Let $A \to S$ be an abelian scheme over an irreducible curve $S$ over complex numbers of relative dimension $g \geq 1$. Let $s: S \to A$ be a non-constant section and denote by $X:=s(S)$. Is it true that $\mathbb{Z}\cdot X:= \bigcup\_{n \in \mathbb{Z}} nX$ is Zariski dense in A?


I expect it is false in general but cannot find a counterexample. Thanks for any guidance.



Zariski dense in abelian scheme
How about if $A = E\_1\times E\_2$ with $E\_i$ non isotrivial elliptic curves and $s: S \to P\times 0$. Then your $\mathbb Z\cdot X$ is contained in $E\_1\times 0$ and so is not Zariski dense. In fact, this is the only thing that can happen.


Let $Z$ be the Zariski closure of $\mathbb Z\cdot X$ in $A$. Then $Z$ is closed under addition and inversion and therefore the connected component of $Z$ is a sub abelian variety. If $X$ is non-torsion, then this connected component will be of positive dimension. Conversely, if we had such a sub abelian variety, any section contained in this subvariety will give rise to a non Zariski dense $\mathbb Z\cdot X$.


