Lemma 34.9.6. Any fppf covering is an fpqc covering, and a fortiori, any syntomic, smooth, étale or Zariski covering is an fpqc covering.

**Proof.**
We will show that an fppf covering is an fpqc covering, and then the rest follows from Lemma 34.7.2. Let $\{ f_ i : U_ i \to U\} _{i \in I}$ be an fppf covering. By definition this means that the $f_ i$ are flat which checks the first condition of Definition 34.9.1. To check the second, let $V \subset U$ be an affine open subset. Write $f_ i^{-1}(V) = \bigcup _{j \in J_ i} V_{ij}$ for some affine opens $V_{ij} \subset U_ i$. Since each $f_ i$ is open (Morphisms, Lemma 29.24.10), we see that $V = \bigcup _{i\in I} \bigcup _{j \in J_ i} f_ i(V_{ij})$ is an open covering of $V$. Since $V$ is quasi-compact, this covering has a finite refinement. This finishes the proof.
$\square$

## 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Your email address will not be published.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: