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. $\square$

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