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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 $\operatorname{Disc}(x^4-x^3-2x^2-2x-1)=-5\times95$.
Michael Hardy
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Davood
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    Do you get something from the Conductor-discriminant_formula – reuns Jun 11 '20 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. –  Jun 12 '20 at 04:17
  • 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 $D=2$. –  Jun 12 '20 at 04:19
  • @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.

     – Davood Jun 12 '20 at 09:35
  • @TokenToucan (1) That ring class field is a field extension of $\mathbb{Q}(\sqrt{D})$, so we can speak about its discriminant. (2) In the direct procedure of this MO post, we assumed that $D$ is a discriminant, so $\mathcal{O}=\mathbb{Z} [\frac{D+\sqrt{D}}{2}]$ is definitely an order. (3) In the reverse procedure, when we speak about the order $\mathcal{O}=\mathbb{Z} [\frac{D+\sqrt{D}}{2}]$, we assume the Implicit condition that we are looking for a Discriminant $D$. – Davood Jun 12 '20 at 09:50
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    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. – reuns Jun 12 '20 at 18:50
  • @reuns I am not proficient in the subject, and maybe I am mistaken; but as far as my knowledge allows, your point about the second question just says about the existence, and does not give a practical and effective method to construct that order explicitly. – Davood Jun 12 '20 at 22:08

1 Answers1

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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.

Davood
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