# Proving that $F[x] / \langle f(x) g(x) \rangle \cong F[x] / \langle f(x) \rangle \oplus F[x] / \langle g(x) \rangle$

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 $F[x]$ ?

• The surjectivity, in fact, the whole statement, is a corollary of Chinese Remainder Theorem. Did you encounter that before, or is this an exercise as a stepping stone to C.R.T.? – Cave Johnson Dec 29 '16 at 9:50
• I haven't seen the Chinese Remainder Theorem yet in the context of rings. This is just a problem from a book on abstract algebra I'm studying from. – Kamil Dec 29 '16 at 9:55

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 $\Phi(q(x)g(x)h(x))=(\overline{h(x)},0)$.
• Thank you for the hint. What is your $h(x)$ though? Do you multiply Bezout identity with $h(x)$ ? – Kamil Dec 29 '16 at 10:19
• @Kamil $h(x)$ is an arbitrary polynomial. And yes, the evalutation of $\Phi(p(x)f(x)h(x))$ involves multiplying the both sides of Bezout by $h(x)$, the same for $\Phi(q(x)g(x)h(x))$. – Cave Johnson Dec 29 '16 at 10:21
• I proved the surjectivity. Do I now have to prove that $\langle f(x) \rangle \cap \langle g(x) \rangle = \langle f(x) g(x) \rangle$? I think this is needed to prove the kernel is $\langle f(x) g(x) \rangle$. – Kamil Dec 29 '16 at 10:39
• @Kamil Yes, you need to prove that. This is not difficult once we have Bezout's identity. Suppose $a(x)f(x)=b(x)g(x)$. We find $f(x)\mid b(x)g(x)$, hence $f(x)\mid b(x)q(x)g(x)=b(x)-b(x)p(x)f(x)$. So $f(x)\mid b(x)$. – Cave Johnson Dec 29 '16 at 11:01