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

Statement
Suppose $$R$$ is a commutative unital ring and $$N$$ is a fact about::Euclidean norm on $$R$$ -- in particular, $$R$$ is a fact about::Euclidean ring. Suppose $$b$$ is a nonzero element of $$R$$ that is not a unit, and such that $$N(b) \le N(r)$$ for all nonzero non-units $$r$$ of $$R$$. Then, $$b$$ is a fact about::universal side divisor in $$R$$.

Universal side divisor
A nonzero non-unit $$b$$ in a commutative unital ring $$R$$ is termed a universal side divisor in $$R$$ if for any $$a \in R$$, either $$b | a$$ or there exists a unit $$r \in R$$ such that $$b | a - r$$.

Related facts

 * Element of minimum norm in Euclidean ring is a unit
 * Every Euclidean ring has a unique smallest Euclidean norm: With respect to this Euclidean norm, the units are precisely the elements of norm zero and the universal side divisors are precisely the elements of norm one.

Proof
Given: A commutative unital ring $$R$$ with a Euclidean norm $$N$$. A nonzero non-unit element $$b$$ of $$R$$ such that $$N(b) \le N(r)$$ for all nonzero non-units $$r$$ in $$R$$.

To prove: $$b$$ is a universal side divisor in $$R$$.

Proof: Pick any $$a \in R$$. Then, by the Euclidean algorithm, we can write:

$$a = bq + r$$

where either $$r = 0$$ or $$N(r) < N(b)$$. If $$r = 0$$, $$b | a$$, and we are done. Otherwise, $$N(r) < N(b)$$. By assumption, $$b$$ has smallest norm among the non-units, so $$r$$ must be a unit, hence $$bq = a - r$$, so $$b | a - r$$ for a unit $$r$$, and we are done.