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\textbf{Application:} Assume $\char K \nmid n$.
There is a short exact sequence of $\Gal(\ol{K} /
K)$-modules
\begin{align*}
  0
  \to \mu_n
  \to \ol{K}^*
  &\to \ol{K}^*
  \to 0 \\
  x
  &\mapsto x^n
\end{align*}
Long exact sequence:
\begin{align*}
  K^*
  &\to K^*
  \to \Hone(\Gal(\ol{K} / K), \mu_n)
  \to \ob{\Hone(\Gal(\ol{K} / K), \ol{K}^*)}^{= 0
  \text{ by \gls{hilb90}}} \\
  x
  &\mapsto x^n
\end{align*}
Therefore $\Hone(\Gal(\ol{K} / K), \mu_n) \cong
K^* / (K^*)^n$.

If $\mu_n \subset K$ then
\[
  \Hom_{\text{cts}}(\Gal(\ol{K} / K), \mu_n)
  \cong K^* / (K^*)^n
\]
Finite subgroups of $\Hom_{\text{cts}}(\Gal(\ol{K}
/ K), \mu_n)$ are of the form $\Hom(\Gal(L / K),
\mu_n)$ for $L / K$ a finite abelian extension of
$K$ of exponent dividing $n$.

This gives another proof of \cref{thm:11.2}.

\begin{notation*}
  $\Hone(K, -)$ means $\Hone(\Gal(\ol{K} / K),
  -)$.
\end{notation*}

Let $\phi : E \to E'$ be an \gls{isog} of
\glspl{ellc} over $K$. Short exact sequence of
$\Gal(\ol{K} / K)$-modules
\[
  0
  \to E[\phi]
  \to E
  \stackrel{\phi}{\to} E'
  \to 0
\]
has long exact seqeucne
\[
  E(K)
  \stackrel{\phi}{\to} E'(K)
  \stackrel{\delta}{\to} \Hone(K, E[\phi])
  \to \Hone(K, E)
  \stackrel{\phi_*}{\to} \Hone(K, E')
.\]
We get a short exact sequence
\begin{picmath}
  \begin{tikzcd}
    0 \ar[r]
    & \frac{E'(K)}{\phi E(K)}
    \ar[r, "\delta"] \ar[d]
    & \Hone(K, E[\phi])
    \ar[r] \ar[d, "\res_v"] \ar[rd]
    & \Hone(K, E)[\phi_*]
    \ar[r] \ar[d, "\res_v"]
    & 0 \\
    0 \ar[r]
    & \prod_v \frac{E'(K_v)}{\phi E(K_v)}
    \ar[r, "\delta_V"]
    & \prod_v \Hone(K_v, E[\phi])
    \ar[r]
    & \prod_v \Hone(K_v, E)[\phi^*]
    \ar[r]
    & 0
  \end{tikzcd}
\end{picmath}
Now take $K$ a number field.

For each place $v$, fix an embedding $\ol{K}
\subset \ol{K}_v$. Then $\Gal(\ol{K}_v / K_v)
\subset \Gal(\ol{K} / K)$.

\begin{fcdefnstar}[Selmer-group]
  \glssymboldefn{selmer}%
  We define the \emph{$\phi$-Selmer group}
  \begin{align*}
    S^{(\phi)}(E / K)
    &= \ker\left(\Hone(K, E[\phi]) \to \prod_v
    \Hone(K_v, E)\right) \\
    &= \{\alpha \in \Hone(K, E[\phi]) \st
    \res_v(\alpha) \in \Im(\delta_v) ~\forall v\}
  \end{align*}
  (the map $\Hone(K, E[\phi]) \to \prod_v
  \Hone(K_v, E)$ is as in the commutative diagram
  above).
\end{fcdefnstar}

\begin{fcdefnstar}[Tate-Shafarevich group]
  \glssymboldefn{TS}%
  The \emph{Tate-Shafarevich group} is
  \[
    \Sha(E / K)
    = \ker(\Hone(K, E) \to \prod_v \Hone(K_v, E))
  .\]
\end{fcdefnstar}

We get a short exact sequence
\[
  0
  \to \frac{E'(K)}{\phi E(K)}
  \to \Selmer(E / K)
  \to \TS(E / K)[\phi_K]
  \to 0
.\]
Taking $\phi = [n]$ gives
\[
  0
  \to \frac{E(K)}{n E(K)}
  \to \Selmer[n](E / K)
  \to \TS(E / K)\tors[n]
  \to 0
.\]
Reorganising the proof of \gls{mwt} gives

\begin{fcthm}[]
  \label{thm:15.3}
  $\Selmer[n](E / K)$ is \cloze{finite.}
\end{fcthm}

\begin{proof}
  For $L / K$ a finite Galois extension,
  \begin{picmath}
    \begin{tikzcd}
      0 \ar[r]
      & \ob{\Hone(\Gal(L / K),
      E(L)\tors[n])}^{\text{finite}}
      \ar[r, "\inf"]
      & \Hone(K, E\tors[n])
      \ar[r, "\res"]
      \ar[d, phantom, sloped, "\supset"]
      & \Hone(L, E\tors[n])
      \ar[d, phantom, sloped, "\supset"]
      \\
      &
      & \Selmer[n](E / K)
      \ar[r]
      & \Selmer[n](E / L)
    \end{tikzcd}
  \end{picmath}
  Therefore by extending our field, we may assume
  $E\tors[n] \subset E(K)$ and hence by the Weil
  pairing $\mu_n \subset K$.

  Therefore $E\tors[n] \cong \mu_n \times \mu_n$
  as a $\Gal(\ol{K} / K)$-module. Then
  \begin{align*}
    \Hone(K, E\tors[n])
    &\cong \Hone(K, \mu_n) \times \Hone(K, \mu_n)
    \\
    &\cong K^* / (K^*)^n \times K^* / (K^*)^n
  \end{align*}
  Let
  \[
    S
    = \{\text{primes of \gls{bred} for $E$}\}
    \cup \{v \mid n \infty\}
  \]
  (a finite set of places).

  Define the subgroup of $\Hone(K, A)$ unramified
  outside of $S$ as
  \[
    \Hone(K, A; S)
    = \ker(\Hone(K, A) \to \prod_{v \notin S}
    \Hone(K_v^{\text{nr}}, A))
  .\]

  There is a commutative diagram with exact rows:
  \begin{picmath}
    \begin{tikzcd}
      E(K_v)
      \ar[r, "\times n"] \ar[d]
      & E(K_v)
      \ar[r, "\delta_v"] \ar[d]
      & \Hone(K_v, E\tors[n])
      \ar[d, "\res"]
      \\
      E(K_v^\nr)
      \ar[r, "\times n"]
      & E(Kv^\nr)
      \ar[r, "0"]
      & \Hone(K_v^\nr, E\tors[n])
    \end{tikzcd}
  \end{picmath}
  The bottom map $\times n$ is surjective $\forall
  v \notin S$ (see \cref{thm:9.8}).

  Therefore
  \begin{align*}
    \Selmer[n](E / K)
    &\subset \Hone(K, E\tors[n]; S) \\
    &\cong \Hone(K, \mu_n; S) \times \Hone(K,
    \mu_n; S)
  \end{align*}
  But
  \begin{align*}
    \Hone(K, \mu_n; S)
    &= \ker \left( \frac{K^*}{(K^*)^n} \to
    \prod_{v \notin S}
    \frac{(K_v^\nr)^*}{(K_v^\nr)^{*n}} \right) \\
    &\subset K(S, n)
  \end{align*}
  which is finite by \cref{lemma:11.4}.
\end{proof}

\begin{remark*}
  $\Selmer[n](E / K)$ is finite and effectively
  computable.

  It is conjectured that $|\TS(E / K)| < \infty$.
  This would imply that $\rank E(K)$ is
  effectively computable.
\end{remark*}

\newpage
\section{Descent by Cyclic Isogeny}

Let $E, E'$ be \glspl{ellc} over a number field
$K$, and $\phi : E \to E'$ an \gls{isog} of degree
$n$.

Suppose $E'[\hat{\phi}] \cong \Zbb / n\Zbb$
generated by $T \in E'(K)$. Then $E[\phi] \cong
\mu_n$ as a $\Gal(\ol{K} / K)$-module
\[
  S \mapsto \ephi(S, T)
.\]
Have a short exact sequence of $\Gal(\ol{K} /
K)$-modules
\[
  0
  \to \mu_n
  \to E
  \stackrel{\phi}{\to} E'
  \to 0
.\]
Get long exact sequence
\begin{picmath}
  \begin{tikzcd}
    E(K)
    \ar[r, "\phi"]
    & E'(K)
    \ar[r, "\delta"]
    \ar[rd, "\alpha"]
    & \Hone(K, \mu_n)
    \ar[r] \ar[d, phantom, sloped, "\cong"]
    & \Hone(K, E) \\
    &
    & K^* / (K^*)^n
  \end{tikzcd}
\end{picmath}
(the $\cong$ is by \gls{hilb90}).

\begin{fcthm}[]
  \label{thm:16.1}
  Assuming:
  - $f \in K(E')$ and $g \in K(E)$
  - $\div(f) = n(T) - n(0)$
  - $\phi^* f = g^n$
  Then:
  $\alpha(P) = f(P) \pmod{(K^*)^n}$ for all $P \in
  E;(K) \setminus \{0, T\}$.
\end{fcthm}
