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% Example Sheet 3:
% Questions 1, 3, 8

  So we have a map
  \begin{align*}
    \ker \left( \frac{E(K)}{nE(K)} \to
    \frac{E(L)}{n E(L)} \right)
    &\stackrel{(*)}{\to}
    \Maps(\Gal(L / L), E\tors[n]) \\
    P + nE(K)
    &\mapsto (\sigma \mapsto \sigma Q - Q)
    &&\text{where $nQ = P$}
  \end{align*}
  It remains to show ($*$) is injective.

  So suppose $P_1, P_2 \in E(K)$, $P_i = nQ_i$ for
  $i = 1, 2$, and suppose $\sigma Q_1 - Q_1 =
  \sigma Q_2 - Q_2$ for all $\sigma \in \Gal(L /
  K)$. Then $\sigma(Q_1 - Q_2) = Q_1 - Q_2$ for
  all $\sigma \in \Gal(L / K)$, hence $Q_1 - Q_2
  \in E(K)$, so $P_1 - P_2 \in n E(K)$. Hence $P_1
  + n E(K) = P_2 + n E(K)$ as desired.
\end{proof}

\begin{fcthmstar}[Weak Mordell-Weil Theorem]
  \glsnoundefn{wmwt}{Weak Mordell-Weil
  Theorem}{NA}%
  Assuming:
  - $K$ a number field
  - $E / K$ a \gls{ellc}
  - $n \ge 2$ an integer
  Then:
  $\frac{E(K)}{n E(K)}$ is finite.
\end{fcthmstar}

\begin{proof}
  \cref{lemma:12.1} tells us that we may replace
  $K$ by a finite Galois extension.

  So without loss of generality $\mu_n \subset K$
  and $E\tors[n] \subset E(K)$. Let
  \[
    S
    = \{\mathfrak{p} \st n\}
    \cup \{\text{primes of \gls{pbred} for $E / K$}\}
  .\]
  For each $P \in E(K)$, the extension $K([n]^{-1}
  P) / K$ is unramified outisde $S$ by
  \cref{thm:9.8}. Since $\Gal(\ol{K} / K)$ acts on
  $[n]^{-1} P$, it follows that $\Gal(\ol{K} /
  K([n]^{-1} P))$ is a normal subgroup of
  $\Gal(\ol{K} / K)$ and hence $K([n]^{-1} P) / K$
  is a Galois extension.

  Let $Q \in [n]^{-1} P$. Since $E\tors[n] \subset
  E(K)$, we have $K(Q) = K([n]^{-1} P)$. Consider
  \begin{align*}
    \Gal(K(Q) / K)
    &\to E\tors[n] \cong (\Zbb / n\Zbb)^2 \\
    \sigma
    &\mapsto \sigma Q - Q
  \end{align*}
  \textbf{Group homomorphism:} $\sigma \tau Q - Q =
  \cancel{\sigma} (\tau Q - Q) + (\sigma Q - Q)$.

  \textbf{Injective:} If $\sigma Q = Q$ then
  $\sigma$ fixes $K(Q)$ pointwise, i.e. $\sigma =
  1$.

  Therefore $K(Q) / K$ is an abelian extension of
  exponent $n$, unramified outside $S$.

  \cref{prop:11.3} shows that as we vary $P \in
  E(K)$ there are only finitely many possibilities
  for $K(Q)$.

  Let $L$ be the composite of all such extensions
  of $K$. Then $L / K$ is finite and Galois, and
  $\frac{E(K)}{n E(K)} \to \frac{E(L)}{n E(L)}$ is
  the zero map. \cref{lemma:12.1} gives $\left|
  \frac{E(K)}{n E(K)} \right| < \infty$.
\end{proof}

\begin{remark*}
  If $K = \Rbb$ or $\Cbb$, or $[K : \Qbb_p] <
  \infty$ then $\left| \frac{E(K)}{n E(K)} \right|
  < \infty$, yet $E(K)$ is uncountable, so not
  finitely generated.
\end{remark*}

\textbf{Fact:} If $K$ is a number field, then
there exists a \gls{qf} (= canonical height)
$\hat{h} : E(K) \to \Rbb_{\ge 0}$ with the
property that for any $B \ge 0$,
\[
  \{P \in E(K) \mid \hat{h}(P) \le B\}
  \tag{$*$}
  \label{lec18eq1}
\]
is finite.

\begin{fcthmstar}[Mordell-Weil Theorem]
  \glsnoundefn{mwt}{Mordell-Weil}{NA}%
  Assuming:
  - $K$ a number field
  - $E / K$ an \gls{ellc}
  Then:
  $E(K)$ is a finitely generated abelian group.
\end{fcthmstar}

\begin{proof}
  Fix an integer $n \ge 2$.

  \gls{wmwt} implies that $\left| \frac{E(K)}{n
  E(K)} \right| \infty$. Pick coset
  representatives $P_1, P_2, \ldots, P_m$. Let
  \[
    \Sigma
    = \{P \in E(K) \st \hat{(P)} \le \max_{1 \le i
    \le m} \hat{h}(P_i)\}
  .\]
  
  \textbf{Claim:} $\Sigma$ generates $E(K)$.

  If not, then there exists $P \in E(K) \setminus
  (\text{subgroup generated by $\Sigma$})$ of
  minimal height (exists by \eqref{lec18eq1}).
  Then $P = P_i + nQ$ for some $i$ and $Q \in
  E(K)$. Note that $Q \in E(K) \setminus
  (\text{subgroup generated by $\Sigma$})$.

  Minimal choice of $P$ gives
  \begin{align*}
    4\hat{h}(P)
    &\le 4\hat{h}(Q) \\
    &\le n^2 \hat{h}(Q) \\
    &= \hat{h}(nQ) \\
    &= \hat{h}(P - P_i) \\
    &\le \hat{h}(P - P_i) + \hat{h}(P + P_i) \\
    &= 2\hat{h}(P) + 2\hat{h}(P_i)
    &&\text{parallelogram law}
  \end{align*}
  Therefore $\hat{h}(P) \le \hat{h}(P_i)$. Hence
  $P \in \Sigma$ (by definition of $\Sigma$),
  which contradicts the choice of $P$.

  This proves the claim.

  By \eqref{lec18eq1}, $\Sigma$ is finite.
\end{proof}

\newpage
\section{Heights}

For simplicity, take $K = \Qbb$. Write $P \in
\Pbb^n(Q)$ as $P = (a_0 : a_1 : \cdots : a_n)$
where $a_0, \ldots, a_n \in \Zbb$ and $\gcd(a_0,
a_n) = 1$.

\begin{fcdefnstar}[Height of a point]
  \glssymboldefn{pointheight}%
  $H(P) = \max_{0 \le i \le n} |a_i|$.
\end{fcdefnstar}

\begin{fclemma}[]
  \label{lemma:13.1}
  Assuming:
  - $f_1, f_2 \in \Qbb[X_1, X_2]$ coprime
  homogeneous polynomials of degree $d$
  - let
  \begin{align*}
    F : \Pbb^1
    &\to \Pbb^1 \\
    (x_1 : x_2)
    &\mapsto (f_1(x_1, x_2) : f_2(x_1, x_2))
  \end{align*}
  Then:
  there exist $c_1, c_2 > 0$ such that
  for all $P \in \Pbb^1(\Qbb)$,
  \[
    c_1 \height(P)^d
    \le \height(F(P))
    \le c_2 \height(P)^d
  .\]
\end{fclemma}

\begin{proof}
  Without loss of generality $f_1, f_2 \in
  \Zbb[X_1, X_2]$.

  \textbf{Upper bound:} Write $P = (a_1 : a_2)$,
  $a_1, a_2 \in \Zbb$ coprime.
  \begin{align*}
    \height(F(P))
    &\le \max(|f_1(a_1, a_2)|, |f_2(a_1, a_2)|) \\
    &\le c_2 \max(|a_1|^d, |a_2|^d)
  \end{align*}
  where
  \[
    c_2 = \max_{i = 1, 2} \text{(sum of absolute
    values of coefficients of $f_i$)}
  .\]
  Therefore $\height(F(P)) \le c_2 \height(P)^d$.

  \textbf{Lower bound:} We claim that there exists
  $g_{ij} \in \Zbb[X_1, X_2]$ homogeneous of
  degree $d - 1$ and $\kappa \in \Zbb_{> 0}$ such that
  \[
    \sum_{j = 1}^2 g_{ij} f_j = \kappa X_i^{2d - 1}
    , \quad i = 1, 2
    \tag{$\dag$}
    \label{lec18eq2}
  .\]
  Indeed, running Euclid's algorithm on $f_1(X,
  1)$ and $f_2(X, 1)$ gives $r, s \in \Qbb[X]$ of
  degree $< d$ such that
  \[
    r(X) f_1(X, 1) + s(X) f_2(X, 1)
    = 1
  .\]
  Homogenising and clearing denominators gives
  \eqref{lec18eq2} for $i = 2$. Likewise for $i =
  1$.

  Write $P = (a_1 : a_2)$ with $a_1, a_2 \in \Zbb$
  coprime. \eqref{lec18eq2} gives
  \[
    \sum_{j = 1}^2 g_{ij}(a_1, a_2) f_j(a_1, a_2)
    = \kappa a_i^{2d - 1}
    , \quad i = 1, 2
  .\]
  Therefore $\gcd(f_1(a_1, a_2), f_2(a_1, a_2))$
  divides $\gcd(\kappa a_1^{2d - 1}, \kappa
  a_2^{2d - 1}) = \kappa$.