Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $A$ be an absolutely simple abelian variety over a number field $K$. Assume that, for some prime $p$, the Tate module $T_p A$ has a submodule of rank one, invariant under the absolute Galois group of $K$. Does it follow that $A$ is has CM?

For elliptic curves, I guess this follows from Serre's open image theorem. That's all I know. I would be surprised if there was a counterexample as it would be a way of constructing abelian extensions of $K$ using non-CM abelian varieties, which would be surprising.

share|improve this question
    
This seems very true. Let $L_{\mathfrak p}$ be a finite extension of $\mathbb Q_{p}$ containing all the traces of the Frobenius morphisms acting on $T_{p}A$. Assume that $A$ has no CM. Then there are no quadratic character $\eta$ such that $V=T_{p}A\otimes L_{\mathfrak p}$ is isomorphic to $V\otimes\eta$. This implies that $\operatorname{End}_{L_{\mathfrak p}[G_{K}]}$ is equal to $L_{\mathfrak p}$ by Frobenius reciprocity and this in turn implies that $V$ is irreducible. Does that sound good to you or am I missing something? Didn't Bogomolov proved the open image theorem you want anyway? –  Olivier Feb 16 '11 at 22:11
1  
@Olivier: What's the result of Bogomolov you've alluded to? I'd be interested to see. –  Felipe Voloch Feb 17 '11 at 0:24
    
Dear Felipe, I was referring to Sur l'algébricité des représentations l-adiques (C.R.A.S 290 F.Bogomolov). There are several results of Serre from the 80s, mostly found in letters to other people, which also cover these kind of results. –  Olivier Feb 17 '11 at 9:42
    
I thought Bogomolov proved something about homotheties and not a full open image theorem. I'll have a look, thanks. –  Felipe Voloch Feb 18 '11 at 1:37
add comment

1 Answer

up vote 12 down vote accepted

Yes. This follows from the main result of the following paper of Zarhin.


MR0885780 (88h:14046) Zarhin, Yu. G. Endomorphisms and torsion of abelian varieties. Duke Math. J. 54 (1987), no. 1, 131–145.


His result, specialized to the $K$-simple case, is the following (fantastic) theorem.

Let $A$ be a $K$-simple abelian variety defined over a number field $K$. The following are equivalent:
(i) $A(K^{\operatorname{ab}})[\operatorname{tors}]$ is infinite.
(ii) $A$ is of CM-type over $K$.

Your hypotheses imply that there is infinite torsion over the abelian extension cut out by the action of Galois on the one-dimensional subspace (the Galois group is contained in $\mathbb{Z}_{p}^{\times}$), so Zarhin's theorem applies.

share|improve this answer
    
This is very pretty. In (i) did you mean 'infinite'? –  Keerthi Madapusi Pera Feb 16 '11 at 22:33
    
@Keerthi: thanks. Yes, I either meant "infinite" in (i) or was missing a "not" in (ii) (and not both!). I fixed it as you suggested. –  Pete L. Clark Feb 16 '11 at 22:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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