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?