Let $O$ be a valuation ring with fraction field $K$ of characteristic zero and residue field $O/m=k$ of characteristic $p>0$, and $X$ be a proper smooth scheme over $O$. Then can we control the mod $p$ etale cohomology of the special fiber by that of the generic fiber? Namely, do we have $\dim_{\Bbb F_p}H^i_{et}(X_k,\Bbb F_p) \leq\dim_{\Bbb F_p}H^i_{et}(X_K,\Bbb F_p)$ in general?

I am interested in the case $O=O_{\Bbb C_p}$, and it turns out mod p etale cohomology vanish for $i>\text{dim}(X_k)$ by Artin-schreier sequence while the $2\text{dim}(X_K)$'s etale cohomology can be nonzero for the generic fiber, see this answer. So they are not of the same dimension in general and it's natural to ask for a bound.

Also, smooth base change theorem is good for $\Bbb Z/\ell$ ($\ell \not=p$) but don't hold if the coefficient sheaf has $p$-torion. From BMS we know $\text{dim}_{\Bbb F_p}H^i_{et}(X_K,\Bbb F_p) \leq \text{dim}_{k}H^i_{dR}(X_k)$, so maybe a lower bound is also possible in our setting.

Edit: The case $i=0,1$ is true. What about the case $i=2$? Moreover, $\text{dim} _{\Bbb F_p}H^i_{et}(X,\Bbb F_p)$ is finite and less than $\text{dim}_k H^i(X_k, O_k)$, see Lemma 0A3L for a related semi-linear algebra result (so maybe we need to understand the Frobenius action).

Edit: Note that $LHS \leq \text{dim}_{k}H^i_{k}(X_k,O_{X_k})\geq \text{dim}_{C}H^i(X_C,O_{X_C}) \leq \text{dim}_{C}H^i_{sing}(X_K^{an},K) \leq RHS. $ The only $\geq$ is due to upper semi-continuities, and we use an abstract isomorphism between $\Bbb C$ and $K$ to apply singular cohomologies, Hodge-de Rham decomposition and the universal coefficient theorem. So if $\text{dim}_{k}H^i_{k}(X_k,O_{X_k})=\text{dim}_{C}H^i(X_C,O_{X_C})$ then what we want holds, however this is not true in general. For example, one considers lifting of a singular Enriques surface in char $2$, but even in this case the inequality still holds (because the second Betti number is $10$, which is much larger than $h^{2,0}=1$).

  • $\begingroup$ What's your argument for $i=1$? $\endgroup$ – gdb Oct 1 '18 at 4:29
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    $\begingroup$ @gdb Possible argument: $\pi_1(X_{\bar K})\to \pi_1(X_{\bar k})$ is surjective (SGA1), so $\mathrm{Hom}( \pi_1(X_{\bar k}), \mathbf{F}_p)\to \mathrm{Hom}( \pi_1(X_{\bar K}), \mathbf{F}_p)$ is injective, but $\mathrm{Hom}(\pi_1, \mathbf{F}_p) = H^1(\pi_1, \mathbf{F}_p) = H^1(X, \mathbf{F}_p)$? $\endgroup$ – Piotr Achinger Oct 1 '18 at 11:44

(Not an answer, just a suggestion as to where to look.)

When $O$ is a Henselian DVR, the groups $\mathrm{H}^i(X_K, \mathbb{Z}/p)$ were studied by Bloch and Kato in "p-adic étale cohomology", Pub. math. IHÉS 63 (1986). They use Leray + proper base change to get a spectral sequence

$E_2^{p,q} = \mathrm{H}^p(X_k, i^* \mathrm{R}^q j_* \mathbb{Z}/p) \Rightarrow \mathrm{H}^{p+q}(X_K, \mathbb{Z}/p)$

where, as usual, $i$ and $j$ are the inclusions of the special and generic fibres, respectively.

They then study the sheaves $i^* \mathrm{R}^q j_* \mathbb{Z}/p(q)$ by putting a filtration on them and relating the graded pieces of the filtration to various spaces of differentials on the special fibre.

I suppose that to answer your question you'd need to understand not only the groups in this spectral sequence but also the maps...


There was a nonsense answer here earlier. It is now removed.

  • $\begingroup$ Why does Artin-Schreier theory imply that $H^i(Y,\mathbb F_p) \otimes_{\mathbb F_p} k \to H^i(Y,\mathcal O_Y)$ is injective? I guess I don't understand how $\operatorname{id} - \operatorname{Frob}$ acts on $H^i(Y,\mathcal O_Y)$ (although the $-\otimes_{\mathbb F_p} k$ step also feels dangerous). $\endgroup$ – R. van Dobben de Bruyn Sep 29 '18 at 0:49
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    $\begingroup$ @R.vanDobbendeBruyn By a semilinear lemma 0A3L the Artin-Schreier long exact sequence splits, see stacks.math.columbia.edu/tag/0A3J $\endgroup$ – sawdada Sep 29 '18 at 1:23
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    $\begingroup$ While the argument is clever, and you get that the special fiber's etale cohomology can be controlled by the de Rham cohomology in special fiber, buy how do you control the latter term by the etale cohomology of the generic fiber? (BMS is for the opposite direction) $\endgroup$ – sawdada Sep 29 '18 at 1:28
  • $\begingroup$ Good point. I removed the wrong answer, and failed to find a correct answer. $\endgroup$ – Boa Sep 29 '18 at 2:41

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