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

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Localisation commutes with quotient rings: $S^{- 1} R / S^{- 1} I ≅ {\overset{ˉ}{S}}^{- 1} (R / I)$

Statement

Let $I$ be an ideal of a commutative ring $R$ , let $S \subseteq R$ be multiplicative, and let $\overset{ˉ}{S}$ be the image of $S$ in $R / I$ . There is a canonical isomorphism $(S^{- 1} R) / (S^{- 1} I) ≅ {\overset{ˉ}{S}}^{- 1} (R / I), r / s + S^{- 1} I ⟼ (r + I) / (s + I) .$ This includes the case $S \cap I \neq \emptyset$ , when both sides are the zero ring.

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $I$ , and a multiplicative subset $S$ with image $\overset{ˉ}{S}$ in $R / I$ .

[F1]

A map that sends a multiplicative subset to units factors uniquely through the corresponding localisation (Universal property of localisation: maps that invert $S$ factor uniquely through $S^{- 1} R$ ).

[F2]

A homomorphism killing an ideal factors uniquely through the quotient by that ideal (A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring).

[F3]

The extended ideal $S^{- 1} I$ consists of the fractions with a numerator in $I$ (Ideals of $S^{- 1} R$ correspond to $S$ -saturated ideals of $R$ , and prime ideals correspond to primes disjoint from $S$ ).

Proof

technique · construction of inverse maps

1.1

The map $R \to {\overset{ˉ}{S}}^{- 1} (R / I)$ given by $r \mapsto (r + I) / 1$ sends $S$ to units, so [F1] gives $Φ : S^{- 1} R \to {\overset{ˉ}{S}}^{- 1} (R / I)$ with $Φ (r / s) = (r + I) / (s + I)$ . It kills $S^{- 1} I$ by [F3], so [F2] gives a homomorphism $\overline{Φ} : (S^{- 1} R) / (S^{- 1} I) \to {\overset{ˉ}{S}}^{- 1} (R / I)$ .

F1F2F3construct

1.2

The map $R \to (S^{- 1} R) / (S^{- 1} I)$ given by $r \mapsto r / 1 + S^{- 1} I$ kills $I$ , so [F2] induces $R / I \to (S^{- 1} R) / (S^{- 1} I)$ . Every $s + I \in \overset{ˉ}{S}$ maps to the unit $s / 1 + S^{- 1} I$ , so [F1] extends this to $Ψ : {\overset{ˉ}{S}}^{- 1} (R / I) \to (S^{- 1} R) / (S^{- 1} I)$ .

F1F2F3

2.1

The composites $Ψ \overline{Φ}$ and $\overline{Φ} Ψ$ fix, respectively, every class $r / s + S^{- 1} I$ and every fraction $(r + I) / (s + I)$ by the formulas in steps 1.1 and 1.2. Hence the maps are inverse isomorphisms. If $s \in S \cap I$ , then $1 = s / s \in S^{- 1} I$ , so the left side is zero; also $s + I = 0$ lies in $\overset{ˉ}{S}$ , so the right localisation is zero.

step 1.1step 1.2F3algebradischarge-construct∎

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Dependency tree · two levels

16 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, Proposition 10.9.14 (standard reference, not scraped)

Localisation commutes with quotient rings: S−1R/S−1I≅Sˉ−1(R/I)S−1R/S−1I≅Sˉ−1(R/I)