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Let $K$ be an imaginary quadratic field with Hilbert class field $H$, and let $E$ be an elliptic curve defined over $H$ with complex multiplication by the ring of integers $O_K$ of $K$. It is known that for an integral ideal $\mathfrak{m}$ of $O_K$, $K(j(E),h(E[\mathfrak{m}]))$ is the ray class field of K modulo $\mathfrak{m}$, where $h$ is the Weber function for $E/H$. (This is stated, for example, on page 135 of Silverman's Advanced Topics in the Arithmetic of Elliptic Curves.)

My question is this: what if $E$ has CM by an arbitrary order? Can any generalization of this statement be made? I've read that if $E$ has CM by an order of conductor $\mathfrak{f}$, then $K(j(E))$ is the ring class field of $K$ with conductor $\mathfrak{f}$, but I'm wondering if anything more can be said.

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There is an isogeny from $E$ to a regular CM curve naturally defined by the subgroup of the fundamental group which has full CM, so the isogeny is defined over $\mathbb Q$. It has a cyclic kernel of degree $k$, so for $(m,k)=1$ it induces an isomorphism on $m$-torsion points and so preserves the CM theory for those torsion points. – Will Sawin Aug 30 '12 at 16:52
Beyond that, you can use the induced map on Tate modules to determine the Galois representation of $E$ and so the action on torsion points. – Will Sawin Aug 30 '12 at 16:56
I don't think this characterization of $K(j(E))$ is correct. I believe that it is a subfield of that ray class field. Instead of modding out by principal ideals with a generator $1$ mod $k$, I think you should mod out by ideals with a generator integral mod $k$, that is, ideals with a generator in the order. The reason is because you can define a CM curve by an order using a $\mathbb Z/k$-subgroup of a CM curve by the full ring, and the defining equation for a subgroup doesn't get you the full ray class field, but the subfield I described. – Will Sawin Aug 30 '12 at 18:00
Thanks for your comments! About what I said concerning K(j(E))--this is coming specifically from the first page of Kwon's "Degree of Isogenies of Elliptic Curves with Complex Multiplication." Other places (for example, Silverman's The Arithmetic of Elliptic Curves, page 427) at least say that [K(j(E)):K] is equal to the class number of the order by which E has CM. Am I misunderstanding something? There could be some assumptions I've missed, or perhaps I've misinterpreted. – abourdon Aug 30 '12 at 20:06
@Will Sawin: you are in fact both correct, $K(j(E))$ is the subfield of the ray class field that corresponds to the ring class field; I provided more details in my answer below. – Damien Robert Apr 14 '14 at 12:22
up vote 9 down vote accepted

In fact Shimura handled the case of an abelian variety $A$ with complex multiplication by an order $O$ inside the maximal order $O_K$ of the CM field $K$. A very good modern reference is the following article by Marco Streng: An explicit version of Shimura's reciprocity law for Siegel modular functions. I also recommend his PhD thesis. Both are available here:

Now for the statement (see Theorem 2.2 of the paper above): let $\tau$ be the element of the Siegel space that corresponds to the abelian variety $A$ above. The Siegel modular functions of level $N$ with $q$-expansion in $\mathbb{Q}(\zeta_N)$ evaluated at $\tau$ generate an abelian extension $H(N)$ of the reflex field $K^r$ (for the type norm $\Phi$ associated to $A$).

This abelian extension is associated by class field theory to the class group $I_K(NF)/H_{\Phi,O}(N)$ where

  • $F$ is the conductor of $O$
  • $I_K(NF)$ are the fractional ideals of $K$ prime to $NF$
  • $H_{\Phi,O}(N)$ are the ideals $\mathfrak{a} \in I_K(NF)$ such that $\exists \mu \in K$ with $N_{\Phi^r,O}(\mathfrak{a})=\mu O$, $\mu \overline{\mu}=N(\mathfrak{a})\in \mathbb{Q}$ and $\mu \equiv 1 {\bmod^\times} NO$. (Here $N_{\Phi^r,O}$ is the type norm from the reflex field $K^r$ to $K$.)

Now specializing to elliptic curves, we get that

  • If $E$ has CM by $O_K$, then $K(j(E),h(E[𝔪]))$ is the ray class field associated to the class group $I_K(m)/\{ \mu O_K \mid \mu \equiv 1 {\bmod^\times} m\}$
  • If $E$ has CM by $O$ where the conductor of $O$ is $F \in \mathbb{Z}$, then $K(j(E))$ is the ring class field of $O$, meaning the extension associated to the class group $I_K(F)/\{ \text{principal ideals of $O$ primes to $F$}\}$. And finally, $K(j(E),h(E[𝔪]))$ will correspond to $I_K(mF)/\{ \mu O \mid \text{$\mu$ is prime to $F$ and}\ \mu \equiv 1 {\bmod^\times} mO\}$. (At least we have the inclusion, but I am pretty sure that in the elliptic curve case, the evaluation of Weber functions at the points of $m$-torsion over the $j$-invariant give the field generated by level-$m$ modular forms evaluated at $\tau$.) So in this case we have something intermediate between a ring class field and a ray class field. [I have not checked carefully but it should be the compositium of the ring class field of conductor $F$ and the ray class field of modulus $m$ when $m$ is prime to $F$].
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Thanks very much for your answer. This will actually be used for a paper that the OP and I are writing together, so in the near future I'll look up the reference in detail. I certainly intend to award you the bounty. – Pete L. Clark Apr 14 '14 at 22:12
Well all credits should go to Shimura for proving all the results (but for elliptic curves I guess everything was already known by Deuring!), and Streng for a nice exposition. – Damien Robert Apr 15 '14 at 11:04
"I guess everything was already known by Deuring." I am almost sure of that; what I wonder is how much of this was already known by Weber. In any case, it is un/fortunately quite true that many things which were already known to the great mathematicians of the 19th and early 20th centuries are nevertheless not known to me. :) (To be more honest: in this case I think I am being a little lazy. I was quite sure that the result I was asking about was true, and if necessary I think I could have made appropriate modifications in the proofs I knew. But I didn't want to...) – Pete L. Clark Apr 15 '14 at 18:08

Now as for an intuitive explanation concerning Pete Clark's question:

I would be interested in, at least, a reference for the fact that K(j(E),h(E[N])) contains the N-ray class field of K for arbitrary orders.

Here is my intuitive point of view on it (I put it in another answer because I won't try to be perfectly rigorous), in terms of moduli.

Let $E_1$ be a an elliptic curve with maximal CM by $O_K$, and $E_2$ be a curve with CM by $O$, an order $O$ of conductor $F$ in $O_K$. $E_2$ is defined over $K(j(E_2)) \supset K(j(E_1))$, so there is a rational isogeny $E_2 \to E_1$ of degree $F$. If we add the field of definition of the points of $m$-torsion of $E_2$ then it is clear that if $F$ is prime to $m$, these points get transported to the points of $m$-torsion of $E_1$, so we already have the $m$-ray class field.

What is more surprising is that it works also when $m$ is not prime to $F$. So let's assume that $m \mid F$ and see why we can still transport the $m$-torsion from $E_2$ to $E_1$.

The reason is as follow: when we are in the ring class field of $O$, all isogenies of degree $F$ between $E_1$ and an elliptic curve with endomorphism by $O$ are already rational; this means that the Galois action on $E_1[F]$ is given by a diagonal matrix. (And of course being in the $F$-ray class field means that the Galois action on $E_1[F]$ is the identity.) So in particular the kernel $K_1$ of the isogeny $E_1 \to E_2$ has a rational complement $K_2$.

Now if we have the $m$-torsion on $E_2$, pushing it through the dual isogeny $E_2 \to E_1$, we have that at least the points of $m$-torsion of $K_1$ are all rationals. But because the points of $m$-torsion are rationals in $E_2$, the $m$-roots of unity are rationals, and so by looking at the Weil pairing we see that the points of $m$-torsion in $K_2$ are rationals. So all points of $m$-torsion in $E_1$ are rationals.

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Upon further thought, I am having some trouble filling in the details of this answer in the general case. I see that if $E_2$ has full $N$-torsion over a field $F$ containin the CM field, then (i) $F$ contains the $N$th roots of unity and (ii) $E_1$ has at least one $F$-rational point of order $N$. This means that the mod $N$ Galois rep on $E_1$ has the shape $[[1 * ][0 1]]$. That last $*$ means that we get full $N$-torsion via a cyclic extension of degree dividing $N$. If $N$ is unramified in the CM field, this forces the torsion to be rational over $F$. But if $N$ is ramified....? – Pete L. Clark Apr 24 '14 at 18:06
Sorry, I missed your comment! It comes from the fact that all isogenies of degree $m$ starting from $E_1$ will give you an elliptic curve $E_3$ with endomorphism ring of conductor a divisor of $m$, so in particular $\mathrm{End}(E_3) \supset \mathcal{O}$. This means that $E_3$ is rational over $F$, so the isogeny is rational over $F$. In particular the Galois action is of the form $\lambda \mathrm{Id}$ on the $m$-torsion; but since we have a rational point $\lambda=1$. – Damien Robert Oct 28 '14 at 14:36

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