Lemma 98.27.15. 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 65.13.6. Let $\mathcal{Z} = \mathcal{X} \times _ X Z_ x$. By Lemma 98.27.3 the algebraic stack $\mathcal{Z}$ is a gerbe over $Z_ x$. Hence $|\mathcal{Z}| = |Z_ x|$ (Lemma 98.27.13) is a singleton. Since $\mathcal{Z} \to Z_ x$ is locally of finite presentation as a base change of $\pi$ (see Lemmas 98.27.8 and 98.26.3) we see that $\mathcal{Z}$ is locally Noetherian, see Lemma 98.17.5. 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). $\square$

There are also:

• 2 comment(s) on Section 98.27: Gerbes

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).