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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