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.

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