Lemma 100.50.4. 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 100.49.3. 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 99.10.4. Then $\mathcal{Z}$ is an integral algebraic stack because it is decent (Lemma 100.48.3), 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 100.5.6), $\mathcal{Z}$ is reduced, and $|\mathcal{Z}|$ is irreducible.
$\square$

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