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

Question

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 $D$?

2. 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:

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

4. 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:

5. 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 $\operatorname{Disc}(x^4-x^3-2x^2-2x-1)=-5\times95$.

Answer 1

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.