Loading web-font TeX/Main/Regular

The Stacks project

Lemma 17.17.2. Let (X, \mathcal{O}_ X) be a ringed space. An \mathcal{O}_ X-module \mathcal{F} is flat if and only if the stalk \mathcal{F}_ x is a flat \mathcal{O}_{X, x}-module for all x \in X.

Proof. Assume \mathcal{F}_ x is a flat \mathcal{O}_{X, x}-module for all x \in X. In this case, if \mathcal{G} \to \mathcal{H} \to \mathcal{K} is exact, then also \mathcal{G} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{H} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{K} \otimes _{\mathcal{O}_ X} \mathcal{F} is exact because we can check exactness at stalks and because tensor product commutes with taking stalks, see Lemma 17.16.1. Conversely, suppose that \mathcal{F} is flat, and let x \in X. Consider the skyscraper sheaves i_{x, *} M where M is a \mathcal{O}_{X, x}-module. Note that

M \otimes _{\mathcal{O}_{X, x}} \mathcal{F}_ x = \left(i_{x, *} M \otimes _{\mathcal{O}_ X} \mathcal{F}\right)_ x

again by Lemma 17.16.1. Since i_{x, *} is exact, we see that the fact that \mathcal{F} is flat implies that M \mapsto M \otimes _{\mathcal{O}_{X, x}} \mathcal{F}_ x is exact. Hence \mathcal{F}_ x is a flat \mathcal{O}_{X, x}-module. \square


Comments (1)

Comment #779 by Anfang Zhou on

Typo. There is an extra ')' and it should be


Your email address will not be published.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.