Lemma 2

Let $(x_{\lambda})_{\lambda\in\Lambda}$ be a family of pairwise commuting elements of A. The full subalgebra generated by $(x_{\lambda})$ is the set of elements of the form $R((x_{\lambda}))$, where $R\in K((X_{\lambda}))$ runs through the set of rational fractions in which the family $(x_{\lambda})$ is substitutable.

Let B be the full subalgebra of A generated by the family $(x_{\lambda})$, and let us denote by $B_1$ the set of elements of the form $R((x_{\lambda}))$, where $R((X_{\lambda}))$ is a rational fraction in which $(x_{\lambda})$ can be substituted. Explicitly, $B_1$ is the set of elements of A of the form $P((x_{\lambda}))Q((x_{\lambda}))^{-1}$, where $P,Q\in K[(X_{\lambda})]$ and $Q((x_{\lambda}))$ is invertible in A. The set $B_1$ is a unital subalgebra of A containing the family $(x_{\lambda})$. It is a full subalgebra: if $P((x_{\lambda}))Q((x_{\lambda}))^{-1}$ is invertible in A, then $P((x_{\lambda}))$ is invertible in A and the inverse $Q((x_{\lambda}))P((x_{\lambda}))^{-1}$ of $P((x_{\lambda}))Q((x_{\lambda}))^{-1}$ belongs to $B_1$. Hence $B\subset B_1$. On the other hand, if $P,Q\in K[(X_{\lambda})]$, and if $Q((x_{\lambda}))$ is invertible in A, then $P((x_{\lambda}))\in B$ and $Q((x_{\lambda}))\in B$, hence $Q((x_{\lambda}))^{-1}\in B$ and $P((x_{\lambda}))Q((x_{\lambda}))^{-1}\in B$, hence $B_1\subset B$.