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
Comments (0)
There are also: