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% 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.
