Etale cohomology with coefficients in $\mathbb{Q}$

Question

Let $X$ be a smooth variety of a field $k$. Then is $$H_{et}^i(X, \mathbb{Q}) = 0$$ for all $i > 0$?

• The result is true for $i=1$. This follows from the same argument given for $\mathbb{Z}$-coefficients given here: Etale cohomology with coefficients in the integers. Note that one needs to at the very least assume that $X$ is normal; otherwise $H_{et}^1(X, \mathbb{Q})$ can be non-trivial.
• The result is true if $X = \mathrm{Spec}(k)$. This follows from standard properties of Galois cohomology. Note that the analogous result is not true with $\mathbb{Z}$-coefficients; $H^2(k, \mathbb{Z})$ is often non-trivial.
• If it helps, I mostly care about the case $i=2$.

Answer 1

The following is surely expressing whatever is in the core non-formal aspect of Joe Berner's answer (which is above my pay grade); it is offered as an alternative version of the same ideas.

Let $X$ be a normal noetherian scheme. We'll show the higher etale cohomology with coefficients in any flat $\mathbf{Z}$-module $M$ is torsion, and hence vanishes when $M$ is a $\mathbf{Q}$-vector space. Via the exact sequence $$0 \rightarrow M \rightarrow \mathbf{Q} \otimes_{\mathbf{Z}} M \rightarrow T \rightarrow 0$$ with $T$ visibly torsion, we just have to show the higher etale cohomology of the middle term vanishes; i.e., we are back to the original question of vanishing of higher etale cohomology with coefficients in a $\mathbf{Q}$-vector space $V$.

We can assume $X$ is connected, so it is irreducible (as $X$ is noetherian and normal); let $\eta$ be its unique generic point. For any quasi-compact etale $E \rightarrow X$, the scheme $E$ is normal and noetherian, so its connected components are all irreducible and as such have their unique generic point over $\eta$ (due to flatness of $E$ over $X$). Thus, $E \mapsto E_{\eta}$ does not lose any information about connected components, so for the natural map $j:\eta \rightarrow X$ we see that the natural map of etale sheaves $V_X \rightarrow j_{\ast}(V_{\eta})$ induces an isomorphism upon evaluating on any such $E$, whence it is an isomorphism of sheaves. Thus, our task is the same as proving the vanishing of the higher etale cohomology on $X$ of $j_{\ast}(V_{\eta})$.

Consider the Leray spectral sequence $$E_2^{n,m} = {\rm{H}}^n(X, {\rm{R}}^mj_{\ast}(V_{\eta})) \Rightarrow {\rm{H}}^{n+m}(\eta,V_{\eta}).$$ The sheaf ${\rm{R}}^m j_{\ast}(V_{\eta}$ is the etale sheafification of the presheaf on $X_{\rm{et}}$ given by $U \mapsto {\rm{H}}^m(U_{\eta},V)$ (with quasi-compact etale $U$ over $X$). The latter vanishes for $m>0$ since it is Galois cohomology (for the finitely many points of $U_{\eta}$) and $V$ is a $\mathbf{Q}$-vector space. Hence, $E_2^{n,m}=0$ for $m>0$, so $${\rm{H}}^n(X, j_{\ast}(V_{\eta})) \simeq {\rm{H}}^n(\eta,V_{\eta})$$ for all $n \ge 0$. But the right side vanishes for $n>0$ for the same Galois-cohomological reasons, so we are done.

Answer 2

Yes. It is a theorem that for a quasi-compact quasi-separated locally noetherian and normal scheme $X$ that its protruncated etale homotopy type is already profinite. You can look at DAG XIII for details, but this is stated in slightly more/less generality in Artin and Mazur's book. That means that its truncated etale homotopy type is represented by some cofiltered diagram of $\pi$-finite spaces $\operatorname{Ver}_X : D \rightarrow \mathcal{S}$ (roughly, the Verdier functor in Artin-Mazur). The point here is that for any constant sheaf of abelian groups $A$ we have that the sheaf cohomology $H^i(X,\underline{A})$ is isomorphic to $\lim_{d \in D} H^i(\operatorname{Ver}_X(d),A)$. This reduces our question to showing that $H^i(Z,\mathbb{Q})$ is zero for all $i>0$ and all '$\pi$-finite spaces' $Z$, which means that $Z$ has finitely many connected components, and finitely many non-zero homotopy groups (for arbitrary basepoints $z \in Z$) which are all finite groups.

We can do this pretty explicitly. Cohomology commutes with disjoint union, so assume that $Z$ is connected, its universal cover $\tilde{Z}$ is of course simply connected, and the rational Hurewicz theorem implies it has trivial rational homology in positive degrees. The universal coefficient theorem then implies it has trivial rational cohomology in positive degrees. We know that the rational cohomology of $Z$ and $\tilde{Z}$ differ by a spectral sequence $E_2^{p,q} = H^p(\pi_1(Z,z),H^q(\tilde{Z},\mathbb{Q})) \Rightarrow H^{p+q}(Z,\mathbb{Q})$. This is just a row spectral sequence, and so the question reduces to the statement that finite groups have torsion cohomology in positive degrees, which seems to be well known (although I don't have a source).

Answer 3

You can find this for normal schemes in Deninger, A proper base change theorem for non-torsion sheaves in étale cohomology, Journal of Pure and Applied Algebra 50 (1988), 231–235. http://www.sciencedirect.com/science/article/pii/0022404988901028 (2.1) (proof by reduction to Galois cohomology using the Leray spectral sequence)