If $p$ is an irreducible element of an integral domain $D$, and if $e$ belongs to $D^\ast$ prove that $ep$ is also ireducible

Question

If $p$ is an irreducible element of an integral domain $D$, and if $e$ belongs to $D^\ast$, prove that $ep$ is also ireducible .

To me it seems so profound, but I cannot get the proof . I'm trying to say that it is not irreducible, so there exists $a,b$ so that $ep=ab$ where $a,b$ don't belong to $D^\ast$....

Answer 1

This is, as has been mentioned, ill-written, but that said, if $ep=ab$ then $p=e^{-1}ab$. And if $a$ is not a unit, neither is $e^{-1}a$.

Answer 2

Hint $\, p\,$ isn't irreducible iff $\, p=0\,$ or $\,p$ is a unit or $\,p = ab\,$ for nonunits $\,a,b.\,$ All are preserved by scaling by a unit $\,e,\,$ e.g. $\ (ab)e = a(be)\,$ stays a product of nonunits, by $\,be\,$ unit $\!\iff\! b,e\,$ units. Therefore $\,pe\,$ isn't irreducible $\iff p\,$ isn't irreducible. Negating yields the sought result.