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.

`$\operatorname{End}_{L_{\mathfrak p}[G_{K}]}$`

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 '11 at 22:11