Lemma 99.10.2. Let $\mathcal{X}$ be an algebraic stack. If $\mathcal{X}' \subset \mathcal{X}$ is a closed substack, $\mathcal{X}$ is reduced and $|\mathcal{X}'| = |\mathcal{X}|$, then $\mathcal{X}' = \mathcal{X}$.

**Proof.**
Choose a presentation $[U/R] \to \mathcal{X}$ with $U$ a scheme. As $\mathcal{X}$ is reduced, we see that $U$ is reduced (by definition of reduced algebraic stacks). By Lemma 99.9.11 $\mathcal{X}'$ corresponds to an $R$-invariant closed subscheme $Z \subset U$. But now $|Z| \subset |U|$ is the inverse image of $|\mathcal{X}'|$, and hence $|Z| = |U|$. Hence $Z$ is a closed subscheme of $U$ whose underlying sets of points agree. By Schemes, Lemma 26.12.7 the map $\text{id}_ U : U \to U$ factors through $Z \to U$, and hence $Z = U$, i.e., $\mathcal{X}' = \mathcal{X}$.
$\square$

## 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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)