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\begin{flashcard}[div-class-rOmega-well-def-prop]
\begin{proposition*}
  Let $\omega, \omega' \in \rOmega_{\K(C) / \KK}$.
  Then $\rdiv(\omega)$ and $\rdiv(\omega')$ are
  \gls{linequiv}.
\end{proposition*}

\begin{proof}
  \cloze{For $t$ a local parameter at some point $p \in
  C$, $\omega = f \dd t$, $\omega' = f' \dd t$,
  then $\omega = \frac{f'}{f} \cdot \omega'$. Then
  \[ \rdiv(\omega) = \rdiv(\omega') +
  \left( \frac{f'}{f} \right) . \qedhere \]}
\end{proof}
\end{flashcard}

\begin{flashcard}[canonical-class-defn]
\begin{definition*}[Canonical class]
  \glsnoundefn{cancls}{canonical class}{canonical
  classes}
  \glssymboldefn{KC}{$K_C$}{$K_C$}
  \cloze{The \emph{canonical class} of a projective
  non-\gls{sing} \gls{curve} $C$ is the linear
  equivalence class of $\rdiv \omega$ in $\Cl C$,
  for any $0 \neq \omega \in \rOmega_{\K(C) /
  \KK}$. We write the canonical class as $K_C$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[genus-defn]
\begin{definition*}[Genus]
  \glsadjdefn{genus}{genus}{\gls{curve}}
  \cloze{The \emph{genus} of $C$ is $\dim_{\KK}
  \LD(\KC)$.

  If $\KK = \CC$ and we use the Euclidean topology
  rather than the Zariski topology, then this is
  the usual notion of genus!}
\end{definition*}
\end{flashcard}

\begin{example*}
  $C = \P^1$, $\K(C) = \KK(t)$, $t = x_0 / x_1$.
  Note when $x_1 = 1$, $t = p_0$ is a local
  parameter for $C$ at $p_0 = (p_0 : 1) \in \P^1$.
  Thus $\dd t = \dd (t - p_0)$ and $\nu_{p_0} (\dd
  (t - p_0)) = 0$. Thus $\nu_{p_0}(\dd t) = 0$ for
  all $p_0 \in \P^1 \setminus \Z(x_i)$. At $t =
  \infty$, look at $\AA^1 = \P^1 \setminus
  \Z(x_0)$, so $s = x_1 / x_0$ is a local
  parameter at $q = (1 : 0)$. Note $t = s^{-1}$,
  so
  \[ \dd t = \dd (1 / s) = -\frac{\dd s}{s^2} \]
  so $\nu_q(\dd t) = -2$. So $\KC \sim -2q$ where
  $\sim$ means \gls{linequiv}. Thus $\LD(\KC) =
  \LD(-2q) = 0$. Thus
  \[ g(C) = \dim \LD(\KC) = 0 .\]
\end{example*}

\begin{example*}
  Plane cubic
  \[ y^2 = (x - \lambda_1)(x - \lambda_2)(x -
  \lambda_3) \]
  in $\AA^2$ or
  \[ y^2z = (x - \lambda_1 z)(x - \lambda_2 z)(x -
  \lambda_3 z) \]
  $\lambda_1, \lambda_2, \lambda_3 \in \KK$
  distinct. $\omega = \frac{\dd x}{y}$, $2y \dd y
  = f'(x) \dd x$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/fc1c52e4078a4166.png}
  \end{center}
  so
  \[ \frac{2 \dd y}{f'(x)} = \frac{\dd x}{y} .\]
  In fact, $\rdiv(\omega) = 0$. Hardest part: $q
  = (0 : 1 : 0)$. Thus $\KC \sim 0$, and $\LD(\KC)
  = \LD(0)$, so $g(C) = \dim \LD(0) = 1$.
\end{example*}

\begin{flashcard}[riemann-roch-thm]
\begin{theorem*}[Riemann-Roch Theorem]
  \label{rroch}
  \cloze{Write $l(D) \defeq \dim_{\KK} \LD(D)$ for $D \in
  \Div(C)$. Then
  \[ l(D) - l(\KC - D) = \deg D + 1 - g \]
  where $g$ is the genus of $C$.}
\end{theorem*}
\end{flashcard}

\begin{proof}
  Omitted. This is far beyond the scope of this course;
  this theorem is not even proved in part III.
\end{proof}

Consequences:
\begin{enumerate}[(1)]
  \item If $D = 0$ then $l(D) = 1$, so $1 - l(\KC)
    = 0 + 1 - g$ or $l(\KC) = g$, the definition
    of $g$.
  \item If $D = \KC$, then
    \[ \ub{l(\KC) - l(0)}_{= g - 1} = \deg \KC + 1 - g \]
    so $\boxed{\deg \KC = 2g - 2}$.
  \item If $\deg D > 2g - 2$, then $\deg \KC - D =
    2g - 2 - \deg D , 0$. Thus $l(\KC - D) = 0$
    and
    \[ \boxed{l(D) = \deg D + 1 - g} .\]
    \begin{remark*}
      \vspace{1em}
      For $0 \le \deg D \le 2g - 2$, behaviour of
      $l(D)$ can be complicated and unpredictable.
    \end{remark*}
  \item If $\deg D > 2g$, then $\forall P, Q \in
    C$,
    \[ l(D - P - Q) = l(D) - 2 \]
    by (3). Hence $\dbar|D|$ induces an embedding
    of $C$ in some $\P^n$.
    \begin{example*}
      \vspace{1em}
      If $C$ has genus $0$, then every positive degree
      \gls{div} induces an embedding.

      For example, if $P \in C$, $\dbar|P|$ is
      \gls{va}, $l(P) = 2$, so we get an
      embedding of $C$ in $\P^1$. Thus $C \cong
      \P^1$.
    \end{example*}

    \begin{example*}
      \vspace{1em}
      $g = 1$. If $\deg D = 3$, then $D$ is
      \gls{va}, and $l(D) = 3 + 1 - 1 = 3$. So
      $\dbar|D|$ induces an embedding of $C$ in
      $\P^2$. Thus in particular $C$ is isomorphic
      to a \gls{curve} of degree $3$ in $\P^2$.
      Can show $C \cong \Z(f)$ for some
      homogeneous polynomial of degree $3$. More
      specifically, fix $P_0 \in C$, and embed
      using $\dbar|3P_0|$. Let $D \in \Div C$ be
      degree $0$. Then
      \[  l(D + P_0) - l(\KC - D - P_0) = \deg(D +
      P_0) + 1 - g .\]
      The second term of RHS is $0$ since $\deg
      \KC - D - P_0 = -1$. Then since $\deg(D +
      P_0) = 1$ and $g = 1$, we get $l(D + P_0) =
      1$. So there exists an \gls{eff} \gls{div}
      \gls{linequiv} to $D + P_0$, necessarily $D
      + P_0 \sim P$ for some $P \in C$. Thus $P -
      P_0 \sim D$. Note $P$ is unique: if $P - P_0
      \sim P' - P_0$, then $P \sim P'$, so if $P
      \neq P'$, $\dim\dbar|P| \ge 1$, so $l(P) \ge
      2$. But $l(P) = 1$ by \nameref{rroch}.

      Conclusion: every divisor class on $C$ of
      degree $0$ can be represented uniquely by $P
      - P_0$ for some $P \in C$, i.e. $C
      \to \Ker(\deg : \Cl C \to \ZZ))$, $p \mapsto
      p - p_0$ is a bijection. This gives a group
      structure on $C$, i.e. $P + Q = R$ for $P,
      Q, R \in C$ if
      \[ (P - P_0) + (Q - P_0) \sim R - P_0 .\]
      Geometric description: $P, Q \in C
      \stackrel{i}{\hookrightarrow} \P^2$. Let $L$ be the line
      joining $P$ and $Q$ (tangent line to $C$ at
      $P$ if $P = Q$). Then
      \[ \text{``}L \cap C\text{''} = \fstar i L =
      P + Q + S .\]
      (possibly $S = P$ or $S = Q$). Now $P + Q +
      S \sim 3P_0$, or
      \[ (P - P_0) + (Q - P_0) + (S - P_0) \sim 0
      .\]
      Next let $L'$ be the line joining $S$ with
      $P_0$. Then
      \[ \text{``}L' \cap C\text{''} = \fstar i L'
      = S + P_0 + R \sim 3P_0 .\]
      So $(S - P_0) + (R - P_0) \sim 0$ or $(S -
      P_0) \sim -(R - P_0)$. Thus
      \[ (P - P_0) + (Q - P_0) \sim (R - P_0) \]
      so $P + Q = R$.
    \end{example*}
\end{enumerate}