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\section{Extensions of complete valued fields}

\begin{fcthm}[]
  \label{thm_6_1}
  % Theorem 6.1
  Assuming:
  - $(K, \absval|\bullet|)$ is a complete
  \gls{dvf}
  - $L / K$ a finite extension of degree $n$
  Then:
  \begin{enumerate}[(i)]
    \item $\absval|\bullet|$ extends uniquely to
      an \gls{absval} $\absval|\bullet|_L$ on $L$
      defined by
      \[
        \absval|y|_L = \absval|N_{L / L}(y)|^{\frac{1}{n}}
        \qquad \forall y \in L
      .\]
    \item $L$ is complete with respect to
      $\absval|\bullet|_L$.
  \end{enumerate}
\end{fcthm}

\textbf{Recall:} If $L / K$ is finite, $N_{L / K}
: L \to K$ is defined by $N_{L / K}(y) =
\det_K(\mathrm{mult}(y)$ where $\mathrm{mult}(y) :
L \to L$ is the $K$-linear map induced by
multiplication by $y$.

\textbf{Facts:}
\begin{itemize}
  \item $N_{L / K}$ is multiplicative.
  \item Let $X^n + a_{n - 1} X^{n - 1} + \cdots +
    a_0 \in K[X]$ be the minimal polynomial of $y
    \in L$. Then $N_{L / K}(y) = \pm a_0^m$ for
    some $m \ge 1$ (in fact, $m$ is the degree of
    $L / K[y]$).
  \item $N_{L / K}(y) = 0 \iff y = 0$.
\end{itemize}

\begin{fcdefn}[Norm]
  \glsnoundefn{norm}{norm}{norms}%
  \glssymboldefn{norm}%
  Let $(K, \absval|\bullet|)$ be a \gls{nonarch}
  \gls{valf}, $V$ a vector space over $K$. A
  \emph{norm}on $V$ is a function $\|\bullet\| : V
  \to \Rbb_{\ge 0}$ satisfying:
  \begin{enumerate}[(i)]
    \item $\|x\| = 0 \iff x = 0$.
    \item $\|\lambda x\| = \|\lambda\| \|x\|$ for
      all $\lambda \in K$, $x \in V$.
    \item $\|x + y\| \le \max(\|x\|, \|y\|)$ for
      all $x, y \in V$.
  \end{enumerate}
\end{fcdefn}

\begin{example*}
  If $V$ is finite dimensional and $e_1, \ldots,
  e_n$ is a basis of $V$. The supremum
  $\norm|\bullet|_{\sup}$ on $V$ is defined by
  \[
    \norm|x|_{\sup} = \max_i |x_i|
  ,\]
  where $x = \sum_{i = 1}^n x_i e_i$.

  \textbf{Exercise:} $\norm|\bullet|_{\sup}$ is a
  \gls{norm}.
\end{example*}

\begin{fcdefn}[Equivalent norms]
  \glsadjdefn{nequ}{equivalent}{norms}%
  Two \glspl{norm} $\norm|\bullet|_1$ and
  $\norm|\bullet|_2$ on $V$ are equivalent if
  there exists $C, D \in \Rbb_{> 0}$ such that
  \[
    C\norm|x|_1 \le \norm|x|_2 \le D\norm|x|_1
    \qquad \forall x \in V
  .\]
\end{fcdefn}

\textbf{Fact:} A \gls{norm} defines a topology on
$V$, and equivalent \glspl{norm} induce the same
topology.

\begin{fcprop}[]
  \label{prop_6_4}
  % Proposition 6.4
  Assuming:
  - $(K, \absval|\bullet|)$ is a complete
  \gls{nonarch} \gls{valf}
  - $V$ a finite dimensional vector space over $K$
  Then:
  $V$ is complete with respect to
  $\norm|\bullet|_{\sup}$.
\end{fcprop}

\begin{proof}
  Let $(v_i)_{i = 1}^{\infty}$ be a Cauchy
  sequence in $V$, and let $e_1, \ldots, e_n$ be a
  basis for $V$.

  Write $v_i = \sum_{j = 1}^{n} x_j^i e_j$. Then
  $(x_j^i)_{i = 1}^{\infty}$ is a Cauchy sequence
  in $K$. Let $x_j^i \to x_j \in K$, then $v_i \to
  v \defeq \sum_{j = 1}^{n}  x_j e_j$.
\end{proof}

\begin{fcthm}[]
  \label{thm_6_5}
  % Theorem 6.5
  Assuming:
  - $(K, \absval|\bullet|)$ is a complete
  \gls{nonarch} \gls{valf}
  - $V$ a finite dimensional vector space over $K$
  Then:
  any two \glspl{norm} on $K$ are equivalent. In
  particular, $V$ is complete with respect to any
  \gls{norm} (using \cref{prop_6_4}).
\end{fcthm}

\begin{proof}
  Since equivalence defines an equivalence
  relation on the set of \glspl{norm}, it suffices
  to show that any \gls{norm} $\norm|\bullet|$ is
  equivalent to $\norm|\bullet|_{\sup}$.

  Let $e_1, \ldots, e_n$ be a basis for $V$, and
  set $D \defeq \max_i \norm|e_i| > 0$. Then for
  $x = \sum_{i = 1}^n x_i e_i$, we have
  \[
    \norm|x| \le \max_i \norm|x_i e_i| = \max_i
    \absval|x_i| \norm|e_i| \le D \max_i \absval|x_i| =
    D\norm|x|_{\sup}
  .\]
  To find $C$ such that $C\norm|\bullet|_{\sup}
  \le \norm|\bullet|$, we induct on $n = \dim V$.

  For $n = 1$: $\norm|x| = \norm|x_1 e_1| =
  \absval|x_1| \norm|e_1|$, so take $C =
  \norm|e_1|$.

  For $n > 1$: set $V_i = \Span\langle e_1,
  \ldots, e_{i - 1}, e_{i + 1}, \ldots, e_n
  \rangle$. By induction, $V_i$ is complete with
  respect to $\norm|\bullet|$, hence closed.

  Then $e_i + V_i$ is closed for all $i$, and
  hence
  \[
    S \defeq \bigcup_{i = 1}^n e_i + V_i
  \]
  is a closed subset not containing $0$. Thus
  there exists $c > 0$ such that $B(0, C) \cap S =
  \emptyset$ where $B(0, C) = \{x \in V \st
  \norm|x| < C\}$.

  Let $0 \neq x = \sum_{i = 1}^{n} x_i e_i$ and
  suppose $\absval|x_j| = \max_i \absval|x_i|$.
  Then $\norm|x|_{\sup} = |x_j|$, and
  $\frac{1}{x_j} \in S$. Thus $\left\|
  \frac{x_i}{x_j} \right\| \ge C$, and hence
  \[
    \norm|x| \ge C\absval|x_j| = C\norm|x|_{\sup}
  .\]
  $V$ is complete since it is complete with
  respect to $\norm|\bullet|_{\sup}$ (see
  \cref{prop_6_4}).
\end{proof}

\begin{fcdefn}[Integral closure]
  \glsadjdefn{int}{integral}{element}%
  \glsnoundefn{intc}{integral closure}{integral
  closures}%
  \glsadjdefn{intlyc}{integrally closed}{subring}%
  \glssymboldefn{int}%
  % Definition 6.6
  Let $R$ be a subring of $S$. We say $s \in S$ is
  \emph{integral} over $R$ if there exists a monic
  polynomial $f(X) \in R[X]$ such that $f(s) = 0$.

  The \emph{integral closure}
  $R^{\mathrm{int}(S)}$ of $R$ inside $S$ is
  defined to be
  \[
    R^{\mathrm{int}(S)} = \{s \in S \st \text{$s$
    integral over $R$}\}
  .\]
  
  We say $R$ is \emph{integrally closed} in $S$ if
  $R^{\mathrm{int}(S)} = R$.
\end{fcdefn}

\begin{proposition}[]
  % Proposition 6.7
  $R\intc S$ is a subring of $S$. Moreover, $R
  \intc S$ is \gls{intlyc} in $S$.
\end{proposition}

\begin{proof}
  Example Sheet 2.
\end{proof}

\begin{fclemma}[]
  \label{lemma_6_8}
  % Lemma 6.8
  Assuming:
  - $(K, \absval|\bullet|)$  is \gls{nonarch}
  \gls{valf}
  Then:
  $\O_K$ is \gls{intlyc} in $K$.
\end{fclemma}

\begin{proof}
  Let $x \in K$ be \gls{int} over $\O_K$. Without
  loss of generality, $x \neq 0$. Let $f(X) = X^n
  + a_{n - 1} X^{n - 1} + \cdots + a_0 \in
  \O_K[X]$ such that $f(x) = 0$. Then
  \[
    x = -a_{n - 1} \frac{1}{x} - \cdots - a_0
    \frac{1}{x_{n - 1}}
  .\]
  If $\absval|x| > 1$, we have $\left| -a_{n - 1}
  \frac{1}{x} - \cdots - a_0 \frac{1}{x_{n -
  1}}\right| < 1$. Thus $\absval|x| \le 1 \implies
  x \in \O_K$.
\end{proof}

\begin{lemma}
  \label{lemma_6_9}
  % Lemma 6.9
  $\O_L$ is the \gls{intc} of $\O_K$ inside $L$.
\end{lemma}

\begin{proof}
  Let $0 \neq y \in L$ and let
  \[
    f(X) = X^d + a_{d - 1} X^{d - 1} + \cdots +
    a_0 \in K[X]
  \]
  be the minimal (monic) polynomial of $y$.

  \textbf{Claim:} $y$ integral over $\O_K$ if and
  only if $f(X) \in \Qbb_K[X]$.

  \begin{enumerate}[$\Rightarrow$]
    \item[$\Rightarrow$] Clear.
    \item[$\Leftarrow$] Let $g(X) \in \O_K[X]$
      monic such that $g(y) = 0$. Then $f \mid g$
      (in $K[X]$), and hence every root of $f$ is
      a root of $g$. So every root of $f$ in $\ol
      K$ is \gls{int} over $\O_K$, so $a_i$ are
      \gls{int} over $\O_K$ for $i = 0, \ldots , d
      - 1$.
  \end{enumerate}
  Hence $a_i \in \O_k$ (by
  \cref{lemma_6_8}). By \cref{coro_4_5},
  $|a_i| \le \max(|a_0|, 1)$ for $i = 0,
  \ldots, d - 1$. By property of $N_{L / K}$,
  we have $N_{L / K}(y) = \pm a_0^m$ for $m
  \ge 1$.

  Hence
  \begin{align*}
    y \in \O_L
    &\iff |N_{L / K}(y)| \le 1 \\
    &\iff |a_0| \le 1 \\
    &\stackrel{\cref{coro_4_5}}{\iff} |a_i| \le 1
    \qquad \forall i, \text{i.e. $a_i \in \O_K$}
  \end{align*}
  Thus $\O_K\intc L = \O_L$ and proves the Lemma.
\end{proof}


\begin{proof}[Proof of \cref{thm_6_1}]
  We first show $\absval|\bullet|_L = \absval|N_{L /
  K}(\bullet)|^{\frac{1}{n}}$ satisfies the three
  axioms in the definition of \gls{absval}.
  \begin{enumerate}[(i)]
    \item \phantom{} \\[-2.5\baselineskip]
      \begin{align*}
        \absval|y|_L = 0
        &\iff \absval|N_{L / K}(y)|^{\frac{1}{n}} = 0 \\
        &\iff N_{L / K}(y) = 0 \\
        &\iff y = 0
      \end{align*}
    \item \phantom{} \\[-2.5\baselineskip]
      \begin{align*}
        \absval|y_1 y_2|_L
        &= \absval|N_{L / K}(y_1,
        y_2)|^{\frac{1}{n}} \\
        &= \absval|N_{L / K}(y_1)
        N_{L / K}(y_2)
        |^{\frac{1}{n}} \\
        &= \absval|N_{L / K}(y_1)
        |^{\frac{1}{n}}
        \absval|N_{L / K}(y_2)
        |^{\frac{1}{n}}
        \\
        &= \absval|y_1|_L \absval|y_2|_L
      \end{align*}
    \item Set $\O_L = \{y \in L \st \absval|y|_L
      \le 1\}$.

      \textbf{Claim:} $\O_L$ is the \gls{intc} of
      $\O_K$ inside $L$.

      Assuming this, we prove (iii). Let $x, y \in
      L$, and without loss of generality assume
      $\absval|x|_L \le \absval|y|_L$. Then
      $\left| \frac{x}{y} \right|_L$ hence
      $\frac{x}{y} \in \O_L$. Since $1 \in \O_L$
      and $\O_L$is a ring, we have $1 +
      \frac{x}{y} \in \O_L$ and hence $\left| 1 +
      \frac{x}{y}\right|_L \le 1$. Hence
      $\absval|x + y|_L \le \absval|y|_L =
      \max(\absval|x|_L, \absval|y|_L)$ thus (iii)
      is satisfied.
  \end{enumerate}
