# Why are torsion points dense in an abelian variety?

Hi everyone,

let $A$ be an abelian variety of dimension $g$ over an algebraically closed field $k$ of characteristic $p\geqslant 0$. I'm trying to prove that the subgroup $A'$ which is the union of all torsion points $a\in A(k)$ of order prime to $p$ is Zariski dense in $A$.

The statement would follow if the Zariski closure $B$ (which by construction is a group variety) of $A'$ in $A$ would again be an abelian variety of dimension $d$, because assuming $d<g$, the $\ell$-primary part of $B$ would still be $(\mathbf Q_\ell/\mathbf Z_\ell)^{2 g}$, while it SHOULD be of rank $2d<2g$, contradiction.

However, I fail to see why $B$ should be irreducible. Does anyone see a way to salvage the argument, or a different, (simpler) argument?

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If p>0, and assuming A ordinary, then the p-power torsion points of A(k) should also be Zariski dense in A, if I'm right. –  Tommaso Centeleghe Jun 3 at 13:42

Let $C$ be the connected component of the identity in $B$. Then $C$ is a projective group variety, hence an abelian variety; let it have dimension $d$. Let $B/C=G$, a finite group. Then the number of $\ell$-torsion points of $B$ is at most $\left|G\right|\cdot\ell^{2d}$. For large $\ell$, this is less than $\ell^{2n}$ if $d$ is not $n$.