Definition 99.6.1. Let $\mathcal{X}$ be an algebraic stack. We say $\mathcal{X}$ is *quasi-compact* if and only if $|\mathcal{X}|$ is quasi-compact.

## 99.6 Quasi-compact algebraic stacks

The following definition is equivalent with the definition for algebraic spaces by Properties of Spaces, Lemma 65.5.2.

Lemma 99.6.2. Let $\mathcal{X}$ be an algebraic stack. The following are equivalent:

$\mathcal{X}$ is quasi-compact,

there exists a surjective smooth morphism $U \to \mathcal{X}$ with $U$ an affine scheme,

there exists a surjective smooth morphism $U \to \mathcal{X}$ with $U$ a quasi-compact scheme,

there exists a surjective smooth morphism $U \to \mathcal{X}$ with $U$ a quasi-compact algebraic space, and

there exists a surjective morphism $\mathcal{U} \to \mathcal{X}$ of algebraic stacks such that $\mathcal{U}$ is quasi-compact.

**Proof.**
We will use Lemma 99.4.4. Suppose $\mathcal{U}$ and $\mathcal{U} \to \mathcal{X}$ are as in (5). Then since $|\mathcal{U}| \to |\mathcal{X}|$ is surjective and continuous we conclude that $|\mathcal{X}|$ is quasi-compact. Thus (5) implies (1). The implications (2) $\Rightarrow $ (3) $\Rightarrow $ (4) $\Rightarrow $ (5) are immediate. Assume (1), i.e., $\mathcal{X}$ is quasi-compact, i.e., that $|\mathcal{X}|$ is quasi-compact. Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Then since $|U| \to |\mathcal{X}|$ is open we see that there exists a quasi-compact open $U' \subset U$ such that $|U'| \to |X|$ is surjective (and still smooth). Choose a finite affine open covering $U' = U_1 \cup \ldots \cup U_ n$. Then $U_1 \amalg \ldots \amalg U_ n \to \mathcal{X}$ is a surjective smooth morphism whose source is an affine scheme (Schemes, Lemma 26.6.8). Hence (2) holds.
$\square$

Lemma 99.6.3. A finite disjoint union of quasi-compact algebraic stacks is a quasi-compact algebraic stack.

**Proof.**
This is clear from the corresponding topological fact.
$\square$

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