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\begin{proof}
  \cref{lemma:6.3} + induction gives $[n]^* \omega
  = n\omega$.

  $\char K \nmid n$ implies $[n]$ is separable.
  So $\#[n]^{-1} (Q) = \deg[n]$ for all but
  finitely many $Q \in E$.
  But $[n]$ is a group homomorphism, so
  $[n]^{-1}(Q) = \#E\tors[n]$ for all $Q \in E$.

  Putting these two statements together gives
  \[
    \#E\tors[n]
    = \deg[n]
    \stackrel{\text{\cref{coro:5.9}}}{=}
    n^2
  .\]
  Group theory (structure theorem) gives that
  \[
    E\tors[n]
    \cong \Zbb / d_1 \Zbb \times \cdots \times
    \Zbb / d_t \Zbb
  \]
  for some $1 < d_1 \mid d_2 \mid \cdots \mid d_t
  \mid n$.

  Let $p$ be a prime with $p \mid d_1$. Then
  $E\tors[p]
  \cong (\Zbb / p\Zbb)^t$. But $\#E\tors[p] = p^2$, so
  $t = 2$. But $\#E\tors[p] = p^2$, so $t = 2$, i.e.
  $E\tors[n] \cong \Zbb / d_1 \Zbb \times \Zbb / d_2
  \Zbb$. Since $d_1 \mid d_2 \mid n$ and $d_1 d_2
  = n^2$, we get $d_1 = d_2 = n$.

  Thus $E\tors[n] \cong (\Zbb / n\Zbb)^2$.
\end{proof}

\begin{remark*}
  If $\char K = p$ then $[p]$ is inseparable. It
  can be shown that:
  \begin{align*}
    \text{either } E\tors[p^r]
    &\cong \Zbb / p^r \Zbb ~\forall r \ge 1
    &&\text{``ordinary''}
    \text{or } E\tors[p]
    &= 0
    &&\text{``supersingular''}
  \end{align*}
  Do not use this remark on \es{2}!
\end{remark*}

\newpage
\section{Elliptic Curves over Finite Fields}

\begin{fclemma}[]
  \label{lemma:7.1}
  Assuming:
  - $A$ an abelian group
  - $q : A \to \Zbb$ a positive definit quadratic
  form
  Then:
  \[
    |\ub{q(x + y) - q(x) - q(y)}_{=\langle x,
    y\rangle}|
    \le 2\sqrt{q(x) q(y)} \qquad \forall x, y \in
    A
  .\]
\end{fclemma}

\begin{proof}
  We may assume $x \neq 0$, otherwise the result
  is clear. So $q(x) \neq 0$.

  Let $m, n \in \Zbb$. Then
  \begin{align*}
    0
    &\le q(mx + ny) \\
    &= \half \langle mx + ny, mx + ny \rangle \\
    &= m^2 q(x) + mn \langle x, y\rangle + n^2
    q(y) \\
    &= q(x) \left( m + \frac{\langle x,
    y\rangle}{2q(x)} n \right)^2 + \left( q(y) -
    \frac{\langle x, y\rangle^2}{4q(x)} \right) n^2
  \end{align*}
  Take $m = -\langle x, y\rangle$, $n = 2q(x) \neq
  0$ to deduce
  \[
    q(y) - \frac{\langle x, y\rangle^2}{4q(x)}
    \ge 0
  \]
  hence $\langle x, y\rangle^2 \le 4q(x)q(y)$ and
  hence
  \[
    |\langle x, y\rangle|
    \le 2\sqrt{q(x)q(y)}
  .\]
\end{proof}

\begin{fcthm}[Hasse's Theorem]
  \label{thm:7.2}
  Assuming:
  - $E / \Fbb_q$ an \gls{ellc}
  Then:
  \[
    |\#E(\Fbb_q) - (q + 1)|
    \le 2\sqrt{q}
  .\]
\end{fcthm}

\begin{proof}
  Recall $\Gal(\Fbb_{q^r} / \Fbb_q)$ is cyclic of
  order $r$ and generated by Frobenius $x \mapsto
  x^q$.

  Let $E$ have Weierstrass equation with
  coefficients $a_1, \ldots, a_6 \in \Fbb_q$ (so
  $a_i^q = a_i$). Define the Frobenius
  endomorphism
  \begin{align*}
    \phi : E
    &\to E \\
    (x, y)
    &\mapsto (x^q, y^q)
  \end{align*}
  This is an \gls{isog} of degree $q$.

  Then
  \[
    E(\Fbb_q)
    = \{P \in E \mid \phi(P) = P\}
    = \ker(1 - \phi)
  .\]
  \begin{align*}
    \phi^* \omega
    &= \phi^* \left( \frac{\dd x}{y} \right) \\
    &= \frac{\dd (x^q)}{y^q} \\
    &= \frac{qx^{q - 1} \dd x}{y^q} \\
    &= 0
    &&\text{($q \equiv 0 \pmod{p}$)}
  \end{align*}
  \cref{lemma:6.3} tells us that
  \[
    (1 - \phi)^* \omega
    = \omega - \phi^*(\omega)
    = \omega
    \neq 0
  .\]
  Hence $1 - \phi$ is separable.

  By \cref{thm:2.8} and the fact that $1 - \phi$
  is a group homomorphism, we argue as in the
  proof of \cref{thm:6.5} that
  \[
    \ub{\#\ker(1 - \phi)}_{= \#E(\Fbb_q)}
    = \deg(1 - \phi)
  .\]
  $\deg : \Hom(E, E) \to \Zbb$ is a positive
  definite \gls{qf} (\cref{thm:5.7}, and
  positive definiteness is obvious since
  non-constant morphisms have positive degree).

  \cref{lemma:7.1} gives
  \[
    |\deg(1 - \phi) - 1 - \deg\phi|
    \le 2\sqrt{\deg\phi}
  .\]
  Hence
  \[
    |\#E(\Fbb_q) - 1 - q|
    \le 2\sqrt{q}
    .
    \qedhere
  \]
\end{proof}

\begin{fcdefnstar}[]
  \glssymboldefn{isogip}%
  For $\phi, \psi \in \End(E) = \Hom(E, E)$, we
  put $\langle \phi, \psi\rangle = \deg(\phi +
  \psi) - \deg \phi - \deg \psi$ and $\trinternal(\phi) =
  \langle \phi, 1\rangle$.
\end{fcdefnstar}

\begin{fccoro}[]
  \label{coro:7.3}
  Assuming:
  - $E / \Fbb_q$ an \gls{ellc}
  - $\phi \in \End(E)$ the $q$-power Frobenius
  Then:
  $\#E(\Fbb_q) = q + 1 - \isogtr(\phi)$ and
  $|\isogtr(\phi)| \le 2\sqrt{q}$.
\end{fccoro}

\subsection{Zeta functions}

For $K$ a number field, let
\[
  \zeta_K(s)
  = \sum_{\mathfrak{a} \subset \mathcal{O}_K}
  \frac{1}{(N \mathfrak{a})^s}
  = \prod_{\substack{\mathfrak{p} \subset
  \mathcal{O}_K \\ \text{prime}}} \left( 1 -
  \frac{1}{(N \mathfrak{p})^s} \right)^{-1}
.\]
For $K$ a function field, i.e. $K = \Fbb_q(C)$
where $C / \Fbb_q$ is a smooth projective curve,
\[
  \zeta_K(s)
  = \prod_{x \in |C|} \left( 1 -
  \frac{1}{(Nx)^s} \right)^{-1}
,\]
where
\[
  |C| = \{\text{closed points on $C$}\}
\]
(closed points are orbits for action of
$\Gal(\ol{\Fbb_q} / \Fbb_q)$ on
$C(\ol{\Fbb_q})$) and $Nx = q^{\deg x}$, $\deg
x$ is the size of orbit.

We have $\zeta_K(s) = F(q^{-s})$ for some $F \in
\Qbb[[T]]$,
\begin{align*}
  F(T)
  &= \prod_{x \in |C|} (1 - T^{\deg x})^{-1}
  \\
  \log F(T)
  &= \sum_{x \in |C|} \sum_{m = 1}^\infty
  \frac{1}{m} T^{m \deg x}
  &&-\log(1 - x) = \sum_{m = 1}^\infty
  \frac{x^m}{m}
  \\
  T \frac{\dd }{\dd T} (\log F(T))
  &= \sum_{x \in |C|} \sum_{m = 1}^\infty \deg x
  T^{m \deg x} \\
  &= \sum_{n = 1}^{\infty} \left(
  \sum_{\substack{x \in |C| \\ \deg x \mid n}}
  \deg x \right) T^n
  &&n = m \deg x \\
  &= \sum_{n = 1}^{\infty}  \#C(\Fbb_{q^n}) T^n
\end{align*}
Therefore
\[
  F(T)
  = \exp \left( \sum_{n = 1}^{\infty}
  \frac{\#C(\Fbb_{q^n})}{n} T^n \right)
.\]

\begin{fcdefnstar}[Zeta function]
  \glssymboldefn{zetafunction}%
  The zeta function $Z_C(T)$ of a smooth
  projective curve $C / \Fbb_q$ is defined by
  \[
    Z_C(T)
    = \exp \left( \sum_{n = 1}^{\infty}
    \frac{\#C(\Fbb_{q^n})}{n} T^n \right)
  .\]
\end{fcdefnstar}