R} M}$ . Hence tensoring over $S$ with $S \otimes_{R} M$ preserves every injection.

step 2.1L1

4.1

By [L2] over the ring $S$ , the extended module $S \otimes_{R} M$ is flat.

step 3.1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 35 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Stacks Project, Lemma 10.39.7 (standard reference, not scraped)