Lemma 10.122.8 (00PP)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.122: Quasi-finite maps
Lemma 10.122.8 (cite)

Lemma 10.122.8. Let $R \to S$ be a ring map of finite type. Let $R \to R'$ be any ring map. Set $S' = R' \otimes _ R S$ .

The set $\{ \mathfrak q' \mid R' \to S' \text{ quasi-finite at }\mathfrak q'\}$ is the inverse image of the corresponding set of $\mathop{\mathrm{Spec}}(S)$ under the canonical map $\mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(S)$ .
If $\mathop{\mathrm{Spec}}(R') \to \mathop{\mathrm{Spec}}(R)$ is surjective, then $R \to S$ is quasi-finite if and only if $R' \to S'$ is quasi-finite.
Any base change of a quasi-finite ring map is quasi-finite.

Proof. Let $\mathfrak p' \subset R'$ be a prime lying over $\mathfrak p \subset R$ . Then the fibre ring $S' \otimes _{R'} \kappa (\mathfrak p')$ is the base change of the fibre ring $S \otimes _ R \kappa (\mathfrak p)$ by the field extension $\kappa (\mathfrak p) \to \kappa (\mathfrak p')$ . Hence the first assertion follows from the invariance of dimension under field extension (Lemma 10.116.6) and Lemma 10.122.1. The stability of quasi-finite maps under base change follows from this and the stability of finite type property under base change. The second assertion follows since the assumption implies that given a prime $\mathfrak q \subset S$ we can find a prime $\mathfrak q' \subset S'$ lying over it. $\square$

The Stacks project

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