Theorem 60.26.4. In Situation 60.26.1 let $(\mathcal{E}, F_\mathcal {E})$ be a nondegenerate $F$-crystal. Assume $A$ is a $p$-adically complete Noetherian ring and that $X \to S_0$ is proper smooth. Then the canonical map

$F_\mathcal {E} \circ (F_ X)_{\text{cris}}^* : R\Gamma (\text{Cris}(X/S), \mathcal{E}) \otimes ^\mathbf {L}_{A, \sigma } A \longrightarrow R\Gamma (\text{Cris}(X/S), \mathcal{E})$

becomes an isomorphism after inverting $p$.

Proof. We first write the arrow as a composition of three arrows. Namely, set

$X^{(1)} = X \times _{S_0, F_{S_0}} S_0$

and denote $F_{X/S_0} : X \to X^{(1)}$ the relative Frobenius morphism. Denote $\mathcal{E}^{(1)}$ the base change of $\mathcal{E}$ by $\mathop{\mathrm{Spec}}(\sigma )$, in other words the pullback of $\mathcal{E}$ to $\text{Cris}(X^{(1)}/S)$ by the morphism of crystalline topoi associated to the commutative diagram

$\xymatrix{ X^{(1)} \ar[r] \ar[d] & X \ar[d] \\ S \ar[r]^{\mathop{\mathrm{Spec}}(\sigma )} & S }$

Then we have the base change map

60.26.4.1
$$\label{crystalline-equation-base-change-sigma} R\Gamma (\text{Cris}(X/S), \mathcal{E}) \otimes ^\mathbf {L}_{A, \sigma } A \longrightarrow R\Gamma (\text{Cris}(X^{(1)}/S), \mathcal{E}^{(1)})$$

see Remark 60.24.8. Note that the composition of $F_{X/S_0} : X \to X^{(1)}$ with the projection $X^{(1)} \to X$ is the absolute Frobenius morphism $F_ X$. Hence we see that $F_{X/S_0}^*\mathcal{E}^{(1)} = (F_ X)_{\text{cris}}^*\mathcal{E}$. Thus pullback by $F_{X/S_0}$ is a map

60.26.4.2
$$\label{crystalline-equation-to-prove} F_{X/S_0}^* : R\Gamma (\text{Cris}(X^{(1)}/S), \mathcal{E}^{(1)}) \longrightarrow R\Gamma (\text{Cris}(X/S), (F_ X)^*_{\text{cris}}\mathcal{E})$$

Finally we can use $F_\mathcal {E}$ to get a map

60.26.4.3
$$\label{crystalline-equation-F-E} R\Gamma (\text{Cris}(X/S), (F_ X)^*_{\text{cris}}\mathcal{E}) \longrightarrow R\Gamma (\text{Cris}(X/S), \mathcal{E})$$

The map of the theorem is the composition of the three maps (60.26.4.1), (60.26.4.2), and (60.26.4.3) above. The first is a quasi-isomorphism modulo all powers of $p$ by Remark 60.24.9. Hence it is a quasi-isomorphism since the complexes involved are perfect in $D(A)$ see Remark 60.24.13. The third map is a quasi-isomorphism after inverting $p$ simply because $F_\mathcal {E}$ has an inverse up to a power of $p$, see Remark 60.26.3. Finally, the second is an isomorphism after inverting $p$ by Lemma 60.25.6. $\square$

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