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Let $f:X\rightarrow Y$ be an abelian galois map (not necessarily unramified) of nonsingular complete curves over algebraically closed $k$, where the order of the galois group $A$ is coprime to the characteristic of $k$. We can view the function field $K(X)$ as a $K(Y)$ vector space, and by galois theory we know its structure as a $k[A]$ module is given by $K(Y)[A]$.

So given our assumptions, we see that $K(X)$ has a basis of eigenvectors for $A$, corresponding the the distinct one dimensional representations of $A$ over $k$. Explicitly, it has a basis of $f_\lambda\in K(X)$ such that $g.f_\lambda=\lambda(g).f$ for $\lambda(g)$ an element of $K(Y)$. So the divisors associated to these $f_\lambda$ are canonical up to principal divisors from $Y$, and we have one for each character of $A$.

My question is, can you describe these divisors more geometrically?

I am vaguely aware that all coverings of this form should come from isogenies of the jacobian, but my knowledge of abelian covers of curves is not great, so apologies if this question is something really simple.

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  • $\begingroup$ Assume that $Gal(K(X)/K(Y))$ is cyclic of degree $n$. Then $K(X)=K(Y)[f^{1/n}]$ and $K(X)/K(Y)$ is unramified (so $X\to Y$ is a covering) iff every zeros and poles of $f$ are of order $ln$. Thus the unramified coverings of $Y$ correspond to the $D\in Div^0(Y)$ such that $nD\in Prin(Y)$ through $nD=Div(f)$ ie. to the points of order $n$ in $Pic^0(Y)$. The non-cyclic case is to generate an abelian extension from any finite subgroup of $Pic^0(Y)$. $\endgroup$
    – reuns
    Commented Feb 28, 2020 at 3:02
  • $\begingroup$ Sorry, I'll edit the question, I used covering forgetting the technical meaning, I just mean a nonconstant map of curves, possibly with ramification. I'll think about your comment further, though I don't see how the noncyclic case works, but that's probably because I'm not comfortable with $Pic^0$. $\endgroup$
    – Chris H
    Commented Feb 28, 2020 at 3:51
  • $\begingroup$ For non-cyclic abelian unramified coverings let $K(X)=K(Y)[f_1^{1/n},\ldots,f_j^{1/n}]$ where $Div(f_j)$ correspond as above to cyclic coverings. I don't think the ramified cyclic extensions correspond to some points of the Jacobian (which is isomorphic to $Pic^0$) try with $K(Y)$ an elliptic curve, it is its own Jacobian and the finite subgroups correspond to isogenies which are unramified. It might work if you replace $Pic^0$ by the class group of $O_Y(Y-\{p_1,\ldots,p_l\})$ where $p_l$ contain the ramified points of your branched covering. $\endgroup$
    – reuns
    Commented Feb 28, 2020 at 4:00

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