TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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

Statement

Let $R$ be a commutative ring and $S \subseteq R$ multiplicative. For an ideal $I \subseteq R$ , write $S^{- 1} I = {r / s : r \in I, s \in S}, I^{s a t} = {r \in R : s r \in I for some s \in S} .$ An ideal is $S$ -saturated when $I = I^{s a t}$ . Extension $I \mapsto S^{- 1} I$ and contraction $J \mapsto λ_{S}^{- 1} (J)$ give inverse inclusion-preserving bijections between $S$ -saturated ideals of $R$ and ideals of $S^{- 1} R$ . Moreover, they restrict to inverse bijections ${p \in Spec R : p \cap S = \emptyset} ⟷ Spec (S^{- 1} R), p ⟼ S^{- 1} p .$

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S$ , and the localisation map $λ_{S}$ .

[F1]

Localisation has the usual fraction arithmetic, equality is detected by a witness in $S$ , and every $s / 1$ is a unit (The localisation relation is an equivalence relation and fraction arithmetic is well defined, Equality, vanishing, and the kernel of the localisation map).

[F2]

Ideals are additive subgroups absorbing multiplication; a prime ideal is proper and contains one factor whenever it contains a product (Left, right and two-sided ideals, Prime ideals and maximal ideals in a commutative ring).

[L1]

Each ideal of $S^{- 1} R$ is $S^{- 1} I$ for its inverse-image ideal $I$ in $R$ (The Stacks Project, Lemma 10.9.16).

[L2]

The map $R \to S^{- 1} R$ induces a bijection from the prime ideals of $S^{- 1} R$ to the prime ideals of $R$ disjoint from $S$ , with inverse $p \mapsto S^{- 1} p$ (The Stacks Project, Lemma 10.17.5).

Proof

technique · direct extension and contraction

1.1

The displayed $S^{- 1} I$ is an ideal: common-denominator addition preserves numerator membership in $I$ , and multiplication by an arbitrary fraction does likewise. It is also the ideal generated by the elements $i / 1$ with $i \in I$ , since $i / s = (i / 1) (1 / s)$ .

F1F2

1.2

For completeness, let $p$ be prime and disjoint from $S$ . It is saturated, since $s r \in p$ with $s \notin p$ forces $r \in p$ . Its extension is proper: $1 = i / s$ with $i \in p$ would give $u (s - i) = 0$ , hence $u s = u i \in p$ , contradicting $u, s \in S$ .

F1F2

2.1

An element $r$ lies in the contraction of $S^{- 1} I$ exactly when $r / 1 = i / s$ for some $i \in I$ . Equality gives $u (r s - i) = 0$ , hence $(u s) r = u i \in I$ , so $r \in I^{s a t}$ . Conversely, if $t r \in I$ for $t \in S$ , then $r / 1 = t r / t \in S^{- 1} I$ . Thus the contraction is exactly $I^{s a t}$ .

F1step 1.1

3.1

If $J \subseteq S^{- 1} R$ is an ideal and $I = λ_{S}^{- 1} (J)$ , then $r / s \in J$ implies $r / 1 = (s / 1) (r / s) \in J$ , while $r / 1 \in J$ implies $r / s = (1 / s) (r / 1) \in J$ . Hence $J = S^{- 1} I$ , as in [L1]. Step 2.1 also makes $I$ saturated.

F1F2L1

3.2

If $(a / s) (b / t) \in S^{- 1} p$ , contraction and saturation give $u a b \in p$ for some $u \in S$ . Since $u \notin p$ , primality gives $a \in p$ or $b \in p$ , so one factor lies in $S^{- 1} p$ . Thus the extension is prime.

F1F2step 2.1step 1.2

4.1

Steps 2.1 and 3.1 prove that extension and contraction are inverse on the stated ideals. Both operations visibly preserve inclusion.

step 2.1step 3.1

5.1

Conversely, the contraction $q^{c}$ of a prime $q \subseteq S^{- 1} R$ is proper because $1 / 1 \notin q$ . If $a b \in q^{c}$ , then $(a / 1) (b / 1) \in q$ , so primality gives $a \in q^{c}$ or $b \in q^{c}$ . It also misses $S$ , because each $s / 1$ is a unit and no proper ideal contains a unit. Together with steps 3.2 and 4.1, this proves the prime bijection stated in [L2].

F1F2L2step 3.2step 4.1∎

Depends on

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11 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, Lemmas 10.9.16 and 10.17.5 (standard reference, not scraped)
The Stacks Project, Lemma 10.17.5 (standard reference, not scraped)

Ideals of S−1RS−1R correspond to SS-saturated ideals of RR, and prime ideals correspond to primes disjoint from SS