Is there an example of an abelian variety $X$ defined over a number field $K$, with $\dim X > 1$, and a $K$-rational point $P$ on $X$, such that no curve $C$ on $X$ (say defined over a number field) contains all integer multiples of $P$? I'm particularly interested in the case $\dim X = 2$. Can we make a statement to the effect that for a "generic" $P$ and $X$, this always happens?
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Every $K$-rational point that's not torsion or contained in an elliptic curve through the origin has this property by Falting's theorem: Since $P$ is not torsion, there are infinitely many integer multiples of $P$, so if $C$ contains them all, $C$ has infinitely many rational points, hence genus $\leq 1$, but can't have genus $0$ as it lives in an abelian variety, thus must be an elliptic curve, and contains the origin since the origin is an integer multiple of $P$.
This can be made to work even for points that are not $K$-rational using the Mordell-Lang conjecture.
answered Nov 2, 2023 at 20:20
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3$\begingroup$ I don't think you need Faltings's theorem to answer the question: the Zariski closure of all multiples of $P$ must be a subgroup, positive dimensional if $P$ is not torsion. $\endgroup$– nafCommented Nov 3, 2023 at 3:59
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$\begingroup$ @naf That's true, Falting's theorem instead gives the stronger statement that you can't have infinitely many multiples of $P$ in $C$ unless $P$ lies in a torsion translate of an elliptic curve. $\endgroup$ Commented Nov 3, 2023 at 14:05
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2$\begingroup$ Actually this can also be proved in a more elementary way using Mahler's theorem, see the article "The dynamical Mordell-Lang problem for etale maps" by Bell, Ghioca and Tucker. $\endgroup$– nafCommented Nov 4, 2023 at 4:44