The Stacks project

Lemma 97.8.2. Let $S$ be a scheme. Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$. If

  1. $\mathcal{Y}$ is an algebraic stack, and

  2. $F$ is algebraic (see above),

then $\mathcal{X}$ is an algebraic stack.

Proof. By assumption (1) there exists a scheme $T$ and an object $\xi $ of $\mathcal{Y}$ over $T$ such that the corresponding $1$-morphism $\xi : (\mathit{Sch}/T)_{fppf} \to \mathcal{Y}$ is smooth an surjective. Then $\mathcal{U} = (\mathit{Sch}/T)_{fppf} \times _{\xi , \mathcal{Y}} \mathcal{X}$ is an algebraic stack by assumption (2). Choose a scheme $U$ and a surjective smooth $1$-morphism $(\mathit{Sch}/U)_{fppf} \to \mathcal{U}$. The projection $\mathcal{U} \longrightarrow \mathcal{X}$ is, as the base change of the morphism $\xi : (\mathit{Sch}/T)_{fppf} \to \mathcal{Y}$, surjective and smooth, see Algebraic Stacks, Lemma 94.10.6. Then the composition $(\mathit{Sch}/U)_{fppf} \to \mathcal{U} \to \mathcal{X}$ is surjective and smooth as a composition of surjective and smooth morphisms, see Algebraic Stacks, Lemma 94.10.5. Hence $\mathcal{X}$ is an algebraic stack by Algebraic Stacks, Lemma 94.15.3. $\square$


Comments (0)


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Your email address will not be published.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 05XY. Beware of the difference between the letter 'O' and the digit '0'.