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: Fix $s$ irreducible. Now consider $p,q$ polynomials such that $pq = rs$ for some other polynomial $r$. Now, since $\mathbb{Z}[X]$ is a unique factorization domain, the polynomials $p,q$ and $r$ have a decomposition in irreducibles, 
$$
p = f_1^{\alpha_1} \cdots f_n^{\alpha_n}, \quad q = g_1^{\beta_1} \cdots g_m^{\beta_m}, \quad r = h_1^{\gamma_1} \cdots h_k^{\gamma_k}. 
$$
So the former equality says that
$$
f_1^{\alpha_1} \cdots f_n^{\alpha_n} \cdot g_1^{\beta_1} \cdots g_m^{\beta_m} = h_1^{\gamma_1} \cdots h_k^{\gamma_k} \cdot s. 
$$
Once again, by the uniqueness of factorization, one of the irreducibles of the left must coincide with $s$ up to a unit. Without loss of generality, suppose that $f_i = \pm s$ for some $1 \leq n \leq n$. Then, 
$$
f = \pm s(f_1^{\alpha_1} \cdots \widehat{f_i^{\alpha_i}} \cdots f_n^{\alpha_n})
$$
and so $s$ divides $p$. 
This proves that if $s$ divides $pq$ then it divides either $p$ or $q$, which is to say that $(s)$ is prime. In general $R[X]$ is a UFD if $R$ is one, and that's all we have used. So $(s)$ will be prime if $s$ is irreducible in any UFD. 
In general, an irreducible element $s$ of an UDF is always prime (by the argument above), which is to say that $(s)$ is prime.
A: 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 $\Bbb{Z}[x]$.

A: 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 $\Bbb Z[x]$.)
