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: You don't have to go about building an isomorphism yourself if you make use of the isomorphism theorems.
Use the obvious mapping $\phi:\Bbb Z[x]\to (\Bbb Z/(7))[x]$ where you are just reducing the coefficients of polynomials mod $7$. This is clearly onto the latter ring. The kernel is precisely the polynomials whose coefficients are all divisible by $7$, and that's exactly $(7)\lhd \Bbb Z[x]$.
By an isomorphism theorem, $\Bbb Z[x]/\ker(\phi)\cong (\Bbb Z/(7))[x]$, and you already know what $\ker(\phi)$ is.
A: HINT: 
Consider the quotient maps from $\Bbb Z[x]$ to either ring. What can you say about the kernels?
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 $\mathbb Z/(n)[x]$.
