The Stacks project

Lemma 101.48.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $f$ is quasi-compact and surjective and $\mathcal{X}$ is decent, then $\mathcal{Y}$ is decent.

Proof. Let $x : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism where $k$ is a field and denote $y = f \circ x$. Since $f$ is surjective, every point of $|\mathcal{Y}|$ arises in this manner, see Properties of Stacks, Lemma 100.4.4. Consider an affine scheme $T$ and morphism $T \to \mathcal{Y}$. It suffices to show that $T \times _{\mathcal{Y}, y} \mathop{\mathrm{Spec}}(k)$ is quasi-compact, see Lemma 101.7.10. We have

\[ (T \times _{\mathcal{Y}} \mathcal{X}) \times _{\mathcal{X}, x} \mathop{\mathrm{Spec}}(k) = T \times _{\mathcal{Y}, y} \mathop{\mathrm{Spec}}(k) \]

The morphism $T \times _{\mathcal{Y}} \mathcal{X} \to T$ is quasi-compact hence $T \times _\mathcal {Y} \mathcal{X}$ is quasi-compact. Since $x$ is a quasi-compact morphism as $\mathcal{X}$ is decent we see that the displayed fibre product is quasi-compact. $\square$


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Your email address will not be published.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 0GW4. Beware of the difference between the letter 'O' and the digit '0'.