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Suppose $v$ is a \gls{dv} on $K$, $\pi \in
\O_K$ a \gls{unif}. For $x \in K^\times$, let $n
\in \Zbb$ such that $v(x) = nv(\pi)$. Then $u =
\pi^{-n} x \in \O_K^\times$ and $x = u\pi^n$. In
particular, $K = \O_K \left[ \frac{1}{\pi}
\right]$ and hence $K = \Frac(\O_K)$.

\begin{fcdefn}[Discrete valuation ring]
  % Definition 2.2
  \glsnoundefn{dvr}{discrete valuation
  ring}{discrete valuation rings}%
  A ring $R$ is called a \emph{discrete valuation
  ring} (DVR) if it is a PID with exactly one
  non-zero prime ideal (necessarily maximal).
\end{fcdefn}

\begin{lemma}
  % Lemma 2.8
  \phantom{}
  \begin{enumerate}[(i)]
    \item Let $v$ be a \gls{dv} on $K$. Then
      $\O_K$ is a \gls{dvr}.
    \item Let $R$ be a \gls{dvr}. Then there
      exists a \gls{valt} on $K \defeq \Frac(R)$
      such that $R = \O_K$.
  \end{enumerate}
\end{lemma}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $\O_K$ is a PID by \cref{lemma_2_6}.
      Hence any non-zero prime ideal is maximal
      and hence $\O_K$ is a \gls{dvr} since it is
      a local ring.
    \item Let $R$ be a \gls{dvr}, with maximal
      ideal $m$. Then $m = (\pi)$ for some $\pi
      \in R$. Since PIDs are UFDs, we may write
      any $x \in R \setminus \{0\}$ uniquely as
      $\pi^n u$ with $n \ge 0$, $u \in R^\times$.
      Then any $y \in K^\times$ can be written
      uniquely as $\pi^m u$ with $u \in R^\times$,
      $m \in \Zbb$. Define $v(\pi^m u) = m$; check
      $v$ is a \gls{valt} and $\O_K = R$. \qedhere
  \end{enumerate}
\end{proof}

\begin{example*}
  $\Zbb_{(p)}$, $k[[t]]$ ($k$ a field) are
  \glspl{dvr}.
\end{example*}

\newpage
\section{The $p$-adic numbers}

Recall that $\Qp$ is the completion of $\Qbb$
with respect to $\absp|\bullet|$. On Example Sheet
1, we will show that $\Qp$ is a field. We also
show that $\absp|\bullet|$ extends to $\Qp$ and
the associated \gls{valt} is \gls{disc}.

\begin{fcdefn}[]
  % Definition 3.1
  \glssymboldefn{Zp}%
  The ring of $p$-adic integers $\Zbb_p$ is the
  \gls{valt} ring
  \[
    \Zbb_p = \{x \in \Qp \st \absp|x| \le 1\}
  .\]
\end{fcdefn}

\textbf{Facts:} $\Zp$ is a \gls{dvr}, with maximal
ideal $p\Zp$, and non-zero ideals are given by
$p^n \Zp$.

\begin{proposition*}[]
  $\Zp$ is the closure of $\Zbb$ inside
  $\Qp$. In particular, $\Zp$ is the
  completion of $\Zbb$ with respect to
  $\absp|\bullet|$.
\end{proposition*}

\begin{proof}
  Need to show $\Zbb$ is dense in $\Zp$. Note
  $\Qbb$ is dense in $\Qp$. Since $\Zp \subseteq
  \Qp$ is open, we have that $\Zp \cap \Qbb$ is
  dense in $\Zp$. Now:
  \[
    \Zp \cap \Qbb = \{x \in \Qbb \st \absp|x| \le
    1\} = \left\{ \frac{a}{b} \in \Qbb \Bigg| p
    \nmid b \right\} = \Zbb_{(p)}
  .\]
  Thus it suffices to show $\Zbb$ is dense in
  $\Zbb_{(p)}$.

  Let $\frac{a}{b} \in \Zbb_{(p)}$, $a, b \in
  \Zbb$, $p \nmid b$. For $n \in \Nbb$, choose
  $y_n \in \Zbb$ such that $by_n \equiv a
  \pmod{p^n}$. THen $y_n \to \frac{a}{b}$ as $n
  \to \infty$.

  In particular, $\Zp$ is complete and $\Zbb
  \subseteq \Zp$ is dense.
\end{proof}

\begin{definition*}[Inverse limit]
  \glssymboldefn{invlim}%
  \glsnoundefn{invlim}{inverse limit}{inverse
  limits}%
  Let $(A_n)_{n = 1}^\infty$ be a sequence of sets /
  groups / rings together with homomorphisms
  $\varphi_n : A_{n + 1} \to A_n$ (transition maps).
  Then the \emph{inverse limit} of $(A_n)_{n =
  1}^\infty$ is the set / group / ring defined by
  \[
    \lim_{\stackrel[n]{}{\longleftarrow}} A_n
    = \left\{(a_n)_{n = 1}^\infty
    \in \prod_{n = 1}^{\infty} A_n \Bigg|
    \varphi(a_{n + 1}) = a_n ~\forall n\right\}
  .\]
  Define the group / ring operation componentwise.
\end{definition*}

\begin{notation*}
  \glssymboldefn{limproj}%
  Let $\theta_m : \invlim{n} A_n \to A_m$ denote
  the natural projection.
\end{notation*}

The \gls{invlim}
satisfies the following universal property:

\begin{fcprop}[Universal property of inverse
  limits]
  \label{univ_invlim}
  % Proposition 3.3
  Assuming:
  - $B$ is a set / group / ring
  - $\psi_n$ are homomorphisms $\psi_n : B \to
  A_n$ such that
  \begin{picmath}
    \begin{tikzcd}
      B \ar[r, "\psi_{n + 1}"] \ar[rd, "\psi_n"]
      & A_{n + 1} \ar[d, "\varphi_n"] \\
      & A_n
    \end{tikzcd}
  \end{picmath}
  commutes for all $n$
  Then:
  there exists a unique homomorphism $\psi : B \to
  \invlim n A_n$ such that $\limproj_n \circ \psi
  = \psi_n$.
\end{fcprop}

\begin{proof}
  Define
  \begin{align*}
    \psi : B
    &\to \prod_{n = 1}^{\infty} A_n \\
    b
    &\mapsto \prod_{n = 1}^{\infty} \psi_n(b)
  \end{align*}
  Then $\psi_n = \varphi_n \circ \psi_{n + 1}$
  implies that $\psi(b) \in \invlim n A_n$. The
  map is clearly unique (determined by $\psi_n =
  \theta_n \circ \psi$) and is a homomorphism of
  sets / groups / rings.
\end{proof}

\begin{fcdefn}[$I$-adic completion]
  \glsnoundefn{Iadicc}{$I$-adic
  completion}{$I$-adic completions}%
  \glssymboldefn{Iadicc}%
  % Definition 3.4
  Let $I \subseteq R$ be an ideal ($R$ a ring).
  The $I$-adic completion of $R$ is the
  \[
    \hat{R} \defeq \invlim R / I^n
  \]
  where $R / I^{n + 1} \to R / I^n$ is the natural
  projection.
\end{fcdefn}

\glsadjdefn{adiclyc}{adically complete}{ring}%
Note that there exists a natural map $i : R \to
\iadicc{R}$ by the \nameref{univ_invlim} (there
exist maps $R \to R / I^n$). We say $R$ is
\emph{$I$-adically complete} if it is an isomorphism.

\textbf{Fact:} $\ker(i : R \to \iadicc R) =
\bigcap_{n = 1}^\infty I^n$.

Let $(K,
\absval|\bullet|)$ be a \gls{nonarch}
\gls{valf}and $\pi \in \O_K$ such that $|\pi| <
1$.

\begin{fcprop}[]
  \label{prop_3_5}
  % Proposition 3.5
  Assuming:
  - $K$ is complete with respect to
  $|\bullet|$
  Then:
  \begin{enumerate}[(i)]
    \item Then $\O_K \cong \invlim n \O_K / \pi^n
      \O_K$ ($\O_K$ is $\pi$-\gls{adiclyc})
    \item Every $x \in \O_K$ can be written
      uniquely as $x = \sum_{i = 0}^{n} a_i
      \pi^i$, $a_i \in A$, where $A \subseteq
      \O_K$ is a set of coset representatives for
      $\O_K / \pi \O_K$.
  \end{enumerate}
\end{fcprop}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $K$ is complete and $\O_K$ is closed, so
      $\O_K$ is complete.

      $x \in \bigcap_{n = 1}^\infty \pi^n \O_K$
      impies $v(x) \ge nv(\pi)$ for all $n$, and
      hence $x = 0$. Hence $\O_K \to \invlim \O_K
      / \pi^n \O_K$ is injective.

      Let $(x_n)_{n = 1}^\infty \in \invlim \O_K /
      \pi^n \O_K$ and for each $n$, let $y_n \in
      \O_K$ be a lifting of $x_n \in \O_K / \pi^n
      \O_K$. Then $y_n - y_{n + 1} \in \pi^n \O_K$
      so that $v(y_n - y_{n + 1}) \ge nv(\pi)$.

      Thus $(y_n)_{n = 1}^\infty$ is a Cauchy
      sequence in $\O_K$. Let $y_n \to y \in
      \O_K$. Then $y$ maps to $(x_n)_{n =
      1}^\infty$ in the $\invlim n \O_K / \pi^n
      \O_K$. Thus $\O_K \to \invlim n \O_K / \pi^n
      \O_K$ is surjective.
    \item Exercise on Example Sheet 1. \qedhere
  \end{enumerate}
\end{proof}

\begin{corollary}
  % Corollary 3.6
  \phantom{}
  \begin{enumerate}[(i)]
    \item $\Zp \cong \invlim \Zbb / p^n \Zbb$.
    \item Every element $x \in \Qp$ can be written
      uniquely as
      \[
        x = \sum_{i = n}^{\infty} a_i p^i
      ,\]
      with $n \in \Zbb$, $a_i \in \{0, 1, \ldots,
      - 1\}$.
  \end{enumerate}
\end{corollary}
