# The Stacks Project

## Tag 07LM

Proposition 54.23.1. In Situation 54.7.5. 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 54.15.1 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 54.9.5. Thus Lemma 54.21.2 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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