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.



Non-degenerate points on a Jacobian surface
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.


