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?


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)

  • $\begingroup$ Thanks, but I am not assuming that $char(K) = 0 $ so this argument needs to modified (in an obvious way) to the case of supersingular elliptic curves? $\endgroup$
    – isildur
    Jun 29 '11 at 1:10
  • 5
    $\begingroup$ @isildur: in case of a supersingular elliptic curve, the Galois group is abelian (topologically generated by the Frobenius). This is because such an elliptic curve is defined over a finite field and so its torsion points are all defined over the algebraic closure of a finite field. So the only Galois action is that of the Frobenius! this case is easier than that of char $K = 0$. $\endgroup$
    – SGP
    Jun 29 '11 at 1:54
  • $\begingroup$ Of course, how stupid of me! $\endgroup$
    – isildur
    Jun 30 '11 at 7:23

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