Lemma 10.79.1 (05GE)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.79: Open loci defined by module maps
Lemma 10.79.1 (cite)

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Lemma 10.79.1. Let $R$ be a ring. Let $\varphi : M \to N$ be a map of $R$ -modules with $N$ a finite $R$ -module. Then we have the equality

$\begin{align*} U & = \{ \mathfrak p \subset R \mid \varphi _{\mathfrak p} : M_{\mathfrak p} \to N_{\mathfrak p} \text{ is surjective}\} \\ & = \{ \mathfrak p \subset R \mid \varphi \otimes \kappa (\mathfrak p) : M \otimes \kappa (\mathfrak p) \to N \otimes \kappa (\mathfrak p) \text{ is surjective}\} \end{align*}$

and $U$ is an open subset of $\mathop{\mathrm{Spec}}(R)$ . Moreover, for any $f \in R$ such that $D(f) \subset U$ the map $M_ f \to N_ f$ is surjective.

Proof. The equality in the displayed formula follows from Nakayama's lemma. Nakayama's lemma also implies that $U$ is open. See Lemma 10.20.1 especially part (3). If $D(f) \subset U$ , then $M_ f \to N_ f$ is surjective on all localizations at primes of $R_ f$ , and hence it is surjective by Lemma 10.23.1. $\square$

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