Invertible elements of $A[t, t^{-1}]$ 
Let $A$ be a commutative ring and consider the Laurent polynomial ring:
  $$A[t,t^{-1}]:=\left\{\sum_{i\in\mathbb Z} a_it^i: \text{$a_i=0$ for almost all $i$}\right\}.$$
  What is the group of invertible elements $\left(A[t, t^{-1}]\right)^\times$? 

Is it the set of all polynomials whose lowest degree coefficient is invertible?
 A: When $A$ is a domain, no element of $A[t,t^{-1}]$ with more than one nonzero coefficient can be invertible, by considering the highest and lowest degree terms in a product.  If $f\in A[t,t^{-1}]$ is invertible, then for every prime ideal $P\subset A$, the image of $f$ in $A/P[t,t^{-1}]$ is invertible.  By the remark above, this means every coefficient of $f$ except one must be in $P$.  For each $n\in \mathbb{Z}$, the set $S_n$ of $P$ such that the $t^n$ coefficient of $f$ is not in $P$ is open in $\operatorname{Spec} A$.  So, these sets $S_n$ form a partition of $\operatorname{Spec} A$ into clopen subsets, corresponding to orthogonal idempotents $e_n\in A$ with $\sum_n e_n=1$ (all but finitely many of the $e_n$ are $0$ since $f$ has only finitely many nonzero coefficients).  Looking at the image of $f$ in the localization $A_{e_n}[t,t^{-1}]$, we see that all coefficients of $f$ except the $t^n$ coefficient are in every prime ideal of $A_{e_n}$ and hence are nilpotent.  Also, the $t^n$ coefficient must be a unit mod every prime ideal of $A_{e_n}$ and thus is a unit in $A_{e_n}$ (and thus is the image of some unit in $A$).
So to sum up, we have found that if $f\in A[t,t^{-1}]$ is a unit, there is a system of orthogonal idempotents $e_n\in A$ (all but finitely many of which are $0$) with $\sum_n e_n=1$ and $f=\sum u_ne_nt^n+g$, where each $u_n$ is a unit and each coefficient of $g$ is nilpotent.  Conversely, any element of this form is a unit: it suffices to check that it is a unit after localizing at each $e_n$.  But after localizing at $e_n$, $f$ is just $u_nt^n+g$, which is a unit plus a nilpotent element and hence also a unit.
A: Partial answer: If $A$ is an integral domain, then the units are precisely the elements of the form $ut^j$, where $u \in A$ is a unit and $j$ is some integer. Not sure whether one can give a nice answer for arbitrary rings.
