%! TEX root = LF.tex % vim: tw=50 % 14/11/2024 12PM From now on in this course: if unspecified $L / K$ is a finite separable extension of (\gls{nonarch}) \glspl{lf}. Also, all \glspl{lf} that we consider from now on will be \gls{nonarch}. \begin{fcthm}[] \label{thm_13_3} % Theorem 13.3 Assuming: - $L / K$ a finite separable extension of \gls{nonarch} \glspl{lf} Then: there exists a field $K_0$, $K \subseteq K_0 \subseteq L$ and such that \begin{enumerate}[(i)] \item $K_0$ is \gls{unramext} \item $L / K_0$ is \gls{totramext} \end{enumerate} Moreover $[L : K_0] = e_{L / K}$, $[K_0 : K] = f_{L / K}$ and $K_0 / K$ is Galois. \end{fcthm} \begin{proof} Let $k = \Fbb_q$, so that $k_L = \Fbb_{q^f}$, $f_{L / K} = f$. Set $m = q^f - 1$, $[\blank] : \Fbb_{q^f} \to L$ the \gls{teich} map for $L$. Let $\zeta_m \defeq [\alpha]$ for $\alpha$ a generator of $\Fbb_{q^f}^\times$. $\zeta_m$ a primitive $m$-th root of unity. Set $K_0 = K(\zeta_m) \subseteq L$, then $K_0 / K$ is Galois and has residue field $k_0 = \Fbb_q(\alpha) = k_L$. Hence $f_{L / K_0} = 1$, i.e. $L / K_0$ is \gls{totramext}. Let $\res : \Gal(K_0 / K) \to \Gal(k_0 / k)$ be the natural map. For $\sigma \in \Gal(K_0 / K)$. We have $\sigma(\zeta_m) = \zeta_m$ if $\sigma(\zeta_m) \equiv \zeta_m \bmod m$ (since $\mu_m(K_0) \cong \mu_m(k_0)$ by \nameref{hensel}). Hence $\res$ is injective. Thus $|\Gal(K_0 / K)| \le |\Gal(k_0 / k)| = f_{K_0 / K}$, so $[K_0 : K] = f_{K_0 / K}$. Hence $\res$ is an isomorphism, and $K_0 / K$ is \gls{unramext}. \end{proof} \begin{fcthm}[] \label{thm_13_4} % Theorem 13.4 Assuming: - $k = \Fbb_q$ - $n \ge 1$ Then: there exists a unique \gls{unramext} $L / K$ of degree $n$. Moreover, $L / K$ is Galois and the natural $\Gal(L / K) \to \Gal(k_L / k)$ is an isomorphism. In particular, $\Gal(L / K) \cong \langle \Frob_{L / K} \rangle$ is cyclic, where $\Frob_{L / K}(x) = x^q \bmod m_L$ for all $x \in \O_L$. \end{fcthm} \begin{proof} For $n \ge 1$, take $L = K(\zeta_m)$ where $m = q^n - 1$. As in \cref{thm_13_3}: \[ \Gal(L / K) \stackrel{\sim}{\to} \Gal(k_L / K) \cong \Gal(\Fbb_{q^n} / \Fbb_q) .\] Hence $\Gal(L / K)$ is cyclic, generated by a lift of $x \mapsto x^q$. Uniqueness: $L / K$ of degree $n$ \gls{unramext}. Then \gls{teich} gives $\zeta_m \in L$, so $L = K(\zeta_m)$. \end{proof} \begin{corollary} \label{coro_13_5} % Corollary 13.5 $L / K$ a finite Galois extension. Then $\res : \Gal(L / K) \to \Gal(k_L / k)$ is surjective. \end{corollary} \begin{proof} $\res$ factorises as \[ \Gal(L / K) \surjto \Gal(K_0 / K) \stackrel{\sim}{\to} \Gal(k_L / k) . \qedhere \] \end{proof} \begin{fcdefn}[Inertial subgroup] \glssymboldefn{I}% \glsnoundefn{inert}{inertial subgroup}{inertial subgroups}% \label{defn_13_6} % Definition 13.6 The inertial subgroup is \[ I_{L / K} = \ker(\Gal(L / K) \surjto \Gal(k_L / k)) .\] \end{fcdefn} \begin{itemize} \item Since $e_{L / K} f_{L / K} = [L : K]$, we have $|\I_{L / K}| = e_{L / K}$. \item $\I_{L / K} = \Gal(L / K_0)$ -- $K_0$ as in \cref{thm_13_3}. \end{itemize} \begin{fcdefn}[Eisenstein polynomial] \glsadjdefn{eispoly}{Eisenstein}{polynomial}% \label{defn_13_7} % Definition 13.7 $f(x) = x^n + a_{n - 1} x^{n - 1} + \cdots + a_0 \in \O_K[x]$ is \emph{Eisenstein} if $v_K(a_i) \ge 1$ for all $i$, and $v_K(a_0) = 1$. \end{fcdefn} \textbf{Fact:} $f(x)$ \gls{eispoly} implies $f(x)$ irreducible. \begin{theorem} \label{thm_13_8} % Theorem 13.8 \begin{enumerate}[(i)] \item Let $L / K$ finite \gls{totramext}, $\pi_L \in \mathcal{O}_L$ a \gls{unif}. Then the minimal polynomial of $\pi_L$ is \gls{eispoly} and $\O_L = \O_K[\pi_L]$ (hence $L = K(\pi_L)$) \item Conversely, if $f(x) \in \O_K[x]$ is \gls{eispoly} and a root of if f, then $L \defeq K(\alpha) / K$ is \gls{totramext} and $\alpha$ is a \gls{unif} of $L$. \end{enumerate} \end{theorem} \begin{proof} \phantom{} \begin{enumerate}[(i)] \item $[L : K] = e = e_{L / K}$. Let \[ f(x) = x^m + a_{m - 1} x^{m - 1} + \cdots + a_0 \in \O_K[x] \] the minimal polynomial for $\pi_L$. Then $m \le e$. Since $v_L(K^\times) = e\Zbb$, we have $v_L(a_i\pi^i) \equiv e \bmod e$, for $i < m$. Hence these terms have distinct valuations. As \[ \pi_L^m = -\sum_{i = 0}^{m - 1} a_i \pi_L^i .\] we have \[ m = v_L(\pi_L^m) = \min_{0 \le i \le m - 1} (i + ev_K(a_i)) \] hence $v_K(a_i) \ge 1$ for all $i$. Hence $v_K(a_0) = 1$ and $m = e$. Thus $f(x)$ is \gls{eispoly} and $L = K(\pi_L)$. For $y \in L$, we write $y = \sum_{i = 0}^{e - 1} \pi_L^i b_i$, $b_i \in K$. Then \[ v_L(y) = \min_{0 \le i \le e - 1} (i + ev_K(b_i)) .\] Thus \begin{align*} y \in \O_L &\iff v_L(y) \ge 0 \\ &\iff v_K(b_i) \ge 0 \forall i \\ &\iff y \in \O_K[\pi_L] \end{align*} \item Let $f(x) = x^n + a_{n - 1} x^{n - 1} + \cdots + a_0$ is \gls{eispoly} and $e = e_{L / K}$. Thus $v_L(a_i) \ge e$ and $v_L(a_0) = e$. If $v_L(\alpha) \le 0$, we have \[ v_L(\alpha^n) < v_L \left( \sum_{i = 0}^{n - 1} a_i \alpha^i \right) \] hence $v_L(\alpha) > 0$. For $i \neq 0$, $v_L(a_i \alpha^i) > e = v_L(a_0)$. Therefore \[ v_L(\alpha^n) = v_L \left( -\sum_{i = 0}^{n - 1} a_i \alpha^i \right) = v_L(a_0) = e .\] Hence $nv_L(\alpha) = e$. But $n = [L : K] \ge e$, so $n = e$ and $v_L(\alpha) = 1$. \end{enumerate} \end{proof} \subsection{Structure of Units} Let $[K : \Qbb] < \infty$, $e \defeq e_{K / \Qbb_p}$, $\pi$ a \gls{unif} in $K$. \begin{fcprop}[] \label{prop_13_8} % Proposition 13.8 Assuming: - $r > \frac{e}{p - 1}$ Then: $\exp(x) = \sum_{n = 0}^{\infty} \frac{x^n}{n!}$ converges on $\pi^r \O_K$ and induces an isomorphism \[ (\pi^r \O_K, +) \stackrel{\sim}{\to} (1 + \pi^r \O_K, \times) .\] \end{fcprop} \begin{proof} \begin{align*} v_K(n!) &= e v_p(n!) \\ &= \frac{e(n - s_p(n))}{p - 1} &&\text{Example Sheet 1} \\ &\le e \left( \frac{n - 1}{p - 1} \right) \end{align*} For $x \in \pi^r \O_K$ and $n \ge 1$, \begin{align*} v_K \left( \frac{x^n}{n!} \right) &\ge nr - \frac{e(n - 1)}{p - 1} \\ &= r - (n - 1) \ub{\left( r - \frac{e}{p - 1} \right)}_{> 0} \end{align*} Hence $v_K \left( \frac{x^n}{n!} \right) \to \infty$ as $n \to \infty$. Thus $\exp(x)$ converges.