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

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?

• is it that the sum is within the k summands? Commented Sep 5 at 1:52

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: $$F[x]/(q)\cong F[x]/(p_1)\oplus\cdots \oplus F[x]/(p_k)$$
• small edit: $p_i$ is irreducible. Tried to edit, but edits must be at least 6 characters : ) Commented Sep 5 at 4:09