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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?

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See en.wikipedia.org/wiki/… –  Stopple Jun 28 '11 at 22:48
    
You can take a look to the beginning of "Arithmetic of elliptic curves with complex multiplication", LNM by Gross. This is explained quite well I think. –  Tommaso Centeleghe Feb 9 '12 at 21:06
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1 Answer 1

up vote 4 down vote accepted

Since $E$ has CM over $K$, the ring $F:= End_K(E) \otimes Q$ is an imaginary quadratic field. Suppose $l$ is a prime integer unramified in $F$. Now $F_l:= F \otimes_Q Q_l$ is either two copies of $Q_l$ or a quadratic extension of $Q_l$.

The two dimensional $Q_l$-vector space $V_l:= T_l \otimes_{Z_l} Q_l$ is a rank one free module over $F_l$. The Galois action on $T_l \otimes Q_l$ respects the action of $F_l$ and hence factorises as $Gal(\bar{K}/K) \to F^{\times}_l \hookrightarrow Aut (V_l) = GL_2(Q_l)$. Here $F^{\times}_l $ is the units of $F_l$.

(This is a standard argument, see for example Silverman AEC or Milne EC)

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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? –  isildur Jun 29 '11 at 1:10
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@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 $K = 0$. –  SGP Jun 29 '11 at 1:54
    
Of course, how stupid of me! –  isildur Jun 30 '11 at 7:23
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