Lemma 101.8.4. Let $\mathcal{X}$ be a locally Noetherian algebraic stack. Then $|\mathcal{X}|$ is quasi-sober (Topology, Definition 5.8.6).
Proof. We have to prove that every irreducible closed subset $T \subset |\mathcal{X}|$ has a generic point. Choose an affine scheme $U$ and a smooth morphism $f : U \to \mathcal{X}$ such that $f^{-1}(T) \subset |U|$ is nonempty. Since $U$ is Noetherian, the closed subset $f^{-1}(T)$ has finitely many irreducible components (Topology, Lemma 5.9.2). Say $f^{-1}(T) = Z_1 \cup \ldots \cup Z_ n$ is the decomposition into irreducible components. As $f$ is open, the image of $f|_{f^{-1}(T)} : f^{-1}(T) \to T$ contains a nonempty open subset of $T$. Since $T$ is irreducible, this means that $f(f^{-1}(T))$ is dense. Since $T$ is irreducible, it follows that $f(Z_ i)$ is dense for some $i$. Then if $\xi _ i \in Z_ i$ is the generic point we see that $f(\xi _ i)$ is a generic point of $T$. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)