# 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 $f(x)$.

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.

$af(x)=g(x)\cdot h(x)$ implies $f(x)=a^{-1}g(x)\cdot h(x)$
You're taking the definition the wrong way. To prove that $f$ is irreducible, you must assume $f(x)=g(x)h(x)$ for some $g,h$ and prove that one of them is a unit (i.e. constant).
To do this, observe that $$af(x)=(ag(x))h(x);$$now because $af$ is irreducible, $ag(x)$ or $h(x)$ must be constant. If $h(x)$ is constant, it's over; otherwise $ag(x)$ is constant, and then so is $g(x)=a^{-1}(ag(x))$.