Definition 101.50.1. We say an algebraic stack $\mathcal{X}$ is *integral* if it is reduced, decent, $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is quasi-compact, and $|\mathcal{X}|$ is irreducible.

## 101.50 Integral algebraic stacks

This section is the analogue of Spaces over Fields, Section 72.4. Motivated by the considerations in that section and by the result of Proposition 101.49.3 we define an integral algebraic stack as follows (and it does not conflict with the already existing definitions of integral schemes and integral algebraic spaces).

Note that if $\mathcal{X}$ is quasi-separated, then for it to be integral, it suffices that $\mathcal{X}$ is reduced and that $|\mathcal{X}|$ is irreducible, see Lemma 101.50.3.

Lemma 101.50.2. Let $\mathcal{X}$ be an integral algebraic stack. Then

$|\mathcal{X}|$ is sober, irreducible, and has a unique generic point,

there exists an open substack $\mathcal{U} \subset \mathcal{X}$ which is a gerbe over an integral scheme $U$.

**Proof.**
Proposition 101.49.3 tells us that $|\mathcal{X}|$ is sober. Of course it is also irreducible and hence has a unique generic point $x$ (by the definition of sobriety). Proposition 101.29.1 shows the existence of a dense open $\mathcal{U} \subset \mathcal{X}$ which is a gerbe over an algebraic space $U$. Then $U$ is a decent algebraic space by Lemma 101.48.5 (and the fact that $\mathcal{U}$ is decent by Lemma 101.48.3). Since $|U| = |\mathcal{U}|$ we see that $|U|$ is irreducible. Finally, since $\mathcal{U}$ is reduced the morphism $\mathcal{U} \to U$ factors through $U_{red}$, see Properties of Stacks, Lemma 100.10.3. Now since $\mathcal{U} \to U$ is flat, locally of finite presentation, and surjective (Lemma 101.28.8), this implies that $U = U_{red}$, i.e., $U$ is reduced (small detail omitted). It follows that $U$ is an integral algebraic space, see Spaces over Fields, Definition 72.4.1. Then finally, we may replace $U$ (and correspondingly $\mathcal{U}$) by an open subspace and assume that $U$ is an integral scheme, see discussion in Spaces over Fields, Section 72.4.
$\square$

Lemma 101.50.3. Let $\mathcal{X}$ be an algebraic stack which is reduced and quasi-separated and whose associated topological space $|\mathcal{X}|$ is irreducible. Then $\mathcal{X}$ is integral.

**Proof.**
If $\mathcal{X}$ is quasi-separated, then $\mathcal{X}$ is decent by Lemma 101.48.2. If $\mathcal{X}$ is quasi-separated, then $\Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is quasi-compact, hence $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is quasi-compact as the base change of $\Delta $ by $\Delta $, see Lemma 101.7.3. Thus we see that all the hypotheses of Definition 101.50.1 hold (and we also see that we may replace “quasi-separated” by “$\Delta _\mathcal {X}$ is quasi-compact”).
$\square$

Lemma 101.50.4. Let $\mathcal{X}$ be a decent algebraic stack such that $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is quasi-compact. There are canonical bijections between the following sets:

the set of points of $\mathcal{X}$, i.e., $|\mathcal{X}|$,

the set of irreducible closed subsets of $|\mathcal{X}|$,

the set of integral closed substacks of $\mathcal{X}$.

The bijection from (1) to (2) sends $x$ to $\overline{\{ x\} }$. The bijection from (3) to (2) sends $\mathcal{Z}$ to $|\mathcal{Z}|$.

**Proof.**
Our map defines a bijection between (1) and (2) as $|\mathcal{X}|$ is sober by Proposition 101.49.3. Given $T \subset |\mathcal{X}|$ closed and irreducible, there is a unique reduced closed substack $\mathcal{Z} \subset \mathcal{X}$ such that $|\mathcal{Z}| = T$, namely, $\mathcal{Z}$ is the reduced induced subspace structure on $T$, see Properties of Stacks, Definition 100.10.4. Then $\mathcal{Z}$ is an integral algebraic stack because it is decent (Lemma 101.48.3), the morphism $\mathcal{I}_\mathcal {Z} \to \mathcal{Z}$ is quasi-compact (as the base change of $\mathcal{I}_\mathcal {X} \to \mathcal{X}$, see Lemma 101.5.6), $\mathcal{Z}$ is reduced, and $|\mathcal{Z}|$ is irreducible.
$\square$

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