Prove that $\langle f(x) \rangle $ is prime ideal in $\Bbb Z[x]$ if $f(x)$ is irreducible over $\Bbb Z$ I know $(x^2+1)$ is prime ideal of $\Bbb Z[x]$ without being maximal ideal.  It can be easily proved  as quotient ring isomorphic to integral domain $\Bbb Z[i]$. My question is the generalization of this i. e.  if $(x^2+1)$ is replaced by arbitrary irreducible polynomial $f(x)$  over $\Bbb Z$.  Can anyone suggest me an outline of this proof ? Thank you.

A simple argument could be:


*

*If $f(x)$ is irreducible in $\Bbb{Z}[x]$, including that it has no non-unit constant factor, then it is also irreducible in $\Bbb{Q}[x]$ (the usual argument invokes Gauss's lemma at a key step).

*Therefore $f(x)$ generates a maximal ideal in $\Bbb{Q}[x]$.

*Therefore $\Bbb{Q}[x]/\langle f(x)\rangle$ is a field.

*But $\Bbb{Z}[x]/\langle f(x)\rangle$ is a subring of $\Bbb{Q}[x]/\langle f(x)\rangle$, so it is an integral domain.

*Therefore $\langle f(x)\rangle$ is a prime ideal of $\Bbb{Z}[x]$.

