Tate module of CM elliptic curves This is an exercise in Silverman's book "the arithmetic of elliptic curves". Ex 3.24, page 109.
E/K CM elliptic curve, Prove for $\ell \neq char(K)$, the action of $Gal(\bar{K}/K)$ on $T_{\ell}(E)$ is abelian.
 A: You can find the answer to your question (and learn a whole lot more about complex multiplication) in another book by Joe Silverman, "Advanced Topics in the Arithmetic of Elliptic Curves". See Chapter II, and in particular read the proof of Theorem 2.3.
A: @Emerton So, what if the char($K$) $\ne 0$, and the endomorphism ring is a quaternion algebra? Then you end up getting a map of Galois into $\textrm{GL}_1$ over a noncommutative ring...
Does your proof extend to the case of noncommutative endomorphism rings?
A: Here is the standard argument: you can decide whether it is prettier than the one you had in mind.
Let $V_{\ell}(E)$ be ${\mathbb Q}_{\ell}\otimes_{{\mathbb Z}_{\ell}}T_{\ell}(E)$; it is a two-dimensional ${\mathbb Q}_{\ell}$ vector space.  When $E$ has CM by a quad. imag. field
$F$, it is free of rank one over $F\otimes_{\mathbb Q}{\mathbb Q}_{\ell}$.  Thus 
the image of $Gal(\bar{K}/K)$ acting on $V_{\ell}(E)$ (or equivalently, $T_{\ell}(E)$)
lies in $GL_1(F\otimes_{\mathbb Q}{\mathbb Q}_{\ell})$, and so is abelian.
Note that this argument gets to the very heart of CM theory, and its relation to the class field theory (i.e. to the construction of abelian extensions): the elliptic curve $E$
(or, more precisely, its Tate modules) look 1-dim'l as modules over $F$, and so give
abelian Galois reps.  (Just as the $\ell$-adic Tate modules of the multiplicative group
${\mathbb G}_m$ give 1-dim'l. reps. of $Gal(\bar{\mathbb Q}/{\mathbb Q}).$)
You might also want to compare with Lubin--Tate theory, which is very similar: one uses
formal groups with an action of the ring of integers ${\mathcal O}$ in an extension of 
${\mathbb Q}_p$, and again they are constructed so that the $p$-adic Tate module is free of rank one over ${\mathcal O}$, and hence gives abelian Galois reps.

Added in response to basic's questions below:  To say that $E$ has CM over $K$ is to say that it has an action by an order in a quad. imag. field $F$.  By a standard theorem (in Silverman, say) the ring $F_{\ell} := F\otimes_{\mathbb Q}{\mathbb Q}_{\ell}$ acts faithfully on $V_{\ell}(E)$.  Counting dimensions over ${\mathbb Q_{\ell}},$ we find that $V_{\ell}(E)$ is free of rank 1 over $F_{\ell}$.  The Galois action on $V_{\ell}(E)$ is
being $F_{\ell}$-linear (we have assumed that the action of $F$ is defined over $K$).
Thus we have a group, $Gal(\bar{K}/K)$, acting on a free rank 1 module over a ring
(namely, the $F_{\ell}$-module $V_{\ell}(E)$).  Such an action must be given by $1\times 1$ invertible matrices (just choose a basis for $V_{\ell}(E)$ as an $F_{\ell}$-module),
i.e. is described by a homomorphism $Gal(\bar{K}/K) \to GL_1(F_{\ell})$.
Since the group of invertible $1\times 1$-matrices over any commutative ring is
itself commutative, we see that the $Gal({\bar K}/K)$ action on $V_{\ell}(E)$ is
through an abelian group, as required.
