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  Since $v_K \left( \frac{x^n}{n!} \right) \ge r$
  for all $n \ge 1$, $\exp(x) \in 1 + \pi^r \O_K$.

  Consider $\log$: $1 + \pi^r \O_K \to \pi^r
  \O_K$.
  \[
    \log(1 + x) = \sum_{n = 1}^{\infty}
    \frac{(-1)^{n - 1}}{n} x^n
  \]
  which converges as before.

  Recall identities in $\Qbb[[X, Y]]$:
  \begin{align*}
    \exp(X + Y)
    &= \exp(X) \exp(Y) \\
    \exp(\log(1 + X))
    &= 1 + X \\
    \log(\exp(X))
    &= X
  \end{align*}
  Thus $\exp ; (\pi^r \O_K, +)
  \stackrel{\sim}{\to} (1 + \pi^r \O_K, \times)$
  is an isomorphism.
\end{proof}

$K$ \emph{any} local field: $U_K \defeq
\O_K^\times$, $\pi \in \O_K$ \gls{unif}.

\begin{fcdefn}[$s$-th unit group]
  \label{defn_13_9}
  % Definition 13.9
  For $s \in \Zbb$, the $s$-th unit group
  $U_K^{(s)}$ is defined by
  \[
    U_K^{(s)} = (1 + \pi^s \O_K, \times)
  .\]
  Set $U_K^{(0)} = U_K$. Then we have
  \[
    \cdots \subseteq U_K^{(s)}
    \subseteq U_K^{(s - 1)}
    \subseteq \cdots
    \subseteq U_K^{(0)}
    = U_K
  .\]
\end{fcdefn}

\begin{proposition}
  \label{prop_13_10}
  % Proposition 13.10
  \phantom{}
  \begin{enumerate}[(i)]
    \item $U_K^{(0)} / U_K^{(i)} \cong (k^\times,
      \times)$ ($k \cong \O_K / \pi$)
    \item $U_K^{(s)} / U_K^{(s + 1)} \cong (k,
      +)$ for $s \ge 1$
  \end{enumerate}
\end{proposition}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Reduction modulo $\pi$. $\O_K^\times \to
      k^\times$ is surfective with kernel $1 +
      \pi\O_K = U_K^{(1)}$.
    \item $f ; U_K^{(s)} \to k$, $1 + \pi^s x
      \mapsto x \mod \pi$.
      \[
        (1 + \pi^s x)(1 + \pi^s y) = 1 + \pi^s(x +
        y + \pi^s xy)
      .\]
      $x + y + \pi^s xy \equiv x + y \bmod \pi$,
      hence $f$ is a group homomorphism,
      surjective with kernel $U_K^{(s + 1)}$.
  \end{enumerate}
\end{proof}

\begin{remark*}
  Let $[K : \Qp] < \infty$. \cref{prop_13_8},
  \cref{prop_13_9} implies that there exists
  finite index subgroup of $\O_K^\times$
  isomorphism to $(\O_K, +)$.
\end{remark*}

\begin{example*}
  $\Zp$, $p > 2$, $e = 1$, take $r = 1$. Then
  \begin{align*}
    \Zp^\times
    &\stackrel{\sim}{\to} (\Zbb / p\Zbb)^\times
    \times (1 + p\Zp) \cong \Zbb / (p - 1)\Zbb
    \times \Zp \\
    x
    &\mapsto \left( x \bmod p, \frac{x}{[x \bmod p]} \right)
  \end{align*}
  $p = 2$, take $r = 2$.
  \begin{align*}
    \Zbb_2^\times
    &\stackrel{\sim}{\to} (\Zbb / 4\Zbb)^\times
    \times (1 + p^2\Zp) \cong \Zbb / 2\Zbb
    \times \Zbb_2 \\
    x
    &\mapsto \left( x \bmod 4, \frac{x}{\eps(x)} \right)
  \end{align*}
  where
  \[
    \eps(x) = \begin{cases}
      +1 & x \equiv 1 \pmod 4 \\
      -1 & x \equiv -1 \pmod 4
    \end{cases}
  \]
  So:
  \[
    \Zp^\times / (\Zp^\times)^2
    \cong \begin{cases}
      \Zbb / 2\Zbb & \text{if $p > 2$} \\
      (\Zbb / 2\Zbb)^2 & \text{if $p = 2$}
    \end{cases}
  \]
\end{example*}

\newpage
\section{Higher Ramification Groups}

Let $L / K$ be a finite Galois extension of
\glspl{lf}, and $\pi_L \in \O_L$ a \gls{unif}.

\begin{fcdefn}[$s$-th ramification group]
  \label{defn_14_1}
  % Definition 14.1
  Let $v_L$ be a normalised valuation in $\O_L$.
  For $s \in \Rbb_{\ge -1}$, the $s$-th
  ramification group is
  \[
    G_s(L / K) = \{\sigma \in \Gal(K) \st
    v_L(\sigma(x) - x) \ge s + 1 ~\forall x \in
    O_L\}
  .\]
\end{fcdefn}

\begin{remark*}
  $G_s$ only changes at integers.

  $G_s$, $s \in \Rbb_{\ge -1}$ used to define
  upper numbering.
\end{remark*}

\begin{example*}
  \begin{align*}
    G_{-1}(L / K)
    &= \Gal(L / K) \\
    G_0(L / K)
    &= \{\sigma \in \Gal(L / K) \st \sigma(x)
    \equiv x \bmod \pi_L ~\forall x \in \O_L\} \\
    &= \ker(\Gal(L / K) \surjto \Gal(k_L / k)) \\
    &= \I_{L / K}
  \end{align*}
\end{example*}

\begin{note*}
  For $s \in \Zbb_{\ge 0}$,
  \[
    G_s(L / K) = \ker(\Gal(L / K) \surjto
    \Aut(\O_L / \pi_L^{s + 1} \O_L))
  \]
  hence $G_s(L / K)$ is normal in $G_{-1}$.
  \[
    \cdots \subseteq G_s \subseteq G_{s - 1}
    \subseteq \cdots \subseteq G_{-1} = \Gal(L / K)
  .\]
\end{note*}

\begin{theorem}
  \label{thm_14_2}
  % Theorem 14.2
  \phantom{}
  \begin{enumerate}[(i)]
    \item For $s \ge 1$,
      \[
        G_s = \{\sigma \in G_0 \st v_L(\sigma(\pi_L)
        - \pi_L) \ge s + 1\}
      .\]
    \item $\bigcap_{n = 0}^\infty G_n = \{1\}$.
    \item Let $s \in \Zbb_{\ge 0}$. Then there
      exists an injective group homomorphism
      \[
        G_s / G_{s + 1} \injto U_L^{(s)} / U_L^{(s
        + 1)}
      \]
      induced by $\sigma \mapsto
      \frac{\sigma(\pi_L)}{\pi_L}$. This map is
      independent of the choice of $\pi_L$.
  \end{enumerate}
\end{theorem}

\begin{proof}
  Let $K_0 \subseteq L$ be a maximal
  \gls{unramext} extension of $K$ in $L$. Upon
  replacing $K$ by $K_0$, we may assume that $L /
  K$ is \gls{totramext}.
  \begin{enumerate}[(i)]
    \item \cref{thm_13_8} implies $\O_L /
      \O_K[\pi_L]$. Suppose $v_L(\sigma(\pi_L) -
      \pi_L) \ge s + 1$. Let $x \in \O_L$, then $x
      = f(\pi_L)$, $f(X) \in \O_K[X]$.
      \begin{align*}
        \sigma(x) - x
        &= \sigma(f(\pi_L)) - f(\pi_L) \\
        &= f(\sigma(\pi_L)) - f(\pi_L) \\
        &= (\sigma(\pi_L) - \pi_L) g(\pi_L)
      \end{align*}
      for some $g(X) \in \O_K[X]$, using the fact
      that $X^n - Y^n = (X - Y)(X^{n - 1} +
      \cdots + Y^{n - 1})$. Thus
      \[
        v_L(\sigma(x) - x) = v_L(\sigma(\pi_L) -
        \pi_L) + \ub{v_L(g(\pi_L))}_{\ge 0} \ge s
        + 1
      .\]
    \item Suppose $\sigma \in \Gal(L / K)$,
      $\sigma \neq 1$. Then $\sigma(\pi_L) \neq
      \pi_L$, because $L = K(\pi_L)$ and hence
      $v_L(\sigma(\pi_L) - \pi_L) < \infty$. Thus
      $\sigma \notin G_s$ for some $s \gg 0$ by
      (i).
    \item Note: for $\sigma \in G_s$, $s \in
      \Zbb_{\ge 0}$,
      \[
        \sigma(\pi_L) \in \pi_L + \pi_L^{s + 1}
        \O_L
      \]
      hence
      \[
        \frac{\sigma(\pi_L)}{\pi_L} \in 1 +
        \pi_L^s \O_L = U_L^{(s)}
      .\]
      We claim
      \begin{align*}
        \varphi : G_s
        &\to U_L^{(s)} / U_L^{(s + 1)} \\
        \sigma
        &\mapsto \frac{\sigma(\pi_L)}{\pi_L}
      \end{align*}
      is a group homomorphism with kernel $G_{s +
      1}$. For $\sigma, \tau \in G_s$, let
      $\tau(\pi_L) = u\pi_L$, $u \in \O_L^\times$.
      Then
      \begin{align*}
        \frac{\sigma\tau(\pi_L)}{\pi_L}
        &= \frac{\sigma(\tau(\pi_L))}{\tau(\pi_L)}
        \frac{\tau(\pi_L)}{\pi_L} \\
        &= \frac{\sigma(u)}{u}
        \frac{\sigma(\pi_L)}{\pi_L}
        \frac{\tau(\pi_L)}{\pi_L}
      \end{align*}
      But $\sigma(u) \in u + \pi_L^{s + 1} \O_L$
      since $\sigma \in G_s$. Thus
      $\frac{\sigma(u)}{u} \in U_L^{(s + 1)}$ and
      hence
      \[
        \frac{\sigma\tau(\pi_L)}{\pi_L} \equiv
        \frac{\sigma(\pi_L)}{\pi_L} \cdot
        \frac{\tau(\pi_L)}{\pi_L}
        \bmod U_L^{(s + 1)}
      .\]
      Hence $\varphi$ is a group homomorphism.
      Moreover,
      \[
        \ker(\varphi) = \{\sigma \in G_s \st
        \sigma(\pi_L) \equiv \pi_L \bmod \pi_L^{s
        + 1}\} = G_{s + 1}
      .\]
      If $\pi_L' = a \pi_L$ is another \gls{unif},
      $a \in \O_L^\times$. Then
      \[
        \frac{\sigma(\pi_L')}{\pi_L'}
        = \frac{\sigma(a)}{a} \cdot
        \frac{\sigma(\pi_L)}{\pi_L}
        \equiv \frac{\sigma(\pi_L)}{\pi_L}
        \bmod U_L^{(s + 1)}
        .
        \qedhere
      \]
  \end{enumerate}
\end{proof}

\begin{corollary}
  \label{coro_14_3}
  % Corollary 14.3
  $\Gal(L / K)$ is solvable.
\end{corollary}