Let $R = \mathbb{Z}\left[\frac{1}{2} \right]$ be the ring of all rational numbers of the form $\displaystyle \frac{n}{2^{m}}$. I need to describe the set of units of the polynomial ring $R[x_{1},x_{2}]$
I am assuming that polynomials in this ring look like $f = \frac{n_{3}}{2^{m_{3}}}x_{2}x_{1} +\frac{n_{2}}{2^{m_{2}}}x_{2} + \frac{n_{1}}{2^{m_{1}}}x_{1} + \frac{n_{0}}{2^{m_{0}}}$? Where $n_{i}$ and $m_{i}$ are integers?
Would I consider another polynomial $g = ax_{2}x_{1} + b x_{2} + cx_{1} + d$, multiply $f$ and $g$ together, and then figure out the general form I would need $a$, $b$, $c$, and $d$ to be in in order to get $fg=1$, or is there an easier and/or better way to approach this? If so, can you give me any guidance as to how to proceed?
Thanks in advance
For any integral domain $R$ we have $R[x]^{\times}=R^{\times}$, (mainly because $\mathrm{deg}(fg) = \mathrm{deg}(f)+\mathrm{deg}(g)$). If we apply this twice to $R[x_{1},x_{2}]$, we get that $R[x_{1},x_{2}]^{\times}=\mathbb{Z}[\frac{1}{2}]^{\times}$.
So it suffices to study the units of $R$. Consider $R$ as a subring of $\mathbb{Q}$ if $a = p 2^k \in R$ for $p,k \in \mathbb Z$, assuming $\mathrm{gcd}(p,2)=1$, $p\neq 0$, then $a \in R^\times \Leftrightarrow R \ni\frac{1}{a} = \frac{2^{-k}}{p}$, so $p = \pm1$, thus the units of $R$ are precisely the elements of the form $\pm 2^k$ for $k\in \mathbb Z$.