Element of minimum Dedekind-Hasse norm is a unit

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

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

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

Related facts

 * Element of minimum norm in Euclidean ring is a unit
 * Element of minimum 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 $$N(b) \le N(r) \ \forall \ r \in R$$.

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 $$1 = bq$$.