Theorem 2

Let $A'$ be a ring, $\mathscr{G}$ a finite group operating on $A'$ and $A$ the ring of invariants of $\mathscr{G}$, so that $A'$ is integral over $A$ (§ 1, no. 9, Proposition 22).

(i) Given two prime ideals $p', q'$ of $A'$ lying above the same prime ideal $p$ of $A$, there exists $\sigma \in \mathscr{G}$ such that $q' = \sigma.p'$; in other words, $\mathscr{G}$ operates transitively on the set of prime ideals of $A'$ lying above $p$.

(ii) Let $p'$ be a prime ideal of $A'$, $p = p' \cap A$ and $k$ (resp. $k'$) the field of fractions of $A/p$ (resp. $A'/p'$). Then $k'$ is a quasi-Galois extension () of $k$ and the canonical homomorphism $\sigma \mapsto \bar{\sigma}$ from $\mathscr{G}^{\mathbf{Z}}(p')$ to the group $\Gamma$ of $k$-automorphisms of $k'$ defines, by taking the quotient, an isomorphism of $\mathscr{G}^{\mathbf{Z}}(p')/\mathscr{G}^{\Gamma}(p')$ onto $\Gamma$.*

(*) In order to avoid confusion with other meanings of the word "normal" we henceforth use the term "quasi-Galois extension" as synonymous with the term "normal extension" defined in Algebra, Chapter V, § 6, no. 2, Definition 2.

(i) If $x \in \mathfrak{q}$ then $\prod_{\sigma \in \mathscr{G}} \sigma.x \in q' \cap A = p \subset p'$; then there exists $\sigma \in \mathscr{G}$ such that $\sigma.x \in p'$, that is $x \in \sigma^{-1}.q'$. We conclude that $q' \subset \bigcup_{\sigma \in \mathscr{G}} \mathfrak{s}.\mathfrak{p}'$ and hence (since $\mathscr{G}$ is finite and the $\mathfrak{s}.\mathfrak{p}'$ prime) there exists $\sigma \in \mathscr{G}$ such that $\mathfrak{q}' \subset \mathfrak{s}.\mathfrak{p}'$ (Chapter II, § 1, no. 1, Proposition 2); as $q'$ and $\sigma.\mathfrak{p}'$ both lie above $p$, $q' = \sigma.\mathfrak{p}'$ (no. 1, Corollary 1 to Proposition 1).

(ii) To see that $k'$ is a quasi-Galois extension of $k$, it suffices to prove that every element $\bar{x} \in A'/p'$ is a root of a polynomial $P$ in $k[X]$ all of whose roots are in $A'/p'$ (Algebra, Chapter V, § 6, no. 3, Corollary 3 to Proposition 9). Now, let $x \in A'$ be a representative of the class $\bar{x}$; the polynomial $Q(X) = \prod_{\sigma \in \mathscr{G}} (X - \sigma.x)$ has all its coefficients in $A$; let $P(X)$ be the polynomial in $(A/p)[X]$ whose coefficients are the images of those of $Q$ under the canonical homomorphism $\pi : A \to A/p$. As $\pi$ may be considered as the restriction to $A$ of the canonical homomorphism $\pi' : A' \to A'/p'$, it is seen that $P$ is the product in $(A'/p')[X]$ of the linear factors $X - \pi'(\sigma.x)$ and therefore solves the problem since $\bar{x} = \pi'(x)$.

Clearly, for all $\sigma \in \mathscr{G}^z$, $\hat{\sigma}$ is a $k$-automorphism of $k'$; it remains to verify that $\sigma \mapsto \hat{\sigma}$ maps $\mathscr{G}^z$ onto the group of all $k$-automorphisms of $k'$. Let us write $S = A - p$; $k$ and $k'$ are not changed by replacing $A'$ and $p'$ by $S^{-1}A'$ and $S^{-1}p'$ respectively, by virtue of § 1, no. 9, Proposition 23 and the relation $S^{-1}p' \cap S^{-1}A = S^{-1}(A \cap p') = S^{-1}p$ (Chapter 11, § 2, no. 4); it follows from Lemma 3 that neither $\mathscr{G}^z$ nor its operation on $k'$ is changed; we may therefore restrict our attention to the case where $p$ is maximal, in which case we know that so is $p'$ (no. 1, Proposition 1) and every element of $k'$ is therefore of the form $\pi'(x)$ for some $x \in A'$; it has been seen above that such an element is a root of a polynomial in $k[X]$ of degree $\leq \mathrm{Card}(\mathscr{G})$. As every finite separable extension of $k$ admits a primitive element (Algebra, Chapter V, § 7, no. 7, Proposition 12 and § 11, no. 4, Proposition 4), it is seen that every finite separable extension of $k$ contained in $k'$ is of degree $\leq \mathrm{Card}(\mathscr{G})$, whence it follows that the greatest separable extension $k'_s$ of $k$ contained in $k'$ (Algebra, Chapter V, § 7, no. 6, Proposition 11) is of degree $\leq \mathrm{Card}(\mathscr{G})$ (Algebra, Chapter V, § 3, no. 2, Remark 2). Let $y \in A'$ be an element such that $\pi'(y)$ is a primitive element of $k'_s$. The ideals $\sigma.p'$ for $\sigma \in \mathscr{G} - \mathscr{G}^z$ are maximal and distinct from $p'$ by definition; there therefore exists $x \in A'$ such that $x \equiv y \pmod{p'}$ and $x \in \sigma^{-1}p'$ for $\sigma \in \mathscr{G} - \mathscr{G}^z$ (Chapter 11, § 1, no. 2, Proposition 5). This being so, let $u$ be a $k$-automorphism of $k'$ and let $P(X) = \prod_{\sigma \in \mathscr{G}} (X - \pi'(\sigma.x))$; as $\pi'(x)$ is a root of $P$ and $P \in k[X]$, $u(\pi'(x))$ is also a root of $P$ in $k'$ and hence there exists $\tau \in \mathscr{G}$ such that

but $u(\pi'(x)) \neq 0$ and, for $\sigma \in \mathscr{G} - \mathscr{G}^z$, $\sigma.x \in p'$ and hence $\pi'(\sigma.x) = 0$; we conclude that necessarily $\tau \in \mathscr{G}^z$. But as $u$ and $\tilde{\tau}$ have the same value for the primitive element $\pi'(y) = \pi'(x)$ of $k'_s$, they coincide on $k'_s$ and, as $k'$ is a radicial extension of $k'_s$, they coincide on $k'$.