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 29.25.10), 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.
$\square$

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