The Stacks project

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

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

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). $\square$