Proposition 69.11.7. Let $S$ be a scheme. $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume

1. $f$ is of finite type and separated, and

2. $Y$ is quasi-compact and quasi-separated.

Then there exists a separated morphism of finite presentation $f' : X' \to Y$ and a closed immersion $X \to X'$ over $Y$.

Proof. By Lemma 69.11.6 there is a closed immersion $X \to Z$ with $Z/Y$ of finite presentation. Let $\mathcal{I} \subset \mathcal{O}_ Z$ be the quasi-coherent sheaf of ideals defining $X$ as a closed subscheme of $Y$. By Lemma 69.9.2 we can write $\mathcal{I}$ as a directed colimit $\mathcal{I} = \mathop{\mathrm{colim}}\nolimits _{a \in A} \mathcal{I}_ a$ of its quasi-coherent sheaves of ideals of finite type. Let $X_ a \subset Z$ be the closed subspace defined by $\mathcal{I}_ a$. These form an inverse system indexed by $A$. The transition morphisms $X_ a \to X_{a'}$ are affine because they are closed immersions. Each $X_ a$ is quasi-compact and quasi-separated since it is a closed subspace of $Z$ and $Z$ is quasi-compact and quasi-separated by our assumptions. We have $X = \mathop{\mathrm{lim}}\nolimits _ a X_ a$ as follows directly from the fact that $\mathcal{I} = \mathop{\mathrm{colim}}\nolimits _{a \in A} \mathcal{I}_ a$. Each of the morphisms $X_ a \to Z$ is of finite presentation, see Morphisms, Lemma 29.21.7. Hence the morphisms $X_ a \to Y$ are of finite presentation. Thus it suffices to show that $X_ a \to Y$ is separated for some $a \in A$. This follows from Lemma 69.5.13 as we have assumed that $X \to Y$ is separated. $\square$

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