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\begin{fcthm}[Ostrowski's Theorem]
  \label{thm_7_9}
  % Theorem 7.9
  Assuming:
  - $\absval|\bullet|$ is a non-trivial
  \gls{absval} on $\Qbb$
  Then:
  $\absval|\bullet|$ is equivalent to either the
  usual \gls{absval} $\absval|\bullet|_\infty$ or
  the $p$-adic \gls{absval} $\absp|\bullet|$ for
  some prime $p$.
\end{fcthm}

\begin{proof}
  \textbf{Case:} $\absval|\bullet|$ is \gls{arch}.
  We fix $b > 1$ an integer such that $|b| > 1$
  (exists by \cref{lemma_7_8}). Let $a > 1$ be an
  integer and write $b^n$ in base $a$:
  \[
    b^n = c_m a^m + c_{m - 1} a^{m - 1} + \cdots +
    c_0
  \]
  with $0 \le c_i < a$, $c_m \neq 0$. Let $B =
  \max_{0 \le c < a - 1} (|c|)$, and then we have
  \begin{align*}
    |b^n|
    &\le (m + 1)B \max(|a|^m, 1) \\
    \implies |b|
    &\le \ub{[n (\log_a b + 1)B]^{1 /
    n}}_{\to 1} \max(|a|^{\log_a b}, 1)
    &&m \le \log_a b^n \\
    \implies
    |b|
    &\le \max(|a|^{\log_a b}, 1)
  \end{align*}
  Then $\absval|a| > 1$ and
  \[
    \absval|b| \le \absval|a|^{\log_a b}
    \tag{$*$} \label{9star}
  .\]
  Switching roles of $a$ and $b$, we also obtain
  \[
    \absval|a| \le \absval|b|^{\log_b a}
    \tag{$**$} \label{9sstar}
  .\]
  Then \eqref{9star} and \eqref{9sstar} gives
  (using $\log_a b = \frac{\log b}{\log a}$):
  \[
    \frac{\log\absval|a|}{\log a} =
    \frac{\log\absval|b|}{\log b} = \lambda \in
    \Rbb_{> 0}
  .\]
  Hence $\absval|a| = a^\lambda$ for all $a \in
  \Zbb_{> 1}$, hence $\absval|x| =
  |x|_\infty^\lambda$ for all $x \in \Qbb$.

  \textbf{Case 2:} $\absval|\bullet|$ is
  \gls{nonarch}. As in \cref{lemma_7_8}, we have
  $|n| \le 1$ for all $n \in \Zbb$. Since
  $\absval|\bullet|$ is non-trivial, there exists
  $n \in \Zbb_{> 1}$ such that $\absval|n| < 1$.
  Write $n = p_1^{e_1} \cdots p_r^{e_r}$
  decomposition into prime factors. Then
  $\absval|p| < 1$, for some $p \in \{p_1, \ldots,
  p_r\}$. Suppose $\absval|q| < 1$ for some prime
  $q$, $q \neq p$. Write $1 = rp + sq$ with $r, s
  \in \Zbb$. Then
  \begin{align*}
    1
    &= \absval|rp + sq| \\
    &\le \max(\absval|rp|, \absval|sq|) \\
    &< 1
  \end{align*}
  contradiction. Thus $\absval|p| = \alpha < 1$
  and $\absval|q| = 1$ for all primes $q \neq p$.
  Hence $\absval|\bullet|$ is equivalent to
  $\absp|\bullet|$.
\end{proof}

\begin{fcthm}[]
  % Theorem 7.10
  Assuming:
  - $(K, \absval|\bullet|)$ is a \gls{nonarch}
  \gls{lf} of \gls{mixc}
  Then:
  $K$ is a finite extension of $\Qp$.
\end{fcthm}

\begin{proof}
  $K$ \gls{mixc} implies that $\char K = 0$, hence
  $\Qbb \subseteq K$. $K$ \gls{nonarch} implies
  that $\absval|\bullet| \big|_{\Qbb} =
  \absval|\bullet|_p$ for some prime $p$. Since
  $K$ is complete, $\Qp \subseteq K$. Suffices to
  show that $\O_K$ is finite as a $\Zp$-module.

  Let $\pi \in \O_K$ be a \gls{unif}, $v$
  a normalised \gls{valt} and set $v(p) = e$. Then
  $\O_K / p\O_K \cong \O_K / \pi^e \O_K$ is finite
  since $\pi^i \O_K / \pi^{i + 1} \O_K \cong \O_K
  / \pi\O_K$ is finite. Since $\Fp \cong \Zbb /
  p\Zbb \hookrightarrow \O_K / p\O_K$ we have $\O_K / p\O_K$ a
  finite dimensional vector space over $\Fp$.

  Let $x_1, \ldots, x_n \in \O_K$ be coset
  representatives for $\Fp$-basis of $\O_K /
  p\O_K$. Then
  \[
    \left\{ \sum_{i = 1}^{n} a_i x_i \Bigg| a_i
    \in \{0, \ldots, r - 1\} \right\}
  \]
  is a set of coset representatives for $\O_K /
  p\O_K$. Let $y \in \O_K$. \cref{prop_3_5}(ii)
  tells us that
  \begin{align*}
    y
    &= \sum_{i = 0}^{\infty} \left( \sum_{j =
    1}^{n} a_{ij} x_j \right) p^i
    &&(a_{ij} \in \{0, \ldots, p - 1\}) \\
    &= \sum_{j = 1}^{n} \left( \sum_{i =
    0}^{\infty} a_{ij} p^i \right) x_j \\
    &\in \Zp
  \end{align*}
  Hence $\O_K$ is finite over $\Zp$.
\end{proof}

On Example Sheet 2 we will show that if $K$ is
complete and \gls{arch}, then $K \simeq \Rbb$ or
$\Cbb$. In summary:

If $K$ a \gls{lf}, then either:
\begin{enumerate}[(i)]
  \item $K \cong \Rbb$ or $\Cbb$ (\gls{arch})
  \item $K \cong \Fbb_{p^n}((t))$ (\gls{nonarch}
    \gls{eqc})
  \item $K$ a finite extension of $\Qp$
    (\gls{nonarch} \gls{mixc})
\end{enumerate}

\newpage
\section{Global Fields}

\begin{fcdefn}[Global field]
  \glsnoundefn{gf}{global field}{global fields}%
  % Definition 8.1
  A \emph{global field} is a field which is
  either:
  \begin{enumerate}[(i)]
    \item An algebraic number field
    \item A global function field, i.e. a finite
      extension of $\Fp(t)$.
  \end{enumerate}
\end{fcdefn}

\begin{fclemma}[]
  \label{lemma_8_2}
  % Lemma 8.2
  Assuming:
  - $(K, \absval|\bullet|)$ is a complete \gls{dvf}
  - $L / K$ a finite Galois extension with
  \gls{absval} $\absval|\bullet|_L$ extending
  $\absval|\bullet|$.
  Then:
  for $x \in L$ and $\sigma \in \Gal(L / K)$, we
  have $\absval|\sigma(x)|_L = \absval|x|_L$.
\end{fclemma}

\begin{proof}
  Since $x \mapsto \absval|\sigma(x)|_L$ is
  another \gls{absval} on $L$ extending
  $\absval|\bullet|$ on $K$, the result follows
  from uniqueness of $\absval|\bullet|_L$.
\end{proof}

\begin{fclemma}[Kummer's Lemma]
  \label{kummer}
  % Lemma 8.3
  Assuming:
  - $(K, \absval|\bullet|)$ a complete \gls{dvf}
  - $f(X) \in K[X]$ a separable irreducible
  polynomial with roots $\alpha_1, \ldots,
  \alpha_n \in K^{\text{sep}}$ ($K^{\text{sep}}$
  is the separable closure of $K$)
  - $\beta \in K^{\text{sep}}$ with
  \[
    |\beta - \alpha_1| < |\beta - \alpha_i|
  \]
  for $i = 2, \ldots, n$.
  Then:
  $\alpha_1 \in K(\beta)$.
\end{fclemma}

\begin{proof}
  Let $L = K(\beta)$, $L' = L(\alpha_1, \ldots,
  \alpha_n)$. Then $L' / L$ is a Galois extension.
  Let $\sigma \in \Gal(L' / L)$. We have
  \begin{align*}
    \absval|\beta - \sigma(\alpha_1)|
    &= \absval|\sigma(\beta - \alpha_1)| \\
    &= \absval|\beta - \alpha_1|
  \end{align*}
  using \cref{lemma_8_2}. Hence $\sigma(\alpha_1)
  = \alpha_1$, so $\alpha_1 \in K(\beta)$.
\end{proof}

\begin{fcprop}[]
  \label{prop_8_4}
  % Proposition 8.4
  Assuming:
  - $(F, \absval|\bullet|)$ is a complete \gls{dvf}
  - $f(X) = \sum_{i = 0}^{n} a_i X^i \in \O_K[X]$
  a separable irreducible monic polynomial
  - $\alpha \in K\sep$ a root of $f$
  Then:
  there exists $\eps > 0$ such that for any $g(X)
  = \sum_{i = 0}^{n} b_i X^i \in \O_K[X]$ monic
  with $|a_i - b_i| < \eps$ for all $i$, there
  exists a root $\beta$ of $g(X)$ such that
  $K(\alpha) = K(\beta)$.
\end{fcprop}

``Nearby polynomials define the same extensions''.

\begin{proof}
  Let $\alpha_1, \ldots, \alpha_n \in K\sep$ be
  the roots of $f$ which are necessarily distinct.
  Then $f'(\alpha_1) \neq 0$. We choose $\eps$
  sufficiently small such that $|g(\alpha_1)| <
  |f'(\alpha_1)|^2$ and $|f'(\alpha_1) -
  g'(\alpha_1)| < |f'(\alpha_1)|$. Then we have
  $|g'(\alpha_1)| < |f'(\alpha_1)|^2 =
  |g'(\alpha_1)|^2$ (the equality is by
  \cref{lemma_1_5}).

  By \nameref{hensel} applied to the field
  $K(\alpha_1)$ there exists $\beta \in
  K(\alpha_1)$ such that $g(\beta) = 0$ and
  $\absval|\bullet - \alpha_1| <
  \absval|g'(\alpha_1)|$. Then
  \begin{align*}
    \absval|g'(\alpha_1)|
    &= \absval|f'(\alpha_1)| \\
    &= \prod_{j = 1}^{n} \absval|\alpha_1 -
    \alpha_j| \\
    &\le \absval|\alpha_1 - \alpha_i|
  \end{align*}
  for $i = 2, \ldots, n$. (Use $\absval|\alpha_1 -
  \alpha_i| \le 1$ since $\alpha_i$ integral).
  Since $\absval|\beta - \alpha_1| <
  \absval|\alpha_1 - \alpha_i| = \absval|\beta -
  \alpha_i|$ using \cref{lemma_1_5}, we have that
  \nameref{kummer} gives that $\alpha_1 \in
  K(\beta)$ and hence $K(\alpha_1) = K(\beta)$.
\end{proof}
