%! TEX root = LF.tex
% vim: tw=50
% 24/10/2024 12PM

  So we have proved that $|\bullet|_L$ is an
  \gls{absval} on $L$.

  Since $N_{L / K}(x) = x^n$ for $x \in K$,
  $|x|_L$ extends $|\bullet|$ on $K$.

  If $\absval|\bullet|'_L$ is another \gls{absval}
  on $L$ extending $\absval|\bullet|$, then
  $|\bullet|_L, |\bullet|'_L$ are \glspl{norm} on
  $L$.

  \cref{thm_6_5} tells us that $|\bullet|_L',
  |\bullet|_L$ induce the same topology on $L$.
  Hence $|\bullet|_L' = |\bullet|_L^c$ for some $c
  > 0$ (by \cref{prop_1_3}) since $|\bullet|_L'$
  extends $|\bullet|$, we have $c = 1$.

  Now we show that $L$ is complete with respect to
  $|\bullet|_L$: this is immediate by
  \cref{thm_6_5}.
\end{proof}

Let $(K, |\bullet|)$ be a complete \gls{dvf}.

\begin{corollary}
  % Corollary 6.10
  Let $L / K$ be a finite extension. Then
  \begin{enumerate}[(i)]
    \item $L$ is \gls{dlyv} with respect to
      $|\bullet|_L$.
    \item $\O_L$ is the integral closure of $\O_K$
      in $L$.
  \end{enumerate}
\end{corollary}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $v$ a \gls{valt} on $K$, $v_L$ \gls{valt}
      on $L$ such that $v_L$ extends $v$. Let $n =
      [L : K]$, and let $y \in L^\times$. Then
      $|y|_L = |N_{L / K}(y)|^{\frac{1}{n}}$ hence
      $v_L(y) = \frac{1}{n} v(N_{L / K}(y))$,
      hence $v_L(L^\times) \le
      \frac{1}{n}v(K^\times)$, so $v_L$ is
      \gls{disc}.
    \item \cref{lemma_6_9}.
  \end{enumerate}
\end{proof}

\begin{corollary}
  % Corollary 6.11
  Let $\ol K / K$ be an algebraic closure of $K$.
  Then $|\bullet|$ extends to a unique absolute
  value $|\bullet|_{\ol K}$ on $\ol K$.
\end{corollary}

\begin{proof}
  Let $x \in \ol K$, then $x \in L$ for some $L /
  K$ finite. Define $|x|_{\ol K} = |x|_L$.
  Well-defined, i.e. independent of $L$ by the
  uniqueness in \cref{thm_6_1}.

  The axioms for $|\bullet|_{\ol K}$ to be an
  \gls{absval} can be checked over finite
  extensions.

  Uniqueness: clear.
\end{proof}

\begin{remark*}
  $|\bullet|_{\ol K}$ on $\ol K$ is never
  \gls{disc}. For example $K = \Qp$, $\sqrt[n]{p}
  \in \ol{\Qp}$ for all $n \in \Zbb_{> 0}$. Then
  \[
    v_p(\sqrt[n]{p}) = \frac{1}{n} v(p) =
    \frac{1}{n}
  .\]
  $\ol{\Qp}$ is not complete with respect to
  $|\bullet|_{\Qp}$.

  Example Sheet 2: $\Cbb_p \defeq$ completion of
  $\ol{\Qp}$ with respect to
  $|\bullet|_{\ol{\Qp}}$, then $\Cbb_p$ is
  algebraically closed.
\end{remark*}

\begin{fcprop}[]
  % Proposition 6.12
  Assuming:
  - $L / K$ finite extension of complete
  \glspl{dvf}.
  - (i): $\O_K$ is compact.
  - (ii): The extension of residue fields $k_L /
  k$ is finite and separable.
  Then:
  there exists $\alpha  \in \O_L$ such that $\O_L
  = \O_K[\alpha]$.
\end{fcprop}

Later we'll prove that the (i) implies (ii).

\begin{proof}
  We'll choose $\alpha \in \O_L$ such that:
  \begin{itemize}
    \item there exists $\beta \in \O_L[\alpha]$ a
      \gls{unif} for $\O_L$
    \item $\O_K[\alpha] \to k_L$ surjective
  \end{itemize}
  $k_L / k$ separable tells us that there exists
  $\ol\alpha \in k_L$ such that $k_L =
  k(\ol{\alpha})$.

  Let $\alpha \in \O_L$ a lift of $\ol \alpha$,
  and $g(X) \in \O_K[X]$ a monic lift of the
  minimal polynomial of $\ol \alpha$.

  Fix $\pi_L \in \O_L$ a \gls{unif}. Then
  $\ol{g}(X) \in k[X]$ irreducible and separable,
  hence $g(\alpha) \equiv 0 \bmod \pi_L$ and
  $g'(\alpha) \not\equiv 0 \bmod \pi_L$.

  If $g(\alpha) \equiv 0 \bmod \pi_L^2$, then
  \let\cancel\relax
  \[
    g(\alpha + \pi_L) \equiv \cancel{g(\alpha)} +
    \pi_L g'(\alpha) \bmod \pi_L^2
  .\]
  Thus
  \[
    v_L(g(\alpha + \pi_L)) = v_L(\pi_L g'(\alpha))
    = v_L(\pi_L) = 1
  .\]
  ($v_L$ normalised \gls{valt} on $L$).

  Thus either $v_L(g(\alpha)) = 1$ or
  $v_L(g(\alpha + \pi_L)) = 1$. Upon possibly
  replacing $\alpha$ by $\alpha + \pi_L$, we may
  assume $v_L(g(\alpha)) = 1$.

  Set $\beta = g(\alpha) \in \O_K[\alpha]$ a
  \gls{unif}. Then $\O_K[\alpha] \subseteq L$ is
  the image of a continuous map:
  \begin{align*}
    \O_K^n
    &\to L \\
    (x_0,\ldots, x_{n - 1})
    &\mapsto \sum_{i = 0}^{n} x_i a^i
  \end{align*}
  where $n = [K(\alpha) : K]$. Since $\O_K$ is
  compact, $\O_K[\alpha] \subseteq L$ is compact,
  hence closed. Since $k_L = k(\ol \alpha)$,
  $\O_K[\alpha]$ contians a set of coset
  representatives for $k_L = \O_L / \beta \O_L$.

  Let $y \in \O_L$. Then \cref{prop_3_5} gives us
  \[
    y = \sum_{i = 0}^{\infty}  \lambda_i \beta^i,
    \qquad \lambda_i \in \O_K[\alpha]
  \]
  Then $y_m = \sum_{i = 0}^{m} \lambda_i \beta^i
  \in \O_K[\alpha]$. Hence $y \in \O_K[\alpha]$,
  since $\O_k[\alpha]$ is closed.
\end{proof}

\newpage
\part{Local Fields}

\newpage
\section{Local Fields}

\begin{fcdefn}[Local field]
  \glsnoundefn{lf}{local field}{local fields}%
  % Definition 7.1
  Let $(K, \absval|\bullet|)$ be a \gls{valf}.
  Then $K$ is a \emph{local field} if it is
  complete and locally compact.
\end{fcdefn}

Reminder: locally compact means for all $x \in K$,
there exists $U$ open and $V$ compact such that $x
\in U \subseteq V$.

\begin{example*}
  $\Rbb$ and $\Cbb$ are compact.
\end{example*}

\begin{fcprop}[]
  % Propositoin 7.2
  Assuming:
  - $(K, \absval|\bullet|)$ is a \gls{nonarch}
  complete \gls{valf}
  Then:
  the following are equivalent:
  \begin{enumerate}[(i)]
    \item $K$ is locally compact
    \item $\O_K$ is compact
    \item $v$ is discrete and $k = \O_K / m$ is
      finite.
  \end{enumerate}
\end{fcprop}
