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. 



Is an abelian variety with a Galois invariant, rank one submodule of its Tate module, CM?
Yes. This follows from the main result of [the following paper](https://projecteuclid.org/journals/duke-mathematical-journal/volume-54/issue-1/Endomorphisms-and-torsion-of-abelian-varieties/10.1215/S0012-7094-87-05410-X.short) of Zarhin.




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




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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.


