# Discriminant of the ring class field of the order $\mathcal{O}=\mathbb{Z} \left[\frac{D+\sqrt{D}}{2}\right]$

Let's look at the procedure in the main theorem of this MO post: $$\color{Red}{\text{Starting}}$$ from a discriminant $$D$$, and at the $$\color{Green}{\text{end}}$$, we $$\color{Green}{\text{find}}$$ a polynomial $$f_{D, h}(x)$$. [That procedure just tells us about the existence of the class field, and does not give us an efficient method to compute the class field, so we do not know $$f_{D, h}(x)$$ practically.]

1. What can we say about the discriminant of the ring class field of the order $$\mathcal{O}=\mathbb{Z}\left[\frac{D+\sqrt{D}}{2}\right]$$ in the imaginary quadratic field $$K=\mathbb{Q}(\sqrt{D})$$ comparing with $$D$$?
1. My Question is: I ask the first question because I am looking for something in the reverse order: $$\color{Green}{\text{Starting}}$$ from a polynomial $$f(x)$$ which is a minimal polynomial of some primitive element for some ring class field of a quadratic order, how can I $$\color{Red}{\text{find}}$$ a corresponding Discriminant $$D$$? In other words: What is the relation between discriminant of the ring class field of the order $$\mathcal{O}=\mathbb{Z}\left[\frac{D+\sqrt{D}}{2}\right]$$ in the imaginary quadratic field $$K=\mathbb{Q}(\sqrt{D})$$ and $$D$$?

If one can give a somehow satisfying relation for the first question, then we may have a good restriction for the choices for $$D$$. For instance:

1. If we let $$f(x)=x^3-x-1$$, I do not know how should I reach to $$D=-4\times23$$, note that $$\operatorname{Disc}(x^3-x-1)=-23$$.

2. If we let $$f(x)=x^3-4x-1$$, I do not know how should I reach to $$D=4\times229$$, note that $$\operatorname{Disc}(x^3-4x-1)=229$$.

considering (3) and (4) would lead me to guess that discriminant of the ring class field of the order $$\mathcal{O}=\mathbb{Z} \left[\frac{D + \sqrt{D}}{2}\right]$$ is equal to $$D$$ module $${\mathbb{Q}}/{\mathbb{Q}^{\times 2}}$$, and this is the reason why I asked (1) but the following prevents me from going on:

1. If we let $$f(x)=x^4-x^3-2x^2-2x-1$$, I do not know how should I reach to $$D=-4\times95$$, note that $$\operatorname{Disc}(x^4-x^3-2x^2-2x-1)=-5\times95$$.
• Do you get something from the Conductor-discriminant_formula Jun 11, 2020 at 23:08
• This seems like it should possibly be split into multiple questions. What do you mean by the "discriminant of X in the field Y"? These rings you talk about are $\mathbb Z$-lattices and have a discriminant without having to think about the field in which they live.
– user208649
Jun 12, 2020 at 4:17
• Also, the ring you describe is not necessarily an order as in particular that generator need not be integral for certain choices of $D$, like $D=2$.
– user208649
Jun 12, 2020 at 4:19
• @reuns Thanks for introducing Conductor-discriminant formula, honestly I can not go further easily. As I emphasized in my question, my concern is the second question: to do the reverse procedure in this MO post in the reverse order. I just meant If one can give a somehow satisfying easy relation for the first question, then we may have a good restriction for the choices for D. Jun 12, 2020 at 9:35
• To me your question 1. is given an order $O$ of $O_K ,K=Q(\sqrt{D})$ and $L$ its ring class field, then (class field theory) you know the Galois group and Hecke characters of $L/K$ and the conductor discriminant formula gives $Disc(L)$. Your question 2. is given an abelian extension $F/K$ then you need to find (through class field theory) the order $O$ and (quotient of) $Cl(O)$ which corresponds to $Gal(F/K)$ through the Artin map. Jun 12, 2020 at 18:50

A partial answer to Q($$1$$): If we add the assumption that "$$D$$ is fundamental", then the root discriminant of the Hilbert class field would be equal to the root discriminant of $$\mathbb{Q}(\sqrt{D})$$. Especially in the case of the odd class numbers, the discriminant is the same as $$D$$ module $${\mathbb{Q}}/{\mathbb{Q}^{\times 2}}$$, but this does not help us to give even a partial answer for Q($$2$$).
A partial answer to Q($$2$$): If we add the assumption that "$$\deg(f)=3$$", then $$D=f^2\text{Disc}(f(x))$$ would works for every nonzero integer $$f$$. Aslo, sometimes we can find smaller $$D$$ than $$\text{Disc}(f(x))$$. This would answer Q($$3$$) and Q($$4$$) immediately.