Proposition 101.49.3. Let $\mathcal{X}$ be a decent algebraic stack such that $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is quasi-compact. Then $|\mathcal{X}|$ is sober.

**Proof.**
By Lemma 101.49.1 we know that $|\mathcal{X}|$ is Kolmogorov (in fact we will reprove this). Let $T \subset |\mathcal{X}|$ be an irreducible closed subset. We have to show $T$ has a generic point. Let $\mathcal{Z} \subset \mathcal{X}$ be the reduced induced closed substack corresponding to $T$, see Properties of Stacks, Definition 100.10.4. Since $\mathcal{Z} \to \mathcal{X}$ is a closed immersion, we see that $\mathcal{Z}$ is a decent algebraic stack, see Lemma 101.48.3. Also, the morphism $\mathcal{I}_\mathcal {Z} \to \mathcal{Z}$ is the base change of $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ (Lemma 101.5.6). Hence $\mathcal{I}_\mathcal {Z} \to \mathcal{Z}$ is quasi-compact (Lemma 101.7.3). Thus we reduce to the case discussed in the next paragraph.

Assume $\mathcal{X}$ is decent, $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is quasi-compact, $\mathcal{X}$ is reduced, and $|\mathcal{X}|$ irreducible. We have to show $|\mathcal{X}|$ has a generic point. By Proposition 101.29.1. there exists a dense open substack $\mathcal{U} \subset \mathcal{X}$ which is a gerbe. In other words, $|\mathcal{U}| \subset |\mathcal{X}|$ is open dense. Thus we may assume that $\mathcal{X}$ is a gerbe in addition to all the other properties. Say $\mathcal{X} \to X$ turns $\mathcal{X}$ into a gerbe over the algebraic space $X$. Then $|\mathcal{X}| \cong |X|$ by Lemma 101.28.13. In particular, $X$ is quasi-compact and $|X|$ is irreducible. Also, by Lemma 101.48.5 we see that $X$ is a decent algebraic space. Then $|\mathcal{X}| = |X|$ is sober by Decent Spaces, Proposition 68.12.4 and hence has a (unique) generic point. $\square$

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