The Stacks project

Lemma 67.27.9. Let $S$ be a scheme. Let $f : X \to Y$ be a finite type morphism of algebraic spaces over $S$. Let $y \in |Y|$. There are at most finitely many points of $|X|$ lying over $y$ at which $f$ is quasi-finite.

Proof. Choose a field $k$ and a morphism $\mathop{\mathrm{Spec}}(k) \to Y$ in the equivalence class determined by $y$. The fibre $X_ k = \mathop{\mathrm{Spec}}(k) \times _ Y X$ is an algebraic space of finite type over a field, in particular quasi-compact. The map $|X_ k| \to |X|$ surjects onto the fibre of $|X| \to |Y|$ over $y$ (Properties of Spaces, Lemma 66.4.3). Moreover, the set of points where $X_ k \to \mathop{\mathrm{Spec}}(k)$ is quasi-finite maps onto the set of points lying over $y$ where $f$ is quasi-finite by Lemma 67.27.2. Choose an affine scheme $U$ and a surjective étale morphism $U \to X_ k$ (Properties of Spaces, Lemma 66.6.3). Then $U \to \mathop{\mathrm{Spec}}(k)$ is a morphism of finite type and there are at most a finite number of points where this morphism is quasi-finite, see Morphisms, Lemma 29.20.14. Since $X_ k \to \mathop{\mathrm{Spec}}(k)$ is quasi-finite at a point $x'$ if and only if it is the image of a point of $U$ where $U \to \mathop{\mathrm{Spec}}(k)$ is quasi-finite, we conclude. $\square$


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Your email address will not be published.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