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.

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    $\begingroup$ As Mariano has comment on a different question, straight up "How do I do this exercise in this book?" questions are not so appropriate for MO. But hint: you want to use the fact that the image of Galois in Aut(T_l(E)) lies in the commuting algebra of the image of End(E) \otimes Z_l. Show that this commuting algebra is itself commutative, and you're done. $\endgroup$ Jan 29, 2010 at 8:02
  • $\begingroup$ Thank you, Pete! I finally worked it out! It's interesting to find out that the commuting algebra is commutative! $\endgroup$
    – natura
    Jan 30, 2010 at 9:35
  • $\begingroup$ only I'm wondering if there is a more pretty proof to prove that the commuting algebra ifs commutative. I used down-to-earth linear algebra computation. $\endgroup$
    – natura
    Jan 30, 2010 at 9:36
  • $\begingroup$ @basic: It follows immediately from some basic facts about simple algebras (e.g. the 2 by 2 matrices). Fact 1: The centre of a simple algebra is a field. Fact 2: The centraliser (=commuting algebra) of a simple subalgebra of a simple algebra is again a simple algebra. Fact 3: The dimension of a central simple algebra over its centre is a square. Now use the fact that 2 is not a square! $\endgroup$
    – user1594
    Feb 4, 2010 at 23:23
  • $\begingroup$ A proof is given in proposition 12.13 in Milne's notes on modular forms. $\endgroup$
    – Watson
    Nov 11, 2018 at 20:59

3 Answers 3


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.

  • $\begingroup$ @Emerton: this is indeed a slightly prettier argument than the one I had in mind (which is also standard, I think). $\endgroup$ Feb 4, 2010 at 5:44
  • $\begingroup$ @Emerton. I have some confusing point. 1. the general definition of CM seems to be $End_K(E)$ strictly bigger than $Z$, now when the field is finite field or local field, these fields cannot be embedded into $C$, so we cannot define $End_C(E)$, right? 2.Even when the field $K$ is a number field, which can be embedded in $C$, now your $F$ is $End_C(E) \otimes Q$, but what is the Galois action on it? Or, IF we can prove that $F = End_C(E) \otimes Q = End_K(E) \otimes Q$, then we can define a Galois action on $End_K(E) \otimes Q$, but still, I can't find a proof of such, and even I assume it to $\endgroup$
    – natura
    Feb 4, 2010 at 6:52
  • $\begingroup$ and even I assume it to be true, I still can't figure out a Galois equivariant homomorphism between $V_{\ell}$ and $F \otimes Q_{\ell}$. Thank you! $\endgroup$
    – natura
    Feb 4, 2010 at 6:54
  • $\begingroup$ If E over K has CM then $V_p(E)$ decomposes as direct sum of two Galois invariant lines. But what about $T_p(E)$? Does such a decomposition hold at the "integral" level as well? I suppose that if the two characters describing the action on $V_p(E)$ are congruent to each other mod $p$, then $T_p(E)$ should not necessarily decompose. $\endgroup$ Feb 9, 2012 at 16:30

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.


@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?

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    $\begingroup$ @oxeimon: It would be best to ask this as a separate question, including a link back to this one. $\endgroup$ Mar 2, 2012 at 19:14
  • $\begingroup$ If $K$ is a finite field of characteristic $p$ and $E$ is supersingular with $End_K(E)$ isomorphic to an order in the definite quaternion ramified at $p$ THEN $|K|=p^{2m}$, the Tate module $T_\ell(E)$, $for \ell\neq p$, is a free rank one module over $End_K(E)\otimes Z_\ell$ ($\simeq$ to the ring of two-by-two matrices with entries in $Z_\ell$), and $Frob_K$ acts as multiplication by either $p^{m}$ or $-p^m$, an element in the center of $End_K(E)\otimes Z_\ell$. $\endgroup$ Mar 2, 2012 at 22:25

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