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.

S'pose $g(x)h(x)\in\langle f(x)\rangle $ and that $f(x)$ is irreducible in $\Bbb Z[x]$.
Then $\exists k(x)\in\Bbb Z[x]$ such that $g(x)h(x)=k(x)f(x)$.  So, $f(x)\mid g(x)h(x)$.  But since $f$ is irreducible,  $f(x)\mid g(x)\lor f(x)\mid h(x)$.  Then $g(x)\in\langle f(x)\rangle \lor h(x)\in \langle f(x)\rangle $.
(The key is that primes and irreducibles are the same in the UFD $\Bbb Z[x]$.)
