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\newpage
\part{Basic Theory}

\begin{example*}
  $f(x_1, \ldots, x_r) \in \Zbb[c_1, \ldots,
  x_r]$, $f(x_1, \ldots, x_r) = 0$? This is hard
  to study. It is easier to study
  \begin{align*}
    f(x_1, \ldots, x_r)
    &\equiv 0 \pmod{p} \\
    f(x_1, \ldots, x_r)
    &\equiv 0 \pmod{p^2} \\
    &~~\vdots \\
    f(x_1, \ldots, x_r)
    &\equiv 0 \pmod{p^n}
  \end{align*}
  A local field packages all this information
  together.
\end{example*}

\newpage
\section{Absolute values}

\begin{fcdefn}[Absolute value]
  % Definition 1.1
  \glsnoundefn{absval}{absolute value}{absolute
  values}%
  \glssymboldefn{absval}%
  \glsnoundefn{valf}{valued field}{valued fields}%
  Let $K$ be a field. An \emph{absolute value} on
  $K$ is a function $|\bullet| : K \to \Rbb_{\ge
  0}$ such that
  \begin{enumerate}[(i)]
    \item $|x| = 0$ if and only if $x = 0$.
    \item $|xy| = |x||y|$ for all $x, y \in K$.
    \item $|x + y| \le |x| + |y| ~\forall x, y \in
      K$ (triangly inequality).
  \end{enumerate}
  We say $(K, |\bullet|)$ is a \emph{valued
  field}.
\end{fcdefn}

\begin{example*}
  \phantom{}
  \begin{itemize}
    \item $K = \Qbb, \Rbb, \Cbb$ with usual absolute value
      $|a + ib| = \sqrt{a^2 + b^2}$. Write
      $|\bullet|_\infty$ for this \gls{absval}.
    \item $K$ any field. The trivial \gls{absval}
      is
      \[ \absval|x| = \begin{cases}
        0 & x = 0 \\
        1 & x \neq 0
      \end{cases} \]
      Although this is technically an
      \gls{absval}, it is not useful or
      interesting, so should be ignored.
  \end{itemize}
\end{example*}

\begin{fcdefn}[$p$-adic absolute value]
  % TODO: add star
  \glssymboldefn{absp}%
  Let $K = \Qbb$, and $p$ be a prime. For $0 \neq
  x \in \Qbb$, write $x = p^n \frac{a}{b}$, where
  $(a, p) = 1$, $(b, p) = 1$.
  The \emph{$p$-adic absolute value} is defined to
  be
  \[ |x|_p = \begin{cases}
    0 & x = 0 \\
    p^{-n} & x = p^n \frac{a}{b}
  \end{cases}
  \]
\end{fcdefn}

Verification:
\begin{enumerate}[(i)]
  \item Clear
  \item Write $y = p^m \frac{c}{d}$. Then
    \[ \absp|xy| = \absp|p^{m + n} \frac{ac}{bd}| =
    p^{-m - n} = \absp|x| \absp|y| .\]
  \item Without loss of generality, $m \ge n$.
    Then
    \[ \absp|x + y| = \absp|p^n \frac{ad + p^{m -
    n} bc}{bd}| \le p^{-n} = \max(\absp|x|,
    \absp|y|) .\]
\end{enumerate}

An \gls{absval} $\absval|\bullet|$ on $K$ induces
a metric $d(x, y) = \absval|x - y|$ on $K$, hence
a topology on $K$.

\begin{fcdefn}[Place]
  \glsnoundefn{place}{place}{places}
  Let $\absval|\bullet|$, $\absval|\bullet|'$ be
  absolute values on a field $K$. We say
  $\absval|\bullet|$ and $\absval|\bullet|'$ are
  \emph{equivalent} if they induce the same
  topology. An equivalence class of \glspl{absval}
  is called a \emph{place}.
\end{fcdefn}

\begin{fcprop}[]
  \label{prop_1_3}
  % Proposition 1.3
  Assuming:
  - $\absval|\bullet|$, $\absval|\bullet|'$ are
    (non-trivial) \glspl{absval} on $K$.
  Then:
  the following are equivalent:
  \begin{enumerate}[(i)]
    \item $\absval|\bullet|$ and
      $\absval|\bullet|'$ are equivalent.
    \item $\absval|x| < 1 \iff \absval|\bullet|' <
      1$ for all $x \in K$.
    \item There exists $c \in \Rbb_{> 0}$ such
      that $\absval|x|^c = \absval|\bullet|'$ for
      all $x \in K$.
  \end{enumerate}
\end{fcprop}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(iii) $\implies$]
    \item[(i) $\implies$ (ii)]
      \begin{align*}
        \absval|x| < 1
        &\iff x^n \to 0 \text{ w.r.t
        $\absval|\bullet|$} \\
        &\iff x^n \to 0 \text{ w.r.t
        $\absval|\bullet|'$} \\
        &\iff \absval|x|' < 1
      \end{align*}
    \item[(ii) $\implies$ (iii)] Note:
      $\absval|x|^c = \absval|x|' \iff c \log
      \absval|x| = \log \absval|x|'$. Let $a \in
      K^\times$ such that $\absval|a| > 1$ (exists
      since $\absval|\bullet|$ is non-trivial). We
      need that $\forall x \in K^\times$,
      \[ \frac{\log \absval|x|}{\log \absval|a|} =
      \frac{\log \absval|x|'}{\log \absval|a|'} .\]
      Assume that 
      \[ \frac{\log \absval|x|}{\log \absval|a|} <
      \frac{\log \absval|x|'}{\log \absval|a|'} .\]
      Choose $m, n \in \Zbb$ (with $n > 0$) such that
      \[ \frac{\log \absval|x|}{\log \absval|a|} <
      \frac{m}{n} < \frac{\log \absval|x|'}{\log
      \absval|a|'} .\]
      Then we have
      \begin{align*}
        n \log \absval|x|
        &< m \log \absval|a| \\
        n \log \absval|x|'
        &> m \log \absval|a|'
      \end{align*}
      Hence $\absval|\frac{x^n}{a^m}| < 1$ and
      $\absval|\frac{x^n}{a^m}| > 1$,
      contradiction. Similarly for the case where
      \[ \frac{\log \absval|x|}{\log \absval|a|} >
      \frac{\log \absval|x|'}{\log \absval|a|'} .\]
    \item[(iii) $\implies$ (i)] Clear.
  \end{enumerate}
\end{proof}

\begin{remark*}
  $|\bullet|_\infty^2$ on $\Cbb$ is not an
  \gls{absval} by our definition. Some authors
  replace the triangle inequality by
  \[ \absval|x + y|^\beta \le \absval|x|^\beta +
  \absval|y|^\beta \]
  for some fixed $\beta \in \Rbb_{> 0}$.
\end{remark*}

\begin{fcdefn}[Non-archimedean]
  % Definition 1.4
  \glsadjdefn{arch}{archimedean}{\gls{absval}}%
  \glsadjdefn{nonarch}{non-archimedean}{\gls{absval}}%
  An \gls{absval} $\absval|\bullet|$ on $K$ is
  said to be \emph{non-archimedean} if it
  satisfies the ultrametric inequality:
  \[ \absval|x + y| \le \max(\absval|x|,
  \absval|y|) .\]
  If $\absval|\bullet|$ is not non-archimedean,
  then it is archimedean.
\end{fcdefn}

\begin{example*}
  \phantom{}
  \begin{itemize}
    \item $\absval|\bullet|_\infty$ on $\Rbb$ is
      archimedean.
    \item $\absval|\bullet|_p$ is a \gls{nonarch}
      \gls{absval}.
  \end{itemize}
  
\end{example*}

\begin{fclemma}[]
  \label{lemma_1_5}
  % Lemma 1.5
  Assuming:
  - $(K, \absval|\bullet|)$ is \gls{nonarch}
  - $x, y \in K$
  - $\absval|x| < \absval|y|$
  Then:
  $\absval|x - y| = \absval|y|$.
\end{fclemma}

\begin{proof}
  \[ \absval|x - y| \le \max(\absval|x|,
  \absval|y|) = \absval|y| \]
  and
  \[ \absval|y| \le \max(\absval|x|, \absval|x -
  y|) \le \absval|x - y| .\]
\end{proof}

\begin{fcprop}[]
  Assuming:
  - $(K, \absval|\bullet|)$ is \gls{nonarch}
  - $(x_n)_{n = 1}^\infty$ a sequence in $K$
  - $\absval|x_n - x_{n + 1}| \to 0$
  Then:
  $(x_n)_{n = 1}^\infty$ is Cauchy. In particular,
  if $K$ is in addition complete, then $(x_n)_{n =
  1}^\infty$ converges.
\end{fcprop}

\begin{proof}
  For $\eps > 0$, choose $N$ such that
  $\absval|x_n - x_{n + 1}| < \eps$ for $n > N$.
  Then $N < n < m$,
  \[ \absval|x_n - x_m| = \absval|(x_n - x_{n +
  1}) + \cdots + (x_{n - 1}) - x_m)| < \eps .\]
  The ``In particular'' is clear.
\end{proof}

\begin{example*}
  $p = 5$, construct sequence $(x_n)_{n =
  1}^\infty$ in $\Zbb$ such that
  \begin{enumerate}[(i)]
    \item $x_n^2 + 1 \equiv 0 \pmod{5^n}$
    \item $x_n \equiv x_{n + 1} \pmod{5^n}$
  \end{enumerate}
  Take $x_1 = 2$. Suppose we have constructed
  $x_n$. Let $x_n^2 + 1 = a5^n$ and set $x_{n + 1}
  = x_n + b5^n$. Then
  \begin{align*}
    x_{n + 1}^2 + 1
    &= x_n^2 + 2b x_n 5^n + b^2 5^{2n} + 1 \\
    &= a 5^n + 2b x_n 5^n + b^2 5^{2n}
  \end{align*}
  We choose $b$ such that $a + 2b x_n \equiv 0
  \pmod{5}$. Then we have $x_{n + 1}^2 + 1 \equiv
  0 \pmod{5^{n + 1}}$. Now (ii) implies that
  $(x_n)_{n = 1}^\infty$ is Cauchy. Suppose $x_n
  \to l \in \Qbb$. Then $x_n^2 \to l^2$. But (i)
  tells us that $x_n^2 \to -1$, so $l^2 = -1$, a
  contradiction. Thus $(\Qbb, \absval|\bullet|_5)$
  is not complete.
\end{example*}

\begin{fcdefn}[]
  % Definition 1.7
  \glssymboldefn{Qp}%
  The $p$-adic numbers $\Qbb_p$ is the completion
  of $\Qbb$ with respect to $\absp|\bullet|$.
\end{fcdefn}

Analogy with $\Rbb$:
\begin{center}
  \includegraphics[width=0.2\linewidth]{images/9f2392abf2324852.png}
\end{center}
