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\begin{proof}
  Without loss of generality $K = \ol{K}$. By a
  change of coordinates, we may assume
  \[
    y^2 = x(x - 1)(x - \lambda)
  \]
  for some $\lambda \in K \setminus \{0, 1\}$.
  Suppose $(x, y) \in E(K(t))$. Put $x =
  \frac{u}{v}$, where $u, v \in K[t]$ are coprime.

  Then $w^2 = uv(u - v)(u - \lambda v)$ for some
  $w \in K[t]$.

  Unique factorisation in $K[t]$ gives that $u, v,
  u - v, u - \lambda v$ are squares. Hence by
  \cref{lemma:1.4}, $u, v \in K$, so $x \in K$, so
  $y \in K$.
\end{proof}

\newpage
\section{Some Remarks on Plane Curves}

Work over $K = \ol{K}$.

\begin{fcdefn}[Rational plane affine curve]
  \label{defn:2.1}
  \glsadjdefn{ratc}{rational}{plane affine curve}%
  A plane affine curve $C = \{f(x, y) = 0\}
  \subset \Abb^2$ is \emph{rational} if it has a
  rational parametrisation, i.e. $\exists \phi(t),
  \psi(t) \in K(t)$ such that:
  \begin{enumerate}[(i)]
    \item $\Abb^1 \to \Abb^2$, $t \mapsto
      (\phi(t), \psi(t))$ is injective on $\Abb^1
      \setminus \{\text{finite set}\}$.
    \item $f(\phi(t), \psi(t)) = 0$.
  \end{enumerate}
\end{fcdefn}

\begin{example}
  \label{eg:2.2}
  \phantom{}
  \begin{enumerate}[(a)]
    \item Any (non-singular) plane conic is
      \gls{ratc}.
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/68a70d5ef3d74904.png}
      \end{center}
      Substitute $y = t(x + 1)$. We get $x^2 +
      t^2(x + 1)^2 = 1$, hence $(x + 1)((x - 1) +
      t^2(x + 1)) = 0$, hence $x = -1$ or $x =
      \frac{1 - t^2}{1 + t^2}$.

      Therefore this has a \gls{ratc}
      parametrisation
      \[
        (x, y) = \left( \frac{1 - t^2}{1 + t^2},
        \frac{2t}{1 + t^2} \right)
      .\]
    \item Any singular plane cubic is \gls{ratc}
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/a21320b0be3c4dda.png}
      \end{center}
      \gls{ratc} parametrisation $(x, y) = (t^2,
      t^3)$.

      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/ad0b793066dd4b2b.png}
      \end{center}
      \gls{ratc} parametrisation: $(x, y) =
      (exercise, exercise)$.
    \item \cref{coro:1.6} shows that \glspl{tellc}
      are \emph{not} \gls{ratc}.
  \end{enumerate}
\end{example}

\begin{remark}
  \label{remark:2.3}
  The genus $g(C) \in \Zbb_{\ge 0}$ is an
  invariant of a smooth projective curve $C$.
  \begin{itemize}
    \item If $K = \Cbb$ then $g(C) = \text{genus
      of Riemann surface}$.
    \item A smooth plane curve $C \subset \Pbb^2$
      of degree $d$ has genus $g(C) = \frac{(d -
      1)(d - 2)}{2}$.
  \end{itemize}
\end{remark}

\begin{fcprop}
  \label{prop:2.4}
  Assuming:
  - $K = \ol{K}$
  - $C$ a smooth projective curve
  Then:
  \begin{enumerate}[(i)]
    \item $C$ is rational (see \cref{defn:2.1})
      if and only if $g(C) = 0$.
    \item $C$ is an \gls{tellc} (see
      \cref{defn:1.5}) if and only if $g(C) = 1$.
  \end{enumerate}
\end{fcprop}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Omitted.
    \item $\Rightarrow$: Check $C$ a smooth plane
      curve (exercise). Then use
      \cref{remark:2.3}.

      $\Leftarrow$: See later.
  \end{enumerate}
\end{proof}

\subsubsection*{Order of vanishing}

$C$ algebraic curve, function field $K(C)$, $P \in
C$ smooth point.

We write $\ord_p(f)$ for the order of vanishing of
$f \in K(C)^\times$ at $P$ (negative of $f$ has a pole).

\textbf{Fact:} $\ord_p : K(C)^\times \to \Zbb$ is
a discrete valuation, i.e. $\ord_p(f_1 f_2) =
\ord_p(f_1) + \ord_p(f_2)$ and $\ord_p(f_1 + f_2)
\ge \min(\ord_p(f_1), \ord_p(f_2))$.

\begin{definition*}[Uniformiser]
  $t \in K(C)^\times$ is a uniformiser at $P$ if
  $\ord_p(t) = 1$.
\end{definition*}

\begin{example}
  \label{eg:2.5}
  $C = \{g = 0\} \subset \Abb^2$, $g \in K[x, y]$
  irreducible.

  $K(C) = \Frac \frac{K[x, y]}{(g)}$.
  \[
    g = g_0 + g_1(x, y) + g_2(x, y) + \cdots
  \]
  where $g_i$ are homogeneous of degree $i$.

  Suppose $P = (0, 0) \in C$ is a smooth point,
  i.e. $g_0 = 0$, $g_1(x, y) = \alpha x + \beta y$
  with $\alpha, \beta$ \emph{not} both zero.

  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/8c83429a86804baf.png}
  \end{center}
  \textbf{Fact:} $\gamma x + \delta y \in K(C)$ is
  a uniformiser at $P$ if and only if $\alpha
  \delta - \beta \gamma \neq 0$.
\end{example}

\begin{example}
  \label{eg:2.6}
  \[
    \{y^2 = x(x - 1)(x - \lambda)\} \subset \Abb^2
  \]
  where $\lambda \neq 0, 1$. Projective closure
  ($x = \frac{X}{Z}$, $y = \frac{Y}{Z}$):
  \[
    \{Y^2 Z = X(X - Z)(X - \lambda Z)\} \subset
    \Pbb^2
  .\]
  Let $P = (0 : 1 : 0)$.

  \textbf{Aim:} Compute $\ord_p(x)$ and
  $\ord_p(y)$.

  Put $w = \frac{Z}{Y}$, $t = \frac{X}{Y}$. So
  \[
    w = t(t - w)(t - \lambda w)
    \tag{$*$}
    \label{eg:2.6_eq}
  \]
  Now $P$is the point $(t, w) = (0, 0)$. This is a
  smooth point with
  \[
    \ord_p(t) = \ord_p(t - w) = \ord_p(t - \lambda
    w) = 1
  .\]
  \eqref{eg:2.6_eq} implies $\ord_p(w) = 3$.
  Therefore $\ord_p(x) = \ord_p(t / w) =1 - 3 =
  -2$ and $\ord_p(y) = \ord_p(1 / w) = -3$.
\end{example}

\subsubsection*{Riemann Roch Spaces}

Let $C$ be a smooth projective curve.

\begin{fcdefnstar}[Divisor]
  A \emph{divisor} is a formal sum of points on
  $C$, say
  \[
    D = \sum_{P \in C} n_p P
  \]
  where $n_p \in \Zbb$ and $n_p = 0$ for all but
  finitely many $P \in C$.

  We write $\deg D = \sum n_P$.

  We say $D$ is effective (written $D \ge 0$) if
  $n_P \ge 0$ for all $P$.
\end{fcdefnstar}

\glssymboldefn{div}%
If $f \in K(C)^\times$, then $\div(f) = \sum_{P
\in C} \ord_p(f) P$.

The Riemann Roch space of $D \in \Div(C)$ is
\[
  \mathcal{L}(D) = \{f \in K(C)^\times \st \div(f)
  + D \ge 0\} \cup \{0\}
,\]
i.e. the $K$-vector space of rational functions on
$C$ with ``poles no worse than specified by $D$''.

We quote: Riemann Roch for genus $1$:
\[
  \dim \mathcal{L}(D) = \begin{cases}
    \deg D & \text{if $\deg D > 0$} \\
    0 \text{ or } 1 & \text{if $\deg D = 0$} \\
    0 & \text{if $\deg D < 0$}
  \end{cases}
\]
For example, in \cref{eg:2.6}:
\begin{align*}
  \mathcal{L}(2P)
  &= \langle 1, x \rangle \\
  \mathcal{L}(3P)
  &= \langle 1, x, y \rangle
\end{align*}
