Lemma 10.41.9 (037F)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.41: Going up and going down
Lemma 10.41.9 (cite)

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Lemma 10.41.9. Let $k$ be a field, and let $R$ , $S$ be $k$ -algebras. Let $S' \subset S$ be a sub $k$ -algebra, and let $f \in S' \otimes _ k R$ . In the commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}((S \otimes _ k R)_ f) \ar[rd] \ar[rr] & & \mathop{\mathrm{Spec}}((S' \otimes _ k R)_ f) \ar[ld] \\ & \mathop{\mathrm{Spec}}(R) & }$

the images of the diagonal arrows are the same.

Proof. Let $\mathfrak p \subset R$ be in the image of the south-west arrow. This means (Lemma 10.18.6) that

$(S' \otimes _ k R)_ f \otimes _ R \kappa (\mathfrak p) = (S' \otimes _ k \kappa (\mathfrak p))_ f$

is not the zero ring, i.e., $S' \otimes _ k \kappa (\mathfrak p)$ is not the zero ring and the image of $f$ in it is not nilpotent. The ring map $S' \otimes _ k \kappa (\mathfrak p) \to S \otimes _ k \kappa (\mathfrak p)$ is injective. Hence also $S \otimes _ k \kappa (\mathfrak p)$ is not the zero ring and the image of $f$ in it is not nilpotent. Hence $(S \otimes _ k R)_ f \otimes _ R \kappa (\mathfrak p)$ is not the zero ring. Thus (Lemma 10.18.6) we see that $\mathfrak p$ is in the image of the south-east arrow as desired. $\square$

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