LemmaStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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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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Primes of a localization avoid the denominator set

Statement

Let $R$ be a commutative ring, let $S \subseteq R$ be a multiplicative subset, and let $λ : R \to S^{- 1} R$ be the localization map. Contraction sends each prime ideal of $S^{- 1} R$ to a prime ideal of $R$ disjoint from $S$ , and extension sends each prime ideal of $R$ disjoint from $S$ back to a prime ideal of $S^{- 1} R$ . These two operations are inverse and preserve strict inclusion.

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , and the localization map $λ : R \to S^{- 1} R$ .

[L1]

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

Proof

technique · direct

1.1

The prime-ideal part of [L1] says exactly that if $q \in Spec (S^{- 1} R)$ , then $λ^{- 1} (q)$ is a prime ideal of $R$ disjoint from $S$ , and if $p \in Spec (R)$ with $p \cap S = \emptyset$ , then $S^{- 1} p$ is a prime ideal of $S^{- 1} R$ .

1.2

The same statement [L1] asserts that these assignments are inverse inclusion-preserving bijections. In particular they preserve strict inclusion.

2.1

Therefore primes of the localization are exactly the primes of $R$ that avoid the denominator set.

step 1.1step 1.2∎

Depends on

Used by

Primes of a localization at a prime Corollary
Primes of a principal localization Corollary
Prime ideals of a localization are exactly the primes disjoint from the denominator set Theorem

Dependency tree · two levels

12 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

The Stacks Project, Section 10.17: The spectrum of a ring (standard reference, not scraped)
A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., §13 The Spectrum of a Ring (standard reference, not scraped)