%! TEX root = EC.tex
% vim: tw=50
% 14/02/2025 11AM

\begin{fcthm}[]
  \label{thm:7.4}
  Assuming:
  - $E / \Fbb_q$ an \gls{ellc} with $\#E(\Fbb_q) =
  q + 1 - a$
  Then:
  \[
    \Z_E(T)
    = \frac{1 - aT + qT^2}{(1 - T)(1 - qT)}
  .\]
\end{fcthm}

\begin{proof}
  Let $\phi : E \to E$ be the $q$ power Frobenius
  map. By \cref{coro:7.3}
  \[
    \#E(\Fbb_q) = q + 1 - \isogtr(\phi)
  .\]
  Hence $\isogtr(\phi) = a$, $\deg(\phi) = q$.

  \es{2}, Q6(iii) implies $\phi^2 - a\phi = q =
  0$, hence $\phi^{n = 2} - a\phi^{n + 1} +
  q\phi^n = 0$, so
  \[
    \isogtr(\phi^{n + 2}) - a\isogtr(\phi^{n + 1})
    + q\isogtr(\phi^n) = 0
  .\]
  This second order difference equation with
  initial conditions $\isogtr(1) = 2$,
  $\isogtr(\phi) = a$ has solution
  \[
    \isogtr(\phi^n) = \alpha^n + \beta^n
  ,\]
  where $\alpha, \beta$ are roots of $X^2 - aX + q
  = 0$.

  Again by \cref{coro:7.3},
  \begin{align*}
    \#E(\Fbb_q^n)
    &= q^n + 1 - \isogtr(\phi^n) \\
    &= 1 + q^n - \alpha^n - \beta^n
  \end{align*}
  Therefore
  \begin{align*}
    \Z_E(T)
    &= \exp \sum_{n = 1}^\infty \left(
    \frac{T^n}{n} + \frac{(qT)^n}{n} -
    \frac{(\alpha T)^n}{n} - \frac{(\beta T)^n}{n}
    \right) \\
    &= \frac{(1 - \alpha T)(1 - \beta T)}{(1 -
    T)(1 - qT)} \\
    &= \frac{1 - aT + qT^2}{(1 - T)(1 - qT)}
    \qedhere
  \end{align*}
\end{proof}

\begin{remark*}
  \nameref{thm:7.2} tells us that $|a| \le
  2\sqrt{q}$. $\alpha = \ol{\beta}$, and so
  $|\alpha| = |\beta| = \sqrt{q}$ $(*)$.

  Let $K = \Fbb_q(E)$. $\zeta_K(s) = 0$, so
  $\Z_E(q^{-s}) = 0$, so $q^s = \alpha$ or
  $\beta$. Then $q^{\Re(s)} = |\alpha|$ or
  $|\beta|$, so by $(*)$, $\Re(s) = \half$.

  ``This is an analog of the Riemann
  hypothesis.''
\end{remark*}

\newpage
\section{Formal Groups}

\begin{fcdefnstar}[$I$-adic topology]
  Let $R$ be a ring and $I \subset R$ an ideal.
  The \emph{$I$-adic topology} on $R$ has basis $\{r +
  I^n \st r \in R, n \ge 1\}$.
\end{fcdefnstar}

\begin{fcdefnstar}[Cauchy sequence]
  \glsadjdefn{cauchy}{Cauchy}{sequence}%
  A sequence $(x_n)$ in $R$ is \emph{Cauchy} if $\forall
  k, \exists N$ such that $\forall m, n \ge N$,
  $x_m - x_n \in I^k$.
\end{fcdefnstar}

\begin{fcdefnstar}[Complete]
  \glsadjdefn{complete}{complete}{ring}%
  $R$ is \emph{complete} if
  \begin{enumerate}[(i)]
    \item $\bigcap_{n \ge 0} I^n = \{0\}$
    \item every \gls{cauchy} sequence converges
  \end{enumerate}
\end{fcdefnstar}

\textbf{Useful remark:} if $x \in I$ then
\[
  \frac{1}{1 - x} = 1 + x + x^2 + \cdots
\]
so $1 - x \in R^\times$.

\begin{example*}
  $R = \Zbb_p$, $I = p\Zbb_p$.

  $R = \Zbb[[t]]$, $I = (t)$.
\end{example*}

\begin{fclemma}[Hensel's Lemma]
  \label{lemma:8.1}
  Assuming:
  - $R$ is \gls{complete} with respect to an ideal $I$
  - $F \in R[X]$, $s \ge 1$
  - $a \in R$ satisfies $F(a) \equiv 0
  \pmod{I^s}$, $F'(a) \in R^\times$
  Then:
  there exists a unique $b \in R$ such that
  $F(b) = 0$ and $b \equiv 0 \pmod{I^s}$.
\end{fclemma}

\begin{proof}
  Let $u \in R^\times$ with $F(a) = u \pmod{I}$
  (e.g. we could take $u = F'(a)$). Replacing
  $F(X)$ by $\frac{F(X + a)}{u}$, we may assume $a
  = 0$, $F'(0) \equiv 1 \pmod{I}$.

  We put $x_0 = 0$,
  \[
    x_{n + 1} = x_n - F(x_n)
    \tag{1}
    \label{lec10eq1}
  \]
  Easy induction gives
  \[
    x_n \equiv 0 \pmod{I^s} \qquad \forall n
    \tag{2}
    \label{lec10eq2}
  \]
  Let
  \[
    F(X) - F(Y)
    = (X - Y) (F'(0) + XG(X, Y) + YH(X, Y))
    \tag{3}
    \label{lec10eq3}
  \]
  for some $G, H \in R[X, Y]$.

  \textbf{Claim:} $x_{n + 1} \equiv x_n \pmod{I^{n
  + s}}$ for all $n \ge 0$.

  Proof: By induction on $n$. Case $n = 0$ is
  true.

  Suppose $x_n \equiv x_{n = 1} \pmod{I^{n + s -
  1}}$. Then
  \[
    F(x_n) - F(x_{n - 1})
    = (x_n - x_{n - 1})(1 + c)
    \pmod{I^{n + s}}
  \]
  for some $c \in I$. Hence
  \[
    F(x_n) - F(x_{n - 1})
    = x_n - x_{n - 1}
    \pmod{I^{n + s}}
  \]
  Hence
  \[
    x_n - F(x_n)
    \equiv x_{n - 1} - F(x_{n - 1})
    \pmod{I^{n + s}}
  \]
  and hence $x_{n + 1} \equiv x_n \pmod{I^{n +
  s}}$.

  This proves the claim.

  Therefore $(x_n)_{n \ge 0}$ is \gls{cauchy}.
  Since $R$ is \gls{complete}, we have $x_n \to b$
  as $n \to \infty$ for some $b \in R$.

  Taking limit $n \to \infty$ in \eqref{lec10eq1}
  gives $b = b - F(b)$, hence $F(b) = 0$.

  Taking limit $n \to \infty$ in \eqref{lec10eq2}
  gives $b \equiv 0 \pmod{I^s}$.

  Uniqueness is proved using \eqref{lec10eq3} and
  the ``useful remark'' (if $x \in I$ then $1 - x
  \in R^\times$).
\end{proof}

\[
  E:
  Y^2 Z + a_1 XYZ + a_3 YZ^2
  = X^3 + a_2 X^2 Z + a_4 XZ^2 + a_6 Z^3
.\]
Affine piece $Y \neq 0$. $t = -\frac{X}{Y}$,
$w = -\frac{Z}{Y}$.
\[
  w = \ub{t^3 + a_1 t w + a_2 t^2 w + a_3 w^2 + a_4
  tw^2 + a_6 w^3}_{= f(t, w)}
.\]
We apply \cref{lemma:8.1} with
\begin{align*}
  R
  &= \Zbb[a_1, \ldots, a_6][[t]] \\
  I
  &= (t) \\
  F(X)
  &= X - f(t, X) \in R[X]
\end{align*}
$s = 3$, $a = 0$.

Check:
\begin{align*}
  F(0)
  &= -f(t, 0) = -t^3 \equiv 0 \pmod{I^3} \\
  F'(0)
  &= 1 - a_1 t - a_2 t^2 \in R^\times
\end{align*}
Hence there exists a unique $w(t) \in R =
\Zbb[a_1, \ldots, a_6][[t]]$ such that
\begin{align*}
  w(t)
  &= f(t, w(t)) \\
  w(t)
  &\equiv 0 \pmod{t^3}
\end{align*}

\begin{remark*}
  Taking $u = 1$ in the proof of
  \cref{lemma:8.1}, $w(t) = \lim_{n \to
  \infty} w_n(t)$ where $w_0(t) = 0$, $w_{n +
  1}(t) = f(t, w_n(t))$.
