Let $K$ be an imaginary quadratic field with discriminant $d_K$. Suppose that $d_K=gt$, where either $g,t$ are discriminants or have the value $g=1,t=d$. Let $f$ be an additional discriminant of a quadratic field, or $f=1$ and at the same time $g\neq1$. Denote by $G$ the positive discriminant from $fg,ft$, and $T$ the negative one. For a prime $\mathfrak p$ of $K$ such that $(\mathfrak p,f)=1$ let
$$\chi(\mathfrak p)=\begin{cases}\left(\frac{G}{N(\mathfrak p)}\right) \text{ if }\left(\frac{G}{N(\mathfrak p)}\right)\neq 0,\\ \left(\frac{T}{N(\mathfrak p)}\right) \text{ if } \left(\frac{T}{N(\mathfrak p)}\right)\neq 0.\end{cases}$$ Siegel writes that

"Für beliebige zu $f$ teilerfremde Ideale a wird $\chi(\mathfrak a)$ auf multiplikative Art gebildet und erweist sich dann als eigentlicher Charakter der Gruppe der Ringklassen mit dem Führer $\lvert f\rvert$."

What exactly he means by "Gruppe der Ringklassen mit dem Führer $f$"? Is it the class group of the order with conductor $f$? What is more systematic or modern way of viewing these characters (and the "Ringklassen")?

In other words, how to compute the weights in the following formula

$$\varepsilon_G^{h_Gh_T}=\prod_{C\in \text{Cl}_f}h(\tau_C)^{-\chi(C)},$$ where $$h(\tau)=y^{1/2}\lvert \eta(\tau)^2\rvert, \\ \tau=x+iy,\\ \tau_C =\text{CM point representing the class $C$},\\\varepsilon_G = \text{the fundamental unit in $\mathbb Q(G^{1/2})$}.$$ See also my previous questions. According to these, in each class of $\mathcal O_f$ is an ideal of norm equal to $1$. But then is the character above trivial?

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    $\begingroup$ Am I the only one who expected a biographical discussion of Carl Ludwig Siegel? $\endgroup$ – Lior Silberman Apr 2 '20 at 11:59
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    $\begingroup$ "Gruppe der Ringklassen mit dem Führer $f$" means "ring class group of conductor $f$". The modern point of you is that you divide the idele class group by the relevant open subgroup, and you consider the characters of the quotient group. There are many books on the subject, e.g. Weil: Basic number theory, Neukirch: Algebraic number theory etc. $\endgroup$ – GH from MO Apr 2 '20 at 12:17
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    $\begingroup$ @user131781 Relax! $\endgroup$ – Francois Ziegler Apr 2 '20 at 12:31
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    $\begingroup$ The ring class group is associated to an order $\mathcal{O}$ of a number field. It equals the multiplicative group of invertible fractional ideals in $\mathcal{O}$ modulo the subgroup of invertible principal fractional ideals in $\mathcal{O}$. In an imaginary quadratic number field, there is a unique order of conductor $f$, and this gives rise to the ring class group of conductor $f$. $\endgroup$ – GH from MO Apr 2 '20 at 14:52
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    $\begingroup$ @GHfromMO How then do you explain the following puzzling phenomenon: in this question I found that each class contains a proper fractional ideal with norm equal to $1$. But it follows from this that the value of the character is always $1$. This does not make sense to me. $\endgroup$ – Shimrod Apr 2 '20 at 17:17

The ring class group is a special kind of a ray class group, introduced connection with the theory of complex multiplication in complex quadratic number fields. Given an integer $f > 1$, consider the group $D_f$ of all ideals coprime to $f$. The group $P_f$ of principal ideals is the group generated by ideals $(\alpha) \in D_f$ with $\alpha \equiv z \bmod f$ for some integer $z$. The ring class group modulo $f$ is simply the quotient $D_f/P_f$, and the ring class field defined modulo $f$ is obviously a subextension of the ray class field modulo $f$. In Siegel's definition of the quadratic character, ${\mathfrak p}$ is an ideal in the maximal order, not in the order with conductor $f$. It is true, however, that the ring class group modulo $f$ is isomorphic to the ideal class group of the order with conductor $f$, but the isomorphism is not as natural as one might expect. You should definitely look into the books by Cox and Cohn that I have already recommended.

Computing the corresponding CM-points is usually done via quadratic forms: to $Q(x,y) = Ax^2 + Bxy + Cy^2$ you associate the root of the equation $Q(z,1) = 0$ in the upper half plane.


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