Lemma 67.46.1. Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of algebraic spaces over $S$. Then $f$ is finite locally free (in the sense of Section 67.3) if and only if $f$ is affine and the sheaf $f_*\mathcal{O}_ X$ is a finite locally free $\mathcal{O}_ Y$-module.

## 67.46 Finite locally free morphisms

We have already defined in Section 67.3 what it means for a representable morphism of algebraic spaces to be finite locally free.

**Proof.**
Assume $f$ is finite locally free (as defined in Section 67.3). This means that for every morphism $V \to Y$ whose source is a scheme the base change $f' : V \times _ Y X \to V$ is a finite locally free morphism of schemes. This in turn means (by the definition of a finite locally free morphism of schemes) that $f'_*\mathcal{O}_{V \times _ Y X}$ is a finite locally free $\mathcal{O}_ V$-module. We may choose $V \to Y$ to be surjective and étale. By Properties of Spaces, Lemma 66.26.2 we conclude the restriction of $f_*\mathcal{O}_ X$ to $V$ is finite locally free. Hence by Modules on Sites, Lemma 18.23.3 applied to the sheaf $f_*\mathcal{O}_ X$ on $Y_{spaces, {\acute{e}tale}}$ we conclude that $f_*\mathcal{O}_ X$ is finite locally free.

Conversely, assume $f$ is affine and that $f_*\mathcal{O}_ X$ is a finite locally free $\mathcal{O}_ Y$-module. Let $V$ be a scheme, and let $V \to Y$ be a surjective étale morphism. Again by Properties of Spaces, Lemma 66.26.2 we see that $f'_*\mathcal{O}_{V \times _ Y X}$ is finite locally free. Hence $f' : V \times _ Y X \to V$ is finite locally free (as it is also affine). By Spaces, Lemma 65.11.5 we conclude that $f$ is finite locally free (use Morphisms, Lemma 29.48.4 Descent, Lemmas 35.23.30 and 35.37.1). Thus we win. $\square$

This clears the way for the following definition.

Definition 67.46.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. We say that $f$ is *finite locally free* if $f$ is affine and $f_*\mathcal{O}_ X$ is a finite locally free $\mathcal{O}_ Y$-module. In this case we say $f$ is has *rank* or *degree* $d$ if the sheaf $f_*\mathcal{O}_ X$ is finite locally free of rank $d$.

Lemma 67.46.3. 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 finite locally free,

$f$ is finite locally free,

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

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 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 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). $\square$

Lemma 67.46.4. The composition of finite locally free morphisms is finite locally free.

**Proof.**
Omitted.
$\square$

Lemma 67.46.5. The base change of a finite locally free morphism is finite locally free.

**Proof.**
Omitted.
$\square$

Lemma 67.46.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

$f$ is finite locally free,

$f$ is finite, flat, and locally of finite presentation.

If $Y$ is locally Noetherian these are also equivalent to

$f$ is finite and flat.

**Proof.**
In each of the three cases the morphism is representable and you can check the property after base change by a surjective étale morphism $V \to Y$, see Lemmas 67.45.3, 67.46.3, 67.30.5, and 67.28.4. If $Y$ is locally Noetherian, then $V$ is locally Noetherian. Hence the result follows from the corresponding result in the schemes case, see Morphisms, Lemma 29.48.2.
$\square$

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