Strictly multiplicatively monotone norm on Bezout domain is a Dedekind-Hasse norm

Statement
Suppose $$R$$ is a fact about::Bezout domain (i.e., it is an integral domain that is also a fact about::Bezout ring: every finitely generated ideal on $$R$$ is principal).

Suppose, further, that $$N: R \setminus \{ 0 \} \to \mathbb{N}_0$$ is a fact about::strictly multiplicatively monotone norm on $$R$$: in other words, we have that for $$ab \ne 0$$, $$N(ab) \ge N(a)$$, with equality occurring if and only if $$a$$ and $$ab$$ are associate elements.

Then, $$N$$ is a fact about::Dedekind-Hasse norm on $$R$$, and $$R$$ is a fact about::principal ideal domain.

Applications

 * Ring of integers in a number field that is Bezout is a PID with absolute value of algebraic norm a Dedekind-Hasse norm
 * Unique factorization and Bezout iff principal ideal

Proof
Given: A Bezout domain $$R$$ with a strictly multiplicatively monotone norm $$N$$. $$a,b \in R$$ with $$b \ne 0$$.

To prove: Either $$b | a$$ or there exists $$d \in (a,b)$$ with $$N(d) < N(b)$$.

Proof: Since $$R$$ is Bezout, $$(a,b) = (d)$$ for some $$d \in R$$. We have $$d | b$$ and $$d | a$$.


 * 1) If $$b | d$$, then $$b | a$$, and we are done.
 * 2) If $$b$$ does not divide $$d$$, then $$b = dc$$ for some $$c$$, where $$b$$ and $$d$$ are not associates. Thus, by the assumption of strictly multiplicatively monotone, $$N(d) < N(b)$$, and we are done.