If a quadratic integer has a Dedekind-Hasse norm, then the absolute value of the corresponding field norm is Dedekind-Hasse

Let $\mathcal{O}$ be a quadratic integer ring. Suppose that $\mathcal{O}$ is a principal ideal domain. Prove that the absolute value of the field norm on $\mathbb{Q}(\sqrt{D})$ (namely $N(a + b \sqrt{D}) = |a^2 - Db^2|$) is Dedekind-Hasse. Deduce that if a quadratic integer ring has a Dedekind-Hasse norm, then the absolute value of the field norm is Dedekind-Hasse.

Recall that a divisibility norm $N$ on an integral domain is called Dedekind-Hasse if for all nonzero $a,b$, either $a \in (b)$ or there exists $s \in (a,b)$ such that $0 < N(s) < N(b)$.

Suppose $\mathcal{O}$ is a principal ideal domain, and let $a,b \in \mathcal{O}$ be nonzero. Note that $(a,b) = (d)$ is principal for some $d$. In particular, $b = du$ for some $u$, so that $N(b) = N(d)N(u)$. If $N(u) = 1$, then $u$ is a unit, so that $(a,b) = (b)$. Thus $a \in (b)$. If $N(u) > 1$, then $N(d) < N(b)$. Thus $N$ is Dedekind-Hasse.

Let $\mathcal{O}$ be a quadratic integer ring. If $\mathcal{O}$ has a Dedekind-Hasse norm, then it is a principal ideal domain. Thus the absolute value of the field norm is Dedekind-Hasse.