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  If $E'$ is singular then $E$ and $E'$ are
  rational, contradiction.

  So $E'$ is smooth. Then \cref{remark:2.9}
  implies that $\phi^{-1}$ is a morphism. So
  $\phi$ is an isomorphism.
\end{proof}

\begin{fcprop}[]
  \label{prop:3.2}
  Assuming:
  - $K$ is perfect
  - $E$, $E'$ are \glspl{ellc} over $K$
  - $E, E'$ are in \gls{wform}
  Then:
  \begin{iffc}
    \lhs
    $E \cong E'$ over $K$
    \rhs
    the equations are related by a change of
    variables
    \begin{align*}
      x
      &= u^2 x' + r \\
      y
      &= u^3 y' + u^2 s x' + t
    \end{align*}
    where $u, r, s, t \in K$ with $u \neq 0$.
  \end{iffc}
\end{fcprop}

\begin{iffproof}
  \leftimpl
  Obvious.

  \rightimpl
  $\langle 1, x \rangle = \mathcal{L}(2.0_E) =
  \langle 1, x' \rangle$ hence $x = \lambda x' +
  r$ for some $\lambda, r \in K$ with $\lambda
  \neq 0$.

  $\langle 1, x, y \rangle = \mathcal{L}(3.0_E) =
  \langle 1, x', y' \rangle$ implies $y = \mu y' +
  \sigma x' + t$ for some $\mu, \sigma, t \in K$
  with $\mu \neq 0$.

  Looking at coefficients of $x^3$ and $y^2$, we
  get $\lambda^3 = \mu^2$. So $\lambda = u^2$,
  $\mu = u^3$ for some $0 \neq u \in K$.

  Put $s = \frac{\sigma}{u^2}$.
\end{iffproof}

A \gls{wform} equation defines an \gls{ellc} if
and only if it defines a smooth curve, which
happens if and only if $\Delta(a_1, \ldots, a_6)
\neq 0$, where $\Delta \in \Zbb[a_1, \ldots, a_6]$
is a certain polynomial.

If $\char K \neq 2, 3$, we may reduce to the case
$E : y^2 = x^3 + ax + b$, with discriminant
$\Delta = -16(4a^3 + 27b^2)$.

\begin{fccoro}[]
  \label{coro:3.3}
  Assuming:
  - $\char K \neq 2, 3$ and $K$ is perfect
  - \glspl{ellc} $E: y^2 = x^3 + ax + b$, $E' :
  y^2 = x^3 + a'x + b'$
  Then:
  \begin{iffc}
    \lhs
    $E$ and $E'$ are isomorphic over $K$
    \rhs
    $a' = u^4 a$, $b' = u^6 b$ for some $u \in
    K^*$.
  \end{iffc}
\end{fccoro}

\begin{proof}
  $E$ and $E'$ are related by a substitution as in
  \cref{prop:3.2} with $r = s = t = 0$.
\end{proof}

\begin{fcdefnstar}[$j$-invariant]
  \glsnoundefn{jinv}{$j$-invariant}{$j$-invariants}%
  The \emph{$j$-invariant} is
  \[
    j(E) = \frac{1728(4a^3)}{4a^3 + 27b^2}
  .\]
\end{fcdefnstar}

\begin{fccoro}[]
  \label{coro:3.4}
  Assuming:
  - $E \cong E'$
  Then:
  $\jinv(E) = \jinv(E')$.

  Moreover, the converse holds if $K = \ol{K}$.
\end{fccoro}

\begin{proof}
  \begin{align*}
    E \cong E'
    &\iff \begin{cases}
      a' = u^4 a \\
      b' = u^6 b
    \end{cases} \text{ for some $u \in K^*$} \\
    &\implies (a^3 : b^2) = ((a')^3 : (b')^2) \\
    &\iff \jinv(E) = \jinv(E')
  \end{align*}
  and the converse holds if $K = \ol{K}$ (to go
  backwards on the $\implies$ step, we only need
  to take some kind of $n$-th root).
\end{proof}

\newpage
\section{The Group Law}

Let $E \subset \Pbb^2$ be a smooth plane cubic,
and $0_E \in E(K)$.
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/e8d388fd8b2d4f04.png}
\end{center}
$E$ meets any line in 3 points counted with
multiplicity.

Let $S$ be the third point of intersection of $PQ$
with $E$, and $R$ be the third intersection point
of $0_E S$ and $E$.

Define $P \oplus Q = R$.

If $PQ$ then take $T_P E$ instead of $PQ$ etc.

This is called the ``chord and tangent process''.

\begin{fcthm}[]
  \label{thm:4.1}
  $(E, \oplus)$ is an abelian group.
\end{fcthm}

\begin{note*}
  $E$ here means $E(\ol{K})$. As mentioned before,
  we only ever mean ``over $K$'' if it is
  explicitly mentioned (otherwise we are always
  working ``over $\ol{K}$'').
\end{note*}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $\oplus$ is clearly commutative.
    \item $0_E$ is the identity:
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/d2bf7594dc384838.png}
      \end{center}
      $0_E \oplus P = P$.
    \item Inverses:
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/675399dfae894dcf.png}
      \end{center}
      Let $S$ be the third point of intersection
      of $T_{0_E} E$ and $E$, and let $Q$ be the
      third point of intersection of $PS$ and $E$.

      Then $P \oplus Q = 0_E$.
    \item Associativity: much harder! We will
      develop a bit of theory first.
      \qedhere
  \end{enumerate}
\end{proof}

\begin{fcdefnstar}[Linearly equivalent]
  \glsadjdefn{linequiv}{linearly
  equivalent}{divisors}%
  $D_1, D_2 \in \Div(E)$ are \emph{linearly equivalent}
  if there exists $f \in \ol{K}(E)^*$ such that
  $\div(f) = D_1 - D_2$.

  Write $D_1 \sim D_2$ and $[D] = \{D' : D' \sim
  D\}$.
\end{fcdefnstar}

\begin{fcdefnstar}[]
  $\operatorname{Pic}(E) = \frac{\Div(E)}{\sim}$,
  $\Pic^0(E) = \frac{\Div^0(E)}{\sim}$.
  where $\Div^0(E) = \{D \in \Div E \st \deg D =
  0\}$.
\end{fcdefnstar}


\begin{fcprop}[]
  \label{prop:4.2}
  Assuming:
  - we define
  \begin{align*}
    \psi : E
    &\to \Pic^0(E) \\
    P
    &\mapsto [(P) - (0_E)]
  \end{align*}
  Thens:[(i)]
  - $\psi(P \oplus Q) = \psi(P) + \psi(Q)$.
  - $\psi$ is a bijection.
\end{fcprop}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item
      \begin{align*}
        \div(l / m)
        &= (P) + \cancel{(S)} + (Q) -
        (0_E) - \cancel{S} - (R) \\
        &= (P) + (Q) - (0_E) - (P \oplus Q)
      \end{align*}
      Hence $(P) + (Q) \sim (P \oplus Q) + (0_E)$.
      Therefore $(P \oplus Q) - (0_E) \sim (P) -
      (0_E) + (Q) - (0_E)$. So $\psi(P \oplus Q) =
      \psi(P) + \psi(Q)$.
    \item Injectivity: Suppose $\psi(P) = \psi(Q)$
      with $P \neq Q$.

      Then there exists $f \in \ol{K}(E)^*$ such
      that $\div(f) = (P) - (Q)$, so $E
      \stackrel{f}{\to} \Pbb^1$ has degree 1,
      hence $E \cong \Pbb^1$, contradiction.

      Note: We now compute
      \[
        \psi((P \oplus Q) \oplus R) = \psi(P) +
        \psi(Q) + \psi(R) = \psi(P \oplus (Q
        \oplus R))
      \]
      hence $(P \oplus Q) \oplus R = P \oplus (Q
      \oplus R)$ for all $P, Q, R \in E$ (using
      injectivity). So $(E, \oplus)$ is
      associative.