I am trying to calculate the group of units of the power series ring $\mathbb{Z}_3[[x]]$. I know that all the unit elements are of the form $u+\sum_1^{\infty} a_nx^n$ where $u$ is a unit in $\mathbb{Z}_3$, where $\mathbb{Z}_3$ are the $3$-adic integers. However I am not sure about the group structure.
For example, we have many subgroups $U_i$ which are the set of elements of the form $u+\sum_i^{\infty} a_nx^n$ i.e the power series where the first power of $x$ is $x^i$. These groups form a filtration on the group of units. Is there a way to relate these groups to the group of units of $\mathbb{Z}_3[[x]]$ similar to the result for the ring $\mathbb{F}_3[[x]]$? For the result for $\mathbb{F}_3[[x]]$ look at Thm 4.4 here.
Thanks in advance.
For $n\in \Bbb{Z}_{\ge 0}$, the map $$a\mapsto (1+x^n)^a=\sum_{k\ge 0} x^{nk} {a\choose k},\qquad\Bbb{Z}_3\to\Bbb{Z}_3[[x]]^\times$$ is well-defined as $a\mapsto {a\choose k},\Bbb{Z}_{\ge 0}\to \Bbb{Z}_{\ge 0}$ is an integer valued polynomial which is thus 3-adically continuous.
If $f=1+ax^n+O(x^{n+1})$ then $f= (1+x^n)^a g$ with $g=1+O(x^{n+1})$,
Whence for $h\in \Bbb{Z}_3[[x]]^\times$ there is a unique $u\in \Bbb{Z}_3^\times$ and a unique sequence $a_n$ such that $$h=u\prod_{n\ge 1} (1+x^n)^{a_n}$$
Where the convergence is in the topology on $\Bbb{Z}_3[[x]]^\times$ inherited from the discrete valuation on $\Bbb{Q}_3[[x]]$. With the product topology on $\Bbb{Z}_3^\times \times \prod_{n\ge 1} (1+x^n)^{\Bbb{Z}_3}$ we get that as topological groups $$\Bbb{Z}_3[[x]]^\times = \Bbb{Z}_3^\times \times \prod_{n\ge 1} (1+x^n)^{\Bbb{Z}_3}$$ Of course we can replace $1+x^n$ by $1+u_n x^n+ r_n x^{n+1} $ for any sequence of $u_n\in \Bbb{Z}_3^\times,r_n\in \Bbb{Z}_3[[x]]$.