The Stacks project

Lemma 99.11.13. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $x \in |\mathcal{X}|$ with image $y \in |\mathcal{Y}|$. Assume the residual gerbes $\mathcal{Z}_ x \subset \mathcal{X}$ and $\mathcal{Z}_ y \subset \mathcal{Y}$ of $x$ and $y$ exist and that there exists a morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ in the equivalence class of $x$ such that

\[ \mathop{\mathrm{Spec}}(k) \times _\mathcal {X} \mathop{\mathrm{Spec}}(k) \longrightarrow \mathop{\mathrm{Spec}}(k) \times _\mathcal {Y} \mathop{\mathrm{Spec}}(k) \]

is an isomorphism. Then $\mathcal{Z}_ x \to \mathcal{Z}_ y$ is an isomorphism.

Proof. Let $k'/k$ be an extension of fields. Then

\[ \mathop{\mathrm{Spec}}(k') \times _\mathcal {X} \mathop{\mathrm{Spec}}(k') \longrightarrow \mathop{\mathrm{Spec}}(k') \times _\mathcal {Y} \mathop{\mathrm{Spec}}(k') \]

is the base change of the morphism in the lemma by the faithfully flat morphism $\mathop{\mathrm{Spec}}(k' \otimes k') \to \mathop{\mathrm{Spec}}(k \otimes k)$. Thus the property described in the lemma is independent of the choice of the morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ in the equivalence class of $x$. Thus we may assume that $\mathop{\mathrm{Spec}}(k) \to \mathcal{Z}_ x$ is surjective, flat, and locally of finite presentation. In this situation we have

\[ \mathcal{Z}_ x = [\mathop{\mathrm{Spec}}(k)/R] \]

with $R = \mathop{\mathrm{Spec}}(k) \times _\mathcal {X} \mathop{\mathrm{Spec}}(k)$. See proof of Lemma 99.11.5. Since also $R = \mathop{\mathrm{Spec}}(k) \times _\mathcal {Y} \mathop{\mathrm{Spec}}(k)$ we conclude that the morphism $\mathcal{Z}_ x \to \mathcal{Z}_ y$ of Lemma 99.11.12 is fully faithful by Algebraic Stacks, Lemma 93.16.1. We conclude for example by Lemma 99.11.11. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 99.11: Residual gerbes

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Your email address will not be published.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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DTI. Beware of the difference between the letter 'O' and the digit '0'.