Lemma 34.9.7. 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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