Q:

Easier way to show $(\mathbb{Z}/(n))[x]$ and $\mathbb{Z}[x] / (n)$ are isomorphic

$$(\mathbb{Z}/(n))[x] \simeq \mathbb{Z}[x] / (n)$$
I've shown this by showing that the map that sends $\overline{1} \mapsto [1+(n)]$ (where the bar denotes the congruence class mod $n$) and $x \mapsto [x+(n)]$ is a homomorphism that is injective and bijective, which is a bit cumbersome. Is there an easier way to show this? I can't tell if this is trivial or not.

A:

The homomorphism in the other direction is maybe easieer to see.
From the $\mathbb Z\to \mathbb Z/(n)\hookrightarrow \mathbb Z/(n)[x]$ (canonical projection and canonical inclusion) and $x\mapsto x$ we obtain a ring homomorphism $\mathbb Z[x]\to \mathbb  Z/(n)[x]$ (universal property of polynomial ring). The kernel is quite clearly the set of polynomials with coefficients multiples of $n$, i.e. $n\mathbb Z[x]$ or $(n)$. Hence we obtain a homomorphism $\mathbb Z[x]/(n)\to\mathbb Z/(n)[x]$.
The homomorphism is clearly onto as it is easy to find an inverse image in $\mathbb Z[x]$ for any element of $\mathbb Z/(n)[x]$.

