Do the Laurent polynomials over $\mathbb{Z}$ form a principal ideal domain? 
I'm trying to prove whether or not the Laurent polynomials $\mathbb{Z}[x, x^{-1}]$ with coefficients in $\mathbb{Z}$ form a principal ideal domain. 

I know that $\mathbb{F}[x, x^{-1}]$ is a PID when $\mathbb{F}$ is a field, but clearly $\mathbb{Z}$ is not a field so I cannot appeal to this result. And my intuitions are not serving me very well at the moment. Can anyone provide a hint or direction to take?
 A: Hint: Consider the ideal $I = (2, 1+x)$.  Can you find a single generator for $I$?
Full solution:

Consider the ideal $(2, 1+x)$; I claim that it is not principal.  Note that
$$(0) \subsetneq (2) \subsetneq (2, 1+x)$$ is a chain of prime ideals of length $2$, so $(2, 1+x)$ has height $\geq 2$.  A principal ideal has height at most $1$ by Krull's Hauptidealsatz, which shows that $(2, 1+x)$ is not principal.

In terms of intuition: the ring $\mathbb{Z}[x]$ is "too big" to be a PID.  A PID that is not a field has Krull dimension $1$, and $\mathbb{Z}[x]$, much like $k[x,y]$, is $2$-dimensional.  Localizing at $x$ gets rid of the single irreducible element $x$ (making it into a unit), but there are plenty of other irreducible polynomials one can use instead to build a chain of primes of length $2$.
EDIT: To respond to your question: since
$$
\frac{\mathbb{Z}[x,x^{-1}]}{(2, 1+x)} \cong \frac{\mathbb{F}_2[x,x^{-1}]}{(1+x)} \cong \mathbb{F}_2[-1, (-1)^{-1}] = \mathbb{F}_2
$$
which is a field, so $(2, 1+x)$ is maximal, hence prime.
