Element of minimum norm in Euclidean ring is a unit

Statement
Suppose $$R$$ is a Euclidean ring, i.e., $$R$$ is a commutative unital ring with a Euclidean norm $$N$$. Suppose $$b \in R$$ is a nonzero element such that:

$$N(b) \le N(r) \ \forall \ r \ne 0$$.

Then, $$b$$ is a unit in $$R$$.

In particular, all elements of norm zero are units.

Related facts

 * Unit in Euclidean ring need not have minimum norm
 * Element of minimum norm among non-units in Euclidean ring is a universal side divisor
 * Every Euclidean ring has a unique smallest Euclidean norm: With respect to this Euclidean norm, the units are precisely the elements of norm zero.

Proof
Given: A commutative unital ring $$R$$ with Euclidean norm $$N$$, such that $$N(b) \le N(r)$$ for all $$r \ne 0$$.

To prove: $$b$$ is a unit in $$R$$.

Proof: By the Euclidean algorithm, we can write:

$$1 = bq + r, \qquad q,r \in R$$

where $$r = 0$$ or $$N(r) < N(b)$$. By the assumption, $$N(r) < N(b)$$ is impossible, so we are forced to have $$r = 0$$. Thus:

$$1 = bq$$,

and we obtain that $$b$$ is a unit.