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. $\square$

Comment #779 by Anfang Zhou on

Typo. There is an extra ')' and it should be $\left(i_{x, *} M \otimes_{\mathcal{O}_X} \mathcal{F}\right)_x$

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