Prove that $(x^2+1)\mathbb Z[x]$ is a prime ideal of $\mathbb Z[x]$, but not maximal Prove that $(x^2+1)\mathbb Z[x]$ is a prime ideal of $\mathbb Z[x]$, but not maximal.


I'm supposed to show this for my homework. My first thought is to show that $\mathbb Z[x]/(x^2+1)\mathbb Z[x]$ is an integral domain but not a field, but I don't know how to do that. Please don't give me the answer but if you can give me a hint I'd really appreciate it. 

Update: I used the unique factorization of $\mathbb Z[x]$ and the irreducibility of $x^2+1$ in $\mathbb Z[x]$ to show that $\mathbb Z[x]/(x^2+1)\mathbb Z[x]$ is an integral domain. I'm still working on the other part.
You have the right idea: the quotient ring is a domain but isn't a field. It might be helpful to note that the quotient is isomorphic to $\mathbb{Z}[i]$, and then show that $2$ (for instance) isn't invertible.
You could define a natural homomorphism which maps $\mathbb{Z}$ to its quotient ring $\mathbb{Z}[x] / &lt;x^2+1&gt;$ which can be defined as $\{[a] | a\in \mathbb{Z}\}$ and we let the homomorphism be $\pi$; that is $\pi:\mathbb{Z}[x]/&lt;x^2+1&gt; \mapsto \mathbb{Z}, \pi([a]) = a$. It is trivial to show that is an isomorphism, since it is a bijection as well as a homomorphism.

We know that $\mathbb{Z}$ is an integral ring but not a field (because $\forall z \in \mathbb{Z} \setminus \{-1, 1\}$, we cannot find an inverse of $z$); therefore $&lt;x^2+1&gt;$ is a prime ideal of $\mathbb{Z}[x]$ but not maximal.