Universal side divisor implies irreducible

Statement
In an commutative unital ring, any fact about::universal side divisor is an fact about::irreducible element.

Converse
The converse is not true, even in a Euclidean domain.

Other related facts

 * Universal side divisor not implies prime
 * Associate implies same orbit under multiplication by group of units in integral domain
 * Associate not implies same orbit under multiplication by group of units
 * Element of minimum norm among non-units in Euclidean ring is a universal side divisor
 * Euclidean ring that is not a field has a universal side divisor

Proof
Given: A commutative unital ring $$R$$, a universal side divisor $$x \in R$$ such that $$x = ab$$.

To prove: $$x$$ is neither zero nor a unit, and either $$a$$ is a unit or $$b$$ is a unit.

Proof: The fact that $$x$$ is neither zero nor a unit follows form the definition of universal side divisor, so it remains to show that if $$x = ab$$, then either $$a$$ is a unit or $$b$$ is a unit.

Since $$x$$ is a universal side divisor, we obtain that either $$x | a$$ or there exists a unit $$u$$ such that $$x | a - u$$.


 * 1) Case $$x | a$$: In this case we have $$x | a$$ and $$a | x$$, so $$x$$ and $$a$$ are associates, and we are done.
 * 2) Case there exists a unit $$u$$ such that $$x | a - u$$: We have $$a - u = xy$$ for some $$y$$, yielding $$u = a - xy = a - aby = a(1 - by)$$. Since $$u$$ is a unit, there exists $$v$$ such that $$uv = 1$$, yielding $$a(1-by)v = 1$$, so $$a$$ is a unit with inverse $$(1-by)v$$. Thus, $$x$$ is associate with $$b$$, and we are done.