What is the class number of $\mathcal{O}_{\sqrt[3]{18}}$?

Question

I accept it without proof that $\mathcal{O}_{\sqrt[3]{2}}$ and $\mathcal{O}_{\sqrt[3]{3}}$ both have class number $1$. Also, I've been told that $\mathcal{O}_{\sqrt[3]{m^2}} = \mathcal{O}_{\sqrt[3]{m}}$; I presume this means that the two are identical, and not that one is a nontrivial subdomain of the other (you know, like $\mathbb{Z}[\sqrt{-8}]$ is a nontrivial subdomain of $\mathbb{Z}[\sqrt{-2}]$). This would mean that $\mathcal{O}_{\sqrt[3]{9}} = \mathcal{O}_{\sqrt[3]{3}}$ and therefore has class number $1$. Even if I am correct in all of this so far, I feel I would be jumping to a conclusion if I just assume that $\mathcal{O}_{\sqrt[3]{18}}$ has class number $1$.

P.S. I have barely begun trying to factorize numbers in $\mathcal{O}_{\sqrt[3]{2}}$ and $\mathcal{O}_{\sqrt[3]{3}}$. Obviously $2 = (\sqrt[3]{2})^3$ and $3 = (\sqrt[3]{3})^3$. But how do you factorize $2$ in $\mathcal{O}_{\sqrt[3]{3}}$, or $3$ in $\mathcal{O}_{\sqrt[3]{2}}$?

Answer 1

Yes, $K=\mathbb{Q}(\sqrt[3]{18})$ has class number $1$. For methods to compute the class number for pure cubic number fields, see also here. As to the last question, $(3)=(1+\sqrt[3]{2})^3$ in $\mathbb{Q}(\sqrt[3]{2})$, so $3$ ramifies completely.

Answer 2

If you want to factorize $3$ in $\mathcal{O}_{\sqrt[3]2}$, render

$3=2+1^3=(\sqrt[3]2+1)[(\sqrt[3]2)^2-\sqrt[3]2+1]$

and similarly

$2=3-1^3=(\sqrt[3]3-1)[(\sqrt[3]3)^2+\sqrt[3]3+1].$