Let $R$ be a commutative ring with unit, $R[[t]]$ the ring of formal power series over $R$ and $R((t))$ the ring of formal Laurent series of $R$.
It is easy to see (and well known) that the group of units $R[[t]]^{\times}$ equals $R^{\times}+tR[[t]]$. On the other hand, the group of units $R((t))^{\times }$ is a less familiar object. In this paper (page 7 example 2.9) it is stated without proof that
$$
R((t))^\times = \{\sum a_n t^n \mid \exists n_0 \in \mathbb{N}, a_{n_0}\in R^\times, a_n \mbox{ nilpotent for } n<n_0 \}
$$
While it is clear that all the elements on the RHS are units, I don't see why all units are of this form. Actually, it seems to me just wrong. Over $R=R_1 \times R_2$ we have
$$
((1,0)\cdot 1+(0,1)\cdot t)((0,1)\cdot t^{-1}+(1,0)\cdot 1)=1
$$
Which seems to be a counter example.
Question 1: Is this indeed wrong, or am I missing something? If it is correct, how to prove it? If not, How can one describe explicitly $R((t))^\times?$
Right after that, there is a claim about $R((t))^{\times}/R[[t]]^\times$, but it also seems incorrect or at least I don't see why it is true.
Question 2: Is there a nice description of the group $R((t))^{\times}/R[[t]]^\times$?
Best Answer
Here is the statement concerning question (1):
$a_0$ is invertible on $U_0$ and nilpotent on $U_1\sqcup \cdots \sqcup U_n$;
$a_1$ is invertible on $U_1$ and nilpotent on $U_2\sqcup \cdots \sqcup U_n$...
$a_n$ is invertible on $U_n$.
In particular, on each $U_k$, $u$ is of the form described in Görtz's paper.
Proof. We have $R((t))=R[[t]][1/t]$. So an element $a(t)/t^N$ with $a(t)=a_0+a_1t+\cdots\in R[[t]]$ is invertible in $R((t))$ if and only if there exists $b(t)=b_0+b_1t+\cdots\in R[[t]]$ such that $a(t)b(t)=t^m$ for some $m\ge 0$.
Developping the product $a(t)b(t)$, we get $$a_0b_0=0, \ a_1b_0+a_0b_1=0, ..., \ a_{m-1}b_0+\cdots +a_0b_{m-1}=0; \ a_mb_0+\cdots +a_0b_m=1.$$ Multiply these equalities by suitable powers of $a_0$ and we obtain $a_0^{k+1}b_k=0$ for all $k\le m-1$ and $a_0^{m+1}b_m=a_0^m$. Therefore $$ a_0^{m}(1-a_0b_m)=0.$$ Let $U_0=D(a_0)$ be the principal open subset of $\mathrm{Spec}(R)$ defined by $a_0$. Then $a_0$ is invertible on $U_0$. Let $V_0=D(1-a_0b_m)$. Then $\mathrm{Spec}(R)$ is the disjoint union of $U_0$ and $V_0$, and $a_0$ is nilpotent on $V_0$. As $t^{n-1}(1-a_0b(t)t^{-n})=(a_1+a_2t+\cdots)b(t)$ and $a_0$ is nilpotent on $V_0$, we see that $a_1+a_2t+\cdots$ is invertible in $O(V_0)((t))$ (see KotelKanim's comment). So we can start again with $a_1+a_2t+\cdots$ to find that $V_0=(D(a_1)\cap V_0)\sqcup V_1$ for some principal open subset $V_1$ of $V_0$ such that $a_1$ is nilpotent on $V_1$. As $D(a_1)\cap V_1\subseteq D(a_1)\cap V_0\cap V_1=\emptyset$, we have $$\mathrm{Spec}(R)=(D(a_0)\cup D(a_1))\sqcup V_1.$$ By induction, we find $V_k=(D(a_{k+1})\cap V_k)\sqcup V_{k+1}$ with $a_{k+1}$ nilpotent on $V_{k+1}$ and $$\mathrm{Spec}(R)=(D(a_0)\cup ... \cup D(a_{k+1}))\sqcup V_{k+1}.$$ Note that the equality $a_0b_m+\cdots+a_mb_0=1$ implies that $\mathrm{Spec}(R)=\cup_{0\le i\le m} D(a_i)$. So the above construction must stop at some step $n\le m$, we find the partition with $U_1=D(a_1)\cap V_0$, ..., $U_n=D(a_n)\cap V_{n-1}$.
It remains to shwo that if $a(t)$ satisfies the condition of partition, then $a(t)$ is a unit in $R((t))$. But it is enough to notice that on each of $U_i$, it is a unit.