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The spectrum of a quotient is a closed subspace

Statement

Let $R$ be a commutative ring, let $I ⊴ R$ , and let $π : R \to R / I$ be the quotient map. Then contraction along $π$ is a homeomorphism from $Spec (R / I)$ onto the closed subset $V (I) \subseteq Spec (R)$ .

Facts & Assumptions

Given: A commutative ring $R$ , an ideal $I ⊴ R$ , and the quotient map $π : R \to R / I$ .

[L1]

Contraction along $π$ is an inclusion-preserving bijection from $Spec (R / I)$ onto $V (I)$ , with inverse $p \mapsto p / I$ (Prime ideals of a quotient ring are exactly the prime ideals containing the ideal).

[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]

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

Proof

technique · direct

1.1L1

By [L1], contraction gives a bijection $c : Spec (R / I) \to V (I)$ .

1.2L1givenalgebra

Let $J$ be an ideal of $R$ containing $I$ . If $q \in Spec (R / I)$ has contraction $p = π^{- 1} (q)$ , then $q \supseteq J / I ⟺ p \supseteq J .$ Therefore $c (V_{R / I} (J / I)) = V_{R} (J) .$ Since $V_{R} (J) \subseteq V_{R} (I)$ , this is closed in the subspace $V (I)$ .

2.1L2step 1.2

By [L2], every closed subset of $Spec (R / I)$ has the form $V_{R / I} (K)$ for some ideal $K ⊴ R / I$ . Writing $J = π^{- 1} (K)$ , one has $K = J / I$ , so step 1.2 shows that $c$ sends every closed subset of $Spec (R / I)$ to a closed subset of $V (I)$ .

2.2A1step 1.2

Conversely, let $C \subseteq V (I)$ be closed. By [A1], $C = V (I + J)$ for some ideal $J ⊴ R$ . Since $I + J$ contains $I$ , step 1.2 gives $C = V_{R} (I + J) = c (V_{R / I} ((I + J) / I)),$ so $c^{- 1}$ also carries closed sets to 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 $Spec (R / I)$ onto the closed subspace $V (I)$ .

The spectrum of a quotient is a closed subspace

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources