Lemma 10.30.5 (00FK)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.30: More on images
Lemma 10.30.5 (cite)

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Lemma 10.30.5. Let $R \subset S$ be an injective ring map. Then $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ hits all the minimal primes.

Proof. Let $\mathfrak p \subset R$ be a minimal prime. In this case $R_{\mathfrak p}$ has a unique prime ideal. Hence it suffices to show that $S_{\mathfrak p}$ is not zero. And this follows from the fact that localization is exact, see Proposition 10.9.12. $\square$

Comments (6)

Comment #4211 by Aaron on May 08, 2019 at 17:44

I think it should be: $\text{Spec}(S)\to\text{Spec}(R)$ hits all the minimal primes.

Comment #4393 by Johan on August 30, 2019 at 14:00

Thanks and fixed here.

Comment #6009 by Ivan on March 25, 2021 at 15:12

could you please explain why it suffices to show that $S_p$ is not zero and we get a $q \in Spec(S)$ such that $q \cap R=p$ ? in additon, here $S_p$ is a ring?

Comment #6010 by Fan on March 25, 2021 at 17:41

See the section on localisation (00CM)

Comment #6012 by Ivan on March 26, 2021 at 02:36

but 00CM doesn't mention about it. even if $0 \rightarrow R_p \rightarrow S_p$ is exact and $S_p$ is non-zero, is it possible that $(p)S_p=S_p$ ? why $S_p$ is non-zero and then there exist $q \in Spec(S)$ such that $q \cap R = p$ ?

Comment #6014 by Ivan on March 26, 2021 at 10:23

I think I got it, thanks

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