How do I know if an ideal $I$ in $\mathbb Z[x]$ is principal? maximal? I have just finished the first section on rings in Artin's Algebra and would like to know if there are necessary and sufficient conditions for when to know if an ideal of $\mathbb Z[x]$ is principal, and/or maximal.
It seems there are big results about polynomial rings in a field but not so much when the base is not a field.
 A: To tell if an ideal is maximal, take the quotient and see if it is a field! They will look something like $J = (p, f(x))$ where $p$ is a prime and $f(x) \in \mathbb{Z}[x]$ is a polynomial. For example the ideal $(2, x)$ is maximal because $\mathbb{Z}[x]/(2, x) = \mathbb{F}_2[x]/(x) = \mathbb{F}_2$. More generally if you are given the ideal in this form, you just need to check whether or not the reduction of $f$ mod $p$ is irreducible. These are in fact all of the examples of maximal ideals in $\mathbb{Z}[x]$ (or more geometrically, the closed points of Spec $\mathbb{Z}[x]$).
Deciding whether or not it is principal depends on what form you are given the ideal. The ideals described above are never principal, since by Krull's Hauptidealsatz the principal ideals have height at most one. 
A: $\Bbb Z[x]$ is not a PID, and maximal ideals are $(p,f(x))$, where $f(x)$ is an irreducible polynomial over the field $\Bbb F_p$.  
For example: $(2,x)=\{2f(x)+xg(x); f(x),g(x)\in \Bbb Z[x] \}$ is not a principal ideal in $\Bbb Z[x]$. Assume $(2,x)=(a(x))$ for some $a(x)\in \Bbb Z[x]$, then $a(x)\mid2$ and $2=a(x)b(x); b(x)\in \Bbb Z[x]$. Since $2$ is a prime number, $a(x),b(x)\in \{\pm1, \pm2 \}$ . 
if $a(x)=\pm1$ ,then $(a(x))=\Bbb Z[x]$ contrary to $(a(x))$ being a proper ideal. 
if $a(x)=\pm2$, then $(a(x))=(2)=(-2). $ We know $x\in(a(x))=(2)$ and so $x=2f(x); f(x)\in \Bbb Z[x] $, clearly impossible. 
