The Stacks project

Proof. Let $f : X \to Y$ be a morphism of schemes. Let $\{ Y_ i \to Y\} $ be an fpqc covering. Write $X_ i = Y_ i \times _ Y X$ and $f_ i : X_ i \to Y_ i$ the base change of $f$. Also denote $q_ i : Y_ i \to Y$ the given flat morphisms. Assume each $f_ i$ is a quasi-compact immersion. By Schemes, Lemma 26.23.8 each $f_ i$ is separated. By Lemmas 35.23.1 and 35.23.6 this implies that $f$ is quasi-compact and separated. Let $X \to Z \to Y$ be the factorization of $f$ through its scheme theoretic image. By Morphisms, Lemma 29.6.3 the closed subscheme $Z \subset Y$ is cut out by the quasi-coherent sheaf of ideals $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ Y \to f_*\mathcal{O}_ X)$ as $f$ is quasi-compact. By flat base change (see for example Cohomology of Schemes, Lemma 30.5.2; here we use $f$ is separated) we see $f_{i, *}\mathcal{O}_{X_ i}$ is the pullback $q_ i^*f_*\mathcal{O}_ X$. Hence $Y_ i \times _ Y Z$ is cut out by the quasi-coherent sheaf of ideals $q_ i^*\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_{Y_ i} \to f_{i, *}\mathcal{O}_{X_ i})$. By Morphisms, Lemma 29.7.7 the morphisms $X_ i \to Y_ i \times _ Y Z$ are open immersions. Hence by Lemma 35.23.16 we see that $X \to Z$ is an open immersion and hence $f$ is a immersion as desired (we already saw it was quasi-compact). $\square$