Lemma 3

Let $A'$ be a ring, $\mathscr{G}$ a group operating on $A'$, $A$ the ring of invariants of $\mathscr{G}$, $p'$ a prime ideal of $A'$ and $S$ a multiplicative subset of $A$ not meeting $p'$. Then $\mathscr{G}^{\mathbf{Z}}(S^{-1}p') = \mathscr{G}^{\mathbf{Z}}(p')$, $\mathscr{G}^{\Gamma}(S^{-1}p') = \mathscr{G}^{\Gamma}(p')$ and, if $\mathscr{G}$ is locally finite, $S^{-1}A^{\mathbf{Z}}(p') = A^{\mathbf{Z}}(S^{-1}p')$ and $S^{-1}A^{\Gamma}(p') = A^{\Gamma}(S^{-1}p')$.

As the elements of $S$ are invariant under $\mathscr{G}$, clearly, if $\sigma.p' = p'$, also $\sigma.(S^{-1}p') = S^{-1}p'$. Conversely, suppose that $\sigma \in \mathscr{G}$ is such that $\sigma.(S^{-1}p') = S^{-1}p'$; then, if $x \in p'$, $(\sigma.x)/1 \in S^{-1}p'$ and there therefore exists $s \in S$ such that $s(\sigma.x) \in p'$, whence $\sigma.x \in p'$ since $p'$ is prime and $s \notin p'$; this proves that $\sigma.p' \subset p'$ and it can be similarly shown that $\sigma^{-1}.p' \subset p'$, hence $\sigma.p' = p'$ and $\sigma \in \mathscr{G}^{\mathbf{Z}}(p')$. If $\sigma \in \mathscr{G}^{\Gamma}(p')$, then $\sigma.x - x \in p'$ for all $x \in A'$, hence also, for all $s \in S$,

and therefore $\sigma \in \mathscr{G}^{\Gamma}(S^{-1}p')$. Conversely, suppose that $\sigma \in \mathscr{G}^{\Gamma}(S^{-1}p')$; then, for all $x \in A'$, $\sigma.(x/1) - (x/1) \in S^{-1}p'$ and therefore there exists $s \in S$ such that $s(\sigma.x - x) \in p'$, whence as above $\sigma.x - x \in p'$, which proves that $\sigma \in \mathscr{G}^{\Gamma}(p')$. The last two assertions follow from § 1, no. 9, Proposition 23.