Universal side divisor implies irreducible

Statement

In an integral domain, any universal side divisor is an irreducible element.

Related facts

Converse

The converse is not true, even in a Euclidean domain. Further information: Irreducible not implies universal side divisor

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 smallest 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: An integral domain $R$, 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 $x|a$: We have $a=xy$ for some $y$, and we get $x=xyb$, yielding $x(1-yb)=0$. Since $x\ne 0$ and $R$ is an integral domain, we obtain $by=1$, so $b$ is a unit.
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 $(1-by)v$.