Lemma 66.49.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The following are equivalent

1. there is a surjective étale morphism $U \to X$ where $U$ is a scheme such that every quasi-compact open of $U$ has finitely many irreducible components,

2. for every scheme $U$ and every étale morphism $U \to X$ every quasi-compact open of $U$ has finitely many irreducible components,

3. for every quasi-compact algebraic space $Y$ étale over $X$ the set of codimension $0$ points of $Y$ (Properties of Spaces, Definition 65.10.2) is finite, and

4. for every quasi-compact algebraic space $Y$ étale over $X$ the space $|Y|$ has finitely many irreducible components.

If $X$ is representable this means that every quasi-compact open of $X$ has finitely many irreducible components.

Proof. The equivalence of (1) and (2) and the final statement follow from Descent, Lemma 35.16.3 and Properties of Spaces, Lemma 65.7.1. It is clear that (4) implies (1) and (2) by considering only those $Y$ which are schemes. Similarly, (3) implies (1) and (2) since for a scheme the codimension $0$ points are the generic points of its irreducible components, see for example Properties of Spaces, Lemma 65.11.1.

Conversely, assume (2) and let $Y \to X$ be an étale morphism of algebraic spaces with $Y$ quasi-compact. Then we can choose an affine scheme $V$ and a surjective étale morphism $V \to Y$ (Properties of Spaces, Lemma 65.6.3). Since $V$ has finitely many irreducible components by (2) and since $|V| \to |Y|$ is surjective and continuous, we conclude that $|Y|$ has finitely many irreducible components by Topology, Lemma 5.8.5. Thus (4) holds. Similarly, by Properties of Spaces, Lemma 65.11.1 the images of the generic points of the irreducible components of $V$ are the codimension $0$ points of $Y$ and we conclude that there are finitely many, i.e., (3) holds. $\square$

Comment #7686 by Anonymous on

In (1), the phrase "is as scheme" should be "is a scheme".

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