4
$\begingroup$

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 $\operatorname{Disc}(x^4-x^3-2x^2-2x-1)=-5\times95$.
$\endgroup$
7
  • 1
    $\begingroup$ Do you get something from the Conductor-discriminant_formula $\endgroup$
    – reuns
    Jun 11, 2020 at 23:08
  • $\begingroup$ 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. $\endgroup$
    – user208649
    Jun 12, 2020 at 4:17
  • $\begingroup$ 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 $D=2$. $\endgroup$
    – user208649
    Jun 12, 2020 at 4:19
  • $\begingroup$ @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. $\endgroup$
    – Davood
    Jun 12, 2020 at 9:35
  • 1
    $\begingroup$ 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. $\endgroup$
    – reuns
    Jun 12, 2020 at 18:50

1 Answer 1

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.