> [!NOTE] Lemma > Let $R$ be an [[Integral Domain|integral domain]] and $a,b\in R$. Then $a\mid b$ and $b\mid a$ if, and only if there exists a [[Unit in a Ring|unit]] $u\in R^*$ such that $a=ub$ where $a\mid b$ denotes $a$ [[Divisibility|divides]] $b$. ###### Proof Suppose there exists $u\in R^*$ such that $a=ub$. Since $u$ is a unit, there exists $v\in R^*$ such that $uv=1.$ Now $a=ub$ tells us that $b\mid a$ and multiplying both sides by $v$ yields $b=va$, so $a\mid b$. Conversely, suppose $a\mid b$ and $b \mid a$. Then $a=ub$ and $b=va$ for some $u,v\in R$. Suppose both $a,b$ are non-zero since if WLOG $a=0$ then $a=b=0$ and $a=1\cdot b$. Multiplying both equations with each other yields $ab=uvab$ and so applying the [[Cancellation Law for Integral Domains|cancellation law]], we conclude that $uv=1$. That is, $u$ is a unit and $a=ub$, as required.