The Stacks project

Lemma 73.11.18. The property $\mathcal{P}(f) =$“$f$ is separated” is fpqc local on the base.

Proof. A base change of a separated morphism is separated, see Morphisms of Spaces, Lemma 66.4.4. Hence the direct implication in Definition 73.10.1.

Let $\{ Y_ i \to Y\} _{i \in I}$ be an fpqc covering of algebraic spaces over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume each base change $X_ i := Y_ i \times _ Y X \to Y_ i$ is separated. This means that each of the morphisms

\[ \Delta _ i : X_ i \longrightarrow X_ i \times _{Y_ i} X_ i = Y_ i \times _ Y (X \times _ Y X) \]

is a closed immersion. The base change of a fpqc covering is an fpqc covering, see Topologies on Spaces, Lemma 72.9.3 hence $\{ Y_ i \times _ Y (X \times _ Y X) \to X \times _ Y X\} $ is an fpqc covering of algebraic spaces. Moreover, each $\Delta _ i$ is the base change of the morphism $\Delta : X \to X \times _ Y X$. Hence it follows from Lemma 73.11.17 that $\Delta $ is a closed immersion, i.e., $f$ is separated. $\square$

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