In this section we prove that Frobenius pullback induces a quasi-isomorphism on crystalline cohomology after inverting the prime $p$. But in order to even formulate this we need to work in a special situation.
Situation 60.26.1. In Situation 60.7.5 assume the following
$S = \mathop{\mathrm{Spec}}(A)$ for some divided power ring $(A, I, \gamma )$ with $p \in I$,
there is given a homomorphism of divided power rings $\sigma : A \to A$ such that $\sigma (x) = x^ p \bmod pA$ for all $x \in A$.
In Situation 60.26.1 the morphism $\mathop{\mathrm{Spec}}(\sigma ) : S \to S$ is a lift of the absolute Frobenius $F_{S_0} : S_0 \to S_0$ and since the diagram
is commutative where $F_ X : X \to X$ is the absolute Frobenius morphism of $X$. Thus we obtain a morphism of crystalline topoi
see Remark 60.9.3. Here is the terminology concerning $F$-crystals following the notation of Saavedra, see [Saavedra].
Definition 60.26.2. In Situation 60.26.1 an $F$-crystal on $X/S$ (relative to $\sigma $) is a pair $(\mathcal{E}, F_\mathcal {E})$ given by a crystal in finite locally free $\mathcal{O}_{X/S}$-modules $\mathcal{E}$ together with a map An $F$-crystal is called nondegenerate if there exists an integer $i \geq 0$ a map $V : \mathcal{E} \to (F_ X)_{\text{cris}}^*\mathcal{E}$ such that $V \circ F_{\mathcal{E}} = p^ i \text{id}$.
\[ F_\mathcal {E} : (F_ X)_{\text{cris}}^*\mathcal{E} \longrightarrow \mathcal{E} \]
Remark 60.26.3. Let $(\mathcal{E}, F)$ be an $F$-crystal as in Definition 60.26.2. In the literature the nondegeneracy condition is often part of the definition of an $F$-crystal. Moreover, often it is also assumed that $F \circ V = p^ n\text{id}$. What is needed for the result below is that there exists an integer $j \geq 0$ such that $\mathop{\mathrm{Ker}}(F)$ and $\mathop{\mathrm{Coker}}(F)$ are killed by $p^ j$. If the rank of $\mathcal{E}$ is bounded (for example if $X$ is quasi-compact), then both of these conditions follow from the nondegeneracy condition as formulated in the definition. Namely, suppose $R$ is a ring, $r \geq 1$ is an integer and $K, L \in \text{Mat}(r \times r, R)$ are matrices with $K L = p^ i 1_{r \times r}$. Then $\det (K)\det (L) = p^{ri}$. Let $L'$ be the adjugate matrix of $L$, i.e., $L' L = L L' = \det (L)$. Set $K' = p^{ri} K$ and $j = ri + i$. Then we have $K' L = p^ j 1_{r \times r}$ as $K L = p^ i$ and It follows that if $V$ is as in Definition 60.26.2 then setting $V' = p^ N V$ where $N > i \cdot \text{rank}(\mathcal{E})$ we get $V' \circ F = p^{N + i}$ and $F \circ V' = p^{N + i}$.
\[ L K' = L K \det (L) \det (M) = L K L L' \det (M) = L p^ i L' \det (M) = p^ j 1_{r \times r} \]
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 becomes an isomorphism after inverting $p$.
\[ F_\mathcal {E} \circ (F_ X)_{\text{cris}}^* : R\Gamma (\text{Cris}(X/S), \mathcal{E}) \otimes ^\mathbf {L}_{A, \sigma } A \longrightarrow R\Gamma (\text{Cris}(X/S), \mathcal{E}) \]
Proof. We first write the arrow as a composition of three arrows. Namely, set
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
Then we have the base change map
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
Finally we can use $F_\mathcal {E}$ to get a map
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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