Lemma 101.49.1. Let $\mathcal{X}$ be a decent algebraic stack. Then $|\mathcal{X}|$ is Kolmogorov (see Topology, Definition 5.8.6).

## 101.49 Points on decent stacks

This section is the analogue of Decent Spaces, Section 68.12. We do not know whether or not the topological space associated to a decent algebraic stack is always sober; see Proposition 101.49.3 for a slightly weaker result.

**Proof.**
Let $x_1, x_2 \in |\mathcal{X}|$ with $x_1 \leadsto x_2$ and $x_2 \leadsto x_1$. We have to show that $x_1 = x_2$. Let $\mathcal{Z} \subset \mathcal{X}$ be the reduced closed substack with $|\mathcal{Z}|$ equal to $\overline{\{ x_1\} } = \overline{\{ x_2\} }$. By Lemma 101.48.3 we see that $\mathcal{Z}$ is decent. After replacing $\mathcal{X}$ by $\mathcal{Z}$ we reduce to the case discussed in the next paragraph.

Assume $|\mathcal{X}|$ is irreducible with generic points $x_1$ and $x_2$. Pick an affine scheme $U$ and $u_1, u_2 \in U$ and a smooth morphism $f : U \to \mathcal{X}$ such that $f(u_ i) = x_ i$. Then we find a third point $u_3 \in U$ which is the generic point of an irreducible component of $U$ whose image $x_3 \in |\mathcal{X}|$ is *also* a generic point of $|\mathcal{X}|$. Namely, we can simply choose $u_3$ any generic point of an irreducible component passing through $u_1$ (or $u_2$ if you like). In the next paragraph we will show that $x_1 = x_3$ and $x_2 = x_3$ which will prove what we want.

By symmetry it suffices to prove that $x_1 = x_3$. Since $x_1$ is a generic point of $|\mathcal{X}|$ we have a specialization $x_1 \leadsto x_3$. By Lemma 101.47.1 we can find a specialization $u'_1 \leadsto u_3$ in $U$ (!) mapping to $x_1 \leadsto x_3$. However, $u_3$ is the generic point of an irreducible component and hence $u'_1 = u_3$ as desired. $\square$

Lemma 101.49.2. Let $\mathcal{X}$ be a decent, locally Noetherian algebraic stack. Then $|\mathcal{X}|$ is a sober locally Noetherian topological space.

**Proof.**
By Lemma 101.8.3 the topological space $|\mathcal{X}|$ is locally Noetherian. By Lemma 101.49.1 the topological space $|\mathcal{X}|$ is Kolmogorov. By Lemma 101.8.4 the topological space $|\mathcal{X}|$ is quasi-sober. This finishes the proof, see Topology, Definition 5.8.6.
$\square$

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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