Lemma 10.29.2 (00F7)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.29: Images of ring maps of finite presentation
Lemma 10.29.2 (cite)

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Lemma 10.29.2. Let $\varphi : R \to S$ be a ring map. The induced continuous map $f : \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is quasi-compact. For any constructible set $E \subset \mathop{\mathrm{Spec}}(R)$ the inverse image $f^{-1}(E)$ is constructible in $\mathop{\mathrm{Spec}}(S)$.

Proof. We first show that the inverse image of any quasi-compact open $U \subset \mathop{\mathrm{Spec}}(R)$ is quasi-compact. By Lemma 10.29.1 we may write $U$ as a finite open of standard opens. Thus by Lemma 10.17.4 we see that $f^{-1}(U)$ is a finite union of standard opens. Hence $f^{-1}(U)$ is quasi-compact by Lemma 10.29.1 again. The second assertion now follows from Topology, Lemma 5.15.3. $\square$

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