Let $ X_0$ be a smooth projective variety over $ \mathbf{F}_q$ and $ {X}$ its base change to an algebraic closure $ k$ of $ \mathbf{F}_q$ .

Crystalline cohomology $ H^*_{\rm cris}(X) := H^*((X/W(k))_{\rm cris},\mathcal{O}_{(X/W)_{\rm cris}})[1/p]$ is known to be a Weil cohomology.

It is also known to be computed as the Zariski hypercohomology of the de Rham-Witt complex $ W\Omega^*_X$ , and the Hodge to de Rham spectral sequence:

$ $ E_2^{i,j} := H^i(X_{\rm Zar},W\Omega^j_X)[1/p]\Rightarrow H^{i+j}_{\rm cris}(X)$ $

degenerates at the second page.

- Do we therefore have a “Hodge decomposition” $ $ \bigoplus_{i+j=n}H^i(X_{\rm Zar},W\Omega^j_X)[1/p]=H^n_{\rm cris}(X)\ \ ?$ $
- When $ n$ is even and $ i=j=n/2$ , does the geometric Frobenius act on $ H^i(X_{\rm Zar},W\Omega^i_X)$ by multiplication by $ q^i$ ?
- Are the Frobenius eigenvalues on $ H^i(X_{\rm Zar},W\Omega^j_X)[1/p]$ (base changed to an algebraic closure of $ \text{Frac}(W)$ )
notintegers when $ i\neq j$ ?- And on a more naive level, where do we crucially use that $ k$ is algebraically closed? We know there is no Weil cohomology with coefficients in $ \mathbf{Q}_p$ , so when $ p = q$ we know crystalline cohomology with $ W(\mathbf{F}_p)[1/p]$ coefficients can’t work. Where do the proofs of the axioms break down when $ k$ is not algebraically closed?