Let $\mathbb{Q}(\sqrt{d})$ be finite extension with square free $d$, with $R_d$ being integral closure of subring $\mathbb{Z}$.
It is known that $R_d=\mathbb{Z}[\sqrt{d}]$ if $d\not\equiv 1\pmod{4}$ and $R_d=\mathbb{Z}[\frac{1+\sqrt{d}}{2}]$ if $d\equiv 1\pmod{4}$ are Dedekind domains.
With tools solely from number theory, is it possible to calculate ideal class group for $R_d$ when $d$ is small?
For instance, show ideal class group for $R_{-5}=\mathbb{Z}[\sqrt{-5}]$ is $\mathbb{Z}/2\mathbb{Z}$; for $R_{-23}=\mathbb{Z}[\frac{1+\sqrt{-23}}{2}]$ is $\mathbb{Z}/3\mathbb{Z}$.
Attempt: For prime ideal $P$ in $\mathbb{Z}[\theta_d]$, $P\cap\mathbb{Z}=(p)$ must be prime. Here $\theta_d$ is the adjoined element in closure.
If $(p)\subseteq\mathbb{Z}[\theta_d]$ is prime, then $P=(p)$, otherwise $(p)=(p,\theta_d+k)(p,\theta_d-k)$ and $P$ is one of them.
$R_{-5}=\mathbb{Z}[\sqrt{-5}]$ has nonprincipal prime ideal $P_2=(2,1+\sqrt{-5})$, so the ideal class group is not trivial.
To show ideal class group is $\mathbb{Z}/2\mathbb{Z}$ it suffices to show for every nonprincipal prime $Q=(p,\sqrt{-5}+k)$, $P_2Q$ principal.
This is where I stuck, is there elementary way to find single generator for $(2,1+\sqrt{-5})(p,\sqrt{-5}+k)$?
$R_{-23}=\mathbb{Z}[\frac{1+\sqrt{-23}}{2}]$ has nonprincipal prime ideal $P_2=(2,\frac{1+\sqrt{-23}}{2})$ and its square $P_2^2=(2,\frac{1-\sqrt{-23}}{2})$.
Similarly, it suffices to show for every nonprincipal prime $Q=(p,\frac{1+\sqrt{-23}}{2}+k)$, either $P_2Q$ or $P_2^2Q$ is principal.
I assume similar argument applies here, but there might be more case analysis.
Many answers I encountered uses (mentions) Minkowski bound, which I would like to avoid.
