In the proof of "maximal ideals of $\Bbb{Z}[x]$ are of the form $(p,f(x))$ " I'm trying to prove the following statement:

Maximal ideals of $R=\Bbb{Z}[x]$ are of the form $(p,f(x))$ where $f(x)$ is irreducible in $\Bbb{F}_p[x]$ and $p$ is prime. 

A quick search on google returns several proofs, which I cannot completely understand. Given a maximal ideal $M\subset {\Bbb Z}[x]$, a key step is to find the prime $p$ first. A(The?) way to do it is as the following. Let
$$
J=M\cap{\Bbb Z},
$$
which turns out to be an ideal in ${\Bbb Z}$ and hence has the form $J=(m)$ where $m$ is some integer. Suppose $m\not=0$ (showing that $J\not=(0)$ is a big part in the proof, which is not what I'm worrying about). My question is: 

how do I show that $m$ must be prime?


It suffices to show that there is an injective homomorphism $\phi:{Z}/(m)\to R/M$. But I don't see a clear way to find $\phi$. (It seems that it can be obtained from some induced homomorphism since $(m)\subset M$).
 A: Here might be an alternative approach without using prime ideals explicitly. 
Consider the inclusion homomorphism $i:\Bbb{Z}\to{\Bbb Z}[x]$ and the canonical homomorphism 
$\pi:{\Bbb Z}[x]\to\Bbb{Z}[x]/M$. It can be checked that the kernel of the homomorphism 
$$
\varphi=\pi\circ i:{\Bbb Z}\to{{\Bbb Z}[x]/M}
$$
is $M\cap {\Bbb Z}=J=(m)$. It follows that the induced homomorphism 
$$
\overline{\varphi}:{\Bbb Z}/(m)\to {\Bbb Z}[x]/M
$$ is injective. But ${\Bbb Z}[x]/M$ is a field. Hence ${\Bbb Z}/(m)$ must be an integral domain. Since it is also finite, it must be a field and hence $m$ is prime.
A: Hint $\ $ If $\, m = ab\in M,\,\ a,b\not\in M\,$ then $M$ is not prime, so not maximal. Generally contractions of primes are prime. Further, in a UFD, prime ideals may be generated by primes (a property which characterizes UFDs).
A: If you consider the map $\varphi : \mathbb Z \to \mathbb Z[X]$ which maps integers to degree zero-degree polynomials (i.e. the canonical injection), then $M \cap \mathbb Z$ is precisely the pullback of the maximal ideal $M$ under this inclusion. Since $M$ is prime, $M \cap \mathbb Z$ is also prime ; since $\mathbb Z$ is a PID, $M \cap \mathbb Z$ is maximal.
Hope that helps,
A: Recall the homomorphism theorem: if $A$ is a subring of $B$ and $M$ is an ideal of $B$, then we have an isomorphism
$$
\frac{A}{M\cap A}\to \frac{A+M}{M}
$$
because we have an obvious homomorphism $\varphi\colon A\to (A+M)/M$ given by $a\mapsto a+M$. The kernel of $\varphi$ is clearly $A\cap M$.
In your case, $A=\mathbb{Z}$, $B=\mathbb{Z}[x]$ and $M$ is maximal, so $(A+M)/M$ is a domain, being a subring of the field $B/M$. Therefore $A\cap M$ is prime.
Note that this doesn't use in any way the properties of $\mathbb{Z}[x]$.
More generally, if $f\colon A\to B$ is a ring homomorphism (the rings are commutative) and $M$ is a prime ideal of $B$, then $f^{-1}(M)$ is a prime ideal of $A$, for the same reasons: if $\pi\colon B\to B/M$ is the projection, then the kernel of $\pi\circ f\colon A\to B/M$ is precisely $f^{-1}(M)$, so we have an injective homomorphism
$$
\frac{A}{f^{-1}(M)}\to \frac{B}{M}
$$
which proves the claim, because $B/M$ is a domain.
