Timeline for Given an abelian galois map of curves, what are the principal divisors on the source fixed by the galois group?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 28, 2020 at 4:00 | comment | added | reuns | 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. | |
| Feb 28, 2020 at 3:55 | history | edited | Chris H | CC BY-SA 4.0 |
content+misleading typo
|
| Feb 28, 2020 at 3:51 | comment | added | Chris H | 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 $Pic^0$. | |
| Feb 28, 2020 at 3:02 | comment | added | reuns | 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 $Pic^0(Y)$. | |
| Feb 27, 2020 at 23:14 | history | asked | Chris H | CC BY-SA 4.0 |