TITLE: Direct product of polynomial rings
QUESTION [5 upvotes]: Let $n = pq$, where $p$ and $q$ are distinct primes. I am trying to show that: $$\mathbb{Z}_n[X] \cong \mathbb{Z}_p[X] \times \mathbb{Z}_q[X].$$
Would it suffice to say that $\rho(np) = \rho(n)\rho(p)$?
I am not sure how to set up the proof, so any help would be appreciated.
Thanks.

REPLY [2 votes]: Hint: Find maps $\mathbb Z[X]\to Z_m[X]$ and $\mathbb Z[x]\to Z_p[X]\times Z_q[X]$. Show that these two maps have the same kernel and are onto. Hence the two ranges are isomorphic.

REPLY [2 votes]: Hints: 

Make sure you understand why for distinct prime numbers $p$ and $q$, $\mathbb{Z}/pq \mathbb{Z} \cong \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/q\mathbb{Z}$.  
Try to show more generally that if for a rings $A,B,C$ you have $A \cong B \times C$, then $A[x] \cong B[x] \times C[x]$.  This should help to focus your efforts.  The actual isomorphism is almost obvious when you write everything down and think a bit.