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Tag 022C

Chapter 33: Topologies on Schemes > Section 33.9: The fpqc topology

Lemma 33.9.6. 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 33.7.2. 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 33.9.1. 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 28.24.9), 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. $\square$

    The code snippet corresponding to this tag is a part of the file topologies.tex and is located in lines 2894–2898 (see updates for more information).

    \begin{lemma}
    \label{lemma-zariski-etale-smooth-syntomic-fppf-fpqc}
    Any fppf covering is an fpqc covering, and a fortiori,
    any syntomic, smooth, \'etale or Zariski covering is an fpqc covering.
    \end{lemma}
    
    \begin{proof}
    We will show that an fppf covering is an fpqc covering, and then the
    rest follows from
    Lemma \ref{lemma-zariski-etale-smooth-syntomic-fppf}.
    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 \ref{definition-fpqc-covering}. 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 \ref{morphisms-lemma-fppf-open}), 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.
    \end{proof}

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