Proposition 23

Let A be a K-algebra, $G$ a locally finite group of operators on A, S a multiplicative subset of A stable under $G$ and $S^G$ the set $S \cap A''$. Then the canonical mapping of $(S^G)^{-1}A$ to $S^{-1}A$ (Chapter II, § 2, no. 1, Corollary 2 to Proposition 2) is an isomorphism which maps $(S^G)^{-1}A^G$ to $(S^{-1}A)^G$.

For all $s \in S$, let $s, s_1, \ldots, s_q$ be the distinct elements of the orbit of s under $G$; as $ss_1 \ldots s_q \in S^G$, the first assertion follows from Chapter 11, § 2, no. 3, Proposition 8. Identifying canonically $(S^G)^{-1}A$ with $S^{-1}A$, clearly every element of $(S^G)^{-1}A^G$ is invariant under $G$. Conversely, let $a/t$ be an element of $(S^G)^{-1}A$ which is invariant under $G$ ($a \in A, t \in S^G$); if $a, (1 \leq j \leq m)$ are the distinct elements of the orbit of $a$ under $G$, then $a_j/t = a/t$ for $1 \leq j \leq m$ and therefore there exists $s \in S^G$ such that $s(a_j - a) = 0$ for $1 \leq j \leq m$; in other words, sa is invariant under $G$ and, as $a/t = (sa)/(st)$, certainly $a/t \in (S^G)^{-1}A^G$.