For a number field F, let E/F be an elliptic curve with CM by a quadratic field K. Let $\rho_\ell: \text{Gal}_F \to \text{Aut}(T_{\ell}E)$ be the $\ell$-adic representation associated to E for some prime $\ell$, and let $\mathcal{G}_{\ell}$ denote its image. According to Theorem 5 in Serre's "Groupes de Lie $\ell$-adiques attaches aux courbes elliptiques" (1966), $\mathcal{G}_{\ell}$ is abelian if and only if K $\subset$ F.

Is it always the case that $\mathcal{G}_{\ell}$ being abelian forces K $\subset$ F, or are there some restrictions on $\ell$ that I haven't been able to find in the paper?

  • $\begingroup$ To be sure I understand the question: you're asking if Serre's theorem 5 is wrong? $\endgroup$
    – Joël
    Jun 26, 2013 at 21:00
  • $\begingroup$ Oh no! I'm just wondering if earlier in the paper Serre has established some conditions on the prime $\ell$. (I couldn't find any, but I may have missed it since the paper is in French.) This result of Serre is referenced in a paper I'm reading and somehow the authors end up with the condition that $\ell$ cannot ramify in K. I thought maybe it was hidden somewhere in Serre's result, since everything else they do seems to follow without this condition. $\endgroup$
    – abourdon
    Jun 26, 2013 at 21:20
  • $\begingroup$ Dear Abbey: are you reading my paper? If so, please feel free to contact me directly, in which case I'll go back and take another look at Serre's paper. I looked back at our paper just now, and the way I read it is: in the ramified case we were able to give a simple enough proof that the $N$-torsion field contains $\mathbb{Q}(\sqrt{D})$, so we didn't need to invoke the "black box of Serre" in this case. So to answer your question: the way I remember it, Serre's result is true with no restriction on $\ell$. $\endgroup$ Jun 27, 2013 at 1:30
  • $\begingroup$ For everyone else: I suspect that the OP may be asking about this paper: math.uga.edu/~pete/torspaper_FINAL.pdf; more specifically the portion from Lemma 15 to Theorem 16. (And apologies to my coauthors for calling it "my" paper.) $\endgroup$ Jun 27, 2013 at 1:32

1 Answer 1


There are no restrictions on $l$. What's going on is that if the $CM$ endomorphisms are defined over $KF$ but not over $F$, there must be elements in the $\operatorname{Gal}(KF|F)$ that conjugate these endomorphisms to different endomorphisms. Since the endomorphisms act faithfully on the Galois representation, the only missing piece is to find an element of the Galois group that has the same action on the Galois-representation as one of the CM endomorphisms. One can check that the Frobenius elements of primes of $F$ that split in $KF$ act this way.

  • 1
    $\begingroup$ +1: this is absolutely right...except there are two instances of $K$ that should be $KF$. $\endgroup$ Jun 27, 2013 at 5:54

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