Dedekind-Hasse norm implies principal ideal ring

Statement

A commutative unital ring that admits a Dedekind-Hasse norm must be a principal ideal ring.

In particular, an integral domain that admits a Dedekind-Hasse norm must be a principal ideal domain.

Definitions used

Dedekind-Hasse norm

Further information: Dedekind-Hasse norm

A Dedekind-Hasse norm on a commutative unital ring $R$ is a function $N$ from the nonzero elements of $R$ to the nonnegative integers with the property that for any elements $a,b\in R$ with $b\ne 0$, it is true that either $a\in (b)$ or there exists an element $r$ in the ideal $(a,b)$ such that $N(r)<N(b)$.

Principal ideal ring

Further information: Principal ideal ring

A commutative unital ring $R$ is termed a principal ideal ring if every ideal of $R$ is principal.

Proof

Given: A commutative unital ring $R$ with a Dedekind-Hasse norm $N$. An ideal $I$ of $R$.

To prove: There exists $b\in I$ such that $I=(b)$.

Proof: If $I=0$, we can take $b=0$, and we are done.

So, suppose $I\ne 0$. Consider the function $N$ on the nonzero elements of $I$. Since $N$ maps to a well-ordered set, there is an element $b\in I\setminus \{0\}$ such that $N(b)\le N(r)$ for all $r\in I\setminus \{0\}$.

Pick any $a\in I$. If $a=0$, then $a\in (b)$ or there exists $r\in (a,b)$ with $N(r)<N(b)$.

Note that in the latter case, we have $r\in (a,b)\subseteq I$, so $r\in I$, and $N(r)<N(b)$, contradicting the assumption that $b$ has minimum norm among the nonzero elements of $I$. Hence, we have the case $a\in (b)$. Thus, $I\subseteq (b)$. Conversely, we clearly have $(b)\subseteq I$, so $I=(b)$.