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\begin{notation*}
  As is usual when working with metric spaces, we
  will be using the notation:
  \begin{align*}
    B(x, r)
    &= \{y \in K \st |x - y| < r\} \\
    \ol{B}(x, r)
    &= \{y \in K \st |x - y| \le r\}
  \end{align*}
\end{notation*}

\begin{fclemma}[]
  % Lemma 1.8
  Assuming:
  - $(K, \absval|\bullet|)$ is a \gls{nonarch}
  \gls{valf}
  Then:
  \begin{enumerate}[(i)]
    \item If $z \in B(x, r)$, then $B(z, r) = B(x,
      r)$ -- so open balls don't have a centre.
    \item If $z \in \ol{B}(x, r)$ then $\ol{B}(x,
      r) = \ol{B}(z, r)$.
    \item $B(x, r)$ is closed.
    \item $\ol{B}(x, r)$ is open.
  \end{enumerate}
\end{fclemma}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Let $y \in B(x, r)$. Then $|x - y| < r$
      hence
      \begin{align*}
        |z - y|
        &= |(z - x) + (x - y)| \\
        &\le \max(|z - x|, |x - y|) \\
        &< r
      \end{align*}
      Thus $B(x, r) \subseteq B(z, r)$.
      $\supseteq$ follows by symmetry.
    \item Same as (i).
    \item Let $y \notin B(x, r)$. If $z \in B(x,
      r) \cap B(y, r)$ then $B(x, r) = B(z, r) =
      B(y, r)$ Hence $y \in B(x, r)$. Hence $B(x,
      r) \cap B(y, r) = \emptyset$.
    \item If $z \in \ol{B}(x, r)$, then $B(z, r)
      \subseteq \ol{B}(z, r) = \ol{B}(x, r)$.
      \qedhere
  \end{enumerate}
\end{proof}

\newpage
\section{Valuation Rings}

\begin{fcdefn}[Valuation]
  % Definition 2.1
  \glsnoundefn{valt}{valuation}{valuations}
  Let $K$ be a field. A valuation on $K$ is a
  function $v : K^\times \to \Rbb$ such that
  \begin{enumerate}[(i)]
    \item $v(xy) = v(x) + v(y)$
    \item $v(x + y) \ge \min(v(x), v(y))$
  \end{enumerate}
\end{fcdefn}

Fix $0 < \alpha < 1$. If $v$ is a \gls{valt} on
$K$, then
\[
  |x| = \begin{cases}
    \alpha^{v(x)} & x \neq 0 \\
    0 & x = 0
  \end{cases}
\]
determines a \gls{nonarch} \gls{absval} on $K$.

Conversely a \gls{nonarch} \gls{absval} determines
a \gls{valt} $v(x) = \log_\alpha |x|$.

\begin{remark*}
  \phantom{}
  \begin{itemize}
    \item Ignore the trivial valuation $v(x) = 0$.
    \item Say $v_1, v_2$ are equivalent if there
      exists $c \in \Rbb{>0}$ such that $v_1(x) =
      cv_2(x)$ for all $x \in K^\times$.
  \end{itemize}
\end{remark*}

\begin{example*}
  \phantom{}
  \begin{itemize}
    \item $K = \Qbb$, $v_p(x) = -\log_p
      \absp|x|$ is known as the $p$-adic
      \gls{valt}.
    \item If $k$ is a field, consider $K = k(t) =
      \Frac(k[t])$
      the rational function field. Then define
      \[
        v \left( t^n \frac{f(t)}{g(t)} \right) = n
      \]
      for $f, g \in k[t]$ with $f(0), g(0) \neq
      0$. We call this the $t$-adic valuation.
    \item $K = k((t)) = \Frac(k[[t]]) = \left\{
      \sum_{i = n}^{\infty} a_i t^i \st a_i \in k,
      n \in \Zbb\right\} $, known as the field of
      formal Laurent series over $k$. Then we can
      define
      \[ v \left( \sum_I a_i t^i \right) = \min
      \{i \st a_i \neq 0\} \]
      is the $t$-adic valuation on $K$.
  \end{itemize}
\end{example*}

\begin{fcdefn}[]
  % Definition 2.2
  \glssymboldefn{O}%
  \glsnoundefn{valr}{valuation ring}{valuation
  rings}%
  Let $(K, \absval|\bullet|)$ be a \gls{nonarch}
  \gls{valf}. The valuation ring of $K$ is defined
  to be
  \begin{align*}
    \mathcal{O}_K
    &= \{x \in K \st |x| \le 1\} (= \ol{B}(0, 1))
    \\
    (&= \{x \in K^\times \st v(x) \ge 0\} \cup \{0\})
  \end{align*}
\end{fcdefn}

\begin{proposition}
  % Proposition 2.3
  \label{OK_facts}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $\O_K$ is an open subring of $K$
    \item The subsets $\{x \in K \st |x| \le r\}$
      and $\{x \in K \st |x| < r\}$ for $r \le 1$
      are open ideals in $\O_K$.
    \item $\O_K^\times = \{x \in K \st |x| =
      1\}$.
  \end{enumerate}
\end{proposition}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $\absval|0| = 0$, $\absval|1| = 1$ so $0, 1 \in \O_K$.
      If $x \in \O_K$, then $\absval|-x| = \absval|x|$ hence $-x
      \in \O_K$. If $x, y \in \O_K$, then
      \[
        \absval|x + y| \le \max(\absval|x|, \absval|y|)
        \le 1
      .\]
      Hence $x + y \in \O_K$. If $x, y \in \O_K$,
      then $\absval|xy| = \absval|x| \absval|y|
      \le 1$, hence $xy \in \O_K$. Thus $\O_K$ is
      a ring. Since $\O_K = \ol{B}(0, 1)$, it is
      open.
    \item Similar to (i).
    \item Note that $\absval|x| \absval|x^{-1}| =
      \absval|xx^{-1}| = 1$. Thus
      \begin{align*}
        \absval|x| = 1
        &\iff \absval|x^{-1}| = 1 \\
        &\iff x, x^{-1} \in \O_K \\
        &\iff x \in \O_K^\times \qedhere
      \end{align*}
  \end{enumerate}
\end{proof}

\begin{notation*}
  \phantom{}
  \begin{itemize}
    \item $m \defeq \{x \in \O_K \st |x| < 1\}$ is a max
      ideal of $\O_K$.
    \item $k \defeq \O_K / m$ is the
      \emph{residue field}.
  \end{itemize}
\end{notation*}

\begin{corollary}[]
  $\O_K$ is a local ring with unique maximal ideal
  $m$ (a local ring is a ring with a unique
  maximal ideal).
\end{corollary}

\begin{proof}
  Let $m'$ be a maximal ideal. Suppose $m' \neq
  m$. Then there exists $x \in m' \setminus m$.
  Using part (iii) of \cref{OK_facts}, we get that
  $x$ is a unit, hence $m' = \O_K$, a
  contradiction.
\end{proof}

\begin{example*}
  $K = \Qbb$ with $\absp|\bullet|$. Then
  \[
    \O_K = \Zbb_{(p)} = \left\{ \frac{a}{b} \in
    \Qbb \Bigg| p \nmid b \right\} 
  ,\]
  and $m = p\Zbb_{(p)}$, $k = \Fbb_p$.
\end{example*}

\begin{fcdefn}[]
  % Definition 2.5
  \glsnoundefn{unif}{uniformiser}{uniformisers}%
  \glsnoundefn{disc}{discrete}{valued field}%
  \glsnoundefn{dv}{discrete valuation}{discrete valuation}%
  \glsnoundefn{dvf}{discretely valued
  field}{discretely valued fields}%
  \glsadjdefn{dlyv}{discretely valued}{field}%
  Let $v : K^\times \to \Rbb$ be a \gls{valt}. If
  $v(K^\times) \cong \Zbb$, we say $v$ is a
  \emph{discrete valuation}. $K$ is said to be a
  discretely valued field. An element $\pi \in
  \O_K$ is uniformiser if $v(\pi) > 0$ and
  $v(\pi)$ generates $v(K^\times)$.
\end{fcdefn}

\begin{example*}
  \begin{itemize}
    \item $K = \Qbb$ with $p$-adic \gls{valt} is a
      \gls{dvr}.
    \item $K = k(t)$ with $t$-adic \gls{valt} is a
      \gls{dvr}.
    \item $K = k(t)(t^{1/2}, t^{1/4}, t{1/8},
      \ldots)$. Here, the $t$-adic \gls{valt} is
      not \gls{disc}.
  \end{itemize}
\end{example*}

\begin{remark*}
  If $v$ is a \gls{dv}, can replace with
  equivalent one such that $v(K^\times) = \Zbb$>
  Call such a $v$ \emph{normalised valuations}
  (then $v(\pi) = 1$ if and only if $\pi$ is a
  unit).
\end{remark*}

\begin{fclemma}[]
  \label{lemma_2_6}
  % Lemma 2.6
  Assuming:
  - $v$ is a \gls{valt} on $K$
  Then:
  the following are equivalent:
  \begin{enumerate}[(i)]
    \item $v$ is \gls{disc}
    \item $\O_K$ is a PID
    \item $\O_K$ is Noetherian
    \item $m$ is principal
  \end{enumerate}
\end{fclemma}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(ii) $\implies$ (ii)]
    \item[(i) $\implies$ (ii)] $\O_K$ is an
      integral domain since it is a subset of $K$,
      which is an integral domain.

      Let $I \subseteq \O_K$ be a non-zero ideal.
      Let $x \in I$ such that $v(x) = \min\{v(a)
      \st a \in I\}$, which exists since $v$ is
      \gls{disc}. Then we claim
      \[
        x \O_K = \{a \in \O_K \st v(a) \ge v(x)\}
      \]
      is equal to $I$.
      \begin{enumerate}[$\subseteq$]
        \item[$\subseteq$] ($I$ is an ideal)
        \item[$\supseteq$] Let $y \in I$. Then
          $v(x^{-1} y) \ge 0$. Hence $y = x(x^{-1}
          y) \in x \O_K$.
      \end{enumerate}
    \item[(ii) $\implies$ (iii)] Clear.
    \item[(iii) $\implies$ (iv)] Write $m =
      x_1\O_K + \cdots + x_n \O_K$. Without loss
      of generality,
      \[
        v(x_1) \le v(x_2) \le \cdots \le v(x_n)
      .\]
      Then $x_2, \ldots, x_n \in x_1 \O_K$. Hence
      $m = x_1 \O_K$.
    \item[(iv) $\implies$ (i)] Let $m = \pi \O_K$
      for some $\pi \in \O_K$ and let $c =
      v(\pi)$. Then if $v(x) > 0$, $x \in m$ hence
      $v(x) \ge c$. Thus $v(K^\times) \cap (0, c)
      = \emptyset$. Since $v(K^\times)$ is a
      subgroup of $(\Rbb, +)$, we deduce
      $v(K^\times) = \Zbb$. \qedhere
  \end{enumerate}
\end{proof}
