## 100.10 Integral and finite morphisms

Integral and finite morphisms of algebraic stacks are defined as follows.

Definition 100.10.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.

1. We say $f$ is integral if $f$ is representable and integral in the sense of Properties of Stacks, Section 99.3.

2. We say $f$ is finite if $f$ is representable and finite in the sense of Properties of Stacks, Section 99.3.

For us it is a little bit more convenient to think of an integral, resp. finite morphism of algebraic stacks as a morphism of algebraic stacks which is representable by algebraic spaces and integral, resp. finite in the sense of Properties of Stacks, Section 99.3. (Recall that the default for “representable” in the Stacks project is representable by schemes.) Since this is clearly equivalent to the notion just defined we shall use this characterization without further mention. We prove a few simple lemmas about this notion.

Lemma 100.10.2. Let $\mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $\mathcal{Z} \to \mathcal{Y}$ be an integral (or finite) morphism of algebraic stacks. Then $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X}$ is an integral (or finite) morphism of algebraic stacks.

Proof. This follows from the discussion in Properties of Stacks, Section 99.3. $\square$

Lemma 100.10.3. Compositions of integral, resp. finite morphisms of algebraic stacks are integral, resp. finite.

Proof. This follows from the discussion in Properties of Stacks, Section 99.3 and Morphisms of Spaces, Lemma 66.45.4. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).