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\begin{corollary}[]
  \label{coro_4_5}
  % Corollary 4.5
  Let $(K, \absval|\bullet|)$ be a complete
  \gls{dvf}. Let
  \[
    f(X) = a_nX^n + \cdots + a_n \in K[X]
  \]
  with $a_0, a_n \neq 0$. If $f(X)$ is
  irreducible, then $|a_i| \le \max(|a_0|, |a_n|)$
  for all $i$.
\end{corollary}

\begin{proof}
  Upon scaling, we may assume $f(X) \in \O_K[X]$
  with $\max_i (|a_i|) = 1$. Thus we need to show
  that $\max(|a_0|, |a_n|) = 1$. If not, let $r$
  minimal such that $|a_r| = 1$, then $0 < r < n$.
  Thus we have
  \[
    \ol{f}(X) = X^r(a_r + \cdots + a_n X^{n -r})
    \pmod m
  .\]
  Then \cref{hensel_2} implies $f(X) = g(X) h(X)$
  with $0 < \deg < n$.
\end{proof}

\newpage
\section{Teichm\"uller lifts}

\begin{fcdefn}[Perfect]
  \glsadjdefn{perf}{perfect}{ring}%
  % Definition 5.1
  A ring $R$ of characteristic $p > 0$ (prime) is a
  \emph{perfect ring} if the Frobenius $x \mapsto
  x^p$ is a bijection. A field of characteristic
  $p$ is a perfect field if it is perfect as a
  ring.
\end{fcdefn}

\begin{remark*}
  Since $\char R = p$, $(x + y)^p = x^p + y^p$, so
  Frobenius is a ring homomorphism.
\end{remark*}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $\Fbb_{p^n}$ and $\ol{\Fp}$ are perfect
      fields.
    \item $\Fp[t]$ is not perfect, because $t
      \notin \Im(\Frob)$.
    \item $\Fp(t^{\frac{1}{p^\infty}}) \defeq \Fp(t,
      t^{\frac{1}{p}}, t^{\frac{1}{p^2}},
      \ldots)$ is a perfect field (called the
      perfection of $\Fp(t)$).
  \end{enumerate}
\end{example*}

\textbf{Fact:} A field of characteristic $p > 0$
is perfect if and only if any finite extension of
$k$ is separable.

\begin{fcthm}[]
  \label{thm_5_2}
  % Theorem 5.2
  Assuming:
  - $(K, \absval|\bullet|)$ is a complete
  \gls{dvf}
  - such that $k \defeq \O_K / m$ is a perfect
  field of characterist $p$
  Then:
  there exists a unique map $[\bullet] : k \to
  \O_K$ such that
  \begin{enumerate}[(i)]
    \item $a \equiv [a] \bmod m$ for all $a \in k$
    \item $[ab] = [a][b]$ for all $a, b \in k$
  \end{enumerate}
  Moreover if $\char \O_K = p$, then $[\bullet]$
  is a ring homomorphism.
\end{fcthm}

\begin{fcdefn}
  \glsnoundefn{teich}{Teichm\"uller}{NA}%
  \glsnoundefn{teichl}{Teichm\"uller
  lift}{Teichm\"uller lifts}%
  \glssymboldefn{lift}%
  % Definition 5.3
  The element $[a] \in \O_K$ constructed in
  \cref{thm_5_2} is the \emph{Teichm\"uller lift}
  of $a$.
\end{fcdefn}

\begin{fclemma}[]
  \label{lemma_5_4}
  % Lemma 5.4
  Assuming:
  - $(K, \absval|\bullet|)$ is a complete
  \gls{dvf}
  - such that $k \defeq \O_K / m$ is a perfect
  field of characterist $p$
  - $\pi \in \O_K$ a fixed \gls{unif}
  - $x, y \in \O_K$ such that $x \equiv y \bmod
  \pi^k$ ($k \ge 1$)
  Then:
  $x^p \equiv y^p \bmod \pi^{k + 1}$.
\end{fclemma}

\begin{proof}
  Let $x = y + u\pi^k$ with $u \in \O_K$. Then
  \begin{align*}
    x^p
    &= \sum_{i = 0}^{p} {p \choose i} y^{p - i}(u
    \pi^k)^i \\
    &= y^p + \sum_{i = 1}^{p} {p \choose i} y^{p -
    1} (u\pi^k)^i
  \end{align*}
  Since $\O_K / \pi \O_K$ has characteristic $p$,
  we have $p \in \pi\O_K$. Thus
  \[
    {p \choose i} (u\pi^k)^i y^{p - i} \in \pi^{k
    + 1} \O_K \qquad \forall i \ge 1
  ,\]
  hence $x^p \equiv y^p \bmod \pi^{k + 1}$.
\end{proof}

\begin{proof}[Proof of \cref{thm_5_2}]
  Let $a \in k$. For each $i \ge 0$ we choose a
  lift $y_i \in \O_K$ of $a^{\frac{1}{p^i}}$, and
  we define
  \[
    x_i \defeq y_i^{p_i}
  .\]
  We claim that $(x_i)_{i = 1}^\infty$ is a Cauchy
  sequence and its limit is independent of the
  choice of $y_i$.

  By construction, $y_i \equiv y_{i + 1}^p \bmod
  \pi$. By \cref{lemma_5_4} and induction on $k$,
  we have $y_i^{p^k} \equiv y_{i + 1}^{p^{k + 1}}$
  and hence $x_i \equiv x_{i + 1} \bmod \pi^{i +
  1}$ (take $i = p$). Hence $(x_i)_{i = 1}^\infty$
  is Cauchy, so $x_i \to x \in \O_K$.

  Suppose $(x_i')_{i = 1}^\infty$ arises from
  another choice of $y_i'$ lifting
  $a_i^{\frac{1}{p^i}}$. Then $(x_i')_{i =
  1}^\infty$ is Cauchy, and $x_i' \to x' \in
  \O_K$. Let
  \[
    x_i'' = \begin{cases}
      x_i & \text{$i$ even} \\
      x_i' & \text{$i$ odd}
    \end{cases}
  .\]
  Then $x_i''$ arises from lifting
  \[
    y_i'' = \begin{cases}
      y_i & \text{$i$ even} \\
      y_i & \text{$i$ odd}
    \end{cases}
  .\]
  Then $x_i''$ is Cauchy and $x_i'' \to x$, $x_i''
  \to x'$. So $x = x'$ and hence $x$ is independet
  of the choice of $y_i$. So we may define $\lift[a] =
  x$.

  Then $x_i = y_i^{p^i} \equiv
  (a^{\frac{1}{p^i}})^{p^i} \equiv a \bmod \pi$.
  Hence $x \equiv a \bmod \pi$. So (i) is
  satisfied.

  We let $b \in k$ and we choose $u_i \in \O_K$ a
  lift of $b^{\frac{1}{p^i}}$, and let $z_i \defeq
  u_i^{p^i}t$. Then $\lim_{i \to \infty} z_i =
  \lift[b]$.

  Now $u_i y_i$ is a lift of
  $(ab)^{\frac{1}{p^i}}$, hence
  \[
    \lift[ab] = \lim_{i \to \infty} x_i z_i = (\lim_{i
    \to \infty} x_i)(\lim_{i \to \infty} z_i) =
    \lift[a]\lift[b]
  .\]
  So (ii) is satisfied.

  If $\char K = p$, $y_i + u_i$ is a lift of
  $a^{\frac{1}{p^i}} + b^{\frac{1}{p^i}} = (a +
  b)^{\frac{1}{p^i}}$. Then
  \begin{align*}
    \lift[a + b]
    &= \lim_{i \to \infty} (y_i + u_i)^{p^i} \\
    &= \lim_{i \to \infty} y_i^{p^i} + u_i^{p^i}
    \\
    &= \lim_{i \to \infty} x_i + z_i \\
    &= \lift[a] + \lift[b]
  \end{align*}
  Easy to check that $\lift[0] = 0$, $\lift[1] =
  1$, and hence $\lift[\bullet]$ is a ring
  homomorphism.

  Uniqueness: let $\phi : k \to \O_K$ be another
  such map. Then for $a \in k$,
  $\phi(a^{\frac{1}{p^i}})$ is a lift of
  $a^{\frac{1}{p^i}}$. It follows that
  \begin{align*}
    \lift[a]
    &= \lim_{i \to \infty}
    \phi(a^{\frac{1}{p^i}})^{p^i} \\
    &= \lim_{i \to \infty} \phi(a) \\
    &= \phi(a) \qedhere
  \end{align*}
\end{proof}

\begin{example*}
  $K = \Qp$, $\lift[\bullet] : \Fp \to \Zp$, $a \in
  \Fp^\times$, $\lift[a]^{p - 1} = \lift[a^{p -
  1}] = \lift[1] = 1$. So $\lift[a]$ is a $(p -
  1)$-th root of unity.
\end{example*}

\begin{fclemma}[]
  % Lemma 5.6
  Assuming:
  - $(K, \absval|\bullet|)$ complete \gls{dvf}
  - $k = \O_K / m \subseteq \ol{\Fp}$
  - $a \in k^\times$
  Then:
  $\lift[a]$ is a root of unity.
\end{fclemma}

\begin{proof}
  \begin{align*}
    a \in k^\times
    &\implies a \in \Fbb_{p^n}^\times \text{ for
    some $n$} \\
    &\implies \lift[a]^{p^n - 1} = \lift[a^{p^n -
    1}] = \lift[1] = 1 \qedhere
  \end{align*}
\end{proof}

\begin{fcthm}[]
  \label{thm_5_7}
  % Theorem 5.7
  Assuming:
  - $(K, \absval|\bullet|)$ complete \gls{dvf}
  - $\char(K) = p > 0$
  - $k$ is \gls{perf}
  Then:
  $K = k((t))$ ($k = \O_K / m$).
\end{fcthm}

\begin{proof}
  Since $K = \Frac(\O_K)$, it suffices to show
  $\O_K \cong k[[t]$. Fix $\pi \in \O_K$ a
  \gls{unif}, and let $\lift[\bullet] : k \to
  \O_K$ be the \glsref[teichl]{Teichm\"uller map}
  and define
  \begin{align*}
    \varphi : k[[t]]
    &\to \O_K \\
    \varphi \left( \sum_{i = 0}^{\infty}  a_i t^i \right)
    &= \sum_{i = 0}^\infty \lift[a_i] \pi^i
  \end{align*}
  Then $\varphi$ is a ring homomorphism since
  $\lift[\bullet]$ is, and it is a bijection by
  \cref{prop_3_5}(ii).
\end{proof}