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Tag 07LM

Chapter 51: Crystalline Cohomology > Section 51.23: Applications

Proposition 51.23.1. In Situation 51.7.5. Let $\mathcal{F}$ be a crystal in quasi-coherent modules on $\text{Cris}(X/S)$. The truncation map of complexes $$ (\mathcal{F} \to \mathcal{F} \otimes_{\mathcal{O}_{X/S}} \Omega^1_{X/S} \to \mathcal{F} \otimes_{\mathcal{O}_{X/S}} \Omega^2_{X/S} \to \ldots) \longrightarrow \mathcal{F}[0], $$ while not a quasi-isomorphism, becomes a quasi-isomorphism after applying $Ru_{X/S, *}$. In fact, for any $i > 0$, we have $$ Ru_{X/S, *}(\mathcal{F} \otimes_{\mathcal{O}_{X/S}} \Omega^i_{X/S}) = 0. $$

Proof. By Lemma 51.15.1 we get a de Rham complex as indicated in the lemma. We abbreviate $\mathcal{H} = \mathcal{F} \otimes \Omega^i_{X/S}$. Let $X' \subset X$ be an affine open subscheme which maps into an affine open subscheme $S' \subset S$. Then $$ (Ru_{X/S, *}\mathcal{H})|_{X'_{Zar}} = Ru_{X'/S', *}(\mathcal{H}|_{\text{Cris}(X'/S')}), $$ see Lemma 51.9.5. Thus Lemma 51.21.2 shows that $Ru_{X/S, *}\mathcal{H}$ is a complex of sheaves on $X_{Zar}$ whose cohomology on any affine open is trivial. As $X$ has a basis for its topology consisting of affine opens this implies that $Ru_{X/S, *}\mathcal{H}$ is quasi-isomorphic to zero. $\square$

    The code snippet corresponding to this tag is a part of the file crystalline.tex and is located in lines 4292–4308 (see updates for more information).

    \begin{proposition}
    \label{proposition-compare-with-de-Rham}
    In Situation \ref{situation-global}.
    Let $\mathcal{F}$ be a crystal in quasi-coherent modules on
    $\text{Cris}(X/S)$. The truncation map of complexes
    $$
    (\mathcal{F} \to
    \mathcal{F} \otimes_{\mathcal{O}_{X/S}} \Omega^1_{X/S} \to
    \mathcal{F} \otimes_{\mathcal{O}_{X/S}} \Omega^2_{X/S} \to \ldots)
    \longrightarrow \mathcal{F}[0],
    $$
    while not a quasi-isomorphism, becomes a quasi-isomorphism after applying
    $Ru_{X/S, *}$. In fact, for any $i > 0$, we have 
    $$
    Ru_{X/S, *}(\mathcal{F} \otimes_{\mathcal{O}_{X/S}} \Omega^i_{X/S}) = 0.
    $$
    \end{proposition}
    
    \begin{proof}
    By Lemma \ref{lemma-automatic-connection} we get a de Rham complex
    as indicated in the lemma. We abbreviate
    $\mathcal{H} = \mathcal{F} \otimes \Omega^i_{X/S}$.
    Let $X' \subset X$ be an affine open
    subscheme which maps into an affine open subscheme $S' \subset S$.
    Then
    $$
    (Ru_{X/S, *}\mathcal{H})|_{X'_{Zar}} =
    Ru_{X'/S', *}(\mathcal{H}|_{\text{Cris}(X'/S')}),
    $$
    see Lemma \ref{lemma-localize}. Thus
    Lemma \ref{lemma-cohomology-is-zero}
    shows that $Ru_{X/S, *}\mathcal{H}$ is a complex of sheaves on
    $X_{Zar}$ whose cohomology on any affine open is trivial.
    As $X$ has a basis for its topology consisting of affine opens
    this implies that $Ru_{X/S, *}\mathcal{H}$ is quasi-isomorphic to zero.
    \end{proof}

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