Irreducible Laurent Polynomial

Question

Is there a general way to determine the irreducible Laurent polynomials in $\Bbb C[x,x^{-1}]$? I already found that $x$, $x^{-1}$ are irreducible, and I suspect that $x+x^{-1}$ is irreducible, but I am not sure if this is the complete set of irreducibles.

Answer 1

$X+X^{-1}$ is not irreducible, since it is associate to $X^{2}+1$ via multiplication by the unit $X$, and $X^{2}+1$ factors as $(X-i)(X+i)$.

In general, $\mathbb{C}[X, X^{-1}]$ is a PID, and its irreducible elements can be described (up to scale) by listing a generator for each nonzero prime ideal. The prime ideals of $\mathbb{C}[X, X^{-1}]$ are in bijection with the prime ideals of $\mathbb{C}[X]$ which don't contain $X$, and the nonzero prime ideals of $\mathbb{C}[X]$ are in bijection with linear monic polynomials.

Therefore, the irreducible elements of $\mathbb{C}[X, X^{-1}]$ are unit multiples of $X - a$, where $a \in \mathbb{C}^{\times}$ (i.e., $a \neq 0$). Intuitively, any element of $\mathbb{C}[X, X^{-1}]$ is associate to some element of $\mathbb{C}[X]$ by multiplying by a suitably high power of $X$. Given an element of $\mathbb{C}[X, X^{-1}]$, we can thus factor it by first multiplying by $X^{k}$ for some $k \in \mathbb{N}$, factoring the resulting element of $\mathbb{C}[X]$ over $\mathbb{C}$, and multiplying one of the factors by $X^{-k}$.