# Do we need the Weber function to generate ray class fields of imaginary quadratic fields of class number one?

I'm a bit confused by the role of the Weber function in generating ray class fields of imaginary quadratic fields of class number one. More specifically, let $$K$$ be such a field and $$E$$ an elliptic curve defined over $$\mathbf{Q}$$ with CM by $$\mathscr{O}_K$$. Let $$\mathfrak{m}$$ be a modulus for $$K$$. To get the ray class field for $$\mathfrak{m}$$, we consider the torsion group $$E[\mathfrak{m}]$$. This is a free $$\mathscr{O}_K/\mathfrak{m}$$-module of rank $$1$$, and $$G_K$$ acts on it by $$\mathscr{O}_K$$-linear automorphisms. Thus we have a character $$\alpha \colon \mathrm{Gal}(K(E[\mathfrak{m}])/K) \hookrightarrow \mathscr{O}_K^\times$$. Moreover, the Main Theorem of CM tells us that $$\alpha(\mathrm{rec}(s))$$ acts by $$\lambda(s) s_\mathfrak{m}^{-1}$$ for $$\lambda \colon \mathbf{A}_{K, f}^\times \rightarrow K^\times$$ the CM Großencharacter. Note that if $$s \equiv 1 \pmod {\mathfrak{m}}$$, this tells us that $$\alpha(\mathrm{rec}(s)) = \lambda(s) \pmod {\mathfrak{m}}$$, and that if $$s = (\gamma)$$ is a principal idele, that $$\lambda(s) = s_\mathfrak{m} \pmod{\mathfrak{m}}$$.

Now, at least if $$\mathfrak{m}$$ is divisible by the conductor of $$\lambda$$ (i.e. $$\mathfrak{m}$$ is divisible by the primes of bad reduction of $$E$$), we see that $$\alpha(\mathrm{rec}(s))$$ only depends on the image of $$s$$ in $$\mathrm{Cl}_\mathfrak{m}(K)$$. Thus, we have a map $$\alpha \circ \mathrm{rec} \colon \mathrm{Cl}_{\mathfrak{m}}(K) \rightarrow \mathrm{Gal}(K(E[\mathfrak{m}])/K) \hookrightarrow (\mathscr{O}_K/\mathfrak{m})^\times$$.

If we compose $$\alpha \circ \mathrm{rec}$$ with the map $$(\mathscr{O}_K/\mathfrak{m})^\times \rightarrow (\mathscr{O}_K/\mathfrak{m})^\times/\mathscr{O}^\times = \mathrm{Cl}_\mathfrak{m}(K)$$, we certainly get an isomorphism, since we can see that $$\alpha \circ \mathrm{rec}(\pi_v) = \lambda(\pi_v)$$ is a generator of $$v$$ for all but finitely many primes $$v$$ of $$K$$ which are split over $$\mathbf{Q}$$.

In the sources I've read on this, it seems that one replaces $$\alpha \colon \mathrm{Gal}(K(E[\mathfrak{m}])/K) \hookrightarrow \mathscr{O}_K^\times$$ with $$\mathrm{Gal}(K(h(E[\mathfrak{m}]))/K) \hookrightarrow (\mathscr{O}_K/\mathfrak{m})^\times/\mathscr{O}_K^\times$$ where $$h$$ is a Weber function. This shows that $$K(h(E[\mathfrak{m}]))$$ is the ray class field of $$K$$ with conductor $$\mathfrak{m}$$. One sometimes comments that the field $$K(E[\mathfrak{m}])$$ might not be abelian over $$K$$ in general, since $$E$$ is only defined over the Hilbert class field of $$K$$. But in the class number one case, this problem goes away.

So what happens to $$\alpha$$? We've shown that when $$\mathfrak{m}$$ is divisible by the conductor of $$\lambda$$, $$K(E[\mathfrak{m}])$$ is an abelian extension of $$K$$ such that the reciprocity map kills ideals with generators which are congruent to $$1$$ mod $$\mathfrak{m}$$, so it must be contained in the ray class field with conductor $$\mathfrak{m}$$, i.e. the subfield $$K(h(E[\mathfrak{m}])$$. Thus, these are the same field. Is $$\alpha$$ surjective, or is the image some index $$2$$ (or $$4$$ or $$6$$ for the cases of extra automorphisms, I suppose) subgroup of $$(\mathscr{O}_K/\mathfrak{m})^\times$$ mapping isomorphically onto the ray class group?

If $$\mathfrak{m}$$ is not divisible by the conductor of $$\lambda$$, $$K(h(E[\mathfrak{m}]))$$ is still the ray class field of conductor $$\mathfrak{m}$$, but I don't think my proof shows the reciprocity map into $$\mathrm{Gal}(K(E[\mathfrak{m}])/K)$$ kills principal ideals with a generator which is congruent to $$1$$ mod $$\mathfrak{m}$$. Are these still the same field? If not, $$K(E[\mathfrak{m}])$$ is some abelian extension of $$K$$, so it must sit inside $$K(h(E[\mathfrak{n}]))$$ for some $$\mathfrak{n}$$ - which one?

A quick comment (which I have to post as an answer, since I was locked out of my old account): The construction of $$K(j(E), E[\mathfrak{m}])$$ you describe in the general gives abelian extensions of the Hilbert class field $$K(j(E))$$ which might not be abelian over $$K$$. The role of the Weber function $$h$$ is to select out the subfields which are abelian over $$K$$. Geometrically, fixing a model for $$E$$, this $$h$$ can be defined as any finite map $$h: E \longrightarrow E/\operatorname{Aut}(E)\approx E/\mathcal{O}_K^{\times}$$, and does not depend on the choice of model. (Analytically, it is defined as $$z \in {\bf{C}}/\Lambda \approx E({\bf{C}}) \longrightarrow (\wp(z, \Lambda), \wp'(z, \Lambda))$$, and does not depend on the choice of lattice $$\Lambda$$). Hence, the way to construct the ray class field of conductor $$\mathfrak{m}$$ of $$K$$ in general is to adjoin to the coordinates of the images in $$X := E/\operatorname{Aut}(E)$$ of the torsion points $$E[\mathfrak{m}]$$. If $$O_K^{\times}$$ has order 2, then the homomorphism $$E \rightarrow X$$ is given by the coordinate $$x$$; if the order is 4 it is given by $$x^2,$$ and if the order is 6 then it is given $$x^3$$. In other words, the ray class extension of conductor $$\mathfrak{m}$$ of $$K$$ is generated by $$j(E)$$ and the coordinates $$x$$, $$x^2$$, or $$x^3$$ respectively, which to be explicit is $$K(j(E), h(E[\mathfrak{m}]))$$ rather than $$K(j(E), E[\mathfrak{m}])$$. However, when the class number of $$K$$ is one, then it is usually $$K(E[\mathfrak{m}])$$ as you suggest, but could be a larger abelian extension of $$K$$. Sorry that this comment does not answer the original question about the exact location of $$K(E[\mathfrak{m}])$$. (This should come down to keeping track of action by elements of $$\operatorname{Aut}(E) = \mathcal{O}_K^{\times}$$ on torsion points in the proof of the main theorem). If I have time, then I will try to post an answer later.
• Doesn't the $\mathscr O_K$-linear Galois representation of $G_K$ on $E[\mathfrak{m}]$ depend intrinsically on $E$? I thought that in the class number $1$ case, there was just one isomorphism class of elliptic curve $E$ with CM by $\mathscr O_K$. – pierre de fermat Oct 29 '18 at 19:59
• Yes, indeed — thanks for pointing this out! You are of course right that there is only one isomorphism class when $K$ has class number one. In this case, the maximal abelian extension of $K$ is in fact generated by the torsion of an elliptic curve $E$ with CM by $O_K$. In general, adjoining the torsion of an elliptic curve $E$ with CM by $\mathcal{O}_K$ generates abelian extensions of the Hilbert class field $K(j(E))$ which are not necessarily abelian over $K$. The role of the Weber function $h$ (as defined geometrically as in my comment) is to select out subfields which are abelian over $K$. – yourstruly2095 Oct 30 '18 at 16:59
• I think the question is not asking about the general case, but the particular case of class number 1. In that case, it is known that $K(E^{tors}) = K(h(E^{tors})) = K^{ab}$. The question is whether an analogous equality holds for the $\mathfrak{m}$-torsion, i.e. what is the relationship between $K(E[\mathfrak{m}])$ and $K(h(E[\mathfrak{m}]))$. You say in your last sentence that these fields are equal -- why is this true? I have not seen this stronger statement in standard texts on this. – Jesse Silliman Oct 30 '18 at 18:10