TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01

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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.

Every localization is flat, and localizing a flat module preserves flatness

Statement

Let $R$ be a commutative ring and let $S \subseteq R$ be a multiplicative set.

The localization $S^{- 1} R$ is a flat $R$ -algebra.
If $M$ is an $S^{- 1} R$ -module, then $M$ is flat over $R$ if and only if it is flat over $S^{- 1} R$ .
In particular, if $N$ is a flat $R$ -module, then $S^{- 1} N$ is flat over $S^{- 1} R$ .

Facts & Assumptions

Given: A commutative ring $R$ and a multiplicative subset $S \subseteq R$ .

[L1]

Flatness means exactness of tensoring (Flat and faithfully flat modules and ring homomorphisms).

[L2]

Localization of modules preserves exact sequences (Localisation of modules is exact).

[L3]

Localization of modules is tensor product: $S^{- 1} N ≅ S^{- 1} R \otimes_{R} N$ (Localisation of modules is extension of scalars).

[L4]

Localization at a prime ideal is the special case $S = R ∖ p$ (Localisation at a prime ideal: $R_{p} = (R ∖ p)^{- 1} R$ ).

Proof

technique · direct

1.1

For an exact sequence $A \to B \to C$ of $R$ -modules, [L2] gives an exact sequence $S^{- 1} A \to S^{- 1} B \to S^{- 1} C .$ By [L3] this is $(S^{- 1} R \otimes_{R} A) \to (S^{- 1} R \otimes_{R} B) \to (S^{- 1} R \otimes_{R} C),$ so $S^{- 1} R$ is flat over $R$ by [L1].

L1L2L3given

2.1

Let $M$ be an $S^{- 1} R$ -module. If $M$ is flat over $S^{- 1} R$ , then for any exact sequence over $R$ we first tensor with $S^{- 1} R$ as in step 1.1 and then tensor over $S^{- 1} R$ with $M$ ; the result is exact, so $M$ is flat over $R$ . Conversely, if $M$ is flat over $R$ , then every exact sequence of $S^{- 1} R$ -modules is in particular exact over $R$ , and tensoring it with $M$ over $R$ agrees with tensoring over $S^{- 1} R$ because the scalars already act through the localization. Hence $M$ is flat over $S^{- 1} R$ .

L1L3step 1.1algebra

3.1

If $N$ is flat over $R$ , then $S^{- 1} N ≅ S^{- 1} R \otimes_{R} N$ by [L3]. Applying step 2.1 to the $S^{- 1} R$ -module $S^{- 1} N$ proves it is flat over $S^{- 1} R$ .

L3step 2.1

3.2

Step 2.1 applies in particular to localization at a prime ideal by [L4].

4.1

Therefore all three claims hold.

algebra∎

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 10.39.18 (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, §11 (standard reference, not scraped)