Isn't $x^2+1 $ irreducible in $\mathbb Z$, then why is $\langle x^2+1 \rangle$ not a maximal ideal in $\mathbb Z[x]?$

Question

Isn't $ x^2+1$ irreducible in $\mathbb Z$, then why is $\langle x^2+1 \rangle$ not a maximal ideal in $\mathbb Z[x]$?

$ x^2+1$ cannot be broken down further non trivially in $\mathbb Z[x]$. hence, it's irreducible in $\mathbb Z[x]$. Hence, shouldn't $\mathbb Z[x]/\langle x^2+1 \rangle$ be a field and hence, $ x^2+1$ a maximal ideal in $\mathbb Z[x]$

Thanks for your help.

Answer 1

Because the quotient is not a field, as you can easily check!

For example, the class of $2$ is neither zero nor invertible in $\mathbb Z[x]/(x^2+1)$.

Answer 2

More directly, the ideal $\langle x^2\!+\!1 \rangle$ is properly contained in the ideal $\langle x^2\! +\! 1, \ 3 \rangle$. By definition, it cannot be maximal.

In particular, it is a theorem that no principal ideal in $\mathbb{Z}[x]$ is maximal. In fact, all maximal ideals of $\mathbb{Z}[x]$ are of the form $\langle p, f(x) \rangle$ where $p$ is prime and $f(x)$ is irreducible over $\mathbb{F}_p$.