Proposition 60.23.1. In Situation 60.7.5. 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 60.15.1 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 60.9.5. Thus Lemma 60.21.2 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. $\square$

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