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

Statement

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

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

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

Related facts

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 $b\neq 0$.

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.