The Stacks project

Lemma 79.12.14. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is separated, flat, and locally of finite presentation. In this case

  1. $(X/Y)_{fin} \to Y$ is separated, representable, and étale, and

  2. if $Y$ is a scheme, then $(X/Y)_{fin}$ is (representable by) a scheme.

Proof. Since $f$ is in particular separated and locally of finite type (see Morphisms of Spaces, Lemma 67.28.5) we see that $(X/Y)_{fin}$ is an algebraic space by Proposition 79.12.11. To prove that $(X/Y)_{fin} \to Y$ is separated we have to show the following: Given a scheme $T$ and two pairs $(a, Z_1)$ and $(a, Z_2)$ over $T$ with identical first component satisfying 79.12.0.1 there is a closed subscheme $V \subset T$ with the following property: For any morphism of schemes $h : T' \to T$ we have

\[ h \text{ factors through } V \Leftrightarrow \Big(T' \times _ T Z_1 = T' \times _ T Z_2 \text{ as subspaces of }T' \times _ Y X\Big) \]

In the proof of Lemma 79.12.7 we have seen that $V = T' \setminus E$ is an open subscheme of $T'$ with closed complement

\[ E = \text{pr}_0|_{Z_1}\left(Z_1 \setminus Z_1 \cap Z_2)\right) \cup \text{pr}_0|_{Z_2}\left(Z_2 \setminus Z_1 \cap Z_2)\right). \]

Thus everything comes down to showing that $E$ is also open. By Lemma 79.12.4 we see that $Z_1$ and $Z_2$ are closed in $T' \times _ Y X$. Hence $Z_1 \setminus Z_1 \cap Z_2$ is open in $Z_1$. As $f$ is flat and locally of finite presentation, so is $\text{pr}_0|_{Z_1}$. This is true as $Z_1$ is an open subspace of the base change $T' \times _ Y X$, and Morphisms of Spaces, Lemmas 67.28.3 and Lemmas 67.30.4. Hence $\text{pr}_0|_{Z_1}$ is open, see Morphisms of Spaces, Lemma 67.30.6. Thus $\text{pr}_0|_{Z_1}\left(Z_1 \setminus Z_1 \cap Z_2)\right)$ is open and it follows that $E$ is open as desired.

We have already seen that $(X/Y)_{fin} \to Y$ is étale, see Proposition 79.12.11. Hence now we know it is locally quasi-finite (see Morphisms of Spaces, Lemma 67.39.5) and separated, hence representable by Morphisms of Spaces, Lemma 67.51.1. The final assertion is clear (if you like you can use Morphisms of Spaces, Proposition 67.50.2). $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 79.12: The finite part of a morphism

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Your email address will not be published.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