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\begin{theorem}[]
  \label{thm:5.7}
  $\deg : \Hom(E_1, E_2) \to \Zbb$ is a \gls{qf}.
\end{theorem}

\noproof

\begin{note*}
  We define $\deg(0) = 0$.
\end{note*}

For the proof we assume $\char K \neq 2, 3$.
Write $E_2$: $y^2 = x^3 + ax + b$.

Let $P, Q \in E_2$ with $P, Q, P + Q, P - Q \neq
0$. Let $x_1, \ldots, x_4$ be the
$x$-coordinates of the $4$ points.

\begin{fclemma}[]
  \label{lemma:5.8}
  There exists $W_0, W_1, W_2 \in \Zbb[a, b][x_1,
  x_2]$ of degree in $x_1$ and of degree $\le 2$
  in $x_2$ such that
  \[
    (1 : x_3 + x_4 : x_3 x_4)
    = (W_0 : W_1 : W_2)
  .\]
\end{fclemma}

\begin{proof}
  Two methods.
  \begin{enumerate}[(1)]
    \item Direct calculation:
      \begin{align*}
        W_0
        &= (x_1 - x_2)^2 \\
        W_1
        &= \cdots \\
        W_2
        &= \cdots
      \end{align*}
      see formula sheet.
    \item Let $y = \lambda x + \nu$ be the line
      through $P$ and $Q$.
      \begin{align*}
        x^3 + ax + b - (\lambda x + \nu)^2
        &= (x - x_1)(x - x_2)(x - x_3) \\
        &= x^3 - s_1 x^2 + x_2 x - s_3
      \end{align*}
      where $s_i$ is the $i$-th elementary
      symmetric polynomial in $x_1, x_2, x_3$.

      Comparing coefficients:
      \begin{align*}
        \lambda^2
        &= s_1 \\
        -2\lambda \nu
        &= s_2 - a \\
        \nu^2
        &= s_3 + b
      \end{align*}
      Eliminating $\lambda$ and $\nu$ gives
      \[
        \ub{(s_2 - a)^2 - 4 s_1(s_3 + b) =
        0}_{=F(x_1, x_2, x_3)}
      \]
      where $F$ has degree $\le 2$ in $x_i$.

      $x_3$ is a root of the quadratic
      \[
        W(t) = F(x_1, x_2, t)
      .\]
      Repeating for the line through $P$ and $-Q$
      shows that $x_4$ is the other root.
      Therefore
      \[
        W_0(t - x_3)(t - x_4) = W_0 t^2 - W_1 t +
        W_2
      ,\]
      hence
      \[
        (1 : x_3 + x_4 : x_3 x_4) = (W_0 : W_1 :
        W_2)
        .
        \qedhere
      \]
  \end{enumerate}
\end{proof}

We show that if $\phi, \psi \in \Hom(E_1, E_2)$,
then
\[
  \deg(\phi + \psi) + \deg(\phi - \psi)
  \le 2\deg(\phi)+  2\deg(\psi)
.\]
We may assume $\phi, \psi, \phi + \psi, \phi -
\psi \neq 0$ (otherwise trivial, or use $\deg[-1]
= 1$, $\deg[2] = 4$).
\begin{align*}
  \phi
  &: (x, y) \mapsto (\xi_1(x), \ldots) \\
  \psi
  &: (x, y) \mapsto (\xi_2(x), \ldots) \\
  \phi + \psi
  &: (x, y) \mapsto (\xi_3(x), \ldots) \\
  \phi - \psi
  &: (x, y) \mapsto (\xi_4(x), \ldots)
\end{align*}
\cref{lemma:5.8} implies
\[
  (1 : \xi_3 + \xi_4 : \xi_3 \xi_4)
  = ((\xi_1 - \xi_2)^2 : \cdots)
.\]
Put $\xi_i = \frac{r_i}{s_i}$, $r_i, s_i \in K[t]$
coprime.
\[
  \ub{(s_3 s_4 : r_3 s_4 + r_4 s_4 : r_3
  r_4)}_{\text{coprime}}
  = ((r_1 s_2 - r_2 s_1)^2 : \cdots)
  \tag{$*$}
  \label{lec7eq}
.\]
Therefore
\begin{align*}
  \deg(\phi + \psi) + \deg(\phi - \psi)
  &= \max(\deg(r_3), \deg(s_3)) + \max(\deg(r_4),
  \deg(s_4)) \\
  &= \max(\deg(s_3 s_4), \deg(r_3 s_4 + r_3 s_3),
  \deg(r_3 r_4)) \\
  &\le 2\max(\deg(r_1), \deg(s_1)) +
  2\max(\deg(r_2), \deg(s_2)) \\
  &= 2\deg(\phi) + 2\deg(\psi)
  \label{lec7eq1}
  \tag{1}
\end{align*}
Now replace $\phi, \psi$ by $\phi + \psi, \phi -
\psi$ to get
\begin{align*}
  \deg(2\phi) + \deg(2\psi)
  &\le 2\deg(\phi + \psi) + 2\deg(\phi - \psi) \\
  4 \deg(\phi) + 4\deg(\psi)
  &\le 2\deg(\phi + \psi) + 2\deg(\phi - \psi) \\
  2 \deg(\phi) + 2\deg(\psi)
  &\le \deg(\phi + \psi) + \deg(\phi - \psi)
  \label{lec7eq2}
  \tag{2}
\end{align*}
\eqref{lec7eq1} and \eqref{lec7eq2} give that
$\deg$ satisfies the parallelogram law, hence
$\deg$ is a \gls{qf}.

This proves \cref{thm:5.7}.

\begin{fccoro}[]
  \label{coro:5.9}
  $\deg(n\phi) = n^2 \deg(\phi)$ for all $n \in
  \Zbb$, $\phi \in \Hom(E_1, E_2)$.

  In particular, $\deg[n] = n^2$.
\end{fccoro}

\begin{example}
  \label{eg:5.10}
  Let $E / K$ be an \gls{ellc}. Suppose $\char K
  \neq 2$. Let $0 \neq T \in E(K)[2]$.

  Without loss of generality $E$: $y^2 = x(x^2 +
  ax + b)$, $a, b \in K$, $b(a^2 - 4b) \neq 0$,
  and $T = (0, 0)$.

  If $P = (x, y)$ and $P' = P + T = (x', y')$ then
  \begin{align*}
    x'
    &= \left( \frac{y}{x} \right)^2 - a - x \\
    &= \frac{x^2 + ax + b}{x} - a - x \\
    &= \frac{b}{x} \\
    y'
    &= \left( \frac{y}{x} \right) x' \\
    &= -\frac{by}{x}
  \end{align*}
  Let
  \begin{align*}
    \xi
    &= x + x' + a = \left( \frac{y}{x} \right)^2
    \\
    \eta
    &= y + y' = \frac{y}{x} \left( x - \frac{b}{x}
    \right) \\
    \eta^2
    &= \left( \frac{y}{x} \right)^2 \left[ \left(
    x + \frac{b}{x} \right)^2 - 4b \right] \\
    &= \xi [(\xi - a)^2 - 4b] \\
    &= \xi(\xi^2 - 2a\xi + a^2 - 4b)
  \end{align*}
  Let $E'$: $y^2 = x(x^2 + a'x + b')$, where $a' =
  -2a$, $b' = a^2 - 4b$. There is an \gls{isog}
  \begin{align*}
    \phi : E
    &\to E' \\
    (x, y)
    &\mapsto \left( \left( \frac{y}{x} \right)^2 :
    \frac{y(x^2 - b)}{x^2} : 1 \right) \\
    &~~~~~    -2 ~~~~~~~~   -3 ~~~~~~~ 0 ~~~~~~~ \ord_{0_E}(\blank) \\
    &~~~~~~~~~ 1 ~~~~~~~~~~~ 0 ~~~~~~~ 3 ~~~~~~~
    +3 \\
    0_E
    &\mapsto (0 : 1 : 0) = 0_{E'}
  \end{align*}
  To compute the degree: $\left( \frac{y}{x}
  \right)^2 = \frac{x^2 + ax + b}{x}$ (coprime
  numerator and denominator as $b \neq 0$).
  \cref{lemma:5.4} gives $\deg \phi = 2$.

  We say $\phi$ is a $2$-\gls{isog}.
\end{example}

\newpage
\section{The Invariant Differential}

Let $C$ be an algebraic curve over $K = \ol{K}$.

\begin{fcdefnstar}[Space of differentials]
  The \emph{space of differentials} $\Omega_C$ is
  the $K(C)$-vector space generated by $\dd f$ for
  $f \in K(C)$, subject to the relations
  \begin{enumerate}[(i)]
    \item $\dd(f + g) = \dd f + \dd g$
    \item $\dd (fg) = f \dd g + g \dd f$
    \item $\dd a = 0~~\forall a \in K$
  \end{enumerate}
\end{fcdefnstar}

\textbf{Fact:} $\Omega_C$ is a $1$-dimensional
$K(C)$-vector space.
