## 60.26 Frobenius action on crystalline cohomology

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

\[ \xymatrix{ X \ar[d] \ar[r]_{F_ X} & X \ar[d] \\ S_0 \ar[r]^{F_{S_0}} & S_0 } \]

is commutative where $F_ X : X \to X$ is the absolute Frobenius morphism of $X$. Thus we obtain a morphism of crystalline topoi

\[ (F_ X)_{\text{cris}} : (X/S)_{\text{cris}} \longrightarrow (X/S)_{\text{cris}} \]

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

\[ F_\mathcal {E} : (F_ X)_{\text{cris}}^*\mathcal{E} \longrightarrow \mathcal{E} \]

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 $V \circ F_{\mathcal{E}} = p^ i \text{id}$.

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
\begin{equation} \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)}) \end{equation}

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
\begin{equation} \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}) \end{equation}

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

60.26.4.3
\begin{equation} \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}) \end{equation}

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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