The Stacks project

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)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05GE. Beware of the difference between the letter 'O' and the digit '0'.