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?


1 Answer 1


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.

  • $\begingroup$ 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 $\mathscr O_K$. $\endgroup$ Commented Oct 29, 2018 at 19:59
  • $\begingroup$ 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 $K$. $\endgroup$ Commented Oct 30, 2018 at 16:59
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    $\begingroup$ 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. $\endgroup$ Commented Oct 30, 2018 at 18:10
  • $\begingroup$ I don't think this is true. Even if you have an elliptic curve $E$ for which $K(E[\mathfrak{m}]) =K(h(E[\mathfrak{m}]))$, one can twist $E$ by a quadratic field $K(\sqrt{d}) \not\subseteq K(h(E[\mathfrak{m}]))$ to get a curve $E_d$. Now, $h$ doesn't depend on which twist you pick, so $K(h(E[\mathfrak{m}])) =K(h(E_d[\mathfrak{m}]))$; but $K(E_d[\mathfrak{m}]) \neq K(h(E_d[\mathfrak{m}]))$. In fact, there are infinitely many such $d$. $\endgroup$
    – Rdrr
    Commented Jul 28, 2021 at 21:49

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