Is the set of points on an abelian surface which project to rational points on the Kummer surface a subgroup?

Question

Let $C$ be a hyperelliptic curve of genus 2 defined over $\mathbb{Q}$, let $J$ be its Jacobian, and let $X$ be the Kummer surface associated to $J$ (i. e. $X$ is the singular Kummer surface which results from quotienting $J$ by the map $x \mapsto -x$, not the K3 surface which resolves the singularities of $X$). I have a few questions suggested by computations I have done in a specific instance of this setup. In my situation $X$ is defined over $\mathbb{Q}$ as a quadric hypersurface in $\mathbb{P}^3$. I don't know if it's relevant, but the Jacobian I am working with happens to have Mordell-Weil group $(\mathbb{Z} / 2 \mathbb{Z})^3$.

Let $\pi: J(\overline{\mathbb{Q}}) \to X(\overline{\mathbb{Q}})$ be the projection map. Computational evidence suggests that if $\pi(x) \in X(\mathbb{Q})$, then $\pi(n \cdot x) \in X(\mathbb{Q})$ for all $n \in \mathbb{Z}$. Is this known to be true?

More broadly, if $\pi(x), \pi(y) \in X(\mathbb{Q})$, is $\pi(x + y) \in X(\mathbb{Q})$? If so, is $\pi^{-1}(X(\mathbb{Q}))$ finitely generated? Are there any algorithmic methods (ideally with available software implementations) which can give information about this supposed group in explicit examples?

Answer 1

Varieties over a field $k$ can be understood by their $\bar k$-points with the Zariski topology, together with the $\operatorname{Gal}(\bar k/k)$-action. Any morphism $f \colon X \to Y$ of $k$-varieties induces a $\operatorname{Gal}(\bar k/k)$-equivariant map $X(\bar k) \to Y(\bar k)$.

Thus $\pi^{-1}(X(\mathbf Q))$ is the set of points whose Galois orbit is contained in the fibre, i.e. such that the subset $\{x,-x\}$ is $\operatorname{Gal}(\bar{\mathbf Q}/\mathbf Q)$-invariant. Since $[n] \colon J \to J$ commutes with the Galois action, we see that $\{nx,-nx\}$ has the same property, so $\pi(nx)$ is defined over $\mathbf Q$.

However, for sums the situation is more complicated: the action on $\{x,-x\}$ factors over $\operatorname{Gal}(K/\mathbf Q)$ for a quadratic field $K$, and likewise the action on $\{y,-y\}$ factors over a quadratic field $L$. Then the action on $\{\pm x, \pm y\}$ factors over the compositum $KL$, and if $K \not\cong L$ we see with a little bit of Galois theory that the orbit of $x+y$ is the full set $\{\pm x \pm y\}$, not merely $\{\pm (x+y)\}$.

Answer 2

The answer to your second question is no. If $x\notin J(\mathbb Q)$, then the condition $\pi(x)\in X(\mathbb Q)$ implies that $\mathbb Q(x)/\mathbb Q$ is a quadratic extension. If you take a point $y$ with the same properties, then it is highly unlikely that $\mathbb Q(x)\cong\mathbb Q(y)$, and then $\mathbb Q(x+y)$ is most likely a quartic extension of $\mathbb Q$, so $\pi(x+y)$ will not be in $X(\mathbb Q)$. You might get some insight if you consider the analogous question for an elliptic curve and the map $E\to E/\{\pm1\}\cong\mathbb P^1$.