Elliptic Curves with CM and Class Field Theory

Question

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.

Answer 1

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: http://pub.math.leidenuniv.nl/~strengtc/research.html

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 $F$].

Answer 2

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.