The Stacks project

Lemma 100.3.1. Let $\mathcal{X}$ be an algebraic stack. Let $T$ be a scheme and let $x, y$ be objects of the fibre category of $\mathcal{X}$ over $T$. Then the morphism $\mathit{Isom}_\mathcal {X}(x, y) \to T$ is locally of finite type.

Proof. By Algebraic Stacks, Lemma 93.16.2 we may assume that $\mathcal{X} = [U/R]$ for some smooth groupoid in algebraic spaces. By Descent on Spaces, Lemma 73.11.9 it suffices to check the property fppf locally on $T$. Thus we may assume that $x, y$ come from morphisms $x', y' : T \to U$. By Groupoids in Spaces, Lemma 77.22.1 we see that in this case $\mathit{Isom}_\mathcal {X}(x, y) = T \times _{(y', x'), U \times _ S U} R$. Hence it suffices to prove that $R \to U \times _ S U$ is locally of finite type. This follows from the fact that the composition $s : R \to U \times _ S U \to U$ is smooth (hence locally of finite type, see Morphisms of Spaces, Lemmas 66.37.5 and 66.28.5) and Morphisms of Spaces, Lemma 66.23.6. $\square$

Comments (0)

There are also:

  • 4 comment(s) on Section 100.3: Properties of diagonals

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Your email address will not be published.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 04XR. Beware of the difference between the letter 'O' and the digit '0'.