Lemma 66.46.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

1. $f$ is representable and finite locally free,

2. $f$ is finite locally free,

3. there exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that $V \times _ Y X \to V$ is finite locally free, and

4. there exists a Zariski covering $Y = \bigcup Y_ i$ such that each morphism $f^{-1}(Y_ i) \to Y_ i$ is finite locally free.

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 65.6.1. 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 65.13.1 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 64.11.5 because the class of finite locally free morphisms satisfies all the required properties (see Morphisms, Lemma 29.48.4 Descent, Lemmas 35.23.30 and 35.37.1). The conclusion of applying this lemma is that $f$ is representable and finite locally free, i.e., (1) holds.

The equivalence of (1) and (4) follows from the fact that being finite locally free is Zariski local on the target (the reference above shows that being finite locally free is in fact fpqc local on the target). $\square$

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