Exercise 8, § 8

Let $K'$ be a field, $B$ the ring of a valuation $v$ on $K'$, $G$ a finite group of automorphisms of $K'$, $K$ the subfield of $K'$ consisting of the elements invariant under $G$ and $A = K \cap B$, which is a valuation ring of $K$, corresponding to the valuation $w = v|K$.

(a) The valuations on $K'$ which extend $w$ are the $v \circ \sigma$, where $\sigma \in G$ (no. 6, Corollary 1 to Proposition 6) and the integral closure $A$ of $A$ in $K'$ is the intersection of the $\sigma . B$, where $\sigma$ runs through $G$. If $p(B)$ is the intersection of $A'$ and the maximal ideal $m(B)$, then $\sigma . p(B) = p(\sigma . B)$. Show that the decomposition group $G^Z(p(B))$ (Chapter V, § 2, no. 2) is the subgroup of $G$ consisting of the $\sigma$ such that $\sigma . B = B$; write $G^Z = G^Z(p(B))$ and denote by $K^Z$ the subfield of $K'$ consisting of the elements invariant under $G^Z$ and by $B^Z$ the ring of the valuation induced on $K^Z$ by $v$, which is equal to $B \cap K^Z$; recall that the residue fields of $B^Z$ and $A$ are the same and that the maximal ideal $m(B^Z)$ is generated by $m(A)B^Z$ (Chapter V, § 2, no. 2, Proposition 4).

(b) Let $v^Z$ denote the restriction of $v$ to $K^Z$ and $\Gamma$ and $\Gamma^Z$ the order groups of $w$ and $v^Z$. Show that $\Gamma = \Gamma^Z$. (Let $G^D \supset G^Z$ be the subgroup of $G$ consisting of the $\sigma \in G$ such that $\sigma . B$ is dependent on $B$ (§ 7, no. 2). Argue by induction on the order of $G$. If $(G/G^D) > 1$, let $K''$ be the field of invariants of $G^D$ and $A'' = K'' \cap B$; show that the order group of $v|K''$ is equal to $\Gamma$: for this, use the Approximation Theorem (§ 7, no. 2, Theorem 1) in order to prove that, for all x \in K", there exists y \in K" such that v(y) = v(x) and v(y_i) = 0 for the conjugates y, of y with respect to K, distinct from y; then $\mathscr{G}$ may be replaced by $\mathscr{G}^D$ and the induction hypothesis may be applied. If $\mathscr{G}^D = \mathscr{G}$, let B, be the valuation ring of K' generated by the $\sigma . B$, where $\sigma \in \mathscr{G}$; let $\overline{K}' = \kappa(B_0)$ be its residue field, $x : B, \to \overline{K}'$ the canonical homomorphism, $\overline{B} = \pi(B)$ a valuation ring of $\overline{K}'$ and $\bar{v}$ the corresponding valuation on $\overline{K}'$ such that $v = \bar{v} \circ \pi$ in B, so that the order group $\overline{\Delta}$ of $\bar{v}$ is a subgroup of the order group A of v; if $A_1 = \Delta / \overline{\Delta}$ and $\overline{\omega} : A \to A_1$, is the canonical homomorphism, $v_0 = \overline{\omega} \circ v$ is a valuation on K' corresponding to B_0 and invariant under $\mathscr{G}$. By taking quotients, $\mathscr{G}$ defines a group of automorphisms $\overline{\mathscr{G}}$ of $\overline{K}'$; let $\mathscr{N}$ be the kernel of the canonical homomorphism $\mathscr{G} \to \overline{\mathscr{G}}$; show that $\mathscr{N} \subset \mathscr{G}^z$ and deduce that, if $\mathscr{N} \neq \{ e \}$, $\mathscr{G}$ may be replaced by $\mathscr{G}/\mathscr{N}$ and the induction hypothesis may be applied. Suppose finally that $\mathscr{N} = \{ e \}$; let $A_1 = K \cap B$, and $w_0$ be the restriction of $v_0$ to K; show that the order group of $w_0$ is A, and that $\pi(A_0) = \overline{K}$ is the field of invariants of $\overline{\mathscr{G}}$ (cf. no. 1, Lemma 2); if $\bar{w}$ is the restriction to $\overline{K}$ of $\bar{v}$ and $\overline{\Gamma}$ its order group, A, is canonically isomorphic to $\Gamma/\overline{\Gamma}$. On the other hand, the group $\mathscr{G}^z(p(\overline{B}))$ is the canonical image of $\mathscr{G}^z(p(B))$ in $\overline{\mathscr{G}}$; observe that $\mathscr{G}^D \neq \overline{\mathscr{G}}$ and that the induction hypothesis may therefore be applied to prove that $\overline{\Gamma}$ is equal to the order group $\overline{\Gamma}^z$ of the restriction of $\bar{v}$ to $\overline{K}^z$.

(c) Let $\mathscr{G}^T = \mathscr{G}^T(p(B))$ be the inertia group of $p(B)$ (Chapter V, § 2, no. 2) and let $K^T$ denote the subfield of K' consisting of the invariants of $\mathscr{G}^T$ and $B^T$ the ring of the valuation induced on $K^T$ by $u$, which is equal to $B \cap K^T$; recall that, if k and k' are the residue fields of A and B, the residue field $k^T$ of $B^T$ is the greatest separable extension of k contained in the quasi-Galois extension $k'$ of k and that $\mathscr{G}^z/\mathscr{G}^T$ is canonically isomorphic to the group of k-automorphisms of $k'$ (or $k^T$) (Chapter V, § 2, Theorem 2 and Proposition 5). Let $v^T$ be the restriction of $v$ to $K^T$; show that the order group of $v^T$ is also equal to $\Gamma$ (apply Lemma 2 of no. 1 to the extension $K^T$ of $K^z$).