Let K be a finite field extension of $\mathbb{C}(t)$. Then $K$ is isomorphic to the field of meromorphic functions on a compact Riemann surface $X$ with genius $g$. By an argument similar to the proof of Douady's theorem for $\mathbb{P}^1(\mathbb{C})$ ( cf. chapter3 of Szamuly's Galois groups and fundamental groups) one can show that 
$$\mathrm{Gal}(\bar{K}/K) \cong \widehat{\mathrm{F}}(X\backslash {x\_0} \cup \{ \gamma\_1,\dots,\gamma\_{2g}\}). $$ 


($\widehat{\mathrm{F}}(S)$ is the profinite completion of the free group on generators from a set $S$, $x\_0$ arbitrary point in $X$ and $\gamma\_i$s the standard generators of $\pi\_1(X,x\_0)$.)


It's obvious that the topology and group structure of $\widehat{\mathrm{F}}(S)$ depends only on the cardinality of $S$ and the cardinality of compact Riemann surfaces are the same. So we can conclude that all function fields over $\mathbb{C}$ have isomorphic absolute Galois groups. But are these fields really isomorphic? How one can prove or disprove this? 


( Note that we are interested only in the field structure not in the structure over $\mathbb{C}$. )



Comparison of finite field extensions of $\mathbb{C}(t)$
For any such $K$, we can recover the subfield of constants $\mathbb{C}\subset K$ as the elements of $K$ that have $n$th roots for all $n$. Indeed, if a rational function on a curve has roots of all orders, it must have valuation $0$ at every point and hence be constant. Thus any isomorphism between two such fields $K$ and $K'$ must fix $\mathbb{C}$ setwise, and so can be described as an automorphism of $\mathbb{C}$ followed by an isomorphism of curves. That is, two such fields are isomorphic iff the corresponding curves are conjugate under some automorphism of $\mathbb{C}$. In particular, for example, this means that the genus is invariant under all such isomorphisms (as any of the usual algebro-geometric definitions of genus are preserved by automorphisms of the base field).


