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

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

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
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 $$u$$.