Let $E$ be an elliptic curve with CM by an order in the imaginary quadratic field $K$. Is there some easy way how to prove that the extension $K(j(E))/\mathbb Q$ is abelian?


In general, the extension $K(j(E))/\mathbb Q$ is not abelian. The question therefore is, whether there is an elementary way to see that $K(j(E))/K$ is abelian (or even Galois).

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    $\begingroup$ $K(j(E))/K$ is abelian. There should be an extension $0 \to H \to \mathrm{Gal}(K(j(E))/\mathbb{Q}) \to \mathbb{Z}/(2 \mathbb{Z})$ where $H$ is the class group and $\mathbb{Z}/(2 \mathbb{Z})$ acts on $H$ by $-1$ so, if $H$ is not $2$-torsion, $\mathrm{Gal}(K(j(E))/\mathbb{Q})$ won't be abelian. $\endgroup$ – David E Speyer May 3 '19 at 16:26
  • $\begingroup$ @DavidESpeyer Probably the OP meant to ask about $K(j(E))/K$, since as you say, $K(j(E))/\mathbb{Q}$ tends to be non-abelian. So the question is whether there is an elementary way to see that the former is an abelian extension. It's clear that it's an algebraic extension, since there are only countably many isomorphism classes of CM elliptic curve, so only finitely many Aut$(\mathbb{C}/K)$-conjugates of $K(j(E))$, hence it is an algebraic extension. But it's not even clear (to me) that there's an elementary reason for $K(j(E))/K$ to be Galois, much less abelian. $\endgroup$ – Joe Silverman May 3 '19 at 19:32
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    $\begingroup$ @JoeSilverman I think there is a fairly elementary way, which I started to write up but had to do other things. To you, I can sketch it quickly. For simplicity, assume the order $R$ is the full ring of integers. Let $E_1$, ..., $E_h$ be the set of isomorphism classes over $\mathbb{C}$ of EC's with CM by $R$. For any ideal $\pi \subset R$, let's say that $E_i \stackrel{\pi}{\longrightarrow} E_j$ if there is an isogeny $E_i \to E_j$ whose kernel is the $\pi$-torsion of $E_i$. Any element of $\mathrm{Aut}(\mathbb{C}/K)$ preserves the property of being $\pi$-related (continued) $\endgroup$ – David E Speyer May 3 '19 at 19:59
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    $\begingroup$ and the set of permutations of $E_1$, ..., $E_h$ which preserve the property of being $\pi$-related is the class group. So the action of $\mathrm{Aut}(\mathbb{C}/K)$ on $j(E_1)$, ..., $j(E_h)$ factors through an action of the class group, and thus the action of $\mathrm{Aut}(\mathbb{C}/K)$ on $K(j(E_1), ..., j(E_h))$ factors through an abelian group. That shows $\mathrm{Gal}(K(j(E_1), ..., j(E_h))/K)$ is abelian, and thus so is $\mathrm{Gal}(K(j(E_1))/K)$. $\endgroup$ – David E Speyer May 3 '19 at 20:01
  • $\begingroup$ @alpoge My attempted answer is getting very long (although I think it works), so I'd be glad to see your shorter one. I don't get it yet, but maybe I will! $\endgroup$ – David E Speyer May 4 '19 at 0:15

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