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If $K$ is complete, then $\mathcal{O}_K$ is
\gls{complete} with respect to $\pi^r
\mathcal{O}_K$ (for any $r \ge 1$).

We fix a \gls{min} \gls{weq} for $E /
K$. Get \gls{fgroup} $\hat{E}$ over
$\mathcal{O}_K$.

Taking $I = \pi^r \mathcal{O}_K$ (with $r \ge 1$)
in \cref{lemma:8.2} gives
\begin{align*}
  \hat{E}(I)
  &= \left\{(x, y) \in E(K) ~\Bigg|~ -\frac{x}{y},
  -\frac{1}{y} \in \pi^r \mathcal{O}_K\right\}
  \cup \{0\} \\
  &= \left\{(x, y) \in E(K) ~\Bigg|~ v \left(
  \frac{x}{y} \right) \ge r, v \left( \frac{1}{y}
  \right) \ge r\right\} \cup \{0\} \\
  &= \{(x, y) \in E(K) \st v(x) = -2s, v(y) = -3s
  \text{ for } s \ge r\} \cup \{0\}
  &&\text{(using \cref{lemma:9.1})} \\
  &= \{(x, y) \in E(K) \st v(x) \le -2r, v(y) \le
  -3r\} \cup \{0\}
  &&\text{(using \cref{lemma:9.1})}
\end{align*}
By \cref{lemma:8.2} this is a subgroup of $E(K)$,
say $E_r(K)$.
\[
  \cdots \subset E_2(K) \subset E_1(K) \subset E(K)
.\]
More generally, for $\mathcal{F}$ a
\gls{fgroup} over $\mathcal{O}_K$
\[
  \cdots \subset \mathcal{F}(\pi^2 \mathcal{O}_K)
  \subset \mathcal{F}(\pi \mathcal{O}_K)
.\]
We claim that
\begin{itemize}
  \item $\mathcal{F}(\pi^r \mathcal{O}_K) \cong
    (\mathcal{O}_K, +)$ for $r$ sufficiently
    large.
  \item $\frac{\mathcal{F}(\pi^r
    \mathcal{O}_K)}{\mathcal{F}(\pi^{r + 1}
    \mathcal{O}_K)} \cong (k, +)$ for $r \ge 1$.
\end{itemize}

Reminder: $\char K = 0$, $\char k = p > 0$.

\begin{fcthm}[]
  \label{thm:9.2}
  Assuming:
  - $K$ a local field with $\char K = 0$, $\char k
  = p > 0$
  - $\mathcal{F}$ a \gls{fgroup} over
  $\mathcal{O}_K$
  - $e = v(p)$
  - $r > \frac{e}{p - 1}$
  Then:
  \[
    \log : \mathcal{F}(\pi^r \mathcal{O}_K)
    \congto \hat{\mathbb{G}}_a(\pi^r \mathcal{O}_K)
  \]
  is an isomorphism of groups with inverse
  \[
    \exp : \hat{\mathbb{G}}_a(\pi^r \mathcal{O}_K)
    \congto \mathcal{F}(\pi^r \mathcal{O}_K)
  .\]
\end{fcthm}

\begin{remark*}
  $\hat{\mathbb{G}}_a(\pi^r \mathcal{O}_K) =
  (\pi^r \mathcal{O}_K, +) \cong (\mathcal{O}_K,
  +)$, $x \leftrightarrow \pi^{-r} x$.
\end{remark*}

\begin{proof}
  For $x \in \pi^r \mathcal{O}_K$ we must show the
  power series $\log(x)$ and $\exp(x)$ converge to
  elements in $\pi^r \mathcal{O}_K$.

  Recall
  \[
    \exp(T)
    = T + \frac{b_2}{2!} T^2 + \frac{b_3}{3!} T^3
    + \cdots
  \]
  for some $b_i \in \mathcal{O}_K$.

  Claim: $v_p(n!) \le \frac{n - 1}{p - 1}$.

  Proof of claim:
  \[
    v_p(n!)
    = \sum_{r = 1}^{\infty} \left\lfloor
    \frac{n}{p^r} \right\rfloor
    < \sum_{r = 1}^{\infty} \frac{n}{p^r}
    = \frac{n\cdot \frac{1}{p}}{1 - \frac{1}{p}}
    = \frac{n}{p - 1}
  .\]
  Therefore
  \begin{align*}
    (p - 1) v_p(n!)
    &< n \\
    (p - 1)v_p(n!)
    &\le n - 1
  \end{align*}
  (we go from $<$ to $\le \blank - 1$ by noting
  that the LHS is in $\Zbb$).

  This proves the claim.

  Now
  \begin{align*}
    v \left( \frac{b_n x^n}{n!} \right)
    &\ge nr - e \left( \frac{n - 1}{p - 1} \right)
    \\
    &= (n - 1) \ub{\left( r - \frac{e}{p - 1}
    \right)}_{> 0}
    + r
  \end{align*}
  This is always $\ge r$ and $\to \infty$ as $n
  \to \infty$. Therefore $\exp(x)$ converges to an
  element in $\pi^r \mathcal{O}_K$.

  Same method works for $\log$.
\end{proof}

\begin{fclemma}[]
  \label{lemma:9.3}
  $\frac{\mathcal{F}(\pi^r
  \mathcal{O}_K)}{\mathcal{F}(\pi^{r + 1}
  \mathcal{O}_K)} \cong$ \cloze{$(k, +)$} for all $r \ge 1$.
\end{fclemma}

\begin{proof}
  Definition of \gls{fgroup} gives
  \[
    F(X, Y) = X + Y + XY (\cdots)
  .\]
  So if $x, y \in \mathcal{O}_K$,
  \[
    F(\pi^r x, \pi^r y) \equiv \pi^r (x + y)
    \pmod{\pi^{r + 1}}
  .\]
  Therefore
  \begin{align*}
    \mathcal{F}(\pi^r \mathcal{O}_K)
    &\to (k, +) \\
    \pi^r x
    &\mapsto x \bmod \pi
  \end{align*}
  is a surjective group homomorphism with kernel
  $\mathcal{F}(\pi^{r + 1} \mathcal{O}_K)$.
\end{proof}

\begin{fccorostar}[]
  Assuming:
  - $|k| < \infty$
  Then:
  $\mathcal{F}(\pi \mathcal{O}_K)$ has a subgroup
  of finite index isomorphic to $(\mathcal{O}_K,
  +)$.
\end{fccorostar}

\begin{notation*}
  Reduction modulo $\pi$
  \begin{align*}
    \mathcal{O}_K
    &\to \frac{\mathcal{O}_K}{\pi \mathcal{O}_K} =
    k \\
    x
    &\mapsto \tilde{x}
  \end{align*}
\end{notation*}

\begin{fcprop}[]
  \label{prop:9.4}
  Assuming:
  - $E / K$ be an \gls{ellc}
  Then:
  the reduction mod $\pi$ of any two \gls{min}
  \glspl{weq} for $E$ define isomorphic
  curves over $k$.
\end{fcprop}

\begin{proof}
  Say \glspl{weq} are related by $[u; r,
  s, t]$, $u \in K^*$, $r, s, t \in K$. Then
  $\Delta_1 = u^{12} \Delta_2$.
  Both equations \gls{min} gives us that
  $v(\Delta_1) = v(\Delta_2)$, hence $u \in
  \mathcal{O}_K^*$.

  Transformation formula for the $a_i$ and $b_i +
  \mathcal{O}_K$ is integrally closed, hence $r,
  s, t \in \mathcal{O}_K$.

  The \glspl{weq} obtained by reducing
  mod $\pi$ are now related by $[\tilde{u};
  \tilde{r}, \tilde{s}, \tilde{t}]$, $\tilde{u}
  \in k^*$, $\tilde{r}, \tilde{s}, \tilde{t} \in
  k$.
\end{proof}

\begin{fcdefnstar}[]
  \glsadjdefn{gred}{good reduction}{\gls{ellc}}%
  \glsadjdefn{bred}{bad reduction}{\gls{ellc}}%
  \glssymboldefn{reduction}%
  The reduction $\tilde{E} / k$ of $E / K$ is
  defined by the reduction of a \gls{min}
  \gls{weq}.

  $E$ has \emph{good reduction} if $\tilde{E}$ is
  non-singular (and so an \gls{ellc}) otherwise
  has \emph{bad reduction}.

  For an \gls{int} \gls{weq},
  \begin{itemize}
    \item If $v(\Delta) = 0$ then good reduction.
    \item If $0 < v(\Delta) < 12$ then bad
      reduction.
    \item If $v(\Delta) \ge 12$ then beware that
      the equation might not be \gls{min}.
  \end{itemize}
\end{fcdefnstar}

There is a well-defined map
\begin{align*}
  \Pbb^2(K)
  &\to \Pbb^2(k) \\
  (x : y : z)
  &\mapsto (\tilde{x} : \tilde{y} : \tilde{z})
\end{align*}
(choose a representative with $\min(v(x), v(y),
v(z)) = 0$).

We restrict to give
\begin{align*}
  E(K)
  &\to \tilde{E}(K) \\
  P
  &\mapsto \tilde{P}
\end{align*}
If $P = (x, y) \in E(K)$ then by \cref{lemma:9.1}
either
\begin{itemize}
  \item $x, y \in \mathcal{O}_K$ in which case
    $\tilde{P} = (\tilde{x}, \tilde{y})$.
  \item or $v(x) = -2s$, $v(y) = -3s$ for some $s
    \ge 1$, in which case $P = (x : y : 1) =
    (\pi^{3s} x : \pi^{3s} y : \pi^{3s})$ and
    $\tilde{P} = (0 : 1 : 0)$.
\end{itemize}

Therefore
\[
  E_1(K) = I didn't manage to write this before it
  was rubbed off
\]
``kernel of reduction''.