Lemma about a prime times a unit

Question

I came across this Lemma:

"Let $R$ be an integral domain, and let $a,u\in R$ such that $u$ is invertible. Then $a$ is a prime if and only if $au$ is a prime.

I tried to prove it unsuccessfully, but would appreciate your help with a formal proof.

thanks :)

Answer 1

Just follow the definitions and make use of the multiplicative inverse. Try to completely justify the following line of thought.

$a$ prime $\implies$ $au$ prime:

Let $au|xy$, and we aim to show $au$ divides $x$ or $y$. Then $a|xy$, so $a$ divides $x$ or $y$. If $az=x$, then $au(u^{-1}z)=x$ implies $au|x$. A similar statement holds if $a$ divides $y$.

$au$ prime $\implies$ $a$ prime:

Actually, this is already done! Invoke the previous argument using the prime $au$ and the prime-times-a-unit $(au)u^{-1}=a$.