The Stacks project

Lemma 78.12.14. 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 ├ętale, 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 66.28.5) we see that $(X/Y)_{fin}$ is an algebraic space by Proposition 78.12.11. 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 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 78.12.7 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 78.12.4 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 66.28.3 and Lemmas 66.30.4. Hence $\text{pr}_0|_{Z_1}$ is open, see Morphisms of Spaces, Lemma 66.30.6. 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 ├ętale, see Proposition 78.12.11. Hence now we know it is locally quasi-finite (see Morphisms of Spaces, Lemma 66.39.5) and separated, hence representable by Morphisms of Spaces, Lemma 66.51.1. The final assertion is clear (if you like you can use Morphisms of Spaces, Proposition 66.50.2). $\square$

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