Exercise 11, § 8

Let K be a field, L a finite Galois extension of K and $\mathscr{G}$ its Galois group. Let $v$ be a valuation on L with residue field k and order group $\Gamma$; suppose that $v$ is invariant under $\mathscr{G}$ and that the restriction $v|K$ has the same residue field $k$ as $v$, so that $\mathscr{G} = \mathscr{G}^Z = \mathscr{G}^\Gamma$ (notation of Exercise 8).

(a) Let $x \in L^*, \sigma \in \mathscr{G}$. Show that $x^{-1} \sigma(x)$ is a unit for $v$, that its image $\varepsilon_\sigma(x)$ in $k^*$ depends only on the valuation $v(x)$ and the class of $\sigma$ modulo the commutator group $\mathscr{G}'$ of $\mathscr{G}$ and that there is thus defined a Z-bilinear mapping (also denoted by $\varepsilon$) $(\mathscr{G}/\mathscr{G}') \times \Gamma \to k^*$, which is equal to 1 on $(\mathscr{G}/\mathscr{G}') \times v(K^*)$.

(b) For all $\sigma \in \mathscr{G}$, let $\phi(\sigma)$ be the homomorphism $\Gamma/v(K^*) \to k^*$ mapping $\varepsilon_\sigma(x)$ to $v(x)$ for all $x \in L^*$; thus there is defined a homomorphism

Show that, if $p$ is the characteristic exponent of k, the kernel $\mathscr{N}$ of $\phi$ has order a power of $p$. (In the contrary case, there would exist a $\sigma \neq e$ in $\mathscr{N}$ and a prime number $q \neq p$ such that $\sigma^q = e$; for some integer $x$ in $L - K$, write $\sigma(x) = x + y$, where $v(y) > v(x)$, and calculate $\sigma^q(x)$ to obtain a contradiction.) Deduce that $\mathscr{G}$ is a solvable group.

(c) Suppose that $n = [L : K]$ is prime to $p$; show that $\phi$ is bijective, that $e(L/K) = n$ and that, if $v$ is the $lcm$ of the orders of the elements of $\mathscr{G}, k^*$ contains the $v$-th roots of unity. (Use (b) and Lemma 2 of no. 1.)