Lemma 67.45.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:
$f$ is representable and integral (resp. finite),
$f$ is integral (resp. finite),
there exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that $V \times _ Y X \to V$ is integral (resp. finite), and
there exists a Zariski covering $Y = \bigcup Y_ i$ such that each of the morphisms $f^{-1}(Y_ i) \to Y_ i$ is integral (resp. finite).
Proof. It is clear that (1) implies (2) and that (2) implies (3) by taking $V$ to be a disjoint union of affines étale over $Y$, see Properties of Spaces, Lemma 66.6.1. Assume $V \to Y$ is as in (3). Then for every affine open $W$ of $V$ we see that $W \times _ Y X$ is an affine open of $V \times _ Y X$. Hence by Properties of Spaces, Lemma 66.13.1 we conclude that $V \times _ Y X$ is a scheme. Moreover the morphism $V \times _ Y X \to V$ is affine. This means we can apply Spaces, Lemma 65.11.5 because the class of integral (resp. finite) morphisms satisfies all the required properties (see Morphisms, Lemmas 29.44.6 and Descent, Lemmas 35.23.22, 35.23.23, and 35.37.1). The conclusion of applying this lemma is that $f$ is representable and integral (resp. finite), i.e., (1) holds.
The equivalence of (1) and (4) follows from the fact that being integral (resp. finite) is Zariski local on the target (the reference above shows that being integral or finite is in fact fpqc local on the target). $\square$
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