8
$\begingroup$

Let $C$ be a (hyperelliptic) genus $2$ curve over a number field $K$ with a $K$-rational Weierstrass point $\infty$. We embed $C$ in its Jacobian $J$ via $\infty$.

Question: Is there a quadratic extension $L/K$ and a point $x\in C(L)$ which is non-degenerate in $J$, i.e. such that $\mathbb{Z}x$ is dense in $J$?

If $J$ has only finitely many abelian subvarieties, the answer is obviously yes. What about the case where $J$ is isogeneous to the square of an elliptic curve $E$? There are infinitely many elliptic curves on $J$ given by isogenies of $E$ and a priori, it seems possible to me though quite implausible that they could cover all quadratic points on $C$.

Possible general statement: I would expect that this is part of a much more general "unlikely intersection" statement for any hyperbolic curve $C$ embedded in an abelian variety and demanding $[L:K]<n$ such that points of degree $<n$ are dense in $C$; but I am mainly interested in the above.

$\endgroup$
2
$\begingroup$

The product $(E \times C) / \sigma$, where $\sigma$ acts by inversion on $E$ and the hypereliptic involution on $C$, is an elliptic surface over $C/\sigma = \mathbb P^1$.

This surface has two sections, which are given by the two maps $C \to E$ we get from the Abel-Jacobi map composed with the two projections $J \to E$. (The Abel-Jacobi map of a hyperelliptic curve sends the hyperelliptic involution to negation of the abelian variety, which makes these maps $\sigma$-equivariant, hence descend to sections of the surface).

At the generic point, these two sections are $\mathbb Q$-linearly independent, specifically because they are $\mathbb Q$-linearly independent as maps $C \to E$ (since $C$ generates $J$ as a group).

It follows from Silverman's specialization theorem that, restricted to the fibers over all but finitely many rational points of $\mathbb P^1$, these two sections remain $\mathbb Q$-linearly independent.

This is exactly what you want - a rational point of $\mathbb P^1$ lifts to a quadratic point of $C$, and the two sections are its two projections to $E$, so because these are linearly independent, your quadratic point does not lie inside any copy of $E$ inside $C$.

In particular, if $E$ has rank $0$, so that all but finitely many rational points on $\operatorname{Sym}^2 (C)$ are orbits of the hyperelliptic involution, then all but finitely many quadratic points are non-degenerate in this sense.

$\endgroup$
4
  • $\begingroup$ Dear Will, I tried this approach too but there is one thing I don't understand: Let $\pi_1,\pi_2:C\to E$ be the two projections. Then we know that $\pi_1(x)$ and $\pi_2(x)$ are generically independent, meaning that for any endomorphism $f$ of $E$ $f(\pi_1(x))\neq \pi_2$ generically. By Silverman, this means that $f(\pi_1(x))\neq f(\pi_2(x))$ outside a finite set $S_f$ of $x$. However, there are infinitely many endomorphisms $f$, so how do we know that the union of all $S_f$ doesn't cover $\mathbb{P}^1(K)$? Maybe you are using a stronger version of the specialization theorem? $\endgroup$ – dfn Jan 27 at 14:19
  • $\begingroup$ @dfn Is the issue to do with CM curves $E$? I realize that when I wrote this argument I forgot the case of a CM curve, but you can handle it the same way, now with $4$ sections. $\endgroup$ – Will Sawin Jan 27 at 14:53
  • $\begingroup$ @dfn If not, I don't understand what the problem is. The specialization theorem says that, outside a finite set, the specialization map is injective. That means that any linear combination of $\pi_1(x)$ and $\pi_2(x)$ will be nonzero. The finite set is not dependent on an endomorphism $f$. $\endgroup$ – Will Sawin Jan 27 at 14:54
  • $\begingroup$ Oh ok, yes, I was worrying about CM but since the endomorphism ring is finite over $\mathbb{Q}$, we can just pick a basis and add the images of $\pi_i(x)$ to the mix. Somehow, I only ever used specialization for the naive implication "generic non-torsion->special non-torsion outside a finite set" but of course, it says more than that and is about linear independence. $\endgroup$ – dfn Jan 27 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.