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\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item It suffices by \cref{prop_3_5} to show
      that
      \[
        \Zp / p^n \Zp \cong \Zbb / p^n \Zbb
      .\]
      Let $f_n : \Zbb \to \Zp / p^n \Zp$ be the
      natural map
      \[
        \ker(f_n) = \{x \in \Zbb \st \absp|x| \le
        p^{-n}\} = p^n \Zbb
      ,\]
      hence $\Zbb / p^n \Zbb \to \Zp / p^n \Zp$ is
      injective.

      Let $\tau \in \Zp / p^n \Zp$ and let $c \in
      \Zp$ be a lift. Since $\Zbb$ is dense in
      $\Zp$, there exists $x \in \Zbb$ such that
      $x \in c + p^n \Zp$ is open in $\Zp$. Then
      $f_n(x) = \tau$, hence $\Zbb / p^n \Zbb \to
      \Zp / p^n \Zp$ is surjective.
    \item It follows from \cref{prop_3_5}(ii) to
      $p^{-n} x \in \Zp$ for some $n \in \Zbb$
      \qedhere
  \end{enumerate}
\end{proof}

\begin{example*}
 \[
   \frac{1}{1 - p} = 1 + p + p^2 + p^3 + \cdots
 \]
\end{example*}

\newpage
\part{Complete Valued Fields}

\newpage
\section{Hensel's Lemma}

\begin{fcthm}[Hensel's Lemma version 1]
  \label{hensel}
  % Theorem 4.1
  Assuming:
  - $(K, \absval|\bullet|)$ is a complete \gls{dvf}
  - $f(X) \in \O_K[X]$
  - assume $\exists a \in \O_K$ such that $|f(a)|
  < |f'(a)|^2$
  Then:
  there exists a unique $x \in \O_K$ such that
  $f(x) = 0$ and $|x - a| < |f'(a)|$.
\end{fcthm}

\begin{proof}
  Let $\pi \in \O_K$ be a \gls{unif} and let $r =
  v(f'(a))$, with $v$ the normalised valuation
  ($v(\pi) = 1$). We construct a sequence
  $(x_n)_{n = 1}^\infty$ in $\O_K$ such that:
  \begin{enumerate}[(i)]
    \item $f(x_n) \equiv 0 \pmod{\pi^{n + 2r}}$
    \item $x_{n + 1} \equiv x_n \pmod{\pi^{n + r}}$
  \end{enumerate}
  Take $x_1 = a$: then $f(x_1) \equiv 0
  \pmod{\pi{1 + 2r}}$.

  Now we suppose we have constructed $x_1, \ldots,
  x_n$ satisfying (i) and (ii). Define
  \[
    x_{n + 1} = x_n - \frac{f(x_n)}{f'(x_n)}
  .\]
  Since $x_n \equiv x_1 \pmod{\pi^{r + 1}}$, we
  have
  \[
    v(f'(x_n)) = v(f'(x_i)) = r
  ,\]
  and hence
  \[
    \frac{f(x_n)}{f'(x_n)} \equiv 0 \pmod{\pi^{n +
    r}}
  \]
  by (i).

  It follows that $x_{n + 1} \equiv x_n
  \pmod{\pi^{n + r}}$, so (ii) holds. Note that
  letting $X, Y$ be indeterminates, we have
  \[
    f(X + Y) = f_0(X) + f_1(X) Y + f_2(X) Y^2 +
    \cdots
  ,\]
  where $f_i(X) \in \O_K[X]$ and $f_0(X) = f(X)$,
  $f_1(X) = f'(X)$. Thus
  \[
    f(x_{n + 1}) = f(x_n) + cf'(x_n) + c^2
    f_2(x_N) + \ub{c^2 f_2(x_n) + \cdots}_{\in
    \pi^{n + 2r + 1}}
  \]
  where $c = -\frac{f(x_n)}{f'(x_n)}$.

  Since $c \equiv 0 \pmod{\pi^{n + r}}$ and
  $v(f_i(x_n)) \ge 0$ we have
  \[
    f(x_{n + 1}) \equiv f(x_n) + f'(x_n)c \equiv 0
    \pmod{\pi^{n + 2r + 1}}
  ,\]
  so (i) holds.

  Property (ii) implies that $(x_n)_{n =
  1}^\infty$ is Cauchy, so let $x \in \O_K$ such
  that $x_n \to x$. Then $f(x) = \lim_{n \to
  \infty} f(x_n) = 0$ by (i).

  Moreover, (ii) impies that
  \begin{align*}
    a = x_1
    &\equiv x_n \pmod{\pi^{r + 1}} \qquad \forall
    n \\
    \implies a
    &\equiv x \pmod{\pi^{r + 1}} \\
    \implies |x - a|
    &< |f'(a)|
  \end{align*}
  This proves existence.

  Uniqueness: suppose $x'$ also satisfies $f'(x) =
  0$, $|x' - a| < |f'(a)|$. Set $\delta = x' - x
  \neq 0$. Then
  \[
    |x' - a| < |f'(a)| \qquad |x - a'| < |f'(a)|
  ,\]
  and the ultrametric inequality implies
  \[
    |\delta| = |x - x'| < |f'(a)| = |f'(x)|
  .\]
  But
  \[
    0 = f(x') = f(x + \delta) = \ub{f(x)}_{=0} + f'(x)
    \delta + \ub{\cdots}_{|\bullet| \le
    |\delta|^2}
  .\]
  Hence $|f'(x)\delta| \le |\delta|^2$, so
  $|f'(x)|
  < |\delta|$, a contradiction.
\end{proof}

\begin{corollary}
  \label{coro_4_2}
  % Corollary 4.2
  Let $(K, \absval|\bullet|)$ be a complete
  \gls{dvf}. Let $f(X) \in \O_K[X]$ and $\ol{c}
  \in k \defeq \O_K / m$ a simple root of
  $\ol{f}(X) \defeq f(X) \pmod m \in k[X]$. Then
  there exists a unique $x \in \O_K$ such that
  $f(x) = 0$, $x \equiv \ol{c} \pmod m$.
\end{corollary}

\begin{proof}
  Apply \cref{hensel} to a lift $c \in \O_K$ of
  $\ol{c}$. Then $|f(c)| < 1 = |f'(c)|^2$ since
  $\ol{c}$ is a simple root.
\end{proof}

\begin{example*}
  $f(X) = X^2 - 2$ has a simple root modulo 7.
  Thus $\sqrt2 \in \Zbb_7 \subseteq \Qbb_7$.
\end{example*}

\begin{corollary}
  % Corollary 4.3
  \[
    \Qp^\times / (\Qp^\times)^2 \cong \begin{cases}
      (\Zbb / 2\Zbb)^2 & \text{if $p > 2$} \\
      (\Zbb / 2\Zbb)^3 & \text{if $p = 2$}
    \end{cases}
  \]
\end{corollary}

\begin{proof}
  \textbf{Case $p > 2$:} Let $b \in \Zp^\times$.
  Applying \nameref{coro_4_2} to $f(X) = X^2 - b$,
  we find that $b \in (\Zp^\times)^2$ if and only
  if $b \in (\Fp^\times)^2$. Thus $\Zp^\times /
  (\Zp^\times)^2 \cong \Fp^\times / (\Fp^\times)^2
  \cong \Zbb / 2\Zbb$ ($\Fp^\times \cong \Zbb / (p
  - 1) \Zbb$).

  We have an isomorphism
  \[
    \Zp^\times \times (\Zbb, +) \cong \Qp^\times
  \]
  given by $(u, n) \mapsto up^n$. Thus
  \[
    \Qp^\times / (\Qp^\times)^2 \cong (\Zbb /
    2\Zbb)^2
  .\]
  
  \textbf{Case $p = 2$:} Let $b \in
  \Zbb_2^\times$. Consider $f(X) = X^2 - b$. Note
  $f'(X) = 2X \equiv 0 \pmod 2$. Let $b \equiv 1
  \pmod 8$. Then
  \[
    |f(1)| = 2^{-3} < 2^{-2} = |f'(1)|^2
  .\]
  \nameref{hensel} gives
  \[
    b \in (\Zbb_2^\times)^2 \iff b \equiv 1 \pmod
    8
  .\]
  Then
  \[
    \Zbb_2^\times / (\Zbb_2^\times)^2 \cong (\Zbb
    / 8 \Zbb)^\times \equiv (\Zbb / 2\Zbb)^j
  .\]
  Again using $\Qbb_2^\times \equiv \Zbb_2^\times
  \times \Zbb$, we find that $\Qbb_2^\times \cong
  (\Zbb / 2 \Zbb)^3$.
\end{proof}

\begin{remark*}
  Proof uses the iteration
  \[
    x_{n + 1} = x_n - \frac{f(x_n)}{f'(x_n)}
  ,\]
  which is the \gls{nonarch} analogue of the
  unewton Raphson method.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/2f579e6492324e0c.png}
  \end{center}
\end{remark*}

\begin{fcthm}[Hensel's Lemma version 2]
  \label{hensel_2}
  % Theorem 4.4
  Assuming:
  - $(K, |\bullet|)$ is a complete \gls{dvf}
  - $f(X) \in \O_K[X]$
  - $\ol{f}(X) \defeq f(X) \pmod m \in k[X]$
  factorises as $\ol{f}(X) = \ol{g}(X) \ol{h}(X)$
  in $k[X]$
  - $\ol{g}(X)$ and $\ol{h}(X)$ coprime.
  Then:
  there is a factorisation
  \[
    f(X) = g(X) h(X)
  \]
  in $\O_K[X]$, with $\ol{g}(X) \equiv g(X) \pmod
  m$, $\ol{h}(X) \equiv h(X) \pmod m$ and $\deg
  \ol{g} = \deg g$.
\end{fcthm}

\begin{proof}
  Example Sheet 1.
\end{proof}

