Let $F$ be a field and let $a$ be a nonzero element of $F$. If $af(x)$ is irreducible over $F$, prove that $f(x)$ is irreducible over $f(x)$.
I see that if $af(x)$ is irreducible, then $af(x) = g(x)h(x)$ where $g(x)$ or $h(x)$ is a unit. Since $a$ is a nonzero element of a field and all elements of a field are units, that implies that $a$ is a unit and $a = g(x)$ (suppose it's the unit). Then $f(x) = h(x)$ is not a unit in $F[x]$, but from here I'm stuck.
Best Answer
$af(x)=g(x)\cdot h(x)$ implies $f(x)=a^{-1}g(x)\cdot h(x)$