I need to somehow prove that if $p(x)$ is irreducible and divides $f(x)g(x)$ then $p(x)$ divides $f(x)$ or $g(x)$.
I've been given the hints that I should use the theorems:
$p(x)$ is irreducible over $K$ implies that $J=K[x]\cdot p(x)$ is a maximal ideal in $K[x]$.
If $P$ is a maximal ideal of $A$ then $P$ is a prime ideal of $A$
Well, since $p(x)$ is irreducible we have that $J=K[x]\cdot p(x)$ is a maximal ideal in $K[x]$. Then we have that this maximal ideal is also a prime ideal of $A$. It means that $p_1(x)p_2(x)\in J \implies p_1(x) \in J \mbox{ or } p_2(x) \in J \implies$ $$p_1(x) = k_1(x)\cdot p(x)$$
or
$$p_2(x) = k_1(x)\cdot p(x)$$
which is what we wanted to prove
Am I right?