The Stacks project

Lemma 66.20.4. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The following are equivalent

  1. $X$ is decent and $|X|$ has finitely many irreducible components,

  2. every quasi-compact scheme ├ętale over $X$ has finitely many irreducible components, there are finitely many $x \in |X|$ of codimension $0$ on $X$, and each of these can be represented by a monomorphism $\mathop{\mathrm{Spec}}(k) \to X$,

  3. there exists a dense open $X' \subset X$ which is a scheme, $X'$ has finitely many irreducible components with generic points $\{ x'_1, \ldots , x'_ m\} $, and the morphism $x'_ j \to X$ is quasi-compact for $j = 1, \ldots , m$.

Moreover, if these conditions hold, then $X$ is reasonable and the points $x'_ j \in |X|$ are the generic points of the irreducible components of $|X|$.

Proof. In the proof we use Properties of Spaces, Lemma 64.11.1 without further mention. Assume (1). Then $X$ has a dense open subscheme $X'$ by Theorem 66.10.2. Since the closure of an irreducible component of $|X'|$ is an irreducible component of $|X|$, we see that $|X'|$ has finitely many irreducible components. Thus (3) holds.

Assume $X' \subset X$ is as in (3). Let $\{ x'_1, \ldots , x'_ m\} $ be the generic points of the irreducible components of $X'$. Let $a : U \to X$ be an ├ętale morphism with $U$ a quasi-compact scheme. It suffices to show that $U$ has finitely many irreducible components whose generic points lie over $\{ x'_1, \ldots , x'_ m\} $. It suffices to prove this for the members of a finite affine open cover of $U$, hence we may and do assume $U$ is affine. Note that $U' = a^{-1}(X') \subset U$ is a dense open. The generic points of irreducible components of $U'$ are the points lying over $\{ x'_1, \ldots , x'_ m\} $ and since $x'_ j \to X$ is quasi-compact there are finitely many points of $U$ lying over $x'_ j$ (Lemma 66.4.5). Hence $U'$ has finitely many irreducible components, which implies that the closures of these irreducible components are the irreducible components of $U$. Thus (2) holds.

Assume (2). This implies (1) and the final statement by Lemma 66.20.3. (We also use that a reasonable algebraic space is decent, see discussion following Definition 66.6.1.) $\square$


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