TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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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.

Support under localisation is restriction to primes disjoint from the denominator set

Statement

Let $R$ be a commutative ring, let $S \subseteq R$ be multiplicative, and let $M$ be a left $R$ -module. Under the prime-ideal correspondence

{p \subseteq R : p \cap S = \emptyset} ⟷ Spec (S^{- 1} R), p ⟼ S^{- 1} p,

the support of $S^{- 1} M$ corresponds exactly to

{p \in {Supp}_{R} (M) : p \cap S = \emptyset} .

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , and a left $R$ -module $M$ .

[L1]

A prime ideal lies in the support of a module exactly when the localisation at that prime is nonzero (Support of a module).

[L2]

Prime ideals of $S^{- 1} R$ correspond to prime ideals of $R$ disjoint from $S$ (Ideals of $S^{- 1} R$ correspond to $S$ -saturated ideals of $R$ , and prime ideals correspond to primes disjoint from $S$ ).

[L3]

Localising twice is the same as localising once at the multiplicative set generated by the two denominator sets (Localising twice is localising once at the multiplicative set generated by both denominator sets).

Proof

technique · direct

1.1

Let $q$ be a prime ideal of $S^{- 1} R$ , and let $p \subseteq R$ be its contraction. By [L2], $p \cap S = \emptyset$ and $q = S^{- 1} p$ .

2.1

The localisation $(S^{- 1} M)_{q}$ is obtained by localising $M$ first at $S$ and then at the complement of $q$ . By [L3], this is the same as localising $M$ once at the multiplicative subset $R ∖ p$ , so $(S^{- 1} M)_{q} ≅ M_{p}$ .

step 1.1L3

3.1

By [L1], step 2.1 gives $q \in {Supp}_{S^{- 1} R} (S^{- 1} M)$ if and only if $p \in {Supp}_{R} (M)$ . Together with step 1.1, this proves the stated description of support under localisation.

L1step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., Exercise 13.32 (standard reference, not scraped)
The Stacks Project, Section 10.40: Support (standard reference, not scraped)