PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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.

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

1.1givenL3

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

2.1step 1.1L2

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

3.1step 2.1L1

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.

4.1step 3.1L2∎

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

Depends on

Used by

Dependency tree · two levels

15 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

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