Lemma 101.28.15. Let $\mathcal{X}$ be an algebraic stack. If $\mathcal{X}$ is a gerbe then for every $x \in |\mathcal{X}|$ the residual gerbe of $\mathcal{X}$ at $x$ exists.

**Proof.**
Let $\pi : \mathcal{X} \to X$ be a morphism from $\mathcal{X}$ into an algebraic space $X$ which turns $\mathcal{X}$ into a gerbe over $X$. Let $Z_ x \to X$ be the residual space of $X$ at $x$, see Decent Spaces, Definition 68.13.6. Let $\mathcal{Z} = \mathcal{X} \times _ X Z_ x$. By Lemma 101.28.3 the algebraic stack $\mathcal{Z}$ is a gerbe over $Z_ x$. Hence $|\mathcal{Z}| = |Z_ x|$ (Lemma 101.28.13) is a singleton. Since $\mathcal{Z} \to Z_ x$ is locally of finite presentation as a base change of $\pi $ (see Lemmas 101.28.8 and 101.27.3) we see that $\mathcal{Z}$ is locally Noetherian, see Lemma 101.17.5. Thus the residual gerbe $\mathcal{Z}_ x$ of $\mathcal{X}$ at $x$ exists and is equal to $\mathcal{Z}_ x = \mathcal{Z}_{red}$ the reduction of the algebraic stack $\mathcal{Z}$. Namely, we have seen above that $|\mathcal{Z}_{red}|$ is a singleton mapping to $x \in |\mathcal{X}|$, it is reduced by construction, and it is locally Noetherian (as the reduction of a locally Noetherian algebraic stack is locally Noetherian, details omitted).
$\square$

## 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Your email address will not be published.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)

There are also: