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Consider a cyclic cover $C$ of $\mathbb{C}P^1$ branched over three points. Let $P, Q\in C$ be branch points. When is $P-Q$ torsion in the Jacobian?

If the cover is totally branched at $P$ and $Q$, then $P-Q$ is torsion, because a multiple of $P-Q$ is the pull back of a degree zero divisor on $\mathbb{C}P^1$. But this is the only result I know about the question.

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  • $\begingroup$ If the ramification index at $P$ and $Q$ is the same you can factor the cover into one totally branched an one totally unbranched. If the totally unbranched cover is $\mathbb CP^1$, you're done, it seems like $P-Q$ will almost certainly not be torsion. This gives a slight strengthening of your result. $\endgroup$
    – Will Sawin
    Feb 4, 2012 at 23:17
  • $\begingroup$ Will, why do you think that P-Q will almost certainly not be torsion? $\endgroup$
    – Alex
    Feb 5, 2012 at 0:27
  • $\begingroup$ It's a pair of unbranched points. Almost all points are unbranched, and almost all differences of points are nontorsion, so almost all differences of unbranched points are nontorsion. There's a minor issue where the pullback of the divisor is not the same as the divisor, but I don't think this changes whether it's torsion. $\endgroup$
    – Will Sawin
    Feb 5, 2012 at 4:16
  • $\begingroup$ I don't think I understand your comments Will. It is not true that every cover can be factored into a totally branched cover and a totally unbranched one. For example $y^4=x^2(x−1)^7$, where we could take $P$ and $Q$ to both lie above $x=0$ or both lie above $x=1$, so they have the same branching. If $P$ and $Q$ do not lie over the same point on $CP^1$ but still have the same ramification, then the cover is not connected unless it is totally branched. $\endgroup$
    – Alex
    Feb 5, 2012 at 15:00
  • $\begingroup$ I'm suggesting the factorization into $z^2=(x-1)$, followed by $y^2=x(x-1)^3z=(z^2+1)z^7$. Two points over $0$ would then be factored into one point over $z=i$ and one point over $z=-i$. There is only one point over $1$ since it's totally branched there. While the cover must be totally branched somewhere (which is not quite true, take $y^2=x/(x-1)$, $z^3=(x-2)/(x-3)$), it need not be totally branched everywhere. $\endgroup$
    – Will Sawin
    Feb 6, 2012 at 1:36

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