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% Friday 28th February, MR9, 3:30pm.
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\begin{fclemma}[]
  \label{lemma:8.4}
  Assuming:
  - $f(T) = aT + \cdots \in R[[T]]$ with $a \in
  R^\times$
  Then:
  there exists a unique $g(T) = a^{-1} T + \cdots
  \in R[[T]]$ such that $f(g(T)) = g(f(T)) = T$.
\end{fclemma}

\begin{proof}
  We construct polynomials $g_n(T) \in R[T]$ such
  that
  \begin{align*}
    f(g_n(T))
    &\equiv T \pmod{T^{n + 1}} \\
    g_{n + 1}(T)
    &\equiv g_n(T) \pmod{T^{n + 1}}
  \end{align*}
  Then $g(T) = \lim_{n \to \infty} g_n(T)$
  satisfies $f(g(T)) = T$.

  To start the induction, we set $g_1(T) = a^{-1}
  T$. Now suppose $n \ge 2$ and $g_{n - 1}(T)$
  exists. Then
  \[
    f(g_{n - 1}(T))
    \equiv T + b T^n \pmod{T^{n + 1}}
  .\]
  We put $g_n(T) = g_{n - 1}(T) + \lambda T^n$ for
  some $\lambda \in R$ to be chosen later.

  Then
  \begin{align*}
    f(g_n(T))
    &= f(g_{n - 1}(T) + \lambda T^n) \\
    &\equiv f(g_{n - 1}(T)) + \lambda a T^n
    \pmod{T^{n + 1}} \\
    &\equiv T + (b + \lambda a)T^n
    \pmod{T^{n + 1}}
  \end{align*}
  We take $\lambda = -\frac{b}{a}$ ($a \in
  R^\times, b \in R \implies \lambda \in R$).

  This completes the induction step.

  We get $g(T) = a^{-1}T + \cdots \in R[[T]]$ such
  that $f(g(T)) = T$. Applying the same
  construction to $g$ gives $h(T) = aT + \cdots
  \in R[[T]]$ such that $g(h(T)) = T$.

  Now note $f(T) = f(g(h(T))) = h(T)$.
\end{proof}

\cref{thm:8.3}(ii) now follows by \cref{lemma:8.4}
and Q12 from \es{2}.

\begin{notation*}
  Let $\mathcal{F}$ (e.g. $\hat{\mathbb{G}}_a$,
  $\hat{\mathbb{G}}_m$, $\hat{E}$) be a
  \gls{fgroup} given by a power series $F \in
  R[[X, Y]]$.

  Suppose $R$ is a ring \gls{complete} with
  respect to ideal $I$. For $x, y \in I$, put
  \[
    x \oplus_{\mathcal{F}} y = F(x, y) \in I
  .\]
  Then $\mathcal{F}(I) \defeq (I,
  \oplus_{\mathcal{F}})$ is an abelian group.

  Examples:
  \begin{itemize}
    \item $\hat{\mathbb{G}}_a(I) = (I, +)$
    \item $\hat{\mathbb{G}}_m(I) = (1 + I, \times)$
    \item $\hat{E}(I) = \text{subgroup of $E(K)$
      in \cref{lemma:8.2}}$
  \end{itemize}
\end{notation*}

\begin{fccoro}[]
  \label{coro:8.5}
  Assuming:
  - $F$ a \gls{fgroup} over $R$
  - $n \in \Zbb$ such that $n \in R^\times$
  Thens:[(i)]
  - $[n] : \mathcal{F} \to \mathcal{F}$ is an
  isomorphism of \glspl{fgroup}
  - If $R$ is \gls{complete} with respect to ideal $I$
  then $\mathcal{F}(I) \stackrel{\times n}{\to}
  \mathcal{F}(I)$ is an isomorphism of groups. In
  particular, $\mathcal{F}(I)$ has no $n$-torsion.
\end{fccoro}

\begin{proof}
  We have
  \begin{align*}
    [1](T)
    &= T \\
    [n](T)
    &= F([n - 1](T), T)
  \end{align*}
  (for $n < 0$ use $[-1](T) = \iota(T)$).

  Since
  \[
    F(X, Y)
    = X + Y + XY(\cdots)
  \]
  we get
  \[
    [2](T)
    = F(T, T)
    2T + \cdots \in R[[T]]
  \]
  and by induction we get
  \[
    [n](T)
    = nT + \cdots \in R[[T]]
  .\]
  \cref{lemma:8.4} shows that if $n \in
  R^{\times}$ then $[n]$ is an isomorphism. This
  proves (i), and (ii) follows.
\end{proof}

\newpage
\section{Elliptic Curves over Local Fields}

Let $K$ be a field, complete with respect to
discrete valuation $v : K^* \to \Zbb$.

\begin{notation*}
  Valuation ring (= ring of integers) will be
  denoted by
  \[
    \mathcal{O}_K
    = \{x \in K^* \st v(x) \ge 0\} \cup \{0\}
  .\]
  
  Unit group will be denoted by
  \[
    \mathcal{O}_K^*
    = \{x \in K^* \st v(x) = 0\}
  .\]
  
  The maximal ideal will be denoted by $\pi
  \mathcal{O}_K$ where $v(\pi) = 1$.

  The residue field will be denoted by $k =
  \mathcal{O}_K / \pi \mathcal{O}_K$.

  We assume $\char K = 0$ and $\char k = p > 0$.
  For example, $K = \Qbb_p$, $\mathcal{O}_K =
  \Zbb_p$, $k = \Fbb_p$.
\end{notation*}

Let $E / K$ be an \gls{ellc}.

\begin{fcdefnstar}[Integral / minimal Weierstrass
  equation]
  \glsadjdefn{int}{integral}{equation}%
  \glsadjdefn{min}{minimal}{equation}%
  A \gls{weq} for $E$ with coefficients
  $a_1, \ldots, a_6 \in K$ is \emph{integral} if
  $a_1, \ldots, a_6 \in \mathcal{O}_K$ and
  \emph{minimal} if $v(\Delta)$ is minimal among
  all integral \glspl{weq} for $E$.
\end{fcdefnstar}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Putting $x = u^2 x'$, $y = u^3 y'$ gives
      $a_i = u^i a_i'$. Therefore \gls{int}
      \glspl{weq} exist.
    \item $a_1, \ldots, a_6 \in \mathcal{O}_K
      \implies \Delta \in \mathcal{O}_K \implies
      v(\Delta) \ge 0$. Therefore \gls{min}
      \glspl{weq} exist.
    \item If $\char k \neq 2, 3$ then there exists
      \gls{min} \gls{weq} of the form
      $y^2 = x^3 + ax + b$.
  \end{enumerate}
\end{remark*}

\begin{fclemma}[]
  \label{lemma:9.1}
  Assuming:
  - $E / K$ have \gls{int} \gls{weq}
  \[
    y^2 + a_1 xy + a_3 y
    = x^3 + a_2 x^2 + a_4 x + a_6
  \]
  - $0 \neq P = (x, y) \in E(K)$
  Then:
  either $x, y \in \mathcal{O}_K$ or
  \[
    \begin{cases}
      v(x) = -2s \\
      v(y) = -3s
    \end{cases}
  \]
  for some $s \ge 1$.
\end{fclemma}

(Compare with Q5 from \es{1})

\begin{proof}
  Throughout this proof, LHS and RHS refer to the
  \gls{weq} of the curve.

  \textbf{Case $v(x) \ge 0$:}. If $v(y) < 0$ then
  $v(\LHS) < 0$ and $v(\RHS) \ge 0$. Therefore $x,
  y \in \mathcal{O}_K$.

  \textbf{Case $v(x) < 0$:} $v(\LHS) \ge
  \min(2v(y), v(x) + v(y), v(y))$ and $v(\RHS) =
  3v(x)$. We get 3 possible inequalities from
  this, and each of them gives $v(y) < v(x)$.

  Now
  \begin{align*}
    v(\LHS)
    &= 2v(y) \\
    v(\RHS)
    &= 3v(x)
  \end{align*}
  so
  \[
    \begin{cases}
      v(x) = -2s \\
      v(y) = -3s
    \end{cases}
  \]
  for some $s \ge 1$.
\end{proof}
