The Stacks project

Lemma 64.6.4. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $U$ be a separated scheme and $U \to X$ ├ętale. 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 ├ętale surjection $U \to X'$, see Lemma 64.4.10. If $U \to X'$ is separated, then so is $U \to X$, see Spaces, Lemma 63.5.4 (as the open immersion $X' \to X$ is separated by Spaces, Lemma 63.5.8 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

\[ \xymatrix{ R = U \times _ X U \ar[r] \ar[d] & U \ar[d] \\ U \ar[r] & X } \]

In the proof of Spaces, Lemma 63.13.1 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 63.11.4 and the diagram above we conclude that $U \to X$ is separated. $\square$


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