Proposition 6

Let $A$ be an integrally closed domain, $K$ its field of fractions, $K'$ a quasi-Galois extension of $K$ and $A'$ the integral closure of $A$ in $K'$. Then:

(i) For every prime ideal $p$ of $A$, the group $\mathscr{G}$ of $K$-automorphisms of $K'$ operates transitively on the set of prime ideals of $A'$ lying over $p$.

(ii) For every prime ideal $p'$ of $A'$, the field of fractions $k'$ of $A'/p'$ is a quasi-Galois extension of the field of fractions $k$ of $A/(A \cap p')$ and the canonical homomorphism $\sigma \mapsto \tilde{\sigma}$ from $\mathscr{G}^Z(p')$ to the group $\Gamma$ of $k$-automorphisms of $k'$ defines, by taking the quotient, a bijection of $\mathscr{G}^Z(p')/\mathscr{G}^T(p')$ onto $\Gamma$.

(A) Suppose first that $K'$ is ajinite Galois extension of $K$. Then $A = A \cap K$ since $A$ is integrally closed and $A$ is therefore the ring of invariants of $\mathscr{G}$ in $A'$. As $\mathscr{G}$ is finite, the proposition follows in this case from no. 2, Theorem 2.

(B) Suppose secondly that $K'$ is any Galois extension of $K$. Then $K'$ is the union of a right directed family $(K_\alpha)_{\alpha \in I}$ of finite Galois extensions of $K$. To show (i), consider two prime ideals $p', q'$ of $A'$ lying above $p$. For all $\alpha \in I$, $p' \cap K_\alpha$ and $q' \cap K_\alpha$ are two prime ideals of $A \cap K_\alpha$, lying above $p$. Since $A' \cap K_\alpha$ is the integral closure of $A$ in $K_\alpha$, and the restrictions to $K_\alpha$ of the elements of $\mathscr{G}$ form the group of $K$-automorphisms of $K$, it follows from case (A) that there exists $\sigma \in \mathscr{G}$ such that $(\sigma.(p' \cap K_\alpha)) = q' \cap K_\alpha$. Let $\mathscr{E}_\alpha$ be the set of $\sigma \in \mathscr{G}$ which have the above property. Let $\sigma \in \mathscr{G} - 8$; then, for all $\tau \in \mathscr{G}$ leaving invariant the elements of $K_\alpha$, (QT) $(\sigma.(p' \cap K_\alpha)) = \varrho.(p' \cap K_\alpha) \neq q' \cap K_\alpha$, and hence $\sigma \tau \in \mathscr{G} - b$. It follows that $\mathscr{E}_\alpha$ is closed in the topological Galois group $\mathscr{G}$ (Algebra, Chapter V, Appendix 11, no. 1) and clearly the family $(\mathscr{E}_\alpha)_{\alpha \in I}$ is left directed. As $\mathscr{G}$ is compact (loc. cit., no. 2, Proposition 3) and the $\mathscr{E}_\alpha$ are non-empty, the intersection $\mathscr{E}$ of the family $(8,)$ is non-empty and $\sigma.p' = q'$ for all $\sigma \in \mathscr{E}$, whence (i).

To show (ii), note that $k'$ is the union of the right directed family $(k_\alpha)_{\alpha \in I}$, where $k_\alpha$ is the field of fractions of $(A' \cap K_\alpha)/(p' \cap K_\alpha)$. As each $k_\alpha$ is a quasi-Galois extension of $k$ by (A), so is $k'$ (Algebra, Chapter V, § 6, no. 3, Proposition 8). On the other hand, let $u$ be a $k$-automorphism of $k'$ and let $\pi': A' \to A'/p'$ be the canonical homomorphism. By virtue of no. 2, Theorem 2 applied to $A' \cap K_\alpha$, there exists for all $\alpha$ a non-empty set $F_\alpha$ of elements $\sigma \in \mathscr{G}$ such that $\sigma.(p' \cap K_\alpha) = p' \cap K_\alpha$ and $u(\pi'(x)) = \pi'(\sigma.x)$ for all $x \in A' \cap K_\alpha$. As above it is seen that $\mathscr{F}_\alpha$ is closed in $\mathscr{G}$ and, as $(\mathscr{F}_\alpha)$ is a left directed set, its intersection $\mathscr{F}$ is non-empty. Clearly for $\alpha \in \mathscr{F}$, $\alpha \in \mathscr{G}^\mathbf{Z}(p')$ and $\tilde{\sigma} = u$, which completes the proof of (ii) in this case.

(C) General care. The field of invariants $K_1$ of $\mathscr{G}$ is a radicial extension of $K$ (Algebra, Chapter V, § 10, no. 9, Proposition 14); there therefore exists a single prime ideal $p_1$ of $A_1 = A' \cap K_1$ lying above $p$ (Lemma 4). If $p'$ and $q'$ are two prime ideals of $A'$ lying above $p$, then they lie above $p_1$; as $K'$ is a Galois extension of $K$, and $A' \cap K$, is integrally closed (§ 1, no. 2, Proposition 7 and Corollary to Proposition 8), it follows from (B) that there exists $\sigma \in \mathscr{G}$ such that $(1.p' = q'$; whence (i). On the other hand, clearly the field of fractions $k$, of $A_1/p_1$ is a radicial extension of $k$ (A being integrally closed); as $k'$ is a quasi-Galois extension of $k$, by (B), $k'$ is a quasi-Galois extension of $k$, every $k$-isomorphism of $k'$ to an algebraically closed extension of $k'$ being a $k_1$-isomorphism. This last remark shows also, taking account of (B), that every $k$-automorphism of $k'$ is of the form $\tilde{\sigma}$ where $\alpha \in \mathscr{G}^\mathbf{Z}(p')$, which completes the proof of (ii).

Remark

(2) Suppose that $K$ is a Galois extension of $K$ and let us keep the notation of the proof of Proposition 6; for all $a$ let $\mathscr{G}_a^\mathbf{Z}$ (resp. $\mathscr{G}_a^\mathbf{T}$) be the subgroup of $\mathscr{G}$ consisting of the $\sigma$ whose restriction to $A' \cap K$, belongs to $\mathscr{G}^\mathbf{Z}(p' \cap K_\alpha)$ (resp. $\mathscr{G}^\mathbf{T}(p' \cap K_\alpha)$). The proof of Proposition 6 shows that these subgroups are closed in $\mathscr{G}$ and that

Moreover, the set of restrictions to $A' \cap K$, of the elements of $\mathscr{G}$ (resp. $\mathscr{G}_a^\mathbf{T}$) is the whole of the group $\mathscr{G}^\mathbf{Z}(p' \cap K_\alpha)$ (resp. $\mathscr{G}^\mathbf{T}(p' \cap K_\alpha)$), every $K$-automorphism of $K$, extending to an element of $\mathscr{G}$.

With the same hypotheses, the ring $A^\mathbf{Z}(p')$ (resp. $A^\mathbf{T}(p')$) is the union of the directed family of the $A^\mathbf{Z}(p' \cap K_\alpha)$ (resp. $A^\mathbf{T}(p' \cap K_\alpha)$); in fact, every $x \in A^\mathbf{Z}(p')$ (resp. every $x \in A^\mathbf{T}(p')$) belongs to one of the $K$, and by the above there exists a $\beta$ such that $K_\alpha \subset K_\beta$ and the restrictions to $A' \cap K$, of the elements of $\mathscr{G}^\mathbf{Z}(p')$ (resp. $\mathscr{G}^\mathbf{T}(p')$) are the same as the restrictions to $A' \cap K$, of the elements of $\mathscr{G}^\mathbf{Z}(p' \cap K_\beta)$ (resp. $\mathscr{G}^\mathbf{T}(p' \cap K_\beta)$), the groups $\mathscr{G}^\mathbf{Z}(p' \cap K_\alpha)$ and $\mathscr{G}^\mathbf{T}(p' \cap K_\beta)$ being finite; hence $x$ belongs to $A^\mathbf{Z}(p' \cap K_\beta)$ (resp. $A^\mathbf{T}(p' \cap K_\beta)$).