For a ring $R$, a *polynomial identity* of $R$ is a polynomial (in non-commuting variables) $f(x_1,\ldots,x_n)\in \mathbb{Z}[x_1,\ldots, x_n]$ such that for any choice of $a_i\in R$, $f(a_1,\ldots, a_n)=0$.

For example, for all $n$, the standard (or symmetric) polynomial $S_{2n}(x_1,\ldots,x_{2n})=\sum_{\sigma\in S_{2n}}$sign$(\sigma)x_{\sigma(1)}\cdots x_{\sigma(2n)}$ is a polynomial identity for the ring $R=M_n(\mathbb{C})$.

I would like to understand how this idea might extend to considering, rather than polynomials in the $x_i$, polynomials in the $x_i$ and their inverses, i.e., an element of $\mathbb{Z}\langle x_1,\ldots,x_n\rangle$.

Obviously, to even be defined, one would need to restrict any 'generalized' polynomial identity to the units of $R$. An easy example is:

**Example:** If $R^\times$ is abelian, then $f(x,y) = 1 - xyx^{-1}y^{-1}$ is such a generalized polynomial identity. Similar statements can be made if $R^\times$ is nilpotent, solvable, &c.

I'm primarily interested in the situation of $R=M_n(\mathbb{C})$, with $R^\times=\textrm{GL}_n(\mathbb{C})$. Is anyone aware of previous work on this subject? Are there any 'obvious' such identities?

More optimistically, I wonder if anything might be said by taking the (skew) field of fractions of $\mathbb{Z}[x_1,\ldots,x_n]$ and asking what identities arise here. For example, something of the form $f(x,y)=x(x-y)^{-1}y$. In this case, the domain of the identity $f$ would need to be restricted to those $x,y\in R$ so that $x-y\in R^\times$. More generally, the domain might become even stranger (consider $f(x)=(1+(1+x)^{-1})^{-1}$, $f(x)=(1+(1+(1+x)^{-1})^{-1})^{-1}$, ...).

I'll add that I'm not particularly worried about how nasty this domain of definition may be for a particular identity, just whether such an identity exists with nonempty domain; again, I'm interested in $R=\textrm{GL}_n(\mathbb{C})$, and here (I believe) for any fixed such $f$, the domain will be nonempty unless some inverted term is itself an identity.