I am wondering if the following holds or not.

Let A be an abelian variety of dimension $d\geq 1$ over $\mathbb{Q}$. Then there is a positive number c depending on d and A such that $[\mathbb{Q}(A[n]):\mathbb{Q}]\geq c^{w(n)} n^2$ where $w(n)$ is the number of distinct prime factors of n.

I know that it must hold for d=1(elliptic curve case). I guess that we can even have stronger inequality like $[\mathbb{Q}(A[n]):\mathbb{Q}]\geq c^{w(n)} |GSp_{2d}(Z/nZ)|$ ,but don't know how to prove it, and couldn't find any reference other than d=1 case.

Thanks, Sungjin Kim.

  • $\begingroup$ Perhaps you should change the $n^2$ to $n^{2d}$. $\endgroup$ – S. Carnahan Feb 3 '12 at 3:05
  • $\begingroup$ Conjecturally, for a large enough prime $\ell$, the image of the Galois representation on the points of $\ell$-division is the $\mathbf Z/\ell$-points of the Mumford-Tate group. So such an inequality would not hold unless the M-T group is $\mathop{\rm GSp}_{2d}$. Already for elliptic curves with complex multiplication, the lower bound is $\ell^2$, and not $\ell^3\approx |\mathop{\rm SL}_2(\mathbf Z/\ell)|$. $\endgroup$ – ACL Feb 3 '12 at 7:25
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    $\begingroup$ According to a result of Serre (see "R\'esum\'e des cours de 1985-1986". Coll`ege de France (1986)), for any $\epsilon>0$, there is a constant $C=C(A,\epsilon)$, such that $[{\bf Q}(P):{\bf Q}]\geq C\cdot n^{1-\epsilon}$ if $P\in A[n](\bar{\bf Q})$. If $A_{\bar{\bf Q}}$ does not contain any subvariety of CM type then one even has $[{\bf Q}(P):{\bf Q}]\geq C\cdot n^{2-\epsilon}$. Since $[{\bf Q}(A[n]):{\bf Q}]\geq [{\bf Q}(P):{\bf Q}]$, this applies to your situation. For further references and results, see A. Silverberg's article "Torsion points on abelian var. of CM type", Compositio 68. $\endgroup$ – Damian Rössler Feb 9 '12 at 12:02
  • $\begingroup$ @Damian Rossler Thank you for the answer. However, $[\mathbb{Q}(A[n]):\mathbb{Q}]\geq C\cdot n^{1-\epsilon}$ is not enough for my purpose. I wonder if there is a result with the exponent greater than 1. That is why I specifically put $n^2$ there. For $1-\epsilon$ result, it follows by $\mathbb{Q}(A[n])\supseteq \mathbb{Q}(\zeta_n)$. $\endgroup$ – Sungjin Kim Feb 10 '12 at 0:14
  • $\begingroup$ If $A_{\bar{\bf Q}}$ does not contain any subvariety of CM type then $2-\epsilon$ works (see my last comment). Is even that not enough ? Do you have to deal with abelian varieties of CM type ? (in that case I don't know what to suggest). $\endgroup$ – Damian Rössler Feb 10 '12 at 9:26

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