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\begin{fcprop}[]
  \label{prop:2.7}
  Assuming:
  - $K = \ol{K}$, $\char K \neq 2$
  - $C \subset \Pbb^2$ a smooth plane cubic
  - $P \in C$ a point of inflection
  Then:
  we may change coordinates such that
  \[
    C : Y^2 Z = X(X - Z)(X - \lambda Z)
  \]
  for some $\lambda \neq 0, 1$ and $P = (0 : 1 : 0)$.
\end{fcprop}

\begin{proof}
  We change coordinates such that $P = (0 : 1 :
  0)$, $T_p(C) = \{Z = 0\}$, and $C : \{F(X, Y, Z)
  = 0\} \subset \Pbb^2$.

  $P \in C$ part of inflection implies $F(t, 1,
  0) = \text{const} t^3$, i.e. $F$ has no terms
  $X^2 Y$, $XY^2$ or $Y^3$.

  Therefore
  \[
    F \in \langle Y^2 Z, XYZ, YZ^2, X^3, X^2 Z,
    XZ^2, Z^3 \rangle
  .\]
  The $Y^2 Z$ coefficient must be $\neq 0$
  otherwise $P \in C$ is singular, and the
  coefficient of $X^3$ is $\neq 0$ otherwize $Z
  \mid F$.

  We are free to rescale $X$, $Y$, $Z$ and $F$.
  Then without loss of generality $C$ is defined
  by
  \[
    Y^2 Z + a_1 XYZ + a_3 YZ^2 = X^3 + a_2 X^z Z +
    a_4 XZ^2 + a_6 Z^3
  .\]
  Weierstrass form.

  Substituting $Y \leftarrow Y - \half a_1 X -
  \half a_3 Z$, we may suppose $a_1 = a_3 = 0$.
  Now
  \[
    Y^2 Z = Z^3 f \left( \frac{X}{Z} \right)
  \]
  for some monic cubic polynomial $f$.

  $C$ is smooth, so $f$ has distinct roots.
  Without loss of generality say $0, 1, \lambda$.
  Then $C$ is given by
  \[
    Y^2 Z = X(X - Z)(X - \lambda Z)
    .
    \qedhere
  \]
\end{proof}

\begin{remark*}
  It may be shown that the points of inflection on
  a smooth plane curve
  \[
    C = \{F(X_1, X_2, X_3) = 0\} \subset \Pbb^2
  \]
  are given by
  \[
    F = \ub{\det \left( \frac{\partial^2
    F}{\partial X_i \partial X_j}
    \right)}_{\text{Hessian}} = 0
  ,\]
\end{remark*}

\subsection{The degree of a morphism}

Let $\phi : C_1 \to C_2$ be a non-constant
morphism of smooth projective curves.

Then $\phi^* : K(C_2) \to K(C_1)$, $f \mapsto f
\circ \phi$.

\begin{fcdefnstar}[Degree of a morphism]
  $\deg\phi [K(C_1) : \phi^* K(C_2)]$.
\end{fcdefnstar}

\begin{fcdefnstar}[Separable morphism]
  \glsadjdefn{sep}{separable}{morphism}%
  $\phi$ is separable if $K(C_1) / \phi^* K(C_2)$
  is a separable field extension.
\end{fcdefnstar}

\begin{fcdefnstar}[Ramification index]
  Suppose $P \in C_1$, $Q \in C_2$, $\phi : P
  \mapsto Q$. Let $t \in K(C_2)$ be a uniformiser
  at $Q$.

  The ramification index of $\phi$ at $P$ is
  \[
    e_\phi(P) = \ord_P(\phi^* t)
  \]
  (always $\ge 1$, independent of choice of $t$).
\end{fcdefnstar}

\begin{fcthm}[]
  \label{thm:2.8}
  Assuming:
  - $\phi : C_1 \to C_2$ a non-constant morphism
  of smooth projective curves
  Then:
  \[
    \sum_{P \in \phi^{-1} Q} e_\phi (P)
    = \deg \phi
    \qquad \forall Q \in C_2
  .\]
  Moreover, if $\phi$ is \gls{sep} then $e_\phi(P)
  = 1$ for all but finitely many $P \in C_1$.

  In particular:
  \begin{enumerate}[(i)]
    \item $\phi$ is surjective (on
      $\ol{K}$-points)
    \item $\#\phi^{-1}(Q) \le \deg \phi$
    \item If $\phi$ is \gls{sep} then equality
      holds in (ii) for all but finitely many $Q
      \in C_2$.
  \end{enumerate}
\end{fcthm}

\begin{remark}
  \label{remark:2.9}
  Let $C$ be an algebraic curve. A rational map is
  given $C \ratmap \Pbb^n$, $P \mapsto (f_0(P) :
  f_1(P) : \cdots : f_n(P))$ where $f_0, f_1,
  \ldots, f_n \in K(C)$ are not all zero.
\end{remark}

\textbf{Important Fact:} If $C$ is smooth then $\phi$ is a
morphism.

\newpage
\section{Weierstrass Equations}

In this section, we drop the assumption that $K =
\ol{K}$, but we instead assume that $K$ is a
perfect field.

\begin{fcdefnstar}[Elliptic curve]
  \glsnoundefn{ellc}{elliptic curve}{elliptic
  curves}%
  An \emph{elliptic curve} $E / K$ is a smooth
  projective curve of genus $1$, defined over $K$
  with a specified $K$-rational point $0_E \in
  E(K)$.
\end{fcdefnstar}

\begin{example*}
  $\{X^3 + pY^3 + p^2 Z^3 = 0\} \subset \Pbb^2$ is
  \emph{not} an \gls{ellc} over $\Qbb$, since it
  has no $\Qbb$-rational points.
\end{example*}

\glsnoundefn{wform}{Weierstrass form}{NA}%
\glsnoundefn{weq}{Weierstrass
equation}{Weierstrass equations}%
\begin{fcthm}[]
  \label{thm:3.1}
  Assuming:
  - $E$ is an \gls{ellc}
  Then:
  $E$ is isomorphic over $K$ to a curve in
  Weierstrass form via an isomorphism taking $0_E$
  to $(0 : 1 : 0)$.
\end{fcthm}

\begin{remark*}
  \cref{prop:2.7} treated the special case $E$ is
  a smooth plane cubic and $0_E$ is a point of
  inflection.
\end{remark*}

\textbf{Fact:} If $D \in \Div(E)$ is defined over
$K$ (i.e. fixed by $\Gal(\ol{K} / K)$) then
$\mathcal{L}(D)$ has a basis in $K(E)$ (not just
$\ol{K}(E)$).

\begin{proof}[Proof of \cref{thm:3.1}]
  $\mathcal{L}(2 . 0_E) \subset \mathcal{L}(3 .
  0_E)$. Pick basis $1, x$ for
  $\mathcal{L}(2 . 0_E)$ and $1, x, y$ for
  $\mathcal{L}(3 . 0_E)$.

  Note: $\ord_{0_E}(x) = -2$, $\ord_{0_E}(y) =
  -3$.

  The 7 elements $1, x, y, x^2, xy, x^3, y^2$ in
  the 6-dimensional space $\mathcal{L}(6 . 0_E)$
  must satisfy a dependence relation.

  Leaving out $x^3$ or $y^2$ gives a basis for
  $\mathcal{L}(6 . 0_E)$ since each term has a
  different order pole at $0_E$.

  Therefore the coefficients of $x^3$ and $y^2$
  are non-zero. Rescaling $x$ and $y$ (if
  necessary) we get
  \[
    y^2 + a_1 xy + a_3 y
    = x^3 + a_2 x^2 + a_4 x + a_6
  \]
  for some $a_i \in K$.

  Let $E'$ be the curve defined by this equation
  (or rather its projective closure). There is a
  morphism
  \begin{align*}
    \phi : E
    &\to E' \subset \Pbb^2 \\
    P
    &\mapsto (x(P) : y(P) : 1) = \left(
    \frac{x}{y}(P) : 1 : \frac{1}{y}(P) \right) \\
    0_E
    &\mapsto (0 : 1 : 0)
  \end{align*}
  Then
  \begin{align*}
    [K(E) : K(x)]
    &= \deg(E \stackrel{x}{\to} \Pbb^1) \\
    &= \ord_{0_E} \left( \frac{1}{x} \right) \\
    &= 2
    [K(E) : K(y)]
    &= \deg(E \stackrel{y}{\to} \Pbb^1) \\
    &= \ord_{0_E} \left( \frac{1}{y} \right) \\
    &= 3
  \end{align*}
  This gives us a diagram of field extensions:
  \begin{picmath}
    \begin{tikzcd}
      & K(E) \ar[ldd, no head] \ar[d, no head]
      \ar[rdd, no head] \\
      & K(x, y) \ar[ld, no head] \ar[rd, no head] \\
      K(x)
      &
      & K(y)
    \end{tikzcd}
  \end{picmath}
  By the Tower Law (since $2, 3$ are coprime), we
  get that $[K(E) : K(x, y)] = 1$. Hence $K(E) =
  K(x, y) = \phi^* K(E')$, so $\deg \phi = 1$, so
  $\phi$ is birational.