Timeline for Question on division field of abelian variety
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Feb 10, 2012 at 20:04 | comment | added | Sungjin Kim | @Damian Rossler I found K.Ribet's paper "Division fields of abelian varieties with complex multiplication" I think this can cover the case with complex multiplication. Together with your comment, I think I got what I need. Thank you. | |
| Feb 10, 2012 at 9:54 | comment | added | Damian Rössler | On second thoughts: if $A$ is a CM abelian variety, then the main theorem of complex multiplication provides an explicit description (in terms of ideles) for the action of the Galois group on $A[n]$, which is very similar to the cyclotomic one. See Lang's book on Complex Multiplication (for lack of a better reference), or Silverman, Advanced topics, II, Th. 8.2 for the elliptic curve case. I think it should be possible to adapt the degree estimates for the cyclotomic case (which involves Euler's function) to the CM case. | |
| Feb 10, 2012 at 9:26 | comment | added | Damian Rössler | 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). | |
| Feb 10, 2012 at 0:14 | comment | added | Sungjin Kim | @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 $\mathbb{Q}(A[n])\supseteq \mathbb{Q}(\zeta_n)$. | |
| Feb 9, 2012 at 12:02 | comment | added | Damian Rössler | 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. | |
| Feb 3, 2012 at 7:25 | comment | added | ACL | 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 $\ell^3\approx |\mathop{\rm SL}_2(\mathbf Z/\ell)|$. | |
| Feb 3, 2012 at 3:05 | comment | added | S. Carnahan♦ | Perhaps you should change the $n^2$ to $n^{2d}$. | |
| Feb 2, 2012 at 21:54 | history | asked | Sungjin Kim | CC BY-SA 3.0 |