Units in $R[x,x^{-1}]$ Let $R$ be an integral domain. I am looking for a general way to describe the units in 
$$ R[x,x^{-1}] := R[x,y]/(xy-1).$$
Clearly
$$\{rx^n \mid n \in \mathbb{Z}, r \in R^\times \} \subseteq R[x,x^{-1}]^\times$$
is a subgroup, but how do I know whether it's all? I was trying to argue with degrees or to write down multiplication of general elements with Cauchy sums, but it all becomes pretty messy, having also negative exponents.

Another way to view your ring is as $R[x]$ localised at the multiplicative set $S = \{r\cdot x^k\mid r\in R^\times, k\in \Bbb N\}$. Really, it's just localised at the set $\{1, x, x^2, \ldots\}$, but including the units of $R$ allows us to skip a step.
I claim that this $S$ is saturated, which is to say if $f, g\in R[x]$ are such that $fg\in S$, then $f, g\in S$. This is not difficult to see: 


*

*$fg$ is a monomial, so $f$ and $g$ must be monomials

*$fg$ has an invertible coefficient, so $f$ and $g$ must have invertible coefficients


which is to say $f, g\in S$.
Why does this matter? Because if you localise by a saturated set $S$, then that saturated set contains all elements that become invertible in the localisation.
Let $\frac fs\in S^{-1}R[x]$ be invertible. Then there is a $\frac gt\in S^{-1}R[x]$ such that $\frac fs\cdot \frac gt = \frac11$. By definition of localisation, this means that
$$
fg\cdot 1 = st\cdot 1
$$
(The absence of zero divisors in $R[x]$ means we can simplify the definition of equality in a localisation to this.) Since $st\in S$ because $S$ is multiplicatively closed, we also have $fg\in S$. However, this implies that $f\in S$. So $\frac fs$ is a fraction where both he numerator and the denominator are in $S$, and because of the general form of elements in $S$, this means that $\frac fs$ is either equal to a fraction of the form $\frac{1}{rx^k}$ or to a fraction of the form $\frac{rx^k}1$ where $r\in R^\times, k\in \Bbb N$.
