%! TEX root = EC.tex
% vim: tw=50
% 12/03/2025 11AM
%
      \item Linearity in second argument: Let
        $T_1, T_2 \in E'[\hat{\phi}]$.
        \begin{align*}
          \div(f_1)
          &= n(T_1) - n(0)
          &\phi^* f
          &= g_1^n \\
          \div(f_2)
          &= n(T_2) - n(0)
          &\phi^* f_2
          &= g_2^n
        \end{align*}
        There exists $h \in \ol{K}(E')$ such that
        \[
          \div(h)
          = (T_1) + (T_2) - (T_1 + T_2) - (0)
        .\]
        Then put $f = \frac{f_1 f_2}{h^n}$ and $g
        = \frac{g_1 g_2}{\phi^* h}$.

        Check: $\div(f) = n(T_1 + T_2) - n(0)$.
        Yes.
        \[
          \phi^* f
          = \frac{\phi^* f_1 \phi^* f_2}{(\phi^*
          h)^n}
          = \left( \frac{g_1 g_2}{\phi^*(h)}
          \right)^n
          = g^n
        .\]
        Therefore
        \begin{align*}
          \ephi(S, T_1 + T_2)
          &= \frac{g(X + S)}{g(X)} \\
          &= \frac{g_1(X + S)}{g_1(X)} \frac{g_2(X
          + S)}{g_2(X)} \ub{\frac{h(\phi(X))}{h(\phi(X
          + S))}}_{\substack{= 1 \\ \text{since $S
          \in E[\phi]$}}} \\
          &= \ephi(S, T_1) \ephi(S, T_2)
        \end{align*}
      \item $\ephi$ is non-degenerate. Fix $T \in
        E'[\hat{\phi}]$. Suppose $\ephi(S, T) = 1$
        for all $S \in E[\phi]$. Then $\tau_S^* g
        = g$ for all $S \in E[\phi]$.

        Have $\ol{K}(E) / \phi^* \ol{K}(E')$ is a
        Galois extension with Galois group
        $E[\phi]$ ($S \in E[\phi]$ acts on
        $\ol{K}(E)$ via $\tau_S^*$).

        Therefore $g = \phi^* h$ for some $h \in
        \ol{K}(E')$. So $\phi^* f = g^n =
        \phi^*(h^n)$. So $f = h^n$, and hence
        $\div(h) = (T) - (0)$. So $T = 0$.
        \qedhere
  \end{enumerate}
\end{proof}

We've shown $E'[\phi] \injto \Hom(E[\phi],
\mu_n)$. It is an isomorphism since $\#E[\phi] =
\#E'[\hat{\phi}] = n$.

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(i)]
    \item If $E, E', \phi$ are defined over $K$
      then $\ephi$ is Galois equivariant, i.e.
      $\forall \sigma \in \Gal(\ol{K} / K)$,
      $\forall S \in E[\phi]$, $\forall T \in
      E'[\phi]$,
      \[
        \ephi(\sigma S, \sigma T)
        = \sigma(\ephi(S, T))
      .\]
    \item Taking $\phi = [n] : E \to E$ (so
      $\hat{\phi} = [n]$) gives
      \[
        e_n : E\tors[n] \times E\tors[n] \to
        \cancel{\mu_{n^2}} \mu_n
      .\]
  \end{enumerate}
\end{remark*}

\begin{fccoro}[]
  \label{coro:14.6}
  Assuming:
  - $E\tors[n] \subset E(K)$
  Then:
  $\mu_n \subset K$.
\end{fccoro}

\begin{proof}
  Let $T \in E\tors[n]$ have order $n$. Non
  degeneracy of $e_n$ implies that there exists $S
  \in E\tors[n]$ such that $e_n(S, T)$ is a
  primitive $n$-th root of unity, say $\zeta_n$.

  Then
  \[
    \sigma(\zeta_n)
    = \sigma(e_n(S, T))
    = e_n(\cancel{\sigma} S, \cancel{\sigma} T)
    = \zeta_n
  \]
  for all $\sigma \in \Gal(\ol{K} / K)$. So
  $\zeta_n \in K$.
\end{proof}

\begin{example*}
  There does not exist $E / \Qbb$ with
  $E(\Qbb)_{\text{tors}} \cong (\Zbb / 3\Zbb)^2$.
\end{example*}

\begin{remark*}
  In fact, $e_n$ is alternating, i.e. $e_n(T, T) =
  1$ for all $T \in E\tors[n]$.

  (This implies $e_n(S, T) = e_n(T, S)^{-1}$).
\end{remark*}

\newpage
\section{Galois Cohomology}

Let $G$ be a group and $A$ be a $G$-module (an
abelian group with an action of $G$ via group
homomorphisms). $G$-module means exactly the same
thing as $\Zbb[G]$-module.

\begin{fcdefnstar}[$H^0$]
  Define
  \[
    H^0(G, A)
    = A^G
    = \{a \in A \st \sigma(a) = a~\forall \sigma \in G\}
  .\]
\end{fcdefnstar}

\begin{fcdefnstar}[Cochains]
  \glsnoundefn{cochain}{cochain}{cochains}%
  Define \[
    C^1(G, A)
    = \{\text{maps $G \to A$}\}
  \]
  (called ``cochains'').
\end{fcdefnstar}

\begin{fcdefnstar}[Cocycles]
  \glssymboldefn{Zone}%
  Define \[
    Z^1(G, A)
    = \{(a_\sigma)_{\sigma \in G}
    \st a_{\sigma\tau} = \sigma(a_\tau) + a_\sigma
    ~\forall \sigma, \tau \in G\}
  \]
  (called ``cocycles'').
\end{fcdefnstar}

\begin{fcdefnstar}[Coboundaries]
  \glssymboldefn{Bone}%
  Define
  \[
    B^1(G, A)
    = \{(\sigma b - b)_{\sigma \in G} \st b \in A\}
  \]
  (called ``coboundaries'').
\end{fcdefnstar}

\begin{note*}
  $C^1(G, A) \supset Z^1(G, A) \supset B^1(G, A)$.
\end{note*}

Then we can define:

\begin{fcdefnstar}[$H^1$]
  \glssymboldefn{Hone}%
  Define
  \[
    H^1(G, A)
    = \frac{Z^1(G, A)}{B^1(G, A)}
  .\]
\end{fcdefnstar}

\begin{remark*}
  If $G$ acts trivially on $A$, then
  \[
    \Hone(G, A)
    = \Hom(G, A)
  .\]
\end{remark*}

\begin{fcthm}[]
  \label{thm:15.1}
  Assuming:
  - we have a short exact sequence of $G$-modules
  \[
    0
    \to A
    \stackrel{\phi}{\to} B
    \stackrel{\psi}{\to} C
    \to 0
  .\]
  Then:
  it gives rise to a long exact sequence of
  abelian groups:
  \[
    0
    \to A^G
    \stackrel{\phi}{\to} B^G
    \stackrel{\psi}{\to} C^G
    \stackrel{\delta}{\to} \Hone(G, A)
    \stackrel{\phi^*}{\to} \Hone(G, B)
    \stackrel{\psi^*}{\to} \Hone(G, C)
  .\]
\end{fcthm}

\noproof

\begin{proof}
  Omitted.
\end{proof}

\begin{fcdefnstar}[$delta$]
  Let $c \in C^G$. Then $b \in B$ such that
  $\psi(b) = c$. Then
  \[
    \psi(\sigma b - b)
    = \sigma c - c
    = 0
    \quad \forall \sigma \in G
  \]
  so $\sigma b - b = \phi(a_\sigma)$ for some
  $a_\sigma \in A$.

  Can check $(a_\sigma)_{\sigma \in G} \in
  \Zone(G, A)$.

  Then $\delta(c) =$ class of $(a_\sigma)_{\sigma
  \in G}$ in $\Hone(G, A)$.
\end{fcdefnstar}

\begin{fcthm}[]
  \label{thm:15.2}
  Assuming:
  - $A$ is a $G$-module
  - $H \normalsub G$ a normal subgroup
  Then:
  there is an inflation restriction exact sequence
  \[
    0
    \to \Hone(G / H, A^H)
    \stackrel{\operatorname{inf}}{\to} \Hone(G, A)
    \stackrel{\operatorname{res}}{\to} \Hone(H, A)
  \]
\end{fcthm}

\noproof

\begin{proof}
  Omitted.
\end{proof}

Let $K$ be a perfect field. Then $\Gal(\ol{K} /
K)$ is a topological group with basis of open
subgroups being the $\Gal(\ol{K} / L)$ for $[L :
K] < \infty$.

If $G = \Gal(\ol{K} / K)$ then we modify the
definition of $\Hone(G, A)$ by insisting:
\begin{enumerate}[(1)]
  \item The stabiliser of each $a \in A$ is an
    open subgroup of $G$.
  \item All \glspl{cochain} $G \to A$ are
    continuous, where $A$ is given the discrete
    topology.
\end{enumerate}
Then
\[
  \Hone(\Gal(\ol{K} / K), A)
  = \lim_{\substack{\longrightarrow \\ L \\
  \text{$L / K$ finite Galois}}}
  \Hone(\Gal(L / K), A^{\Gal(\ol{K} / L)})
.\]
(direct limit is with respect to inflation maps).

\begin{fcthmstar}[Hilbert's Theorem 90]
  \glsnoundefn{hilb90}{Hilbert 90}{NA}%
  Assuming:
  - $L / K$ a finite Galois extension
  Then:
  \[
    \Hone(\Gal(L / K), L^*)
    = 0
  .\]
\end{fcthmstar}

\begin{proof}
  Let $G = \Gal(L / K)$. Let $(a_\sigma)_{\sigma
  \in G} \in \Zone(G, L^*)$. Distinct
  automorphisms are linearly independent, so there
  exists $y \in L$ such that
  \[
    \ub{\sum_{\tau \in G} a_\tau^{-1} \tau(y)}_{=
    x}
    \neq 0
  .\]
  Then
  \begin{align*}
    \sigma(x)
    &= \sum_{\tau \in G} \sigma(a_\tau)^{-1}
    \sigma \tau(y) \\
    &= a_\sigma \ub{\sum_{\tau \in G} a_{\sigma
    \tau}^{-1} \sigma \tau(y)}_{=x}
  \end{align*}
  Then $a_\sigma = \frac{\sigma(x)}{x}$ for all
  $\sigma \in G$. ($a_\sigma \tau = \sigma(a_\tau)
  a_\sigma$ $\implies$ $\sigma(a_\tau)^{-1} =
  a_\sigma a_{\sigma \tau}^{-1}$).

  So $(a_\sigma)_{\sigma \in G} \in \Bone(G,
  L^*)$. Therefore $\Hone(G, L^*) = 0$.
\end{proof}

\begin{corollary*}
  $\Hone(\Gal(\ol{K} / K), \ol{K}^*) = 0$.
\end{corollary*}