In Cox's book "Primes of the form $x^2 + ny^2$", he proves that in a quadratic imaginary field $K$, if $\mathcal O$ is an order of conductor $f \in \mathbb Z$, we have that the class group $\mathrm{Cl}(\mathcal O)$ of $\mathcal O$ is isomorphic to the quotient $I_K(f) / P_{K,\mathbb Z}(f)$, where $I_K(f)$ is the free abelian group generated by the prime ideals of $\mathcal O_K$ coprime to $f\mathcal O_K$, and $P_{K,\mathbb Z}(f)$ is the subgroup of principal ideals $\alpha \mathcal O_K$ where $\alpha$ is congruent mod $f\mathcal O_K$ to an integer coprime to $f$ (proposition 7.22). So $\mathrm{Cl}(\mathcal O)$ is a quotient of the ray class group relative to $f\mathcal O_K$.

I suspect a more general result, but I cannot find any reference. Let $K$ be a number field, and $\mathcal O$ an order of conductor $\mathfrak f \subset \mathcal O_K$,

Can $\mathrm{Cl}(\mathcal O)$ be expressed as a quotient of $\mathrm{Cl}_\mathfrak f(K)$, the ray class group of $K$ relative to $\mathfrak f$ ?

The problem actually boils down to proving that the map $$\varphi : I_K(\mathfrak f) \longrightarrow \mathrm{Cl}(\mathcal O) : \mathfrak a \longmapsto \mathfrak [\mathfrak a \cap \mathcal O]$$ is a surjection. Indeed, if it is the case, then $\mathrm{Cl}(\mathcal O) \cong I_K(\mathfrak f) / \ker(\varphi)$, but it is easy to see that $P_K(\mathfrak f) \subset \ker(\varphi)$ (where $P_K(\mathfrak f)$ is from the definition of the ray class group, i.e. $\mathrm{Cl}_\mathfrak f(K) \cong I_K(\mathfrak f) / P_K(\mathfrak f)$), so $\mathrm{Cl}(\mathcal O)$ would be a quotient of $\mathrm{Cl}_\mathfrak f(K)$.

Is $\varphi$ surjective ?

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    $\begingroup$ I think there's something a bit odd here. According to Magma, if $\mathcal{O}_{K} = \mathbb{Z}[i]$ and $f = 5$, then ${\rm Cl}(\mathcal{O}) \cong \mathbb{Z}/2\mathbb{Z}$, but the ray class group associated to the ideal $(5)$ of $\mathbb{Z}[i]$ is $\mathbb{Z}/4\mathbb{Z}$. I seem to remember there was a difference between the "ray class fields" and the "ring class fields", but I've lent my copy of Cox's book to a student. $\endgroup$ May 10, 2014 at 23:35
  • $\begingroup$ You are right there is a mistake. He proves that $\mathrm{Cl}(\mathcal O)$ is isomorphic to $I_K(f) / P_{K,\mathbb Z}(f)$ ($I_K(f)$ is the free abelian group generated by the prime ideals coprime to $f\mathcal O_K$, and $P_{K,\mathbb Z}(f)$ is the subgroup of principal ideals generated by some $\alpha$ congruent to an integer mod $f\mathcal O_K$), which is not exactly the ray class group (we have this extra condition about the integer). I'll update my question, which becomes : can $\mathrm{Cl}(\mathcal O)$ be expressed as a quotient of $\Cl_\mathfrak f(K)$ ? $\endgroup$
    – Calodeon
    May 10, 2014 at 23:48

1 Answer 1


The answer is yes, essentially by weak approximation.

Let $\mathbb{I}_K=\prod'_{v}K_v$ denote the finite Idelles over $K$. Now let $\prod_v\mathcal{O}_v^{\times}=\widehat{\mathcal{O}}^{\times}$ denote the multiplicative group of the completion of $\mathcal{O}$. Then (as Cox explains) you can write

$$Cl(\mathcal{O}) = K^{\times}\backslash\mathbb{I}_K/\widehat{\mathcal{O}}^{\times}.$$

Also letting $\mathbb{I}_{K,f}:=\prod'_{v\not\mid f}K_v$ we have $$I_K(f) = \mathbb{I}_{K,f}/\prod_{v\not\mid f}\mathcal{O}_{K,v}^{\times}.$$

Then the map $I_K(f)\rightarrow Cl(\mathcal{O})$ is induced by the natural inclusion $\mathbb{I}_{K,f}\hookrightarrow \mathbb{I}_K$ so your question amounts to establishing the equality $$\prod_{v\mid f}K^{\times}_v = K^{\times}\cdot\prod_{v\mid f}\mathcal{O}_v^{\times}.$$

Weak approximation says that $K^{\times}$ is dense in $\prod_{v\mid f}K^{\times}_v$, which implies your claim.


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