Proposition 4

*Let $A'$ be a ring, $\mathscr{G}$ a finite group operating on $A'$, $\mathscr{H}$ a subgroup of $\mathscr{G}$, $A$ and $B$ the rings of invariants of $\mathscr{G}$ and $\mathscr{H}$ respectively and $p'$ a prime ideal of $A$; let us write $p = A \cap p'$ and $p(B) = B \cap p'$.

(i) For $\mathscr{H}$ to be contained in the decomposition group $\mathscr{G}^z(p')$, it is necessary and sufficient that $p'$ be the only prime ideal of $A$ lying above $p(B)$.

(ii) If $\mathscr{H}$ contains $\mathscr{G}^z(p')$, the following conditions are satisfied: (a) The rings $A/p$ and $B/p(B)$ have the same field of fractions. (b) The maximal ideal of the local ring $B_{p(B)}$ is equal to $pB_{p(B)}$.

(iii) Suppose further that $A'$ is an integral domain and that $\bigcap_{n \geq 0} p^nA'_p = 0$; then conditions (a) and (b) & (ii) imply that $\mathscr{G}^z(p')$ leaves invariant the elements of $B$.

(i) It follows from Theorem 2 (i) that the prime ideals of $A$ lying above $p(B)$ are the ideals of the form $\sigma.p'$, where $\sigma \in \mathscr{H}$; whence immediately (i).

(ii) We write $S = A - p$; we know that the rings of invariants of $\mathscr{G}$ and $\mathscr{H}$ in $S^{-1}A'$ are respectively $S^{-1}A$ and $S^{-1}B$ (§ 1, no. 9, Proposition 23) and $\mathscr{G}^{\mathbf{Z}}(S^{-1}p) = \mathscr{G}^{\mathbf{Z}}(p')$ (Lemma 3); finally $S^{-1}p(B) = S^{-1}p' \cap S^{-1}B$ (Chapter II, § 2, no. 4), the local ring of the prime ideal $S^{-1}p(B)$ of the ring $S^{-1}B$ is canonically isomorphic to $B_{p(B)}$ and its residue field is isomorphic to the field of fractions of $B/p(B)$ (Chapter II, § 2, no. 5, Proposition 11). We can therefore show (ii) restricting our attention to the case where $p$ is maximal. To establish (a) it will be sufficient to prove that

for this will show that the fields $A/p$ and $B/p(B)$ are canonically isomorphic. By Theorem 2 there is only a finite number of prime ideals of $A'$ lying above $p$ and by Theorem 1 of no. 1 there is at least one prime ideal of $A'$ lying above every prime ideal of $B$; this implies that there is only a finite number of prime ideals of $B$ lying above $p$; let $n_j$ ($1 \leq j \leq r$) denote those of these ideals which are $\neq p(B)$. Let $x$ be an element of $B$; as the ideals $p(B)$ and $n_j$ are maximal (no. 1, Proposition 1), there exists $y \in B$ such that $y \equiv x$ (mod. $p(B)$) and $y \in n_j$ for $1 \leq j \leq r$ (Chapter 11, § 1, no. 2, Proposition 5). Let $y_1 = y, y_2, \ldots, y_q$ be distinct elements of the orbit of $y$ under $\mathscr{G}$; clearly

and to establish (3) it will be sufficient to show that $y_i \in p'$ for $i \geq 2$, for then we shall deduce that $z - y \in p' \cap B = p(B)$, whence $x \in A + p(B)$ since $x \equiv y$ (mod. $p(B)$). Then let $i \geq 2$ and $\sigma \in \mathscr{G}$ such that $\sigma.y = y_i$; we show that $\sigma^{-1}.p'$ does not lie above $p(B)$. For otherwise there would exist $\tau \in \mathscr{H}$ such that $\sigma^{-1}.p' = \tau.p'$ (Theorem 2 (i)), whence $(\tau^{-1}\sigma^{-1}).p' = p'$, in other words $\tau^{-1}\sigma^{-1} \in \mathscr{G}^{\mathbf{Z}} \subset \mathscr{H}$ by hypothesis, whence $\sigma \in \mathscr{H}$; but as $y \in B$ and $\sigma.y \neq y$, this is absurd. We conclude that $\sigma^{-1}.p'$ lies above one of the ideals $n_j$ and, as $y \in n_j$ by construction, certainly $y \in \sigma^{-1}.p'$ or $y_i = \sigma.y \in p'$.

To prove (b) it will suffice to establish that $p(B)$ is contained in the saturation $q$ of the ideal $pB$ with respect to $p(B)$ (Chapter II, § 2, no. 4, Proposition 10); as $p(B)$ is contained in none of the $n_j$ ($1 \leq j \leq r$), it will suffice even to prove that

by Chapter II, § 1, no. 1, Proposition 2. For this, we consider an element $u \in p(B)$ belonging to none of the $n_j$ ($1 \leq j \leq r$) (Chapter 11, § 1, no. 1, Proposition 2); let $u_1 = u, u_2, \ldots, u_m$ be the distinct elements of the orbit of $u$ under $\mathscr{G}$; we write $w = u_1u_2 \ldots u_m, v = u_2 \ldots u_m$; clearly $w \in A$; on the other hand, if $\tau \in \mathscr{H}$, $\tau.u = u$ and hence necessarily $\tau.u_i \neq u$ for $i \geq 2$, which shows that $\tau.v = v$ and hence $v \in B$. It can be shown as in the proof of (a) that, if $\sigma \in \mathscr{G}$ is such that $\sigma.u = u_i$ where $i \geq 2$, $\sigma^{-1}.p'$ lies above one of the $n_j$ and, as $u \notin n_j$, also $u \notin \sigma^{-1}.p'$, in other words $u_i \notin p'$. We conclude that $v \notin p'$ and therefore $v \notin p(B)$. On the other hand clearly $w \in p' \cap A = p$ and the relation $w = uv$ shows that $u$ is in the saturation of $pB$ with respect to $p(B)$ and hence establishes (4).

(iii) Suppose that $A'$ is an integral domain, that $\bigcap_{n \geq 0} p^n A'_p = 0$ and that conditions (a) and (b) of (ii) hold. With the same notation as in (ii), clearly $S^{-1}A'$ is an integral domain and $S^{-1}A'_p = A'_p$; it is therefore possible to replace $A'$ and $p'$ by $S^{-1}A'$ and $S^{-1}p'$, in other words, suppose also that the ideal $p$ is maximal. The hypotheses (a) and (b) then imply that

By induction on $n$ we deduce that $\mathbf{B}_{p(B)} = A + p^n \mathbf{B}_{p(B)}$ for all $n > 0$. Then let $\sigma$ be an element of $\mathscr{G}^\mathbf{Z}$ and $x$ be an element of $B$. For all $n > 0$, there exists $a_n \in A$ such that $x - a_n \in p^n \mathbf{B}_{p(B)} \subset p^n A'_p$; as $\sigma.a_n = a$, and $\sigma.p' = p'$, we deduce that $\sigma.x - x \in p^n A'_p$. Since this relation holds for all $n$, we conclude from the hypothesis that $\sigma.x = x$.

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(3) If $A'$ is an integral domain and Noetherian, the condition $\bigcap_{n \geq 1} p^n A'_p = 0$ always holds (Chapter III, § 3, no. 2, Corollary to Proposition 5). It can be shown that this condition is also satisfied if $A'$ is assumed to be an integral domain and $A$ to be Noetherian.

(4) If $p$ is not a maximal ideal of $A$ relation (3) does not necessarily hold under the hypotheses of (ii) and therefore $A/p$ and $B/p(B)$ are not necessarily isomorphic even if we take $\mathscr{H} = \mathscr{G}^\mathbf{Z}$, whence $B = A''$ (Exercise 10).