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Let $0 \neq \omega \in \Omega_C$. Let $P \in C$ be
a smooth point, and $t \in K(C)$ a uniformiser at
$P$.

Then $\omega = f \dd t$ for some $f \in K(C)^*$.
We define $\ord_p(\omega) = \ord_p(f)$
(independent of choice of $t$).

We assume $C$ is a smooth projective curve.

\begin{definition*}[]
  $\div(\omega) = \sum_{P \in C} \ord_P(\omega) P
  \in \Div(C)$.
\end{definition*}

\begin{note*}
  This is a divisor, i.e. $\ord_p(\omega) = 0$ for
  all but finitely many $P \in C$.
\end{note*}

\begin{fcdefnstar}[Regular differential]
  A differential $\omega \in \Omega_C$ is
  \emph{regular} if $\div(\omega) \ge 0$, i.e. it
  has no poles.
\end{fcdefnstar}

\[
  g(C) = \dim_K \{\omega \in \Omega_C :
  \div(\omega) \ge 0\}
.\]
As a consequence of Riemann-Roch we have:
\begin{quote}
  If $0 \neg \omega \in \Omega_C$, then
  $\deg(\div \omega) = 2g - 2$.
\end{quote}

\textbf{Fact:} Suppose $f \in K(C)^*$, $\ord_p(f)
= n \neq 0$. If $\char K \nmid n$, then
$\ord_p(\dd f) = n - 1$.

\begin{fclemma}[]
  \label{lemma:6.1}
  Assuming:
  - $\char K \neq 2$
  - $E$: $y^2 = (x - e_1)(x - e_2)(x - e_3)$,
  $e_1, e_2, e_3 \in K$ distinct
  Then:
  $\omega = \frac{\dd x}{y}$ is a differential on
  $E$ with no zeroes or poles $\implies g(E) = 1$.

  In particular, the $K$-vector space of regular
  differentials on $E$ is 1-dimensional, spanned
  by $\omega$.
\end{fclemma}

\begin{proof}
  Let $T_i = (e_i, 0)$, $E\tors[2] = \{0, T_1, T_2,
  T_3\}$.
  \[
    \div(y) = (T_1) + (T_2) + (T_3) - 3(0)
    \tag{1}
    \label{lec8eq1}
  .\]
  For $P \in E \setminus \{0\}$,
  \[
    \div(x - x_p) = (P) + (-P) - 2(0)
  .\]
  If $P \in E \setminus E\tors[2]$, then $\ord_p(x -
  x_p) = 1 \implies \ord_p(\dd x) = 0$.

  If $P \in E \setminus E\tors[2]$ then $\ord_p(x -
  x_p) = 1$ $\implies \ord_p(\dd x) = 0$.

  If $P = T_i$ then $\ord_p(x - x_p) = 2$
  $\implies \ord_p(\dd x) = 1$.
  
  If $P = 0$ then $\ord_p(x) = -2$ $\implies
  \ord_p(\dd x) = -3$.

  Therefore
  \[
    \div(\dd x) = (T_1) + (T_2) + (T_3) - 3(0)
    \tag{2}
    \label{lec8eq2}
  .\]
  \eqref{lec8eq1} and \eqref{lec8eq2} implies
  $\div \left( \frac{\dd x}{y} \right) = 0$.
\end{proof}

\begin{definition*}[]
  For $\phi : C_1 \to C_2$ a non-constant morphism
  we define
  \begin{align*}
    \phi^* : \Omega_{C_2}
    &\to \Omega_{C_1} \\
    f \dd g
    &\mapsto (\phi^* f) \dd (\phi^* g)
  \end{align*}
\end{definition*}

\begin{fclemma}[]
  \label{lemma:6.2}
  Assuming:
  - $P \in E$,
  \begin{align*}
    \tau_P : E
    &\to E \\
    X
    &\mapsto P + X
  \end{align*}
  - $\omega = \frac{\dd x}{y}$ as above
  Then:
  $\tau_P^* \omega = \omega$ (we say $\omega$ is
  the \emph{invariant differential}).
\end{fclemma}

\begin{proof}
  $\tau_P^* \omega$ is a regular differential on
  $E$. So $\tau_P^* \omega = \lambda_p \omega$.

  The map $E \to \Pbb^1$, $P \mapsto \lambda_P$ is
  a morphism of smooth projective curves but
  \emph{not} surjective (misses $0$ and $\infty$).
  Therefore it is constant (by \cref{thm:2.8}),
  i.e. there exists $\lambda \in K^*$ such that
  $\tau_P^* \omega = \lambda \omega$ for all $P
  \in E$.

  Taking $P = 0_E$ shows $\lambda = 1$.
\end{proof}

\begin{remark*}
  If $K = \Cbb$, $\Cbb / \Lambda
  \stackrel{\sim}{\to} E(\Cbb)$, $z \mapsto
  (\wp(z), \wp'(z))$,
  \[
    \frac{\dd x}{y} = \frac{\wp'(z) \dd
    z}{\wp'(z)} = \dd z
  .\]
  (invariant under $z \mapsto z +
  (\text{const})$).
\end{remark*}

\begin{fclemma}[]
  \label{lemma:6.3}
  Assuming:
  - $\phi, \psi \in \Hom(E_1, E_2)$
  - $\omega$ an invariant differential on $E_2$
  Then:
  $(\phi + \psi)^* \omega = \phi^* \omega + \psi^*
  \omega$.
\end{fclemma}

\begin{proof}
  Write $E = E_2$
  \begin{align*}
    E \times E
    &\to E \\
    \mu : (P, Q)
    &\mapsto P + Q \\
    \mathrm{pr}_1 : (P, Q)
    &\mapsto P \\
    \mathrm{pr}_2 : (P, Q)
    &\mapsto Q
  \end{align*}
  \textbf{Fact:} $\Omega_{E \times E}$ is a
  2-dimensional $K(E \times E)$-vector space with
  basis $\mathrm{pr}_1^* \omega$ and
  $\mathrm{pr}_2^* \omega$.

  Therefore
  \[
    \mu^* \omega = f \mathrm{pr}_1^* \omega + g
    \mathrm{pr}_2^* \omega
    \tag{1}
    \label{lec8eq3}
  \]
  for some $f, g \in K(E \times E)$.

  For fixed $Q \in E$, let
  \begin{align*}
    \iota_Q : E
    &\to E \times E \\
    P
    &\mapsto (P, Q)
  \end{align*}
  Applying $\iota_Q^*$ to \eqref{lec8eq3} gives
  \begin{align*}
    (\mu \iota_Q)^* \omega
    &= (\iota_Q^* f)(\mathrm{pr}_1 \iota_Q)^*
    \omega
    + (\iota_Q^* g)(\mathrm{pr}_2 \iota_Q)^*
    \omega \\
    \ub{\tau_Q^* \omega}_{= \omega}
    &= \ub{(\iota_Q^* f)}_{\text{by
    \cref{lemma:6.2}}} \omega + 0
  \end{align*}
  Therefore $\iota_Q^* f = 1$ for all $Q \in E$,
  so $f(P, Q) = 1$ for all $P, Q \in E$.

  Similarly $g(P, Q) = 1$ for all $P, Q \in E$.
  \[
    \eqref{lec8eq3} \implies
    \mu^* \omega
    = \mathrm{pr}_1^* \omega + \mathrm{pr}_2^*
    \omega
  .\]
  Now pull back by
  \begin{align*}
    E_1
    &\to E \times E \\
    P
    &\mapsto (\phi(P), \psi(P))
  \end{align*}
  to get
  \[
    (\phi + \psi)^* \omega
    = \phi^* \omega + \psi^* \omega
    .
    \qedhere
  \]
\end{proof}

\begin{fclemma}[]
  \label{lemma:6.4}
  Assuming:
  - $\phi : C_1 \to C_2$ a non-constant morphism
  Then:
  \begin{iffc}
    \lhs
    $\phi$ is separable
    \rhs
    $\phi^* : \Omega_{C_1} \to \Omega_{C_2}$ is
    non-zero
  \end{iffc}
\end{fclemma}

\noproof

\begin{proof}
  Omitted.
\end{proof}

\begin{example*}
  $\mathbb{G}_m = \Abb^1 \setminus \{0\}$
  (multiplicative group).
  \begin{align*}
    \phi : \mathbb{G}_m
    &\to \mathbb{G}_m \\
    x
    &\mapsto x^n
  \end{align*}
  ($n \ge 2$ integer).

  $\phi^*(\dd x) = \dd (x^n) = nx^{n - 1} \dd x$.
  So if $\char K \nmid n$ then $\phi$ is
  separable.

  \cref{thm:2.8} implies $\#\phi^{-1}(Q) = \deg
  \phi$ for all but finitely many $Q \in
  \mathbb{G}_m$.

  But $\phi$ is a group homomorphism, so
  \[
    \#\phi^{-1}(Q) = \#\ker(Q)
  \]
  for all $Q \in \mathbb{G}_m$. Thus $\#\ker \phi
  = \deg \phi = n$ and hence $K (= \ol{K})$
  contains exactly $n$ $n$-th roots of unity.
\end{example*}

\begin{fcthm}[]
  \label{thm:6.5}
  Assuming:
  - $\char K \nmid n$
  Then:
  $E\tors[n] \cong (\Zbb / n\Zbb)^2$.
\end{fcthm}