In an integral domain $R$ associates of an irreducible element are irreducible.

Question

My attempt :

Let $p$ be an irreducible element in $R$.Let $q$ be an associate of $p$.Then there exists a unit $u \in R$ such that $q = pu$.Now if $q$ were reducible then $q$ can be written as $q = ab$ where none of $a$ and $b$ are units in $R$.Then we have $pu = ab$ and so $p = (u^{-1} a) b$.Since $p$ is irreducible one of them should be a unit.Since $b$ is not a unit.So $u^{-1} a$ is a unit.Now I have stuck and can't proceed further.

Please give me a hint to complete it.

Thank you in advance.

Answer 1

Your idea is right. To finish your proof just note that if $u^{-1}a$ is an unit, then $u(u^{-1}a)=a$ would be an unit too, but this contradicts the hypothesis that $a$ is not an unit.