Lemma 73.9.4. Let $S$ be a scheme, and let $X$ be an algebraic space over $S$. Suppose that $\{ f_ i : X_ i \to X\} _{i \in I}$ is a family of morphisms of algebraic spaces with target $X$. Let $U \to X$ be a surjective étale morphism from a scheme towards $X$. Then $\{ f_ i : X_ i \to X\} _{i \in I}$ is an fpqc covering of $X$ if and only if $\{ U \times _ X X_ i \to U\} _{i \in I}$ is an fpqc covering of $U$.

**Proof.**
If $\{ X_ i \to X\} _{i \in I}$ is an fpqc covering, then so is $\{ U \times _ X X_ i \to U\} _{i \in I}$ by Lemma 73.9.3. Assume that $\{ U \times _ X X_ i \to U\} _{i \in I}$ is an fpqc covering. Let $h : Z \to X$ be a morphism from an affine scheme towards $X$. Then we see that $U \times _ X Z \to Z$ is a surjective étale morphism of schemes, in particular open. Hence we can find finitely many affine opens $W_1, \ldots , W_ t$ of $U \times _ X Z$ whose images cover $Z$. For each $j$ we may apply the condition that $\{ U \times _ X X_ i \to U\} _{i \in I}$ is an fpqc covering to the morphism $W_ j \to U$, and obtain a standard fpqc covering $\{ W_{jl} \to W_ j\} $ which refines $\{ W_ j \times _ X X_ i \to W_ j\} _{i \in I}$. Hence $\{ W_{jl} \to Z\} $ is a standard fpqc covering of $Z$ (see Topologies, Lemma 34.9.10) which refines $\{ Z \times _ X X_ i \to X\} $ and we win.
$\square$

## 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Your email address will not be published.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)

There are also: