Is it possible rewrite $\frac{\mathbb{Z}_m [x]}{<f(x)>}$ as a in direct sum for $m$ composite?

Question

I have a doubt: Given $\frac{\mathbb{Z}_m [x]}{<f(x)>}$, where $m$ is composite and $f(x)=\prod_{i=1}^t f_i^{a_i} (x)$ (irreducible factors), can I admit $\frac{\mathbb{Z}_m [x]}{<f(x)>} \cong \bigoplus_{i=1} ^t \frac{\mathbb{Z}_m [x]}{<f_i ^{a_i}(x)>}$ as rings ? Or is it true just for $\mathbb{Z}_m$, for $m$ prime?

Thank you.

Answer 1

Partial answer

$x$ and $x-p$ are not comaximal in $\Bbb{Z}/(p^2)[x]$ as $(x,x-p)=(x,p)$ which is a maximal ideal

so we don't have $$\Bbb{Z}/(p^2)[x]/(x(x-p)) \cong \Bbb{Z}/(p^2)[x]/(x)\times \Bbb{Z}/(p^2)[x]/(x-p)$$

Answer 2

I think Chinese Remainder theorem will do the work.

Let $I_1,...I_n$ be ideals in a ring $R$ which satisfy $I_i+I_j=R$ for $i \neq j$. Then we have $I_1∩···∩I_n=I_1...I_n$ i.e, $\bigcap_{i=1}^{n} I_i=\prod_{i=1}^{n}I_i$ . Then $$R/I_1 \cdots I_n \simeq \oplus_{i=1}^{n} R/I_i.$$ The requirement $I_i+I_j=R$ is not met in the above example.