Lemma 65.6.4. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $U$ be a separated scheme and $U \to X$ étale. Then $U \to X$ is separated, and $R = U \times _ X U$ is a separated scheme.

**Proof.**
Let $X' \subset X$ be the open subscheme such that $U \to X$ factors through an étale surjection $U \to X'$, see Lemma 65.4.10. If $U \to X'$ is separated, then so is $U \to X$, see Spaces, Lemma 64.5.4 (as the open immersion $X' \to X$ is separated by Spaces, Lemma 64.5.8 and Schemes, Lemma 26.23.8). Moreover, since $U \times _{X'} U = U \times _ X U$ it suffices to prove the result after replacing $X$ by $X'$, i.e., we may assume $U \to X$ surjective. Consider the commutative diagram

In the proof of Spaces, Lemma 64.13.1 we have seen that $j : R \to U \times _ S U$ is separated. The morphism of schemes $U \to S$ is separated as $U$ is a separated scheme, see Schemes, Lemma 26.21.13. Hence $U \times _ S U \to U$ is separated as a base change, see Schemes, Lemma 26.21.12. Hence the scheme $U \times _ S U$ is separated (by the same lemma). Since $j$ is separated we see in the same way that $R$ is separated. Hence $R \to U$ is a separated morphism (by Schemes, Lemma 26.21.13 again). Thus by Spaces, Lemma 64.11.4 and the diagram above we conclude that $U \to X$ is separated. $\square$

## 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Your email address will not be published.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.

## Comments (0)