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?



Basic Question on action of Galois group on Tate module
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)


