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  In fact,
  \[
    w(t) = t^3(1 + A_1 t + A_2 t^2 + \cdots)
  \]
  where $A_1 = a_1$, $A_2 = a_1^2 + a_2$, $A_3
  = a_1^3 + 2a_1 a_2 + 2a_3$, \ldots
\end{remark*}

\begin{fclemma}[]
  \label{lemma:8.2}
  Assuming:
  - $R$ an integral domain which is
  \gls{complete} with respect to an ideal $I$
  - $a_1, \ldots, a_6 \in R$
  - $K = \Frac(R)$
  Then:
  \[
    \hat{E}(I) \defeq \{(t, w) \in E(K) \st t, w
    \in I\}
  \]
  is a subgroup of $E(K)$.
\end{fclemma}

\begin{note*}
  By uniqueness in Hensel's lemma:
  \[
    \hat{E}(I) = \{(t, w(t)) \in E(K) \st t \in I\}
  .\]
\end{note*}

\begin{proof}
  Taking $(t, w) = (0, 0)$ show $0_E \in
  \hat{E}(I)$.

  So it suffices to show that $P_1, P_2 \in
  \hat{E}(I)$ then
  \[
    \ub{-P_1 - P_2}_{=P_3} \in \hat{E}(I)
  .\]
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/05271c376cd44a40.png}
  \end{center}
  $P_1, P_2 \in \hat{E}(I)$ implies $t_1, t_2 \in
  I$, $w_1 = w(t_1) \in I$, $w_2 = w(t_2) \in I$.
  \begin{align*}
    w(t)
    &= \sum_{n = 2}^\infty A_{n - 2} t^{n + 1}
    &&(A_0 = 1)
    \\
    \lambda
    &= \begin{cases}
      \frac{w(t_2) - w(t_1)}{t_2 - t_1} & t_1 \neq
      t_2 \\
      w'(t_1) & t_1 = t_2
    \end{cases} \\
    &= \sum_{n = 2}^{\infty} A_{n - 2} (t_1^n +
    t_1^{n - 1} t_2 + \cdots + t_2^n) \in I \\
    \nu
    &= w_1 - \lambda t_1 \in I
  \end{align*}
  Substituting $w = \lambda t + \nu$ into $w =
  f(t, w)$ gives
  \begin{align*}
    \lambda t + \nu
    &= t^3 + a_1 t(\lambda t + \nu) + a_2
    t^2(\lambda t + \nu) + a_3(\lambda t + \nu)^2
    + a_4 t(\lambda t + \nu)^2 + a_6(\lambda t +
    \nu)^3 \\
    A
    &= (\text{coefficient of $t^3$})
    = 1 + a_2 \lambda + a_4 \lambda^2 + a_6
    \lambda^3 \\
    B
    &= (\text{coefficient of $t^2$})
    = a_1 \lambda + a_2 \nu + a_3 \lambda^2 + 2a_4
    \lambda \nu + 3a_6 \lambda^2 \nu
  \end{align*}
  We have $A \in R^\times$ and $B \in I$. Hence
  $t_3 = -\frac{B}{A} - t_1 - t_2 \in I$, $w_3 =
  \lambda t_3 + \nu \in I$.
\end{proof}

Taking $R = \Zbb[a_1, \ldots, a_6][[t]]$, $I =
(t)$, then \cref{lemma:8.2} gives that there
exists $\iota \in \Zbb[a_1, \ldots, a_6][[t]]$
with $\iota(0) = 0$ and
\[
  [-1](t, w(t))
  = (\iota(t), w(\iota(t)))
.\]

Taking $R = \Zbb[a_1, \ldots, a_6][[t_1, t_2]]$,
$I = (t_1, t_2)$, \cref{lemma:8.2} gives that
there exists $F \in \Zbb[a_1, \ldots, a_6][[t_1,
t_2]]$ with $F(0, 0) = 0$ and
\[
  (t_1, w(t_1)) + (t_2, w(t_2))
  = (F(t_1, t_2), w(F(t_1, t_2)))
\]
and
\[
  F(X, Y)
  = X + Y - a_1 XY - a_2 (X^2 Y + XY^2) + \cdots
\]
By properties of the group law, we deduce
\begin{enumerate}[(i)]
  \item $F(X, Y) = F(Y, X)$
  \item $F(X, 0) = X$ and $F(0, Y) = Y$
  \item $F(X, F(Y, Z)) = F(F(X, Y), Z)$
  \item $F(X, \iota(X)) = 0$
\end{enumerate}

\begin{fcdefnstar}[Formal group]
  \glsnoundefn{fgroup}{formal group}{formal
  groups}%
  Let $R$ be a ring. A \emph{formal group} over
  $R$ is a power series $F(X, Y) \in R[[X, Y]]$
  satisfying
  \begin{cenum}[(i)]
    \item $F(X, Y) = F(Y, X)$
    \item $F(X, 0) = X$ and $F(0, Y) = Y$
    \item $F(X, F(Y, Z)) = F(F(X, Y), Z)$
  \end{cenum}
  This looks like it would only define a
  monoid, but in fact inverses are
  guaranteed to exist in this context.
\end{fcdefnstar}

\textbf{Exercise:} Show that for any formal group,
there exists a unique
\[
  \iota(X) = -X + \cdots \in R[[X]]
\]
such that $F(X, \iota(X)) = 0$.

\begin{example*}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $F(X, Y) = X + Y$ (called
      $\hat{\mathbb{G}}_a$)
    \item $F(X, Y) = X + Y + XY = (1 + X)(1 + Y) -
    1$ (called $\hat{\mathbb{G}}_m$)
    \item $F(X, Y) = \text{[as above]}$ (called
      $\hat{E}$)
  \end{enumerate}
\end{example*}

\begin{fcdefnstar}[Morphism / isomorphic (formal groups)]
  \glsnoundefn{fmorph}{morphism}{morphisms}%
  Let $\mathcal{F}$ and $\mathcal{G}$ be
  \glspl{fgroup} over $R$ given by power series
  $F$ and $G$.
  \begin{cenum}[(i)]
    \item A \emph{morphism} $f : \mathcal{F} \to
      \mathcal{G}$ is a power series $f \in
      R[[T]]$ such that $f(0) = 0$ satisfying
      \[
        f(F(X, Y))
        = G(f(x), f(Y))
      .\]
    \item $\mathcal{F} \cong \mathcal{G}$ if there
      exists $\mathcal{F} \stackrel{f}{\to}
      \mathcal{G}$ and $\mathcal{G}
      \stackrel{g}{\to} \mathcal{F}$ morphisms
      such that $f(g(X)) = g(f(X)) = X$.
  \end{cenum}
\end{fcdefnstar}

\begin{fcthm}[All formal groups are isomorphic]
  \label{thm:8.3}
  Assuming:
  - $\char R = 0$
  Then:
  any \gls{fgroup} $\mathcal{F}$ over $R$ is
  isomorphic to $\hat{\mathbb{G}}_a$ over $R
  \otimes \Qbb$.

  More precisely
  \begin{cenum}[(i)]
    \item There is a unique power series
      \[
        \log(T) = T + \frac{a_2}{2} T^2 +
        \frac{a_3}{3} T^3 + \cdots
      \]
      with $a_i \in R$ such that
      \[
        \log(F(X, Y)) = \log(X) + \log(Y)
        \tag{$*$}
        \label{lec11eq1}
      .\]
    \item There is a unique power series
      \[
        \exp(T)
        = T + \frac{b_2}{2!} + \frac{b_3}{3!} T^3
        + \cdots
      \]
      with $b_i \in R$ such that
      \[
        \exp(\log(T)) = \log(\exp(T)) = T
      .\]      
  \end{cenum}
\end{fcthm}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Notation $F_1(X, Y) = \frac{\partial
      F}{\partial x} (X, Y)$.

      \textbf{Uniqueness:} Let
      \[
        p(T)
        = \frac{\dd}{\dd T} \log(T)
        = 1 + a-2 T + a_3 T^2 + \cdots
      .\]
      Differentiating \eqref{lec11eq1} with
      respect to $X$ gives
      \[
        p(F(X, Y)) F_1(X, Y) = p(X)+  0
      .\]
      Putting $X = 0$ gives $p(Y) F_1(0, Y) = 1$,
      hence $p(Y) = F_1(0, Y)^{-1}$, so $p$ is
      uniquely determined by $F$ and hence $\log$
      is too.

      \textbf{Existence:} Let
      \[
        p(T)
        = F_1(0, T)^{-1} = 1 = a_2 T + a_3 T^2 +
        \cdots
      \]
      (say). Let
      \[
        \log(T)
        = T + \frac{a_2}{2} T^2 + \frac{a_3}{3}T^3
        + \cdots
      .\]
      Calculate
      \begin{align*}
        F(F(X, Y), Z)
        &= F(X, F(Y, Z)) \\
        F_1(F(X, Y), Z) F_1(X, Y)
        &= F_1(X, F(Y, Z))
        &&\frac{\dd}{\dd x} \\
        F_1(Y, Z) F_1(0, Y)
        &= F_1(0, F(Y, Z))
        &&\text{put $X = 0$} \\
        \implies
        F_1(Y, Z) p(Y)^{-1}
        &= p(F(Y, Z))^{-1} \\
        \implies F_1(Y, Z) p(F(Y, Z))
        &= p(Y) \\
        \log(F(Y, Z))
        &= \log(Y) + h(Z)
        &&\text{integrate w.r.t. $Y$}
      \end{align*}
      for some power series $h$.

      Symmetry $Y \leftrightarrow Z$ gives $h(Z) =
      \log(Z)$.

      This proves existence.
    \item We prove a lemma first.
      \qedhere
  \end{enumerate}
\end{proof}