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.


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.

  • $\begingroup$ 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

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