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Let $[K : \Qbb_p] < \infty$ and $L / K$ a finite
extension. Let the residue fields be $k'$ and $k$,
and let $f = [k' : k]$.
\begin{picmath}
  \begin{tikzcd}
    K^* \ar[r, "V_K", two heads]
    \arrow[d, phantom, sloped, "\subset"]
    & \Zbb \ar[d, "\times e"] \\
    L^* \ar[r, "V_L", two heads]
    & \Zbb
  \end{tikzcd}
\end{picmath}

\textbf{Facts:}
\begin{enumerate}[(i)]
  \item $[L : K] = ef$.
  \item If $L / K$ is Galois then the natural map
    $\Gal(L / K) \to \Gal(k' / k)$ is surjective
    with kernel of order $e$.
\end{enumerate}

\begin{fcdefnstar}[Unramified]
  \glsadjdefn{unram}{unramified}{extension}%
  $L / K$ is \emph{unramified} if $e = 1$.
\end{fcdefnstar}

\textbf{Fact:} For each $m \ge 1$
\begin{enumerate}[(i)]
  \item $k$ has a unique extension of degree $m$
    (say $k_m$).
  \item $K$ has a unique unramified extension of
    degree $m$ (say $K_m$).
\end{enumerate}
These extensions are Galois, with cyclic Galois
groups.

\begin{fcdefnstar}[Maximal unramified extension]
  \glssymboldefn{nr}%
  $K^{\text{nr}} = \bigcup_{m \ge 1} K_m$ (inside
  $\ol{K}$).
  ``maximal unramified extension''
\end{fcdefnstar}

\begin{fcthm}[]
  \label{thm:9.8}
  Assuming:
  - $[K : \Qbb_p] < \infty$
  - $E / K$ has \gls{gred}
  - $p \nmid n$
  - $P \in E(K)$
  Then:
  $K([n]^{-1} P) / K$ is \gls{unram}.
\end{fcthm}

\begin{notation*}
  \phantom{}
  \[
    [n]^{-1} (P)
    = \{Q \in E(\ol{K}) : nQ = P\}
  ,\]
  \[
    K(\{Q_1, \ldots, Q_r\})
    = K(x_1, \ldots, x_r, y_1, \ldots, y_r)
  ,\]
  where $Q_i = (x_i, y_i)$.
\end{notation*}

\begin{proof}
  For each $m \ge 1$ there is a short exact
  sequence
  \[
    0 \to E_1(K_m)
    \to E(K_m)
   \to \reduction{E}(K_m)
   \to 0
  .\]
  Taking $\bigcup_{m \ge 1}$ gives a commutative
  diagram with exact rows:
  \begin{picmath}
    \begin{tikzcd}
      0
      \ar[r]
      & E_1(K^\nr)
      \ar[r]
      \ar[d]
      & E(K^\nr)
      \ar[r]
      \ar[d]
      & \reduction{E}(\ol{k})
      \ar[r]
      \ar[d]
      & 0 \\
      0
      \ar[r]
      & E_1(K^\nr)
      \ar[r]
      & E(K^\nr)
      \ar[r]
      & \reduction{E}(\ol{k})
      \ar[r]
      & 0
    \end{tikzcd}
  \end{picmath}
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/dcf3d76d0887429e.png}
  \end{center}
  An isomorphism by \cref{coro:8.5} applied over
  each $K_m$ (using $p \nmid n$ here).
  \begin{itemize}
    \item surjective by \cref{thm:2.8}
    \item kernel $\cong (\Zbb / n\Zbb)^2$ by
      \cref{thm:6.5} (again using $p \nmid n$).
  \end{itemize}
  Snake lemma gives
  \[
    \begin{cases}
      E(K^\nr)[n] \cong (\Zbb / n\Zbb)^2 \\
      \frac{E(K^\nr)}{n E(K^\nr)} = 0
    \end{cases}
  \]
  So if $P \in E(K)$ then there exists $Q \in
  E(K^\nr)$ such that $nQ = P$ and 
  \[
    [n]^{-1} P
    = \{Q + T : T \in E[n]\}
    \subset E(K^\nr)
  .\]
  Hence $K([n]^{-1} P) \subset K^\nr$ and so
  $K([n]^{-1} P) / K$ is \gls{unram}.
\end{proof}

\newpage
\section{Elliptic Curves over Number Fields: The
torsion subgroup}

$[K : \Qbb] < \infty$, $E / K$ an \gls{ellc}.

\begin{notation*}
  $\mathfrak{p}$ a prime of $K$ (i.e. a prime)
  ideal in $\mathcal{O}_K$).

  $K_{\mathfrak{p}}$ is the $\mathfrak{p}$-adic
  completion of $K$, valuation ring
  $\mathcal{O}_{\mathfrak{p}}$.

  $k_{\mathfrak{p}} = \mathcal{O}_K /
  \mathfrak{p}$ residue field.
\end{notation*}

\begin{fcdefnstar}[Good reduction (prime)]
  \glsadjdefn{pgred}{good reduction}{of a prime}%
  \glsadjdefn{pbred}{bad reduction}{of a prime}%
  $\mathfrak{p}$ is a prime of \emph{good
  reduction} for $E / K$ if $E / K_{\mathfrak{p}}$
  has \gls{gred}.
\end{fcdefnstar}

\begin{fclemma}[]
  \label{lemma:10.1}
  $E / K$ has only finitely many \cloze{primes of
  \gls{pbred}.}
\end{fclemma}

\begin{proof}
  Take a \gls{weq} for $E$ with $a_1, \ldots, a_6
  \in \mathcal{O}_K$. $E$ non-singular implies
  that $0 \neq \Delta \in \mathcal{O}_K$.

  Write $(\Delta) = \mathfrak{p}^{\alpha_1} \cdots
  \mathfrak{p}_r^{\alpha_r}$ (factorisation into
  prime ideals).

  Let $S = \{\mathfrak{p}_1, \ldots,
  \mathfrak{p}_r\}$. If $\mathfrak{p} \notin S$
  then $v_{\mathfrak{p}}(\Delta) = 0$. Hence $E /
  K_{\mathfrak{p}}$ has good reduction.

  Therefore $\{\textbf{bad primes for $E$}\}
  \subset S$, hence is finite.
\end{proof}

\begin{remark*}
  If $K$ has class number $1$ (e.g. $K = \Qbb$)
  then we can always find a \gls{weq} for $E$ with
  $a_1, \ldots, a_6 \in \mathcal{O}_K$ which is
  \gls{min} at all primes $\mathfrak{p}$.
\end{remark*}

\textbf{Basic group theory:} If $A$ is a finitely
generatead abelian group then $A \cong
\{\text{finite group}\} \times \Zbb^r$. We call
$r$ the ``rank'', and $\{\text{finite subgroup}\}$
is the torsion subgroup.

\begin{fclemma}[]
  \label{lemma:10.2}
  $E(K)_{\text{tors}}$ is finite.
\end{fclemma}

\begin{proof}
  Take any prime $\mathfrak{p}$. We saw that
  $E(K_{\mathfrak{p}})$ has a finite index
  subgroup $A$ (say) with $A \cong
  (\mathcal{O}_{\mathfrak{p}}, +)$.

  In particular, $A$ is torsion free
  \[
    E(K)_{\text{tors}}
    \injto E(K_{\mathfrak{p}})_{\text{tors}}
    \injto
    \ub{\frac{E(K_{\mathfrak{p}})}{A}}_{\text{finite}}
    .
    \qedhere
  \]
\end{proof}

\begin{fclemma}[]
  \label{lemma:10.3}
  Assuming:
  - $\mathfrak{p}$ is a prime of \gls{pgred}
  - $\mathfrak{p} \nmid n$
  Then:
  reduction modulo $\mathfrak{p}$ gives an
  injective group homomorphism
  \[
    E(K)[n]
    \injto \reduction{E}(k_{\mathfrak{p}})
  .\]
\end{fclemma}

\begin{proof}
  \cref{prop:9.5} gives that $E(K_{\mathfrak{p}})
  \to \reduction{E}(k_{\mathfrak{p}})$ is a group
  homomorphism, with kernel
  $E_1(K_{\mathfrak{p}})$.

  \cref{coro:8.5} and $\mathfrak{p} \nmid n$ gives
  that $E_1(K_{\mathfrak{p}})$ has no $n$-torsion.
\end{proof}

\begin{example*}
  $E / \Qbb$: $y^2 + y = x^3 - x$, $\Delta = -11$.

  $E$ has \gls{pgred} at all $p \neq 11$.
  \begin{center}
    \begin{tabular}{c|cccccc}
      $p$
      & $2$
      & $3$
      & $5$
      & $7$
      & $11$
      & $13$ \\
      \hline
      $\#\reduction{E}(\Fbb_p)$
      & $5$
      & $5$
      & $5$
      & $5$
      & $-$
      & $10$
    \end{tabular}
  \end{center}
  \cref{lemma:10.3} gives:
  \[
    \begin{cases}
      \#(\Qbb)_{\text{tors}} \mid 5 \cdot 2^a
      & \text{for some $a \ge 0$} \\
      \#E(\Qbb)_{\text{tors}} \mid 5 \cdot 3^b
      & \text{for some $b \ge 0$}
    \end{cases}
  \]
  Hence $\#(\Qbb)_{\text{tors}} \mid 5$.

  Let $T = (0, 0) \in E(\Qbb)$. Calculation gives
  $5T = 0$. Therefore $E(\Qbb)_{\text{tors}} \cong
  \Zbb / 5\Zbb$.
\end{example*}

\begin{example*}
  $E / \Qbb$: $y^2 + y = x^3 + x^2$, $\Delta = -43$
  \begin{center}
    \begin{tabular}{c|cccccc}
      $p$
      & $2$
      & $3$
      & $5$
      & $7$
      & $11$
      & $13$ \\
      \hline
      $\#\reduction{E}(\Fbb_p)$
      & $5$
      & $6$
      & $10$
      & $8$
      & $9$
      & $19$
    \end{tabular}
  \end{center}
  \cref{lemma:10.3} gives:
  \[
    \begin{cases}
      \#E(\Qbb)_{\text{tors}} \mid 5 \cdot 2^a
      & \text{for some $a \ge 0$} \\
      \#E(\Qbb)_{\text{tors}} \mid 9 \cdot 11^b
      & \text{for some $b \ge 0$}
    \end{cases}
  \]
  Therefore $E(\Qbb)_{\text{tors}} = \{0\}$.

  Therefore $P = (0, 0)$ is a point of infinite order.

  In particular, $E(\Qbb)$ is infinite.
\end{example*}