Element of minimum Dedekind-Hasse norm is a unit

Statement

Suppose $R$ is a commutative unital ring and $N$ is a Dedekind-Hasse norm on $R$. Then, the set:

$\{b\in R\setminus \{0\}\mid N(b)\le N(r)\mathrm{\forall }r\in R\setminus \{0\}\}$

is contained in the set of units of $R$.

Related facts

• Element of minimum norm in Euclidean ring is a unit
• Element of smallest norm among non-units in Euclidean ring is a universal side divisor

Proof

Given: A commutative unital ring $R$ with Dedekind-Hasse norm $N$, an element $b\in R\setminus \{0\}$ such that $N(b)\le N(r)\mathrm{\forall }r\in R$.

To prove: There exists $q\in R$ such that $uq=1$.

Proof: Apply the definition of Dedekind-Hasse norm to the elements $1$ and $b$. We obtain that either $1\in (b)$, or there exists $r\in (1,b)$ such that $N(r)<N(b)$. However, we know that $N(b)\le N(r)$ for all nonzero $r$, so the latter case cannot occur. Thus, $1\in (b)$, so there exists $q\in R$ such that $1=bq$.