%! TEX root = EC.tex
% vim: tw=50
% 24/01/2025 11AM

% Books:
% - Silverman, The arithmetic of elliptic curves,
%   Springer 1986
% - Cassels, Lectures on elliptic curves, CUP 1991
% Introductory reading:
% - Silverman & Tate, Rational points on elliptic
%   curves, Springer 1992
% Other books:
% - Milne, Washington, Knapp, \ldots

\newpage
\section{Fermat's Method of Infinite Descent}

\begin{center}
  \includegraphics[width=0.6\linewidth]{images/d425f9274f544498.png}
\end{center}

\begin{fcdefnstar}[Rational, primitive triangle]
  \glsadjdefn{ratt}{rational}{triangle}%
  \glsadjdefn{primt}{primitive}{triangle}%
  A triangle is \emph{rational} if $a, b, c \in
  \Qbb$.

  A triangle is \emph{primitive} if $a, b, c \in
  \Zbb$ and are coprime.
\end{fcdefnstar}

\begin{fclemma}[]
  \label{lemma:1.1}
  % Lemma 1.1
  Assuming:
  - $\Delta$ is a \gls{primt} triangle
  Then:
  $\Delta$ is of the form
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/56f7deec1b4e4a90.png}
  \end{center}
  for some integers $u > v > 0$.
\end{fclemma}

\begin{proof}
  Without loss of generality $a$ odd, $b$ even
  (work modulo $4$). This then forces $c$ odd.

  Then
  \[
    \left( \frac{b}{2} \right)^2 = \frac{c + a}{2}
    \cdot \frac{c - a}{2}
  ,\]
  and note that all the fractions are integers.
  Also note that the product on the right hand
  side is a product of positive coprime integers.

  Unique prime factorisation in $\Zbb$ gives that
  $\frac{c + a}{2} = u^2$, $\frac{c - a}{2} = v^2$
  for some $u, v \in \Zbb$.

  Then $a = u^2 - v^2$, $b = 2uv$, $c = u^2 +
  v^2$.
\end{proof}

\begin{fcdefnstar}[Congruent number]
  \glsadjdefn{congn}{congruent}{number}%
  $D \in \Qbb_{> 0}$ is a \emph{congruent number}
  if there exists a \gls{ratt} triangle $\Delta$
  with $\area(\Delta) = D$.
\end{fcdefnstar}

\begin{note*}
  It suffices to consider $D \in \Zbb_{> 0}$
  square-free.
\end{note*}

\begin{example*}
  $D = 5, 6$ are \gls{congn}.
\end{example*}

\begin{fclemma}[]
  \label{lemma:1.2}
  % Lemma 1.2
  Assuming:
  - $D \in \Qbb_{> 0}$
  Then:
  \begin{iffc}
    \lhs
    $D$ is \gls{congn}
    \rhs
    $Dy^2 = x^3 - x$ for some $x, y \in \Qbb$, $y
    \neq 0$.
  \end{iffc}
\end{fclemma}

\begin{proof}
  \cref{lemma:1.1} shows
  \[
    D \text{ congruent} \iff D w^2 = uv(u^2 - v^2)
  \]
  for some $u, v, w \in \Qbb$, with $w \neq 0$.
  Put $x = \frac{u}{v}$ and $y = \frac{w}{v^2}$.
\end{proof}

Fermat showed that $1$ is \emph{not} a
\gls{congn} number.

\begin{theorem}[]
  \label{thm:1.3}
  % Theorem 1.3
  There is no solution to
  \[
    w^2 = uv(u + v)(u - v) \tag{$*$}
    \label{thm:1.3_eq}
  \]
  for $u, v, w \in \Zbb$ and $w \neq 0$.
\end{theorem}

\begin{proof}
  Without loss of generality $u, v$ are coprime,
  $u > 0$, $w > 0$. If $v < 0$ then replace $(u,
  v, w)$ by $(-v, u, w)$. If $u, v$ both odd then
  replace $(u, v, w)$
  by $\left( \frac{u + v}{2}, \frac{u -
  v}{2}, \frac{w}{2} \right)$.

  Then $u, v, u + v, u - v$ are pairwise coprime
  positive integers with product a square.

  Unique factorisation in $\Zbb$ gives
  \[
    u = a^2, \quad v = b^2, \quad u + v = c^2,
    \quad u - v = d^2
  \]
  for some $a, b, c, d \in \Zbb_{> 0}$.

  Since $u \not\equiv v \pmod{2}$, both $c$ and
  $d$ are odd.

  Then consider:
  \[
    \left( \frac{c + d}{2} \right)^2 + \left(
    \frac{c - d}{2} \right)^2 = \frac{c^2 +
    d^2}{2} = u = a^2
  .\]
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/9b3821c31d874348.png}
  \end{center}
  This is a primitive triangle. The area is
  $\frac{c^2 - d^2}{8} = \frac{v}{4} = \left( \frac{b}{2}
  \right)^2$.

  Let $w_1 = \frac{b}{2}$. \cref{lemma:1.1} gives
  $w_1^2 = u_1 v_1 (u_1 + v_1)(u_1 - v_1)$ for
  $u_1, v_1 \in \Zbb$. Therefore we have a new
  solution to \eqref{thm:1.3_eq}. But $4w_1^2 =
  b^2 = v \mid w^2$ so $w_1 \le \half w$.

  So by Fermat's method of infinite descent, there
  is no solution to \eqref{thm:1.3_eq}.
\end{proof}

\subsection{A variant for polynomials}

In Section 1, $K$ is a field with $\char K \neq
2$.

Write $\ol{K}$ for the algebraic closure of $K$.

\begin{fclemma}
  \label{lemma:1.4}
  % Lemma 1.4
  Assuming:
  - $u, v \in K[t]$ coprime
  - $\alpha u + \beta v$ is a square for $4$
  distinct $(\alpha : \beta) \in \Pbb^1$
  Then:
  $u, v \in K$.
\end{fclemma}

\begin{proof}
  Without loss of generality $K = \ol{K}$.

  Changing coordinates on $\Pbb^1$, we may assume
  the ratios $(\alpha : \beta)$ are $(1 : 0), (0 :
  1), (1 : -1), (1 : -\lambda)$ for some $\lambda
  \in K \setminus \{0, 1\}$ (M\"obius map).
  \begin{align*}
    u
    &= a^2 \\
    v
    &= b^2 \\
    u - v
    &= (a + b)(a - b) \\
    u - \lambda v
    &= (a + \mu b)(a - \mu b)
  \end{align*}
  (where $\mu$ is a square root of $\lambda$).
  Unique factorisation in $K[t]$ gives that $a +
  b$, $a - b$, $a + \mu b$, $a - \mu b$ are
  squares. But
  \[
    \max(\deg(a), \deg(b)) \le \half \max(\deg(u),
    \deg(v))
  .\]
  So Fermat's method of infinite descent, we get a
  contradiction, unless the degrees of $u$ and $v$
  are zero. So $u, v \in K$.
\end{proof}

\begin{fcdefn}[Elliptic curve (temporary
  definition)]
  \label{defn:1.5}
  \glsnoundefn{tellc}{elliptic curve}{elliptic
  curves}%
  \phantom{}
  \begin{enumerate}[(i)]
    \item An elliptive curve $E / K$ is the
      projective closure of the plane affine
      curve
      \[
        y^2 = f(x)
      \]
      where $f \in K[x]$ is a monic cubic
      polynomial with distinct roots in $\ol{K}$.
      We call this equation ``a Weierstrass
      equation''.
    \item For $L / K$ any field extension
      \[
        E(L) = \{(x, y) \in L^2 \mid y^2 = f(x)\}
        \cup \{0\}
      .\]
      $0$ is the ``point at infinity'' that we get
      because we take the projective closure.
  \end{enumerate}
\end{fcdefn}

\textbf{Fact}: $E(L)$ is naturally an abelian
group.

In this course, we study $E(K)$ for $K$ being a
finite field, local field ($[K : \Qbb_p] <
\infty$) or number field ($[K :
\Qbb] < \infty$).

\cref{lemma:1.2} and \cref{thm:1.3} tells us that
if $E$ is $y^2 = x^3 - x$, then
\[
  E(\Qbb) = \{0, (0, 0), (\pm 1, 0)\}
.\]


\begin{corollary}
  \label{coro:1.6}
  % Corollary 1.6
  Let $E / K$ be an \gls{tellc}. Then $E(K(t)) =
  E(K)$.
\end{corollary}
