I would like to have a precise description of the finite abelian extensions of the field $ K = \Bbb C(X)$. Typically, can we describe the abelianization of its absolute Galois group $G_K$?
Thoughts:
It is known for instance that the abelianization of the absolute Galois group of $\Bbb Q$ is $\widehat{\Bbb Z}^{\times}$. Maybe there is a geometric approach to my problem, via compact Riemann surfaces (equivalently, via smooth projective algebraic curves over $\Bbb C$). As mentioned in this question, the finite abelian extensions of $K$ correspond to "abelian covers" of $\Bbb P^1_{\Bbb C}$. I don't know if this is could be related to geometric class field theory, somehow?
My idea was to use Kummer theory, since all the roots of unity belong to $K$ and $K$ has characteristic $0$. The map $L \mapsto L^{\times, n} \cap K^{\times}$ is a bijection between the set of abelian extensions $L/K$ of exponent $n$ and the subgroups $K^{\times, n} \leq H \leq K^{\times}$. Under this bijection, if $L/K$ is finite, then its Galois group is isomorphic to $H / K^{\times, n}$. Since $K$ is the fraction field of the UFD $\Bbb C[X]$, whose units are well-known, I think that we have $$K^{\times} \cong \Bbb C^{\times} \oplus \Bbb Z^{(\Bbb C)}$$ where$^{(1)}$ $u \prod\limits_{\alpha \in \Bbb C} (X - \alpha)^{n_{\alpha}}$ is sent to $(u, (n_\alpha))$. But then there are too many subgroups... and I don't know how to relate it to $\varprojlim\limits_{L/K \text{abelian finite}} \mathrm{Gal}(L/K)$. It is mentioned here that the maximal pro-$p$ quotient of $G_K^{\mathrm{ab}}$ is $\Bbb Z_p^{(\Bbb C)}$.
Some other notes: the absolute Galois group of $K$ is the profinite completion of the free group of rank $2^{\aleph_0}$, according to the paper cited here. The algebraic closure of $K$ is difficult to describe, actually. If we are interested in the $X$-adic completion of $K$, then we get $\Bbb C((X))$, which is a quasi-finite field (i.e. its absolute Galois group is the pro-cyclic group $\widehat{\Bbb Z}$).
Thank you for your help!
$^{(1)}$ For instance the subgroup $$H = \left\{ u (X-a)^{k_a} \prod_{b \neq a} (X-b)^{n k_b} \mid u \in \Bbb C^{\times}, k_b = 0 \text{ for almost all } b \right\} \leq K^{\times}$$ corresponds to the abelian extension $L = K(\sqrt[n]{H}) = \Bbb C\left( \sqrt[n]{X-a} \right)/K$.
