Problem statement: Let $f(x)$ and $g(x)$ be irreducible polynomials over a field $F$. If $f(x)$ and $g(x)$ are not associates, prove that $F[x] / \langle f(x) g(x) \rangle$ is isomorphic to $F[x] / \langle f(x) \rangle \oplus F[x] / \langle g(x) \rangle$.
Attempt: We know that $F[x] / \langle f(x) \rangle$ is a field, because $\langle f(x) \rangle$ is a maximal ideal of $F[x]$ since $f$ is irreducible, and likewise for $F[x] / \langle g(x) \rangle$. Now, I was just trying to use the First Isomorphism Theorem. I defined a map $$ \Phi: F[x] \to \frac{F[x]}{\langle f(x) \rangle} \oplus \frac{F[x]}{\langle g(x) \rangle}: f(x) \mapsto \big( \overline{f(x)}, \overline{f(x)} \big). $$ I now want to prove this mapping is surjective, and that the kernel is $\langle f(x) g(x) \rangle$. Let $q(x) \in \ker \Phi$. Then $\Phi(q(x)) = ( \overline{0}, \overline{0}) = ( \langle f(x) \rangle, \langle g(x) \rangle)$. This means that $q(x) \in \langle f(x) \rangle$ and $q(x) \in \langle g(x) \rangle$. So there are $p_1(x), p_2(x) \in F[x]$ such that $q(x) = p_1(x) f(x)$ and $q(x) = p_2(x) g(x)$. I now want to conclude from this somehow that $q(x) \in \langle f(x) g(x) \rangle$, but I don't know how. How do I use the information that $f(x)$ and $g(x)$ are not associates?
Also, I'm not sure how to prove subjectivity. An arbitrary element in $F[x] / \langle f(x) \rangle \oplus F[x] / \langle g(x) \rangle$ is of the form $( \overline{f_1(x)}, \overline{f_2(x)})$. How to prove this is the image of some element in $F[x]$ ?
Hint. $f(x)$ and $g(x)$ are coprime. Bézout's identity implies the existence of $p(x),q(x)\in F[x]$ such that $p(x)f(x)+q(x)g(x)=1$. Now $\Phi(p(x)f(x)h(x))=(0,\overline{h(x)})$ and $\Phi(q(x)g(x)h(x))=(\overline{h(x)},0)$.