Lemma 38.32.2. Let $S$ be a quasi-compact and quasi-separated scheme. Let $f : X \to Y$ be a morphism of schemes over $S$ with $Y$ separated and of finite type over $S$ and $X$ compactifyable over $S$. Then $X$ has a compactification over $Y$.

Proof. Let $j : X \to \overline{X}$ be a compactification of $X$ over $S$. Then we let $\overline{X}'$ be the scheme theoretic image of $(j, f) : X \to \overline{X} \times _ S Y$. The morphism $\overline{X}' \to Y$ is proper because $\overline{X} \times _ S Y \to Y$ is proper as a base change of $\overline{X} \to S$. On the other hand, since $Y$ is separated over $S$, the morphism $(1, f) : X \to X \times _ S Y$ is a closed immersion (Schemes, Lemma 26.21.10) and hence $X \to \overline{X}'$ is an open immersion by Morphisms, Lemma 29.6.8 applied to the “partial section” $s = (j, f)$ to the projection $\overline{X} \times _ S Y \to \overline{X}$. $\square$

Comment #4934 by awllower on

Why $(1,f)$ being a closed immersion implies that $X\rightarrow\overline X'$ is an open immersion? I can only tell that $X\times_SY\rightarrow \overline X\times_SY$ is an open immersion, but this does not seem to help.

Comment #4935 by on

You are right! To see this we should appeal to the result stated in Lemma 29.6.8 for the "partial section" $s = (j, f)$ to the projection $\overline{X} \times_S Y \to \overline{X}$. When I last edited this section I tried to add as many references to Lemma 29.6.8 as I could, but I missed this one. Sorry!

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