Let $\mathcal{A}$ be some complex algebra (i.e., algebra over the complex numbers), and let $a_1,...,a_n \in \mathcal{A}$ be pairwise commuting elements (i.e., $a_i a_j = a_j a_i$ for all $i,j=1,...,n$). Call a subalgebra $\mathcal{B}$ inversion closed if $x \in \mathcal{B}$ and $x$ is invertible, then $x^{-1} \in \mathcal{B}$.
Let $\mathcal{B}$ be the smallest inversion closed subalgebra containing $a_1,...,a_n$. I am trying to show that $\mathcal{B}$ must be commutative. Intuitively this is clear. But I am having trouble explicitly describing the elements of $\mathcal{B}$ to show that it is commutative? Does it consist of all rational polynomials in the variables $a_1,...,a_n$?
EDIT
Maybe it's easier than I thought. Here's an idea. Let $\mathcal{C} = \{ \frac{f(a_1,...,a_n)}{g(a_1,...,a_n)} \mid f,g \in \Bbb{C}[x_1,...,x_n], g \neq 0\}$. I think that this is clearly a inversion closed subalgebra which contains $a_1,...,a_n$, so $\mathcal{B} \subseteq \mathcal{C}$ (perhaps equality holds, but it doesn't matter). Then somewhat tedious calculations show that $\mathcal{C}$ is commutative, so $\mathcal{B}$ must be commutative, being a subalgebra of $\mathcal{C}$. Does this sound right?
Hmm...but does $\mathcal{C}$ even "live" inside of $\mathcal{A}$. I'm not even sure the notation $\frac{f(a_1,...,a_n)}{g(a_1,...,a_n)}$ makes sense. I suppose it makes sense if you interpret it as $f(a_1,...,a_n) g(a_1,...,a_n)^{-1}$, but just because $g(a_1,....,a_n) \neq 0$ doesn't mean it is invertible.
If I have understood it correctly, the elements of the subalgebra are ${\Bbb C}$-linear combinations of the form $$a_1^{m_1}a_2^{m_2}\cdots a_n^{m_n},$$ where $m_1,m_2,\ldots,m_n$ are integers.