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\begin{example*}
  $E_D$: $y^2 = x^3 - D^2 x = f(x)$. $D \in \Zbb$
  square-free, $\Delta = 2^6 D^6$. If $p \nmid
  2D$, then
  \[
    \#\reduction{E}_D(\Fbb_p)
    = 1 + \sum_{x \in \Fbb_p} \left( \left(
    \frac{f(x)}{p} \right) + 1 \right) 
  .\]
  If $p \equiv 3 \pmod 4$ then since $f$ is odd,
  \[
    \left( \frac{f(-x)}{p} \right)
    = \left( \frac{-f(x)}{p} \right)
    = \left( \frac{-1}{p} \right) \left(
    \frac{f(x)}{p} \right)
    = - \left( \frac{f(x)}{p} \right)
  .\]
  Hence
  \[
    \#\reduction{E}_D(\Fbb_p)
    = p + 1
  .\]
  \[
    E_D(\Qbb)_{\text{tors}}
    \supset \{0, (0, 0), (\pm D, 0)\}
    \cong (\Zbb / 2\Zbb)^2
  .\]
  Let $m = \#E_D(\Qbb)_{\text{tors}}$.

  We have $4 \mid m \mid p + 1$ forr all
  sufficiently large ($p \nmid 2mD$) primes $p$
  with $p \equiv 3 \pmod 4$. Hence $m = 4$
  (otherwise get contradiction to Dirichlet's
  theorem on primes on arithmetic progressions).

  So
  \begin{align*}
    \rank E_D(\Qbb) \ge 1
    &\iff \exists x, y \in \Qbb, y \neq 0 \wedge
    y^2 = x^3 - D^2 x \\
    &\stackrel{\text{Lecture 1}}{\iff} \text{$D$
    is a congruent number}
  \end{align*}
\end{example*}

\begin{fclemma}[]
  \label{lemma:10.4}
  Assuming:
  - $E / \Qbb$ is given by a \gls{weq} with $a_1,
  \ldots, a_6 \in \Zbb$
  - $0 \neq T = (x, y) \in E(\Qbb)_{\text{tors}}$
  Thens:[(i)]
  - $4x, 8y \in \Zbb$
  - If $2 \mid a_1$ or $2T \neq 0$ then $x, y \in
  \Zbb$
\end{fclemma}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item The \gls{weq} defines a formal group $\hat{E}$
      over $\Zbb$. For $r \ge 1$,
      \[
        \hat{E}(p^r \Zbb_p)
        = \{(x, y) \in E(\Qbb_p) \st v_p(x) \le -2r,
        v_p(y) \le -3r\} \cup \{0\}
      .\]
      Then \cref{thm:9.2} gives $\hat{E}(p^r \Zbb_p)
      \cong (\Zbb_p, +)$ if $r > \frac{1}{p - 1}$.
      Hence $\hat{E}(4 \Zbb_2)$ and $\hat{E}(p\Zbb_p)$
      for $p$ are odd are torsion free.

      So if $0 \neq T = (x, y) \in
      E(\Qbb)_{\text{tors}}$ then
      \begin{align*}
        v_2(x)
        &\ge -2 \\
        v_2(y)
        &\ge -3 \\
        v_p(x)
        &\ge 0 \\
        v_p(y)
        &\ge 0
      \end{align*}
      for all odd primes $p$.
    \item Suppose $T \in \hat{E}(2\Zbb_2)$, i.e.
      $v_2(x) = -2$, $v_2(y) = -3$.

      Since
      \[
        \frac{\hat{E}(2\Zbb_2)}{\hat{E}(4\Zbb_2)}
        \cong (\Fbb_2, +)
      \]
      and $\hat{E}(4\Zbb_2)$ is torsion free, we
      get $2T = 0$.

      Also $(x, y) = T = -T = (x, -y - a_1 x -
      a_3)$. So
      \begin{align*}
        2y + a_1 x + a_3
        &= 0 \\
        \ub{8y}_{\text{odd}}
        + a_1 \ub{(4x)}_{\text{odd}}
        + \ub{4a_3}_{\text{even}}
        &= 0 \\
      \end{align*}
      Hence $a_1$ is odd.

      So if $2T \neq 0$ or $a_1$ is even then $T
      \notin \hat{E}(2\Zbb_2)$, so $x, y \in
      \Zbb$.
      \qedhere
  \end{enumerate}
\end{proof}

\begin{example*}
  $y^2 + xy = x^3 + 4x + 1$, $\left( -\frac{1}{4},
  \frac{1}{8} \right) \in E(\Qbb)\tors[2]$.
\end{example*}

\begin{fcthm}[Lutz Nagell]
  \label{thm:10.5}
  Assuming:
  - $E / \Qbb$: $y^2 = x^3 + ax + b$, $a, b \in
  \Zbb$
  - $0 \neq T = (x, y) \in E(\Qbb)_{\text{tors}}$
  Then:
  $x, y \in \Zbb$ and either $y = 0$ or $y^2 \mid
  (4a^3 + 27b^2)$.
\end{fcthm}

\begin{proof}
  \cref{lemma:10.4} gives that $x, y \in \Zbb$.

  If $2T = 0$ then $y = 0$. Otherwise $0 \neq 2T
  = (x_2, y_2) \in E(\Qbb)_{\text{tors}}$.

  \cref{lemma:10.4} gives $x_2, y_2 \in \Zbb$. But
  \[
    x_2
    = \left( \frac{f'(x)}{2y} \right)^2 - 2x
  \]
  hence $y \mid f'(x)$.

  $E$ non-singular gives that $f(X)$ adn $f'(X)$
  are coprime, so $f(X)$ and $f'(X)^2$ are
  coprime. Therefore there exists $g, h \in
  \Qbb[X]$ satisfying
  \[
    g(X) f(X) + h(X) f'(X)^2 = 1
  .\]
  Doing this and clearing denominators gives
  \[
    (3X^2 + 4a) f'(X)^2 - 27(X^3 + aX - b) f(X)
    = 4a^3 + 27 b^2
  .\]
  Since $y \mid f'(x)$ and $y^2 = f(x)$, we get
  $y^2 \mid (4a^3 + 27b^2)$.
\end{proof}

\begin{remark*}
  Mazur showed that if $E / \Qbb$ is an \gls{ellc}
  then
  \[
    E(\Qbb)_{\text{tors}} \cong
    \begin{cases}
      \Zbb / n \Zbb
      & 1 \le n \le 12, n \neq 11 \\
      \Zbb / 2\Zbb \times \Zbb / 2n \Zbb
      & 1 \le n \le 4
    \end{cases}
  \]
  Moreover, all 15 possibilities occur.
\end{remark*}

\newpage
\section{Kummer Theory}

Let $K$ be a field and $\char K \nmid n$. Assume
$\mu_n \subset K$.

\begin{fclemma}[]
  \label{lemma:11.1}
  Assuming:
  - $\Delta \subset K^* / (K^*)^n$ a finite
  subgroup
  - $L = K(\sqrt[n]{\Delta})$
  Then:
  $L / K$ is Galois and $\Gal(L / K) \cong
  \Hom(\Delta, \mu_n)$.
\end{fclemma}

\begin{proof}
  $\mu_n \subset K$ gives $L / K$ normal, and
  $\char K \nmid n$ gives that $L / K$ is
  separable. So $L / K$ is Galois.

  Define the Kummer pairing
  \begin{align*}
    \langle , \rangle : \Gal(L / K) \times \Delta
    &\to \mu_n \\
    (\sigma, x)
    &\mapsto \frac{\sigma(\sqrt[n]{x})}{\sqrt[n]{x}}
  \end{align*}
  \textbf{Well-defined:} Suppose $\alpha, \beta \in L$
  with $\alpha^n = \beta^n = x$. Then $\left(
  \frac{\alpha}{\beta} \right)^n = 1$, so
  $\frac{\alpha}{\beta} \in \mu_n \subset K$. Then
  $\sigma \left( \frac{\alpha}{\beta} \right) =
  \frac{\alpha}{\beta}$ for all $\sigma \in \Gal(L
  / K)$, hence $\frac{\sigma(\alpha)}{\alpha} =
  \frac{\sigma(\beta)}{\beta}$ for all $\sigma \in
  \Gal(L / K)$.

  \textbf{Bilinear:}
  \begin{align*}
    \langle \sigma\tau, x \rangle
    &= \frac{\sigma(\tau \sqrt[n]{x})}{\tau
    \sqrt[n]{x}} \frac{\tau
    \sqrt[n]{x}}{\sqrt[n]{x}}
    &= \langle \sigma, x \rangle \langle \tau, x
    \rangle \\
    \langle \sigma, xy \rangle
    &= \frac{\sigma \sqrt[n]{xy}}{\sqrt[n]{xy}} \\
    &= \frac{\sigma \sqrt[n]{x}}{\sqrt[n]{x}}
    \frac{\sigma \sqrt[n]{y}}{\sqrt[n]{y}} \\
    &= \langle \sigma, x \rangle \langle \sigma, y
    \rangle
  \end{align*}

  \textbf{Non-degenerate:} Let $\sigma \in \Gal(L
  / K)$. If $\langle \sigma, x \rangle = 1$ for
  all $x \in \Delta$, then
  \[
    \sigma \sqrt[n]{x}
    = \sqrt[n]{x}
    \qquad \forall x \in \Delta
  \]
  adn hence $\sigma$ fixes $L$ pointwise, i.e.
  $\sigma = 1$.

  Let $x (K^*)^n \in \Delta$. If $\langle \sigma,
  x \rangle = 1$ for all $\sigma \in \Gal(L / K)$,
  then
  \[
    \sigma \sqrt[n]{x}
    = \sqrt[n]{x}
    \qquad \forall \sigma \in \Gal(L / K)
  .\]
  So $\sqrt[n]{x} \in K$, so $x \in (K^*)^n$, i.e.
  $x(K^*)^n \in \Delta$ is the identity element.

  We get injective group homomorphisms
  \begin{enumerate}[(i)]
    \item $\Gal(L / K) \injto \Hom(\Delta,
      \mu_n)$.
    \item $\Delta \injto \Hom(\Gal(L / K),
      \mu_n)$.
  \end{enumerate}
