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Hi everyone,

let A be an abelian variety of dimension g over an algebraically closed field k of characteristic p ≥ 0. I'm trying to prove that the subgroup A' which is the union of all torsion points a ∈ 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 (\Q\ell/\Z_\ell)2g, 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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up vote 5 down vote accepted

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 |G|*\ell^{2d}. For large \ell, this is less than \ell^{2n} if d is not n.

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