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There is serious flaw in the following argument, but I have yet to see what it is:

Let $C$ be a genus $g$ hyperelliptic curve in an abelian variety $A$ (over an algebraically closed field with characteristic $\neq$2), such that the automorphism $-1$ of $A$ descends to the hyperelliptic involution on $C$. The 2:1 projection $C\to \mathbb{P}^1$ is just the image of $C$ in the Kummer variety of $A$, that is $\mbox{Km}(A)=A/\pm$, and so $C$ contains $2g+2$ 2-torsion points of $A$. Let $2_A$ be the morphism multiplication by 2 on $A$, and let $C'=2_A(C)$. This is a (possibly singular) curve on $A$, and the image of $C'$ in $\mbox{Km}(A)$ is also rational (since it is the image of $2_{\mbox{Km}(A)}$ restricted to the projection of $C$ in $\mbox{Km}(A)$). If we take the normalization of $C'$, let's say $\pi:N\to C'$, then we get a map $C\to N$, that lifts $2_A:C\to C'$. Now, $N$ is hyperelliptic and if $g\geq 2$, we must have that $g_N$ (the genus of $N$) is less than $g$, unless the map $C\to N$ is an isomorphism. The map can never be an isomorphism, given that the $2g+2$ torsion points on $C$ all go to 0 in $C'$, and above each point in $C'$ there are at most 2 points in $N$.

One very false implication of this reasoning is that if $C$ is of genus 2, for example, and $A$ is the jacobian of $C$, then $N$ would have to be an elliptic curve! This would mean that every 2-dimensional abelian variety is isogenous to the product of elliptic curves, which is very very false!

What's wrong with this argument?

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It is true that $C$ contains some $2$-torsion points of $A$. If $P'\in C'$ is the image of $P\in C$, then for all $2$-torsion points $t\in A[2]$, $P+t\in A$ also maps to $P'$, but $P+t\notin C$ in general. So $C\to C'$ needs not be of degree $>1$.

Suppose $C$ has no other automorphism than the hyperelliptic involution and $C\to A$ is the embedding of $C$ into its Jacobian given by $P\mapsto [P-P_0]$. Let's show that $C\to C'$ is birational.

First for any $t\in A[2]$ non-zero, we have $C+t\ne C$ because otherwise the translation by $t$ would be an automorphism of $C$ of order 2 without fixed point, so a non-trivial automorphism different from the hyperelliptic involution.

So $C\cap (C+t)$ is finite and $\cup_{t\in A[2], t\ne 0 } C\cap (C+t)$ is finite. Let $P\in C$ be a point not in this finite set. If $P, Q$ map to the same point in $C'$, then $2P\sim 2Q$, so $[P-Q]=t\in A[2]$. If $Q\ne P$, then $Q\not\sim P$ and $t\ne 0$. When $C$ is identified with its image in $A$, this means $P\in Q+t \in C\cap (C+t)$, contradiction.

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  • $\begingroup$ Ok, perfect. Now, one thing I'm still not quite getting (from my reasoning) is the following: We know that $C$ has 2g+2 points fixed by the hyperelliptic involution, and so when we identify $C$ with its image in $A$, we get that $C$ has $2g+2$ 2-torsion points of $A$. From what you just said, $C$ is the normalization of $C'$ (assuming $C$ doesn't have non-trivial automorphisms, etc.). So, for example, composing the map $C'\to\mathbb{P}^1$ with the map $C\to C'$ gives us a generically $2:1$ map between non-singular curves, however there is a point in $\mathbb{P}^1$ that has $2g+2$ preimages! $\endgroup$
    – rfauffar
    Feb 26, 2013 at 12:38
  • $\begingroup$ The normalization map may have arbitrary large number of premiages. For example, start with a smooth curve $C$, identify $n>>0$ points in $C$ to get a singular curve $C'$. Then the identified point has $n$ preimages. Also, the image of $C'$ in Kummer is not necessarily a smooth rational curve. $\endgroup$
    – user18119
    Feb 26, 2013 at 12:42
  • $\begingroup$ Ok, great! That's my mistake, assuming that the rational curve was smooth! Thanks so much, terrific answer! $\endgroup$
    – rfauffar
    Feb 26, 2013 at 13:00

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