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\begin{remark*}
  For $K$ a number field and $P = (a_0 : a_1 :
  \cdots : a_n) \in \Pbb^n(K)$, define
  \[
    H(P) = \prod_v \max_{0 \le i \le n} |a_i|_v
  \]
  where the product is over all places $v$, and
  the absolute values are normalised such that
  \[
    \prod_v |\lambda|_v = 1 \qquad \forall \lambda
    \in K^*
  .\]
  Using this definition, all results in this
  section generalise when $\Qbb$ is replaced by a
  number field $K$.
\end{remark*}

\newpage
\section{Dual isogenies and the Weil pairing}

Let $K$ be a perfect field and $E / K$ an
\gls{ellc}.

\begin{fcprop}[]
  \label{prop:14.1}
  Assuming:
  - $\Phi \subset E(\ol{K})$ be a finite
  $\Gal(\ol{K} / K)$-stable subgroup
  Then:
  there exists an \gls{ellc} $E' / K$ and a
  separable \gls{isog} $\phi : E \to E'$ defined
  over $K$, with kernel $\Phi$, such that
  every \gls{isog} $\psi : E \to E''$ with $\Phi
  \subset \psi$ factors uniquely via $\phi$:
  \begin{picmath}
    \begin{tikzcd}
      E \ar[rr, "\psi"] \ar[rd, "\phi", swap]
      &
      & E'' \\
      & E' \ar[ur, dashed,
      "\exists \text{ unique}", swap]
    \end{tikzcd}
  \end{picmath}
\end{fcprop}

\begin{proof}
  Omitted (see Silverman, Chapter III, Proposition
  4.12).
\end{proof}

\begin{fcprop}[]
  \label{prop:4.12}
  Assuming:
  - $\phi : E \to E'$ an \gls{isog} of degree $n$
  Then:
  there exists a unique \gls{isog} $\hat{\phi} :
  E' \to E$ such that $\hat{\phi} \phi = [n]$.
\end{fcprop}

\begin{proof}
  \textbf{Case $\phi$ is separable:} We
  have $|\ker \phi| = n$, hence $\ker \phi \subset
  E\tors[n]$. Apply \cref{prop:14.1} with $\psi =
  [n]$.

  \textbf{Case $\phi$ is inseparable:} omitted.

  \textbf{Uniqueness:} Suppose $\psi_1 \phi =
  \psi_2 \phi = [n]$. Then $(\psi_1 - \psi_2) \phi
  = 0$, so $\deg(\psi_1 - \psi_2) \deg \phi = 0$.
  Then $\deg(\psi_1 - \psi_2) = 0$, hence $\psi_1
  = \psi_2$.
\end{proof}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Write $E_1 \sim E_2$ to mean ``$E_1$ and
      $E_2$ are \gls{isogous}''. Then $\sim$ is an
      equivalence relation.
    \item $\deg[n] = n^2$ gives that
      $\deg\hat{\phi} = \deg\phi$ and $\hat{[n]} =
      [n]$.
    \item Note:
      \[
        \phi \hat{\phi} \phi
        = \phi \circ [n]_E
        = [n]_{E'} \circ \phi
      \]
      Hence $\phi \hat{\phi} = [n]_{E'}$.

      In particular, $\hat{\hat{\phi}} = \phi$.
    \item If $E \stackrel{\psi}{\to} E'
      \stackrel{\phi}{\to} E''$ then $\hat{\phi
      \psi} = \hat{\psi} \hat{\phi}$.
    \item If $\phi \in \End(E)$ then
      \[
        \phi^2 - (\tr\phi) \phi + \deg \phi = 0
      .\]
      So
      \[
        \ub{([\tr \phi] - \phi)\phi}_{= \hat{\phi}}
        = [\deg\phi]
      .\]
      Hence $[\tr\phi] = \phi + \hat{\phi}$.
  \end{enumerate}
\end{remark*}

\begin{fclemma}[]
  \label{lemma:14.3}
  Assuming:
  - $\phi, \psi \in \Hom(E, E')$
  Then:
  $\widehat{\phi + \psi} = \hat{\phi} + \hat{\psi}$.
\end{fclemma}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item If $E = E'$, then this follows from
      $\tr(\phi + \psi) = \tr(\phi) + \tr(\psi)$.
    \item In general, let $\alpha : E' \to E$ be
      any \gls{isog} (e.g. $\hat{\phi}$). Then:
      \begin{align*}
        \widehat{(\alpha \phi + \alpha \psi)}
        &= \widehat{\alpha \phi} + \widehat{\alpha \psi}
        \\
        \widehat{(\phi + \psi)} \hat{\alpha}
        &= (\hat{\phi} + \hat{\psi}) \hat{\alpha}
      \end{align*}
      Hence the result follows.
      \qedhere
  \end{enumerate}
\end{proof}

\begin{remark*}
  In Silverman's book he proves
  \cref{lemma:14.3}first, and uses this to show
  $\deg : \Hom(E, E') \to \Zbb$ is a \gls{qf}.
\end{remark*}

\begin{notation*}
  \begin{align*}
    \summ : \Div(E)
    &\to E \\
    \ub{\sum n_p (P)}_{\text{formal sum}}
    &\mapsto \ub{\sum n_P(P)}_{add using group law}
  \end{align*}
\end{notation*}

Recall
\begin{align*}
  E
  &\congto \Pic^0(E) \\
  P
  &\mapsto [(P) - (0)]
\end{align*}
Therefore $\summ D \mapsto [D]$ for all $D \in
\Div^0(E)$.

We deduce:

\begin{fclemma}[]
  \label{lemma:14.4}
  Assuming:
  - $D \in \Div(E)$
  Then:
  \begin{iffc}
    \lhs
    $D \sim 0$
    \rhs
    $\deg D = 0$ and $\summ D = 0_E$.
  \end{iffc}
\end{fclemma}

We will now discuss Weil pairing.

Let $\phi : E \to E'$ be an \gls{isog} of degree
$n$, with Dual \gls{isog} $\hat{\phi} : E' \to E$.
Assume $\char K \nmid n$ (hence $\phi$,
$\ol{\phi}$ separable).

We define the Weil pairing
\[
  e_\phi : E[\phi] \times E'[\hat{\phi}]
  \to \mu_n
.\]
Let $T \in E'[\phi]$. Then $nT = 0$, so there
exists $f \in \ol{K}(E')^*$ such that
\[
  \div(f) = n(T) - n(0)
.\]
Pick $T_0 \in E(\ol{K})$ with $\phi(T_0) = T$.
Then
\[
  \phi^*(T) - \phi^*(0)
  =\sum_{P \in E[\phi]} (P + T_0)
  - \sum_{P \in E[\phi]} (P)
\]
has sum
\[
  nT_0
  = \hat{\phi} \phi T_0
  = \hat{\phi} T
  = 0
.\]
So there exists $g \in \ol{K}(E)^*$ such that
\[
  \div(g)
  = \phi^*(T) - \phi^*(0)
.\]
Now
\begin{align*}
  \div(\phi^* f)
  &= \phi^* (\div f) \\
  &= n(\phi^* (T) - \phi^*(0)) \\
  &= \div(g^n)
\end{align*}
Therefore $\phi^* f = cg^n$ for some $c \in
\ol{K}^*$.

Rescaling $f$, we can say without loss of generality
$c = 1$, i.e. $\phi^* f = g^n$.

For $S \in E[\phi]$ we get $\tau_S^*(\div g) =
(\div g)$ so $\tau_S^* g = \zeta g$ for some
$\zeta \in \ol{K}^*$, i.e. $\zeta = \frac{g(X +
S)}{g(X)}$ is independent of choice of $X \in
E(\ol{K})$.

Now
\[
  \zeta^n
  = \frac{g(X + S)^n}{g(X)^n}
  = \frac{f(\phi(X + S))}{f(\phi(X))}
  = 1
\]
since $S \in E[\phi]$. Hence $\zeta \in \mu_n$.

\glssymboldefn{ephi}%
We define
\[
  e_\phi(S, T)
  = \frac{g(X + S)}{g(X)}
.\]

\begin{fcprop}[]
  \label{prop:14.5}
  $\ephi$ is \cloze{bilinear and non-degenerate.}
\end{fcprop}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Linearity in first argument:
      \begin{align*}
        \ephi(S_1 + S_2, T)
        &= \frac{g(X + S_1 + S_2)}{g(X + S_2)}
        \frac{g(X + S_2)}{g(X)} \\
        &= \ephi(S, T) \ephi(S_2, T)
      \end{align*}