Lemma 74.11.18. 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 67.4.4. Hence the direct implication in Definition 74.10.1.

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

is a closed immersion. The base change of a fpqc covering is an fpqc covering, see Topologies on Spaces, Lemma 73.9.3 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 74.11.17 that $\Delta $ is a closed immersion, i.e., $f$ is separated. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)