TITLE: Comparison of finite field extensions of $\mathbb{C}(t)$
QUESTION [6 upvotes]: 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 $\mathbb{C}$. )

REPLY [5 votes]: 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).