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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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    $\begingroup$ 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. $\endgroup$ – Will Sawin Aug 30 '12 at 16:52
  • $\begingroup$ 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. $\endgroup$ – Will Sawin Aug 30 '12 at 16:56
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    $\begingroup$ @Asvin I think yes. Because an order in $\mathcal O$ acts on the fundamental group, $\mathcal O$ acts on the fundamental group tensored with $\mathbb Q$. Consider inside the fundamental group the lattice of all elements such that $\mathcal O$ times that element is an fundamental group. It is clearly preserved under addition and multiplication by $\mathcal O$, and has finite index as the order has finite index in $\mathcal O$. So the corresponding finite etale cover has CM by $\mathcal O$. Because it is "canonical", it should be defined over $\mathbb Q$. $\endgroup$ – Will Sawin Oct 5 '17 at 11:59
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    $\begingroup$ @Asvin To check this, apply the same construction to the action of $\mathcal O$ on the rational $\ell$-adic Tate module for each $\ell$, finding the maximal sublattice of the integral Tate module stable under $\mathcal O$, and again consider the isogeny defind by this lattice, which should be defined ovre $\mathbb Q$ because these sublattices are defined over $\mathbb Q$, then check that they are the same isogeny by checking that he integral lattice tensored with $\mathbb Z_\ell$ is the $\ell$-adic lattice foe each $\ell$. $\endgroup$ – Will Sawin Oct 5 '17 at 12:02
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    $\begingroup$ @Asvin Yes, that is my idea. The comparision isomorphism is pretty simple for abelian varieties - tensoring the classical fundamental group with $\hat{\mathbb Z}$ obtains the etale fundamental group. $\endgroup$ – Will Sawin Oct 7 '17 at 11:38
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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 $F$].
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  • $\begingroup$ 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. $\endgroup$ – Pete L. Clark Apr 14 '14 at 22:12
  • $\begingroup$ 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. $\endgroup$ – Damien Robert Apr 15 '14 at 11:04
  • $\begingroup$ "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...) $\endgroup$ – Pete L. Clark Apr 15 '14 at 18:08
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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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  • $\begingroup$ 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....? $\endgroup$ – Pete L. Clark Apr 24 '14 at 18:06
  • $\begingroup$ 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 $\lambda=1$. $\endgroup$ – Damien Robert Oct 28 '14 at 14:36
  • $\begingroup$ I don't follow why $K(j(E_2))\supseteq K(j(E_1))$ implies there is an isogeny $E_2\to E_1$. Using the fact that $E_2$ is isogenous to at most one twist of $E_1$, I can replace, $E_1$ with a twist over $\mathbb{Q}(j(E_1))$, such that we will still have the containment of ring class fields but $E_2$ wont be isogenous to $E_1$. $\endgroup$ – Rdrr Jan 30 at 19:34

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