TITLE: Basic Question on action of Galois group on Tate module
QUESTION [5 upvotes]: I am trying to show that for an elliptic curve $E/K$ with complex multiplication the action of $G_{\overline{K}/ K}$ on the $T_{l}(E)$, the Tate module is abelian. 
An approach: Let $\rho$ denote the Galois representation on the (rational) Tate module $(T_{l} \otimes \mathbb{Q}_{l})$.
Complex Multiplication implies that $dim(End(E) \otimes \mathbb{Q}_{l}) \geq 2 $. Since, each element in $End(E)$ acts as a intertwiner of $\rho$.
By Schur's lemma, the representation $\rho$ is evidently reducible. So the irreducible components of $\rho$ are one dimensional and thus the action of $G_{\overline{K}/ K}$ is abelian. 
While this is a essentially representation theoretic, is there a more arithmetic proof of this 
result?

REPLY [6 votes]: Since $E$ has CM over $K$, the ring $F:= \operatorname{End}_K(E) \otimes \mathbb{Q}$ is an imaginary quadratic field. Suppose $\ell$ is a prime integer unramified in $F$. Now $F_\ell:= F \otimes_\mathbb{Q} \mathbb{Q}_{\ell}$ is either two copies of $\mathbb{Q}_{\ell}$ or a quadratic extension of $\mathbb{Q}_{\ell}$.  
The two dimensional $\mathbb{Q}_{\ell}$-vector space $V_{\ell}:= T_{\ell} \otimes_{\mathbb{Z}_{\ell}} \mathbb{Q}_{\ell}$ is a rank one free module over $F_{\ell}$. The Galois action on $T_{\ell} \otimes \mathbb{Q}_{\ell}$ respects the action of $F_{\ell}$ and hence factorizes as $\operatorname{Gal}(\bar{K}/K) \to F^{\times}_{\ell} \hookrightarrow \operatorname{Aut} (V_{\ell}) = GL_2(\mathbb{Q}_{\ell})$. Here $F^{\times}_{\ell} $ is the units of $F_{\ell}$. 
(This is a standard argument, see for example Silverman AEC or Milne EC)