Spectrum of $\mathbb{Z}[x]$ Can someone point me towards a resource that proves that the spectrum of $\mathbb{Z}[x]$ consists of ideals $(p,f)$ where $p$ prime or zero and $f$ irred mod $p$? In particular I remember this can be proved simply using localizations, but can't quite remember how to do it! I definitely don't want a link to a long involved argument about polynomials, I can find quite enough of those!
Many thanks in advance.
 A: The intersection of a prime ideal with $\mathbb Z$ is again prime, thus we obtain a prime $p$ (or 0).  By localizing at $p$, we make all non-multiples of $p$ invertible and are left with an ideal in the principle ideal ring $\mathbb Z_p[X]$, i.e. $(f)$ with $f\in\mathbb Z_p[X]$. If we had a nontrivial factorization $f\equiv gh\pmod p$, this could be lifted to a factorization in $\mathbb Z_p[X]$, which is impossible. Hence $f$ is irreducible $\bmod p$. This also holds if we replace $f$ with an approximation in $\mathbb Z[X]$. We also see that any $g$ in the ideal becomes a multiple of $f$ in $\mathbb Z_p[X]$, hence can be written as a multiple of $p$ plus a multiple of $f$ in $\mathbb Z[X]$.
A: Here is a geometric (schemes!) way to think about it.
The inclusion $\Bbb Z\to\Bbb Z[x]$ defines a morphism $\operatorname{Spec}(\Bbb Z[x])\to\operatorname{Spec}(\Bbb Z).$ Thus to figure out the primes of $\operatorname{Spec}(\Bbb Z[x]),$ we can simply determine all the fibres of this map. How do we compute the fibres of this map?
For $\langle p\rangle\subseteq\Bbb Z$ a prime ideal, we pull back the morphism given above over the map $\operatorname{Spec}(\kappa(p))\to\operatorname{Spec}(\Bbb Z)$ induced by $\Bbb Z\to\Bbb Z_p/\frak{m_p}$ $=\kappa(p)$, where $\kappa(p)$ is the residue field of $p.$ The residue field $\kappa(0)=\Bbb Q$ and for all other primes $p$ we have $\kappa(p)=\Bbb F_p.$ 
The fibre over $\langle 0\rangle$ is thus $\operatorname{Spec}(\Bbb Q\otimes_{\Bbb Z}\Bbb Z[x])=\operatorname{Spec}(\Bbb Q[x])$ which is all irreducible polynomials over $\Bbb Q$ and the zero ideal. Similarly, the fibre over $\langle p\rangle$ is $\operatorname{Spec}(\Bbb F_p[x]),$ which is just the irreducible polynomials over $\Bbb F_p$ along with its zero ideal. (The zero ideals correspond to those in $\Bbb Z.$)
A: There is a sketch here but I didn't proofread it. I am gambling that it is useful, so I apologize (and will delete this) if it turns out to be useless.
A: The prime ideals of $\mathbb Z[x]$ are of three kinds depending on their heights 


*

*(height $0$): $\{ 0\}$;

*(height $1$): $F(x)\mathbb Z[x]$ with $F(x)$ an irreducible element in $\mathbb Z[x]$. Equivalently: $F(x)$ is a prime number $p$ or is primitive and irreducible in $\mathbb Q[x]$. 

*(height $2$, maximal ideals): the $(p, f(x))$ as you describe.  
A: There are two cases: if a prime ideal $P$ of $Z[x]$ does not contain any constant then considering the ideal generated by $P$ in $Q[x]$, it is a proper ideal of $Q[x]$; so that it is a principal ideal generated by a single polynomial $f(x)\in Q[x]$. But then an irreducible element $g(x)\in P\subset Z[x]$ can be written as $g(x)=f(x)h(x)$ with $h(x)\in Q[x]$. By considering   the common denominators,  $f(x)=f_1(x)/b$ and $h(x)=h_1(x)/c$ where $f_1(x), h_1(x)  \in Z[x]$. Now $bcg(x)=f_1(x)h_1(x)$. Looking at the content of polynomials we can conclude that $P=(f_1(x))$ is a principle ideal of $Z[x]$.
The case where $P$ contains a constant is easy to solve by passing to the quotient ring.  
