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    \item Surjectivity: Let $[D] \in \Pic^0(E)$.
      Then $D + (0_E)$ has degree $1$.

      Riemann Roch gives $\dim \mathcal{L}(D +
      (0_E)) = 1$. So there exists $0 \neq f \in
      \ol{K}(E)$ such that
      \[
        \ub{\div(f) + D + (0_E)}_{\text{degree 1}} \ge 0
      .\]
      So
      \[
        \div(f) + D + (0_E) = (P)
      \]
      for some $P \in E$. Then $(P) - (0_E) \sim
      D$, so $\psi(P) = [D]$.
      \qedhere
  \end{enumerate}
\end{proof}

\subsubsection*{Formulae for $E$ in Weierstrass
form}

\[
  E: y^2 + a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4
  x + a_6
  \tag{$*$}
  \label{lec5eq}
\]
$0_E = (0 : 1 : 0)$.
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/bc82870f25624072.png}
\end{center}
$P_1 \oplus P_2 = P_3$. $\ominus P_1 = (x_1, -(a_1
x + a_3) - y_1)$.

Substituting $y = \lambda x + \nu$ into
\eqref{lec5eq} and looking at coefficient of $x^2$
gives
\[
  \lambda^2 + a_1 \lambda - a_2 = x_1 + x_2 +
  \ub{x'}_{= x_3}
.\]
Therefore
\begin{align*}
  x_3
  &= \lambda^2 + a_1 \lambda - a_2 - x_1 - x_2 \\
  y_3
  &= -(a_1 x' + a_3) - y' \\
  &= -(a_1 x_3 + a_3) - (\lambda x_3 + \nu) \\
  &= -(\lambda + a_1) x_3 - a_3 - \nu
\end{align*}
It remains to find formulae for $\lambda$ and $\nu$.
\begin{enumerate}[{Case }I]
  \item $x_1 = x_2$, $P_1 \neq P_2$. Then $P_1
    \oplus P_2 = 0_E$.
  \item $x_1 \neq x_2$.
    \begin{align*}
      \lambda
      &= \frac{y_2 - y_1}{x_2 - x_1} \\
      \nu
      &= y_1 + \lambda x_1 \\
      &= \frac{y_1 (x_2 - x_1) - x_1(y_2 -
      y_1)}{x_2 - x_1} \\
      &= \frac{x_2 y_1 - x_1 y_2}{x_2 - x_1}
    \end{align*}
  \item $P_1 = P_2$.
    \begin{align*}
      \lambda
      &= \cdots \\
      \nu
      &= \cdots
    \end{align*}
    (Compute equation for tangent line.)    
\end{enumerate}

\begin{fccoro}[]
  \label{coro:4.3}
  $E(K)$ is an abelian group.
\end{fccoro}

\begin{proof}
  It is a subgroup of $(E, \oplus)$. Identity $0_E
  \in E(K)$ by definition.

  Closure / inverses: see formulae above.

  Associative / commutative: inherited.
\end{proof}

\begin{fcthm}[]
  \label{thm:4.4}
  \glsref[ellc]{Elliptic curves} are group
  varieties, i.e. \cloze{$[-1] : E \to E$; $P
  \mapsto \ominus P$ and $\oplus : E \times E \to
  E$; $(P, Q) \mapsto P \oplus Q$ are morphisms of
  algebraic varieties.}
\end{fcthm}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Above formulae imply $[-1] : E \to E$ is
      a rational map, and hence a morphism (by
      \cref{remark:2.9}).
    \item Above formulae imply $\oplus : E \times
      E \to E$ is a rational map regular on
      \[
        U = \{(P, Q) \in E \times E \st P, Q, P
        \oplus Q, P \ominus Q \neq 0_E\}
      .\]
      For $P \in E$, let $\tau_P : E \to E$; $X
      \mapsto X \oplus P$ ``translation by $P$''.

      $\tau_P$ is a rational map, and hence a
      morphism (by \cref{remark:2.9}).

      Take any $A, B \in E$. We factor $\oplus$ as
      \[
        E \times E \stackrel{\tau_{\ominus A}
        \times \tau_{\ominus B}}{\longrightarrow}
        E \times E \stackrel{\oplus}{\to} E
        \stackrel{\tau_{A \oplus
        B}}{\longrightarrow} E
      .\]
      This shows $\oplus$ is regular on $(\tau_A
      \times \tau_B)(U)$ for all $A, B \in E$.
      Therefore $\oplus$ is regular on $E \times
      E$.
      \qedhere
  \end{enumerate}
\end{proof}

\subsubsection*{Statement of Results}

The isomorphisms in (i), (ii), (iv) respect the
relevant topologies.
\begin{enumerate}[(i)]
  \item $K = \Cbb$, $E(\Cbb) = \Cbb / \Lambda
    \cong \Rbb / \Zbb \times \Rbb / \Zbb$
    ($\Lambda$ is a lattice).
  \item $K = \Rbb$. Then
    \[
      E(\Rbb) = \begin{cases}
        \Zbb / 2\Zbb \times \Rbb / \Zbb & \text{if
        $\Delta > 0$} \\
        \Rbb / \Zbb & \text{if $\Delta < 0$}
      \end{cases}
    \]
  \item $K = \Fbb_q$ (field with $q$ elements).
    Then
    \[
      |\#E(\Fbb_q) - (q + 1)| \le 2\sqrt{q}
    .\]
    (Hasse's Theorem).
  \item $[K : \Qbb_p] < \infty$, ring of integers
    $\mathcal{O}_K$. $E(K)$ has a subgroup of
    finite index which is isomorphic to
    $(\mathcal{O}_K, +)$.
  \item $[K : \Qbb] < \infty$. $E(K)$ is a
    finitely generated abelian group (Mordell-Weil
    Theorem).
\end{enumerate}

\subsubsection*{Brief remarks on the case $K =
\Cbb$}

Let
\[
  \Lambda = \{a\omega_1 + b\omega_2 : a, b \in
  \Zbb\}
,\]
where $\omega_1, \omega_2$ a basis for $\Cbb$ as
an $\Rbb$-vector space.

Then
\[
  \{\text{meromorphic functions on the Riemann
  surface $\Cbb / \Lambda$}\}
  \leftrightarrow
  \{\text{$\Lambda$-invariant meromorphic
  functions on $\Cbb$}\}
.\]
The function field of $\Cbb / \Lambda$ is
generated by
\begin{align*}
  \wp(z)
  &= \frac{1}{z^2} + \sum_{0 \neq \lambda \in
  \Lambda} \left( \frac{1}{(z - \lambda)^2} -
  \frac{1}{\lambda^2} \right) \\
  \wp'(z)
  &= -2\sum_{\lambda \in \Lambda} \frac{1}{(z -
  \lambda)^3}
\end{align*}
These satisfy
\[
  \wp'(z)^2 = 4\wp(z)^3 - g_2 \wp(z) - g_3
\]
for some $g_2, g_3 \in \Cbb$ depending only on
$\Lambda$.

One shows $\Cbb / \Lambda = E(\Cbb)$ (isomorphism
as groups and as Riemann surfaces) where $E$:
$y^2 = 4x^3 - g_2 x - g_3$.

\begin{fcthmstar}[Uniformisation Theorem]
  Every \gls{ellc} over $\Cbb$ arises this way
  (one proof uses modular forms).
\end{fcthmstar}

\begin{fcdefnstar}[]
  For $n \in \Zbb$, let $[n] : E \to E$ be defined
  by\cloze{
  \[
    P \mapsto \ub{P \oplus P \oplus \cdots \oplus
    P}_{\text{$n$ copies}}
  \]
  for $n \ge 0$, and $[-n] = [-1] \circ [n]$.}
\end{fcdefnstar}

\begin{fcdefnstar}[$n$-torsion subgroup]
  \glssymboldefn{tors}%
  The $n$-torsion subgroup of $E$ is
  \[
    E[n] = \ker(E \stackrel{[n]}{\to} E)
  .\]
\end{fcdefnstar}

If $K = \Cbb$ then $E(\Cbb) \cong \Cbb / \Lambda$.
Therefore
\[
  \begin{cases}
    E\tors[n] \cong (\Zbb / n\Zbb)^2 & (1) \\
    \deg[n] = n^2 & (2)
  \end{cases}
\]
We'll show (2) holds over any field $K$, and (1)
holds if $\char K \nmid n$.

\begin{fclemma}[]
  \label{lemma:4.5}
  Assuming:
  - $\char K \neq 2$
  - $E$: $y^2 = f(x) = (x - e_1)(x - e_2)(x -
  e_3)$, $e_1, e_2, e_3 \in \ol{K}$
  Then:
  $E\tors[2] = \{0_E, (e_1, 0), (e_2, 0), (e_3, 0)\}
  \cong (\Zbb / 2\Zbb)^2$.
\end{fclemma}

\begin{proof}
  Let $P = (x, y) \in E$. Then
  \begin{align*}
    [2] P = 0
    &\implies P = -P \\
    &\implies (x, y) = (x, -y) \\
    &\implies y = 0
    \qedhere
  \end{align*}
\end{proof}
