Associate implies same orbit under multiplication by group of units in integral domain

Statement

Suppose $R$ is a integral domain and $a,b\in R$ are associate elements in $R$. Then, there exists a unit $u$ in $R$ such that $b=au$.

Definitions used

Unit

Further information: Unit

An element $u$ of a commutative unital ring $R$ is termed a unit if there exists $v\in R$ such that $uv=1$.

Associate elements

Further information: Associate elements

Two elements $a,b$ in a commutative unital ring $R$ are termed associate elements if they both divide each other: $a|b$ and $b|a$. In other words, there exist elements $u,v\in R$ such that $b=au$, $a=bv$.

Related facts

• Elements in same orbit under multiplication by group of units are associate
• Associate not implies same orbit under multiplication by group of units: The statement breaks down for arbitrary commutative unital rings.

Proof

Given: An integral domain $R$. Two associate elements $a,b\in R$.

To prove: There exists a unit $u\in R$ such that $a=bu$.

Proof: By definition, there exist $u,v\in R$ such that $b=au$ and $a=bv$. Thus, we get:

$a=bv=(au)v=a(uv)$.

This yields:

$a(uv-1)=0$.

Since $R$ is an integral domain, either $a=0$ or $uv-1=0$. We consider both cases:

• $a=0$: In this case, $b=0$, and we obtain that $b=a.1$, so $b$ is the product of $a$ and a unit.
• $uv-1=0$: In this case, we get $uv=1$, so $u$ is a unit. Thus, $b=au$ for a unit $u$.