# 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 Euclidean norm on $R$ -- in particular, $R$ is a 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 universal side divisor in $R$.

## Definitions used

### Euclidean norm

Further information: Euclidean norm

### Universal side divisor

Further information: 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 $b | a - r$.

## 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.