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The spectrum of a localisation is the subspace of primes disjoint from the denominator set

Statement

Let $R$ be a commutative ring, let $S \subseteq R$ be multiplicative, and let $λ : R \to S^{- 1} R$ be the localisation map. Then contraction along $λ$ is a homeomorphism from $Spec (S^{- 1} R)$ onto the subspace $X : = {p \in Spec (R) : p \cap S = \emptyset} .$

Facts & Assumptions

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

[L1]

Contraction along $λ$ is an inclusion-preserving bijection from $Spec (S^{- 1} R)$ onto $X$ , with inverse $p \mapsto S^{- 1} p$ (Prime ideals of a localization are exactly the primes disjoint from the denominator set).

[L2]

In a Zariski spectrum, the closed sets are precisely the vanishing sets (The vanishing sets define the Zariski topology on the prime spectrum).

[A1]

If $K ⊴ S^{- 1} R$ and $I = λ^{- 1} (K)$ , then $K = S^{- 1} I$ .

[A2]

A subset of $X$ is closed in the subspace topology exactly when it has the form $X \cap V (I)$ for some ideal $I ⊴ R$ .

Proof

technique · direct

1.1L1

By [L1], contraction gives a bijection $c : Spec (S^{- 1} R) \to X$ .

1.2L1givenalgebra

Let $I ⊴ R$ . If $q \in Spec (S^{- 1} R)$ has contraction $p$ , then the inverse description in [L1] gives $q = S^{- 1} p$ . Therefore $q \supseteq S^{- 1} I ⟺ p \supseteq I,$ and hence $c (V_{S^{- 1} R} (S^{- 1} I)) = X \cap V_{R} (I) .$

2.1L2A1step 1.2

By [L2], every closed subset of $Spec (S^{- 1} R)$ has the form $V_{S^{- 1} R} (K)$ for some ideal $K ⊴ S^{- 1} R$ . With $I = λ^{- 1} (K)$ , assumption [A1] gives $K = S^{- 1} I$ , so step 1.2 shows that $c$ sends every closed subset of $Spec (S^{- 1} R)$ to a closed subset of $X$ .

2.2A2step 1.2

Conversely, if $C \subseteq X$ is closed, then [A2] gives $C = X \cap V_{R} (I)$ for some ideal $I ⊴ R$ . Step 1.2 then yields $C = c (V_{S^{- 1} R} (S^{- 1} I)),$ so $c^{- 1}$ also preserves closed sets.

3.1step 1.1step 2.1step 2.2∎

The bijection $c$ and its inverse both preserve closed sets, so $c$ is a homeomorphism from $Spec (S^{- 1} R)$ onto the subspace $X$ .

The spectrum of a localisation is the subspace of primes disjoint from the denominator set

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources