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\begin{notation*}
  \glssymboldefn{reductionns}%
  \phantom{}
  \[
    \begin{cases}
      \reduction{E}
      & \text{if $E$ has \gls{gred}} \\
      \reduction{E} \setminus \{\text{singular
      point}\}
      & \text{if $E$ has \gls{bred}}
    \end{cases}
  \]
\end{notation*}

The chord and tangent process still defines a
group law on $\reduction{E}_{\text{ns}}$.

In cases of \gls{bred},
$\reductionns{E} \cong \mathbb{G}_m$
(over $k$ or possibly a quadratic extension of
$k$) or $\reductionns{E} \cong \mathbb{G}_a$ (over
$k$).

For simplicity we suppose $\char k \neq 2$.

Then $\reduction E = y^2 = f(x)$, $\deg f = 3$.

\begin{center}
  \includegraphics[width=0.8\linewidth]{images/46cc041b951c40ba.png}
\end{center}
\begin{align*}
  \reductionns{E}
  &\congto \mathbb{G}_a \\
  (x, y)
  &\mapsto \frac{x}{y} \\
  (t^{-2}, t^{-3})
  &\mapsfrom t \\
  (\text{point at $\infty$})
  &\mapsfrom 0
\end{align*}
Let $P_1, P_2, P_3$ lie on the line $ax + by = 1$.
Write $P_i = (x_i, y_i)$, $t_i = \frac{x_i}{y_i}$.
Then $x_i^3 = y_i^2 = y_i^2(ax_i + b y_i)$. So
$t_i^3 - a t_i - b = 0$. So $t_1, t_2, t_3$ are
the roots of $T^3 - aT - b = 0$.

Looking at coefficient of $T^2$ gives $t_1 + t_2 +
t_3 = 0$.

\begin{fcdefnstar}[$E_0(K)$]
  $E_0(K) = \{P \in E(K) \st \tilde{P} \in
  \reductionns{E}(k)\}$.
\end{fcdefnstar}

\begin{fcprop}[]
  \label{prop:9.5}
  $E_0(K)$ is a subgroup of \cloze{$E(K)$} and
  \cloze[1]{reduction modulo $\pi$ is a surjective
  group homomorphism $E_0(K) \to \reductionns
  E(k)$}.
\end{fcprop}

\begin{note*}
  If $E / K$ has \gls{gred}, then this is a
  surjective group homomorphism $E(K) \to
  \reduction E(k)$.
\end{note*}

\begin{proof}
  \textbf{Group homomorphism:} A line $l$ in
  $\Pbb^2$ defined over $K$ has equation
  \[
    l : aX + bY + cZ = 0 \qquad a, b, c \in K
  .\]
  We may assume $\min(v(a), v(b), v(c)) = 0$.
  Reduction modulo $\pi$ gives a line
  \[
    \tilde{l} : \tilde{a} X + \tilde{b} Y +
    \tilde{c} Z = 0
  .\]
  If $P_1, P_2, P_3 \in E(K)$ with $P_1 + P_2 +
  P_3 = 0$ then these points lie on a line $l$. So
  $\tilde{P}_1, \tilde{P}_2, \tilde{P}_3$ lie on
  the line $\tilde{l}$. If $\tilde{P}_1,
  \tilde{P}_2 \in \reductionns(k)$ then
  $\tilde{P}_3 \in \reductionns(k)$.

  So if $P_1, P_2 \in E_0(K)$ then $P_3 \in
  E_0(K)$ and $\tilde{P}_1 + \tilde{P}_2 +
  \tilde{P}_3 = 0$.

  [Exercise: check this still
  works if $\#\{\tilde{P}_1, \tilde{P}_2,
  \tilde{P}_3\} < \infty$]

  \textbf{Surjective:} Let $f(x, y) = y^2 + a_1 xy
  + a_3 y - (x^3 + \cdots)$. Let $\tilde{P} \in
  \reductionns{E}(k) \setminus \{0\}$, say
  $\tilde{P} = (\tilde{x}_0, \tilde{y}_0)$ for
  some $x_0, y_0 \in \mathcal{O}_K$.

  Since $\tilde{P}$ non-singular, either:
  \begin{enumerate}[(i)]
    \item $\frac{\partial f}{\partial x}(x_0, y_0)
      \not\equiv 0 \pmod\pi$.
    \item $\frac{\partial f}{\partial y}(x_0, y_0)
      \not\equiv 0\pmod\pi$.
  \end{enumerate}
  If (i) then put $g(t) = f(t, y_0) \in
  \mathcal{O}_K[t]$. Then
  \[
    \begin{cases}
      g(x_0) \equiv 0
      & \pmod\pi \\
      g'(x_0) \in \mathcal{O}_K^*
    \end{cases}
  \]
  Hensel's lemma gives us that there exists $b \in
  \mathcal{O}_K$ such that
  \[
    \begin{cases}
      g(b) = 0 \\
      b \equiv x_0 \pmod\pi
    \end{cases}
  \]
  Then $(b, y_0) \in E(K)$ has erduction
  $\tilde{P}$. asdfadsf
\end{proof}

Recall that for $r \ge 1$ we put
\[
  E_r(K)
  = \{(x, y) \st v(x) \le -2r, v(y) \le -3r\} \cup
  \{0\}
.\]
If $r > \frac{e}{p - 1}$, these give:
\[
  \ub{E_r(K)}_{\cong (\mathcal{O}_K, +)}
  \subset \cdots
  \subset \ub{E_2(K)}_{= \hat{E}(\pi^2
  \mathcal{O}_K)}
  \subset \ub{E_1(K)}_{= \hat{E}(\pi \mathcal{O}_K)}
  \subset E_0(K)
  \subset E(K)
,\]
where for $i \ge 1$, each $E_{i + 1}(K) \subset
E_i(K)$ gives a quotient isomorphic to $(k, +)$.

We have $\frac{E_0(K)}{E_1(K)} \cong
\reductionns{E}(k)$.

What about $\frac{E(K)}{E_0(K)}$?

\begin{fclemma}[]
  \label{lemma:9.6}
  Assuming:
  - $|k| < \infty$
  Then:
  $E_0(K) \subset E(K)$ has finite index.
\end{fclemma}

\begin{proof}
  $|k| < \infty$ implies that
  $\frac{\mathcal{O}_K}{\pi^r \mathcal{O}_K}$ is
  finite for all $r \ge 1$.

  Hence
  \[
    \mathcal{O}_K
    \cong \invlim{r} \frac{\mathcal{O}_K}{\pi^r
    \mathcal{O}_K}
  \]
  is a profinite group, hence compact.

  Then $\Pbb^n(K)$ is the union of sets
  \[
    \{(a_0 : \cdots : a_{i - 1} : 1 : a_{i + 1} :
    \cdots : a_n) : a_j \in \mathcal{O}_K\}
  \]
  and hence compact (for the $\pi$-adic topology).

  Now note $E(K) \subset \Pbb^2(K)$ is a closed
  subset, hence compact.

  So $E(K)$ is a compact topological group.

  If $\reduction E$ has a singular point
  $(\tilde{x}_0, \tilde{y}_0)$ then
  \[
    E(K) \setminus E_0(K)
    = \{(x, y) \in E(K) \st v(x - x_0) \ge 1, v(y
    - y_0) \ge 1\}
  \]
  is a closed subset of $E(K)$ hence $E_0(K)$ is
  an open subgroup of $E(K)$.

  The cosets of $E_0(K)$ are an open cover of
  $E(K)$. Hence $[E(K) : E_0(K)] < \infty$.
\end{proof}

\begin{fcdefnstar}[Tamagawa number]
  \glsnoundefn{Tnum}{Tamagawa number}{Tamagawa
  numbers}%
  \glssymboldefn{cK}%
  $c_K(E) = [E(K) : E_0(K)]$ is called the
  \emph{Tamagawa number}.
\end{fcdefnstar}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(i)]
    \item If we have \gls{gred} then $\cK(E) = 1$
      (converse is false).
    \item It can be shown that
      \begin{itemize}
        \item either $\cK(E) = v(\Delta)$
        \item or $\cK(E) \le 4$
      \end{itemize}
      for the above statements: it is essential
      that we work with \gls{min} \glspl{weq}.
  \end{enumerate}
\end{remark*}

We deduce:

\begin{fcthm}[]
  \label{thm:9.7}
  Assuming:
  - $[K : \Qbb_p] < \infty$
  Then:
  $E(K)$ contains a subgroup of finite index
  isomorphic to $(\mathcal{O}_K, +)$.
\end{fcthm}

\noproof
