# Non-degenerate points on a Jacobian surface

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 $$[L:K] such that points of degree $$ 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.

• 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? – dfn Jan 27 at 14:19
• @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. – Will Sawin Jan 27 at 14:53
• @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 $f$. – Will Sawin Jan 27 at 14:54
• 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. – dfn Jan 27 at 14:56