# 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.

## 2 Answers

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.

Another approach you could try: Show that the ideal $M=(x+1,2)$ is maximal. Then show that $(x^2+1)$ is contained in $M$.