9
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Jun 3, 2014 at 13:42

1 Answer 1

16
$\begingroup$

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 $n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.