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.