Singular cohomology of fields

Question

Let's define the singular cohomologies of function fields of complex varieties, as the direct limit of the singular cohomologies of Zariski opens of the variety with analytic topology. So for a complex variety $X$ where its function field is $F(X)$, we have defined $H^i(F(X), \mathbb{Z})$. Similarly let's define singular cohomology of the algebraic closure $\overline{F(X)}$, as the direct limit of singular cohomology of finite extensions of $F(X)$. Now my question is: Is the cohomology ring of $\overline{F(X)}$ generated by the first cohomology group and cup product?

Is this true rationally when considering the generic point of the variety? (is $H^i(F(X), \mathbb{Q})$ generated by $H^1$?). Are there any examples that the cohomology ring of the generic point has been calculated (other than curves).

Edit: As a motivation I wanted to add this: this is true if we consider cohomology with finite coefficients and it follows from Bloch-Kato.

Edit2: I had another similar question and did not want to ask a separate question. Assuming there is a counter-example for the question above about the singular cohomology of $\overline{F(X)}$ (I am not aware of it at this point), I was wondering whether the following weaker statement has a chance of being true:

• For a smooth projective complex variety $X$ and a cohomology class $\alpha \in H^i(X, \mathbb{Q})$, there is a quasi-projective variety $Y$ and an etale map (not necessarily a surjective one) $f:Y\rightarrow X$ such that $f^*\alpha\in H^i(Y, \mathbb{Q})$ is generated by $H^1(Y, \mathbb{Q})$ and cup product.

Answer 1

Considering the direction of the arrows, I think it should an inverse limit, but in any case, the answer would be no with $\mathbb{Q}$ coefficients.

Let $X$ be a projective K3 surface over $\mathbb{C}$. By an easy residue calculation, for any Zariski open $U\subseteq X$, the mixed Hodge structure on $H^1(U,\mathbb{Q})$ should be pure of type $(1,1)$, i.e. a sum of $\mathbb{Q}(-1)$'s. On the other hand, by a similar calculation, $H^2(U)$ has a nonzero $(2,0)$ part. So the cup product cannot induce a surjection $\wedge^2 H^1(U)\to H^2(U)$. (Shrinking $U$ won't change the outcome.)