show that $F[x]/q(x)= F[x]/p_1(x)\bigoplus \cdots \bigoplus F[x]/p_k(x)$

Question

This question is from herstein.

Let $p_1(x), \dots, p_k(x)$ be irreducible polynomials in $F[x]$, where $F$ is a field. Let $q(x)=p_1(x)\cdots p_k(x)$. Then show that $F[x]/q(x)= F[x]/p_1(x)\bigoplus \cdots \bigoplus F[x]/p_k(x)$

Any hints for this? I thought that if I know that the ideals are co-maximal, then by chinese remainder theorem, I will be done. But can I assume that ( since they are irreducible, $p_i(x), p_j(x)$ are relatively prime, so by bezouts we get that they are co-maiximal?

Is my idea correct? Any solutions?

Answer 1

Let $q=p_1\cdots p_k$, and note that since $p_i$ is irreducible for all $i$, we have that $(p_i)+(p_j)=F[x]$. Indeed, this holds for any polynomial ring over a field see here for instance.

Since have that the ideals $(p_i)$ are coprime, we have that by the $(p_i\cdots p_k)=(p_i)\cap \cdots \cap (p_j)$, hence by the Chinese remainder theorem we have that there is an isomorphism: $$F[x]/(q)\cong F[x]/(p_1)\oplus\cdots \oplus F[x]/(p_k)$$

Edit: so yes your idea is correct.