TITLE: Proof for maximal ideals in $\mathbb{Z}[x]$
QUESTION [10 upvotes]: I have been trying to prove the following theorem: 

Every maximal ideal in $\mathbb{Z}[x]$ has the form $(p, f(x))$ where $p$ is prime integer and $f$ is primitive integer polynomial that is irreducible modulo $p$. 

Idea: I tried to establish a homomorphism $\phi: \mathbb{Z}[x] \rightarrow \mathbb{F}$. Since $\mathbb{F}$ is a field it has characteristic $p$ and so integer prime p are mapped to 0 in $\mathbb{F}$. Hence $p\in \ker \phi$. Next we consider $\phi': \mathbb{Z}[x] \rightarrow \mathbb{Z_p}[x]$ and pick an arbitrary maximal ideal $M\in \mathbb{Z}[x]$. So, $\phi'(M)$ is maximal as long as $p \in M$ by correspondence. But now I am stuck at this stage and do not know how to proceed. I guess we might have to use primitivity given in problem but dont know how.

REPLY [5 votes]: Since maximal ideals are prime ideals, and according to this post, it suffices to exclude the situation 2.(which is the only non-trivial case), i.e., we need to show that $(f(x))$ is not a maximal ideal when $f(x)$ is irreducible with $\deg(f(x))>1$.
To see this, note that if we assume that $(f(x))$ is a maximal ideal of $\mathbb Z[x]$, then there should not be any non-unit ideal in $\mathbb Z[x]$ containing $f(x)$. 
However, consider $(p,f(x))\supsetneq (f(x))$ where $p$ is a prime such that $p$ does not divide the leading coefficient of $f(x)$. It is trivial to see that $(p,f(x))\ne\mathbb Z[x]$, since $$\frac{\mathbb Z[x]}{(p,f(x))}\cong\frac{\mathbb F_p[x]}{(f(x))}\ne 0.$$ Contradiction!