4
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
  • 4
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.