Cohomology of a constant etale sheaf

Let $X$ be a smooth proper algebraic variety over $\mathbb{C}$.

I know that in the analytic world, there is an isomorphism between the de Rham cohomology and the cohomology of the constant sheaf $\underline{\mathbb{C}}$: $$H^i_{dR}(X^{an}/\mathbb{C}) \cong H^i(X^{an}, \underline{\mathbb{C}}).$$

The proof of this relies on the fact that the sequence of sheaves on $X^{an}$ $$0 \to \underline{\mathbb{C}} \to \Omega^0_{X^{an}} \to \Omega^1_{X^{an}} \to \Omega^2_{X^{an}} \to \cdots$$ is exact.

There is also an isomorphism between $H^i_{dR}(X^{an}/\mathbb{C})$ and $H^i_{dR}(X^{Zar}/\mathbb{C})$. I assume that this also works for $H^i_{dR}(X^{et}/\mathbb{C})$.

However $H^i(X^{Zar}, \underline{\mathbb{C}}) = 0$ because a constant Zariski sheaf is flabby.

What is $H^i(X^{et}, \underline{\mathbb{C}})$?

So far as I can see, neither of the above arguments to calculate this work: a constant étale sheaf is not flabby, because étale open sets need not be connected (actually I am not sure if the definition of flabby sheaves in the Zariski topology makes sense for the étale site).

And the sequence $$0 \to \underline{\mathbb{C}} \to \Omega^0_{X^{et}} \to \Omega^1_{X^{et}} \to \Omega^2_{X^{et}} \to \cdots$$ is not exact because, if $X=\mathbb{P}^1$, its stalks at the origin are $$0 \to \mathbb{C} \to R \to R \, dX \to 0$$ where $R$ is the strict Henselization of $\mathbb{C}[X]_{(X)}$. $R$ is algebraic over $\mathbb{C}[X]$ so does not contain $\log(1+X)$ and hence the 1-form $dX/(1+X)$ is not exact.

• Étale cohomology doesn't work very well with infinite coefficients. I think $H^1$ vanishes in this case. – Zhen Lin May 30 '14 at 9:59
• $H^1$ vanishes, because étale fundamental groups of normal varieties don't have nontrivial maps to the discrete group $\mathbb{C}$. Higher $H^i$ can be complicated. See mathoverflow.net/questions/84414/… – S. Carnahan May 30 '14 at 15:03

The etale cohomology groups $H^i(X_{et},V)$ vanish for $i > 0$ if $X$ is a quasi-compact, quasi-separated, normal scheme, and $V$ is a $\mathbb{Q}$-vector space. (In fact, one may replace "normal" with "geometrically unibranch.") To see this, recall the following presumably classical fact.
Lemma: If $X$ is a normal integral scheme with generic point $j:\eta \to X$, then $\underline{\mathbb{Q}} \simeq R j_* \underline{\mathbb{Q}}$ in $D(X_{et})$.
Proof: We can check this on stalks, so we may assume $X = \mathrm{Spec}(R)$ for a strictly henselian normal local ring $R$. Then $R\Gamma(X_{et},\mathbb{Q}) \simeq \mathbb{Q}$ as $\Gamma(X_{et},-)$ is exact with $X$ connected, so it is enough to show $R\Gamma(\eta_{et},\mathbb{Q}) \simeq \mathbb{Q}$. As etale and Galois cohomology are identified for fields, it is enough to show that $R\Gamma_{cont}(G,\mathbb{Q}) \simeq \mathbb{Q}$ for the absolute Galois group $G$ of $\eta$, which is a standard fact in (profinite) group cohomology.
The lemma implies the claim in the first sentence of this answer for $V = \mathbb{Q}$ after applying the global sections functor $R\Gamma$ (and using the corresponding result for fields). In general, one writes $V$ as a direct sum of copies of $\mathbb{Q}$, and uses that that etale cohomology commutes with arbitrary direct sums (as $X$ is quasi-compact and quasi-separated) to reduce to the previous case.