$R$ is integral domain. Show that $a$ is irreducible iff $au$ is irreducible where $u\in R^*$.
My Try:
Lets assume $a$ is irreducible and $au = bc \implies a=bcu^{-1}$.
We know that $u\in R^*$ so we can conclude $au$ is irreducible?
I think something is wrong. I didn't use the fact that $a$ is irreducible.
The definition of irreducibility: $q=xy \implies x\in R^* \lor y\in R^*$
Can you help me?
OK, assume $au = bc$, then indeed $a = bcu^{-1} = b(cu^{-1})$. Then either $b \in R^\ast$ or $cu^{-1} \in R^\ast$, as $a$ is irreducible. We are done when the former is the case, so suppose $cu^{-1} \in R^\ast$, then also $(cu^{-1})u = c \in R^\ast$, as $R^\ast$ is closed under multiplication. So then $c \in R^\ast$ and you are done.