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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.

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  • $\begingroup$ 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? $\endgroup$
    – Olivier
    Feb 16, 2011 at 22:11
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    $\begingroup$ @Olivier: What's the result of Bogomolov you've alluded to? I'd be interested to see. $\endgroup$
    – Felipe Voloch
    Feb 17, 2011 at 0:24
  • $\begingroup$ 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. $\endgroup$
    – Olivier
    Feb 17, 2011 at 9:42
  • $\begingroup$ I thought Bogomolov proved something about homotheties and not a full open image theorem. I'll have a look, thanks. $\endgroup$
    – Felipe Voloch
    Feb 18, 2011 at 1:37

1 Answer 1

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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.

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  • $\begingroup$ This is very pretty. In (i) did you mean 'infinite'? $\endgroup$
    – Keerthi Madapusi
    Feb 16, 2011 at 22:33
  • $\begingroup$ @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. $\endgroup$
    – Pete L. Clark
    Feb 16, 2011 at 22:36

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