%! TEX root = EC.tex
% vim: tw=50
% 03/03/2025 11AM
%
  (i) implies $\Gal(L / K)$ is abelian and of
  exponent dividing $n$.

  \textbf{Fact:} If $G$ is a finite abelian group
  of exponent dividing $n$ then $\Hom(G, \mu_n)
  \cong G$ (non canonically).

  So
  \[
    |\Gal(L / K)|
    \stackrel{\text{(i)}}{\le} |\Delta|
    \stackrel{\text{(ii)}}{\le} |\Gal(L / K)|
  .\]
  Therefore (i) and (ii) are isomorphisms.
\end{proof}

\begin{example*}
  $\Gal(\Qbb(\sqrt{2}, \sqrt{3}, \sqrt{5}) / \Qbb)
  \cong (\Zbb / 2\Zbb)^3$.
\end{example*}

\begin{fcdefnstar}[Abelian extension]
  We say $L / K$ is \emph{abelian} if it is
  Galois, and has abelian Galois group.

  Similarly for other group terminology (e.g. we
  can say that $L / K$ has exponent dividing $n$
  to mean that it is Galois, with Galois group
  having exponent dividing $n$).
\end{fcdefnstar}

\begin{fcthm}[]
  \label{thm:11.2}
  There is a bijection
  \begin{align*}
    \mathcloze[1]{
      \left\{\substack{
        \text{finite subgroups} \\
        \Delta \subset K^* / (K^*)^n
      }\right\}
    }
    &\leftrightarrow
    \mathcloze[2]{
      \left\{\substack{
        \text{finite abelian extension $L / K$} \\
        \text{of exponent dividing $n$}
      }\right\}
    } \\
    \mathcloze[1]{\Delta}
    &\mapsto \mathcloze[2]{K(\sqrt[n]{\Delta})} \\
    \mathcloze[1]{\frac{(L^*)^n \cap K^*}{(K^*)^n}}
    &\mapsfrom \mathcloze[2]{L}
  \end{align*}
\end{fcthm}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Let $\Delta \subset K^* / (K^*)^n$ be a
      finite subgroup. Let $L =
      K(\sqrt[n]{\Delta})$ and $\Delta' =
      \frac{(L^*)^n \cap K^*}{(K^*)^n}$.

      We must show $\Delta = \Delta'$.

      Clearly $\Delta \subset \Delta'$. So
      \[
        L
        = K(\sqrt[n]{\Delta})
        \subset K(\sqrt[n]{\Delta'})
        \subset L
      .\]
      So $K(\sqrt[n]{\Delta}) =
      K(\sqrt[n]{\Delta'})$. So \cref{lemma:11.1}
      gives $|\Delta| = |\Delta'|$.

      Since $\Delta \subset \Delta'$, it follows
      that $\Delta = \Delta'$.
    \item Let $L / K$ be a finite abelian
      extension of exponent dividing $n$. Let
      $\Delta = \frac{(L^*)^n \cap K^*}{(K^*)^n}$.
      Then $K(\sqrt[n]{\Delta}) \subset L$ and we
      aim to prove this inclusion is an equality.

      Let $G = \Gal(L / K)$. Thu Kummer pairing
      gives an injection $\Delta \injto \Hom(G,
      \mu_n)$.

      \textbf{Claim:} This map is surjective.

      Granted the claim,
      \begin{align*}
        [K(\sqrt[n]{\Delta}) : K]
        &\stackrel{\text{\cref{lemma:11.1}}}{=}
        |\Delta| \\
        &\stackrel{\text{by claim}}{=} |G| \\
        &= [L : K]
      \end{align*}
      Since $K(\sqrt[n]{\Delta}) \subset L$ it
      follows that $L = K(\sqrt[n]{\Delta})$.

      \textbf{Proof of claim:} Let $\chi : G \to
      \mu_n$ be a group homomorphism. Distinct
      automorphisms are linearly independent. So
      there exists $a \in L$ such that
      \[
        \ub{\sum_{\tau \in G} \chi(\tau)^{-1}
        \tau(a)}_{= y}
        \neq 0
      .\]
      Let $\sigma \in G$. Then
      \begin{align*}
        \sigma(y)
        &= \sum_{\tau \in G} \chi(\tau)^{-1}
        \sigma \tau(a) \\
        &= \sum_{\tau \in G} \chi(\sigma^{-1}
        \tau)^{-1} \tau(a) \\
        &= \chi(\sigma) \sum_{\tau \in G}
        \chi(\tau)^{-1} \tau(a) \\
        &= \chi(\sigma) \cdot y
        \tag{$*$}
        \label{lec17eq1}
      \end{align*}
      Therefore $\sigma(y^n) = y^n$ for all
      $\sigma \in G$. Hence $y^n \in K$.

      Let $x = y^n$. Then $x \in K^* \cap
      (L^*)^n$. Then $x(K^*)^n \in \Delta$.

      Also, by \eqref{lec17eq1}, $\chi : \sigma
      \mapsto \frac{\sigma(y)}{y} = \frac{\sigma
      \sqrt[n]{x}}{\sqrt[n]{x}}$. So the map
      $\Delta \injto \Hom(G, \mu_n)$ sends $x
      \mapsto \chi$. This proves the claim.
      \qedhere
  \end{enumerate}
\end{proof}

\begin{fcprop}[]
  \label{prop:11.3}
  Assuming:
  - $K$ a number field
  - $\mu_n \subset K$
  - $S$ a finite set of promes of $K$
  Then there are only finitely many extensions $L
  / K$ such that
  \begin{enumerate}[(i)]
    \item $L / K$ is a finite abelian extension of
      exponent dividing $n$
    \item $L / K$ is unramified at all
      $\mathfrak{p} \notin S$
  \end{enumerate}
\end{fcprop}

\begin{proof}
  \cref{thm:11.2} gives $L = K(\sqrt[n]{\Delta})$
  for some $\Delta \subset K^* / (K^*)^n$ a finite
  subgroup.

  Let $\mathfrak{p}$ be a prime of $K$.
  \[
    \mathfrak{p} \mathcal{O}_L
    = \mathcal{P}_1^{e_1} \cdots
    \mathcal{P}_r^{e_r}
  \]
  for $\mathcal{P}_1, \ldots, \mathcal{P}_r$ some
  distinct primes in $\mathcal{O}_L$.

  If $x \in K^*$ represents an element of $\Delta$
  then
  \[
    nv_{\mathcal{P}_i}(\sqrt[n]{x})
    = v_{\mathcal{P}_i}(x)
    = e_i v_{\mathfrak{p}}(x)
  .\]
  If $\mathfrak{p} \notin S$ then all $e_i = 1$,
  so $v_{\mathfrak{p}}(x) \equiv 0 \pmod n$.
  Therefore $\Delta \subset K(S, n)$ where
  \[
    K(S, n)
    = \{x \in K^* / (K^*)^n \st
    v_{\mathfrak{p}}(x) \equiv 0 \pmod n ~\forall
    \mathfrak{p} \notin S\}
  .\]
  The proof is completed by the next lemma.
\end{proof}

\begin{fclemma}[]
  \label{lemma:11.4}
  $K(S, n)$ is \cloze{finite.}
\end{fclemma}

\begin{proof}
  The map
  \begin{align*}
    K(S, n)
    &\to (\Zbb / n\Zbb)^{|S|} \\
    x
    &\mapsto (v_{\mathfrak{p}}(x) \pmod
    n)_{\mathfrak{p} \in S}
  \end{align*}
  is a group homomorphism with kernel $K(\emptyset,
  n)$.

  Since $|S| < \infty$, it suffices to prove the
  lemma with $S = \emptyset$.

  If $x \in K^*$ represents an element of
  $K(\emptyset, n)$ then $(x) = \mathfrak{a}^n$
  for some fractional ideal $\mathfrak{a}$. There
  is a short exact sequence
  \begin{align*}
    0
    \to \mathcal{O}_K^* / (\mathcal{O}_K^*)^n
    \to K(\emptyset, n)
    &\to \Cl_K \to 0 \\
    x(K^*)^n
    &\mapsto [\mathfrak{a}]
  \end{align*}
  $|\Cl_K| < \infty$ and $\mathcal{O}_K^*$ being a
  finitely generated abelian group (Dirichlet's
  unit theorem) gives us that $K(\phi, n)$ is
  finite.
\end{proof}

\newpage
\section{Elliptic Curves over Number Fields: The
weak Mordell-Weil Theorem}

\begin{fcthm}[]
  \label{lemma:12.1}
  Assuming:
  - $E / K$ an \gls{ellc}
  - $L / K$ a finite Galois extension
  Then:
  the natural map $\frac{E(K)}{n E(K)} \to
  \frac{E(L)}{n E(L)}$ has finite kernel.
\end{fcthm}

\begin{proof}
  For each element in the kernel we pick a coset
  representative $P \in E(K)$ and then $Q \in
  E(L)$ such that $nQ = P$.

  For any $\sigma \in \Gal(L / K)$ we have
  \[
    n(\sigma Q - Q)
    = \sigma P - P
    = 0
  .\]
  So $\sigma Q - Q \in E\tors[n]$.

  Since $\Gal(L / K)$ and $E\tors[n]$ are finite,
  there are only finitely many possibilities for
  the map
  \begin{align*}
    \Gal(L / K)
    &\to E\tors[n] \\
    \sigma
    &\mapsto \sigma Q - Q
  \end{align*}
  (even without requiring it to be a group
  homomorphism!).
