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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?

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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.

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    Rather than invoking Krull's theorem, you can just invoke the fact that all nonzero primes are maximal in a PID, which is more elementary. – 2017-01-14
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    Is there an easy way to see that $(2, 1 + x)$ is a prime ideal? – 2017-01-14
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    @Jason I edited to address your question. – 2017-01-14
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    Suppose $(2,1+X)=(f(X)/X^n), \gcd(f(X),X)=1$. This implies $2=(f(X)/X^n)(g(X)/X^m)\implies f(X)=\pm1,\pm2$. Since $1+X\in(f(X)/X^n)$ we get $f=\pm1$. The last step is left to the reader. – 2017-01-15