Lemma 66.45.1. Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of algebraic spaces over $S$. Then $f$ is integral, resp. finite (in the sense of Section 66.3), if and only if for all affine schemes $Z$ and morphisms $Z \to Y$ the scheme $X \times _ Y Z$ is affine and integral, resp. finite, over $Z$.

## 66.45 Integral and finite morphisms

We have already defined in Section 66.3 what it means for a representable morphism of algebraic spaces to be integral (resp. finite).

**Proof.**
This follows directly from the definition of an integral (resp. finite) morphism of schemes (Morphisms, Definition 29.44.1).
$\square$

This clears the way for the following definition.

Definition 66.45.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.

We say that $f$ is

*integral*if for every affine scheme $Z$ and morphisms $Z \to Y$ the algebraic space $X \times _ Y Z$ is representable by an affine scheme integral over $Z$.We say that $f$ is

*finite*if for every affine scheme $Z$ and morphisms $Z \to Y$ the algebraic space $X \times _ Y Z$ is representable by an affine scheme finite over $Z$.

Lemma 66.45.3. 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 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 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). $\square$

Lemma 66.45.4. The composition of integral (resp. finite) morphisms is integral (resp. finite).

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

Lemma 66.45.5. The base change of an integral (resp. finite) morphism is integral (resp. finite).

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

Lemma 66.45.6. A finite morphism of algebraic spaces is integral. An integral morphism of algebraic spaces which is locally of finite type is finite.

**Proof.**
In both cases the morphism is representable, and you can check the condition after a base change by an affine scheme mapping into $Y$, see Lemmas 66.45.3. Hence this lemma follows from the same lemma for the case of schemes, see Morphisms, Lemma 29.44.4.
$\square$

Lemma 66.45.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent

$f$ is integral, and

$f$ is affine and universally closed.

**Proof.**
In both cases the morphism is representable, and you can check the condition after a base change by an affine scheme mapping into $Y$, see Lemmas 66.45.3, 66.20.3, and 66.9.5. Hence the result follows from Morphisms, Lemma 29.44.7.
$\square$

Lemma 66.45.8. A finite morphism of algebraic spaces is quasi-finite.

**Proof.**
Let $f : X \to Y$ be a morphism of algebraic spaces. By Definition 66.45.2 and Lemmas 66.8.8 and 66.27.6 both properties may be checked after base change to an affine over $Y$, i.e., we may assume $Y$ affine. If $f$ is finite then $X$ is a scheme. Hence the result follows from the corresponding result for schemes, see Morphisms, Lemma 29.44.10.
$\square$

Lemma 66.45.9. 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, and

$f$ is affine and proper.

**Proof.**
In both cases the morphism is representable, and you can check the condition after base change to an affine scheme mapping into $Y$, see Lemmas 66.45.3, 66.20.3, and 66.40.2. Hence the result follows from Morphisms, Lemma 29.44.11.
$\square$

Lemma 66.45.10. A closed immersion is finite (and a fortiori integral).

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

Lemma 66.45.11. Let $S$ be a scheme. Let $X_ i \to Y$, $i = 1, \ldots , n$ be finite morphisms of algebraic spaces over $S$. Then $X_1 \amalg \ldots \amalg X_ n \to Y$ is finite too.

**Proof.**
Follows from the case of schemes (Morphisms, Lemma 29.44.13) by étale localization.
$\square$

Lemma 66.45.12. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of algebraic spaces over $S$.

If $g \circ f$ is finite and $g$ separated then $f$ is finite.

If $g \circ f$ is integral and $g$ separated then $f$ is integral.

**Proof.**
Assume $g \circ f$ is finite (resp. integral) and $g$ separated. The base change $X \times _ Z Y \to Y$ is finite (resp. integral) by Lemma 66.45.5. The morphism $X \to X \times _ Z Y$ is a closed immersion as $Y \to Z$ is separated, see Lemma 66.4.7. A closed immersion is finite (resp. integral), see Lemma 66.45.10. The composition of finite (resp. integral) morphisms is finite (resp. integral), see Lemma 66.45.4. Thus we win.
$\square$

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