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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 $\mathbb{C}$. )

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    $\begingroup$ Two fields can have the same absolute Galois group and not be isomorphic. The function fields of two Riemann surfaces are isomorphic as function fields (i.e. preserving the copy of $\mathbb{C}$) if and only if the surfaces are isomorphic. But you are asking about isomorphism just as fields and this I don't know. $\endgroup$
    – Felipe Voloch
    Commented Aug 6, 2014 at 21:01

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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).

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    $\begingroup$ The isomorphy type of a complex algebraic curve is not invariant under field automorphisms of $\mathbb{C}$. So there are non-isomorphic compact Riemann surfaces $X,Y$ having $\mathbb{C}(X) \cong \mathbb{C}(Y)$ as abstract fields. But you may show that such $X, Y$ must have the same genus. $\endgroup$
    – Vesselin Dimitrov
    Commented Aug 6, 2014 at 22:16
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    $\begingroup$ Good point, but still that an isomorphism preserves $\mathbf{C}$ does not mean it's identity on $\mathbf{C}$. Indeed, since $\mathbf{C}$ has only countably many elements up to automorphism, it's clear that there are at most countably many finite extensions of $\mathbf{C}(t)$ up to abstract automorphism. $\endgroup$
    – YCor
    Commented Aug 6, 2014 at 22:17
  • $\begingroup$ This is also clear: If a complex algebraic curve is defined over $\mathbb{Q}$, then it is isomorphic to all its conjugates under all field automorphisms of $\mathbb{C}$. $\endgroup$
    – Vesselin Dimitrov
    Commented Aug 6, 2014 at 22:57
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    $\begingroup$ The right question, though, would exclude the curves which are definable over $\bar{\mathbb{Q}}$, or more generally, which have algebraically dependent moduli, in the sense that the minimal field of definition has tr. deg. $< \max(3g-3,1)$ over $\mathbb{Q}$. For $g=1$: since any two transcendental numbers are $\mathrm{Aut}(\mathbb{C})$-conjugate, and since $\mathcal{M}_1(\mathbb{C}) = \mathbb{C}$, we see that the original question has a positive answer if we restrict to genus-$1$ function fields with transcendental $j$-invariant. So in some sense to question is not too far from the truth. $\endgroup$
    – Vesselin Dimitrov
    Commented Aug 7, 2014 at 0:03

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