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.
$\square$

## Comments (0)