Proposition 2

*Let K be a commutative field, E a unital associative K-algebra and $x = (x_i)_{i \in I}$ a family of pairwise permutable elements of E. The set $S_x^{-1}B$ of $f \in K((X_i)_{i \in I})$ such that x is substitutable in $f$ is a K-subalgebra of $K((X_i)_{i \in I})$. The mapping $f \mapsto f(x)$ is a unital homomorphism $\varphi$ of $S_x^{-1}B$ into E. The image $\varphi(S_x^{-1}B)$ is the set of all $yz^{-1}$, where y runs over the unital subalgebra $K[x]_E$ of E generated by the family x and where z runs over the set of all invertible elements of $K[x]_E$.

Let $f_1 = \frac{u_1}{v_1},\ f_2 = \frac{u_2}{v_2}$ be two elements of $K((X_i)_{i \in I})$ such that $v_1,\ v_2 \in S_x$. We have $f_1 + f_2 = \frac{u_1v_2 + u_2v_1}{v_1v_2},\ f_1f_2 = \frac{u_1u_2}{v_1v_2}$ and $v_1,\ v_2 \in S_x$. Hence $S_x^{-1}B$ is a K-subalgebra of $K((X_i)_{i \in I})$. The rest of the proposition is clear.