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% Examples Class
% Friday 14th February
% 3:30PM MR9
% 2, 9, 10

\newpage
\section{Isogenies}

Let $E_1, E_2$ be \glspl{ellc}.

\begin{fcdefnstar}[Isogeny]
  \glsnoundefn{isog}{isogeny}{isogenies}%
  \glsnoundefn{isogous}{isogenous}{\glspl{ellc}}%
  An \emph{isogeny} $\phi : E_1 \to E_2$ is a
  nonconstant morphism with $\phi(0_{E_1})
  = 0_{E_2}$ (by \cref{thm:2.8}, a morphism is
  nonconstant if and only if surjective on
  $\ol{K}$-points).

  We say $E_1$ and $E_2$ are isogenous in this
  case.
\end{fcdefnstar}

\begin{fcdefnstar}[]
  $\Hom(E_1, E_2) = \{\text{\glspl{isog} $E_1 \to
  E_2$}\} \cup \{0\}$.
\end{fcdefnstar}

This is an abelian group under
\[
  (\phi + \psi)(P) = \phi(P) + \psi(P)
.\]
If $E_1 \stackrel{\phi}{\to} E_2
\stackrel{\psi}{\to} E_3$ are \glspl{isog} then
$\psi \circ \phi$ is an \gls{isog}.

Tower Law implies $\deg(\psi \phi) =
\deg(\phi)\deg(\psi)$.

\begin{fcprop}[]
  \label{prop:5.1}
  Assuming:
  - $0 \neq n \in \Zbb$
  Then:
  $[n] : E \to E$ is an \gls{isog}.
\end{fcprop}

\begin{proof}
  $[n]$ is a morphism by \cref{thm:4.4}. We must
  show $[n] \neq [0]$.

  Assume $\char K \neq 2$.

  Case $n = 2$: \cref{lemma:4.5} implies $E\tors[2]
  \neq E$ implies $[2] \neq [0]$.

  Case $n$ odd: \cref{lemma:4.5} implies $\exists
  0 \neq T \in E\tors[2]$. Then $nT = T \neq 0$ which
  gives $[n] \neq [0]$.

  Now use $[mn] = [m] \circ [n]$.

  If $\char K = 2$,  then could rpelace
  \cref{lemma:4.5} with an explicit lemma about
  $3$-torsion points.
\end{proof}

\begin{fccoro}[]
  \label{coro:5.2}
  \cloze{$\Hom(E_1, E_2)$} is a torsion free
  $\Zbb$-module.
\end{fccoro}

\noproof

\begin{fcthm}[]
  \label{thm:5.3}
  Assuming:
  - $\phi : E_1 \to E_2$ an \gls{isog}
  Then:
  $\phi(P + Q) = \phi(P) + \phi(Q)$ for all $P, Q
  \in E_1$.
\end{fcthm}

\begin{proof}[Proof (sketch)]
  $\phi$ incudes
  \begin{align*}
    \phi_* : \Div^0(E_1)
    &\to \Div^0(E_2) \\
    \sum_{P \in E_1} n_P P
    &\mapsto \sum_{P \in E_1} n_P \phi(P)
  \end{align*}
  Recall $\phi^* : K(E_2) \hookrightarrow K(E_1)$.
  \begin{picmath}
    \begin{tikzcd}
      K(E_1) \ar[d, no head]
      \ar[d, bend left=50, "\text{norm map}"]
      \\
      K(E_2)
    \end{tikzcd}
  \end{picmath}
  \textbf{Fact:} If $f \in \ol{K}(E_1)^*$ then
  \[
    \div(N_{K(E_1) / K(E_2)} f) = \phi_*(\div f)
  .\]
  So $\phi_*$ sends principal divisors to
  principal divisors.

  Since $\phi(0_{E_1}) = 0_{E_2}$, the following
  diagram commutes:
  \begin{picmath}
    \begin{tikzcd}
      P
      \ar[d, mapsto]
      & E_1 \ar[r, "\phi"]
      \ar[d, "\cong"]
      & E_2
      \ar[d, "\cong"]
      & Q
      \ar[d, mapsto]
      \\
      {[(P) - (0_{E_1})]}
      & \Pic^0(E_1)
      \ar[r, "\phi_*"]
      & \Pic^0(E_2)
      & {[(Q) - (0_{E_2})]}
    \end{tikzcd}
  \end{picmath}
  $\phi_*$ a group homomorphism implies $\phi$ is
  a group homomorphism.
\end{proof}

\begin{fclemma}[]
  \label{lemma:5.4}
  Assuming:
  - $\phi : E_1 \to E_2$ an \gls{isog}
  Then:
  there exists a morphism $\xi$ making the
  following diagram commute:
  \begin{picmath}
    \begin{tikzcd}
      E_1
      \ar[r, "\phi"]
      \ar[d, "x_1"]
      & E_2
      \ar[d, "x_2"]
      \\
      \Pbb^1
      \ar[r, "\xi"]
      & \Pbb^1
    \end{tikzcd}
  \end{picmath}
  ($x_i = $ $x$-coordinate on a Weierstrass
  equation for $E_i$).

  Moreover if $\xi(t) =
  \frac{r(t)}{s(t)}$, $r, s \in K[t]$ coprime,
  then
  \[
    \deg \phi = \deg \xi = \max(\deg(r), \deg(s))
  .\]
\end{fclemma}

\begin{proof}
  For $i = 1, 2$. $K(E_i) / K(x_i)$ is a degree
  $2$ Galois extension, with Galois group
  generated by $[-1]^*$.

  \cref{thm:5.3} implies that $\phi \circ [-1] =
  [-1] \circ \phi$.

  So if $f \in K(x_2)$ then 
  \[
    [-1]^* \phi^* f
    = \phi^* [-1]^* f
    = \phi^* f
  .\]
  Therefore $\phi^* f \in K(x_1)$.
  \begin{picmath}
    \begin{tikzcd}
      & K(E_1) = K(x_1, y_1)
      \ar[ld, no head, "2"]
      \ar[dd, no head, "\deg \phi"]
      \\
      K(x_1)
      \ar[dd, no head, "\deg \xi"]
      \\
      & K(E_2) = K(x_2, y_2)
      \ar[ld, no head, "2"]
      \\
      K(x_2)
    \end{tikzcd}
  \end{picmath}
  In particular, $\phi^* = \xi(x_1)$ for some
  rational function $\xi$.

  Tower Law implies $\deg \phi = \deg \xi$.

  Now $K(x_2) \hookrightarrow K(x_1)$, $x_2
  \mapsto \xi(x_1) = \frac{r(x_1)}{s(x_1)}$, $r, s
  \in K[t]$ coprime.

  We claim the minimal polynomial of $x_1$ over
  $K(x_2)$ is
  \[
    F(t)
    = r(t) - s(t) x_2
    \in K(x_2) [t]
  .\]
  Check:
  \begin{itemize}
    \item $F(x_1) = 0$: true by the definition of
      our embedding.
    \item $F$ is irreducible in $K[x_2, t]$ (since
      $r, s$ coprime), hence irreducible in
      $K(x_2)[t]$ by Gauss's lemma.
  \end{itemize}
  Therefore
  \[
    \deg \xi
    = [K(x_1) : K(x_2)]
    = \deg F
    = \max(\deg r, \deg s)
    .
    \qedhere
  \]
\end{proof}

\begin{fclemma}[]
  $\deg[2] = 4$.
\end{fclemma}

\begin{proof}
  Assume $\char K \neq 2, 3$ (the result is true
  even in the case of $\char K \in \{2, 3\}$, but
  we will only prove the simpler case).

  $E$: $y^2 = x^3 + ax + b = f(x)$.

  If $P = (x, y) \in E$, then
  \begin{align*}
    x(2P)
    &= \left( \frac{3x^2 + a}{2y} \right)^2 - 2x
    \\
    &= \frac{(3x^2 + a)^2 - 8x f(x)}{4f(x)} \\
    &= \frac{x^4 + \cdots}{f(x)}
  \end{align*}
  The numerator and denominator are coprime.
  Indeed, otherwise there exists $\theta \in
  \ol{K}$ with $f(\theta) = f'(\theta) = 0$ and
  hence $f$ has a multiple root (contradiction).

  Now \cref{lemma:5.4} implies $\deg[2] = \max(4,
  3) = 4$.
\end{proof}

\begin{fcdefnstar}[Quadratic form in an abelian
  group]
  \glsnoundefn{qf}{quadratic form}{quadratic
  forms}%
  Let $A$ be an abelian group. We say $q : A \to
  \Zbb$ is a \emph{quadratic form} if
  \begin{enumerate}[(i)]
    \item $q(nx) = n^2 q(x)$ for all $n \in \Zbb$,
      $x \in A$.
    \item $(x, y) \mapsto q(x + y) - q(x) - q(y)$
      is $\Zbb$-bilinear.
  \end{enumerate}
\end{fcdefnstar}

\begin{fclemma}[]
  \label{lemma:5.6}
  \begin{iffc}
    \lhs
    $q : A \to \Zbb$ is a \gls{qf}
    \rhs
    it satisfies the parallelogram law:
    \[
      q(x + y) + q(x - y) = 2q(x) = 2q(y)
      \qquad \forall x, y \in A
    .\]
  \end{iffc}
\end{fclemma}

\begin{iffproof}
  \rightimpl
  Let $\langle x, y \rangle = q(x + y) - q(x) -
  q(y)$. Then
  \[
    \langle x, x \rangle = q(2x) - 2q(x) = 2q(x)
  \]
  (the equality in the last step is using property
  (i) with $n = 2$).

  But by (ii),
  \[
    \langle x + y, x + y \rangle + \langle x - y,
    x - y \rangle
    = 2\langle x, x \rangle + 2\langle y, y
    \rangle
  .\]
  Hence (upon dividing by $2$),
  \[
    q(x + y) + q(x - y)
    = 2q(x) + 2q(y)
  .\]

  \leftimpl
  See \es{2}.
\end{iffproof}
