The Stacks project

Lemma 73.4.1. The morphism $\epsilon $ of (73.4.0.1) is a flat morphism of ringed sites. In particular the functor $\epsilon ^* : \textit{Mod}(\mathcal{O}_ X) \to \textit{Mod}(\mathcal{O}_{\acute{e}tale})$ is exact. Moreover, if $\epsilon ^*\mathcal{F} = 0$, then $\mathcal{F} = 0$.

Proof. The second assertion follows from the first by Modules on Sites, Lemma 18.31.2. To prove the first assertion we have to show that $\mathcal{O}_{\acute{e}tale}$ is a flat $\epsilon ^{-1}\mathcal{O}_ X$-module. To do this it suffices to check $\mathcal{O}_{X, x} \to \mathcal{O}_{{\acute{e}tale}, \overline{x}}$ is flat for any geometric point $\overline{x}$ of $X$, see Modules on Sites, Lemma 18.39.3, Sites, Lemma 7.34.2, and Étale Cohomology, Remarks 58.29.11. By Étale Cohomology, Lemma 58.33.1 we see that $\mathcal{O}_{{\acute{e}tale}, \overline{x}}$ is the strict henselization of $\mathcal{O}_{X, x}$. Thus $\mathcal{O}_{X, x} \to \mathcal{O}_{{\acute{e}tale}, \overline{x}}$ is faithfully flat by More on Algebra, Lemma 15.45.1. The final statement follows also: if $\epsilon ^*\mathcal{F} = 0$, then

\[ 0 = \epsilon ^*\mathcal{F}_{\overline{x}} = \mathcal{F}_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{\acute{e}tale} \]

(Modules on Sites, Lemma 18.36.4) for all geometric points $\overline{x}$. By faithful flatness of $\mathcal{O}_{X, x} \to \mathcal{O}_{{\acute{e}tale}, \overline{x}}$ we conclude $\mathcal{F}_ x = 0$ for all $x \in X$. $\square$


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Your email address will not be published.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 08H8. Beware of the difference between the letter 'O' and the digit '0'.