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Let $C$ be a projective non-\gls{sing}
\gls{curve}, and $f \in \K(C)^\times$. $f : C \to
\P^1$, $p \mapsto (f(p) : 1)$, or writing $f =
\frac{g}{h}$, $g, h$ homogeneous polynomials of
the same degree, then $f : C \to \P^1$, $p \mapsto
(g(p) : h(p))$.

Then
\[ \divzp(f) = \sum_{p \in f^{-1}((0 : 1))} e_p p -
\sum_{q \in f^{-1}((1 : 0))} e_q q .\]
Thus, if we define
\[ \deg \sum_{p \in C} a_p p = \sum_{p \in C} a_p
,\]
then $\deg \divzp(f) = \deg f - \deg f = 0$. Thus
every \gls{prin} \gls{div} is degree $0$.

Thus the homomorphism $\deg : \Div C \to \ZZ$
descends to $\deg : \Cl C \to \ZZ$, and this is
surjective as $\deg p = 1$.

\subsubsection*{Linear systems}

\begin{flashcard}[effective-divisor-defn]
\begin{definition*}[Effective divisor]
  \glsadjdefn{eff}{effective}{\gls{div}}
  \glssymboldefn{effvs}{$L(D)$}{$L(D)$}
  Let $D \in \Div C$, $D = \sum_i n_i p_i$. We say
  $D$ is \emph{effective} if $n_i \ge 0$ for all
  $i$. Define
  \[ \LD(D) = \{f \in \K(C)^\times \st D +
  \divzp(f) \text{ is \gls{eff}}\} \cup \{0\} .\]
\end{definition*}
\end{flashcard}

\begin{flashcard}[LD-vec-space-lemma]
\begin{lemma*}
  $\LD(D)$ is a vector space.
\end{lemma*}

\begin{proof}
  \cloze{$f \in \LD(D)$ implies $cf \in \LD(D)$ for $c
  \in \KK$, $c \neq 0$, since $\divzp(f) = (cf) =
  (c) + (f)$. If $f, g \in \LD(D)$, $f, g$
  non-zero, $f + g \neq 0$, then
  \[ \divzp(f + g) = \sum_p \nuf_p(f + g) p \]
  and $\nuf_p(f + g) \ge \min\{\nuf_p(f),
  \nuf_p(g)\}$. Thus if $D + \divzp(f)$, $D +
  \divzp(g)$ are effective, then so is $D +
  \divzp(f + g)$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[LD-finite-dimensional-thm]
\begin{theorem*}
  $\LD(D)$ is a finite dimensional vector space
  and $\LD(0) = \KK$. Furthermore, $\dim_{\KK}
  \LD(D) \le \deg D + 1$.
\end{theorem*}

\begin{proof}
  \cloze{Induction on $\deg D$. If $\deg D < 0$, then
  there are no \gls{eff} \glspl{div}
  \gls{linequiv} to $D$, so $\LD(D) = 0$. Suppose
  $\deg D \ge 0$, write $D = \sum_i n_i p_i$ and
  pick $p \in C \setminus \{p_1, \ldots, p_n\}$.
  Consider the map
  \[ \lambda : \LD(D) \to \KK, \qquad f \mapsto
  f(p) .\]
  which makes sense since $\nuf_p(f) \ge 0$ for $f
  \in \LD(D)$, since otherwise the coefficient of
  $p$ in $D + \divzp(f)$ is negative. If $f \in
  \ker \lambda$, then $f \in m_p \subseteq
  \OXp_{C, p}$, so $\nuf_p(f) \ge 1$. Thus $f \in
  \LD(D - P)$. Note also $\LD(D - D) \subseteq
  \LD(D)$, since if $D - P + \divzp(f)$ is
  \gls{eff} then so is $D - \divzp(f)$. Thus
  $\LD(D - P) = \ker\lambda$, and
  $\frac{\LD(D)}{\LD(D - P)} \subseteq \KK$. Thus
  $\dim_{\KK} \LD(D) \le \dim \LD(D - P) + 1$.
  Thus by induction, $\dim_{\KK} \LD(D) \le \deg D
  + 1$.

  Thus $\dim\LD(0) \le 1$, but $\KK \subseteq
  \LD(D)$ since $0 + \divzp(c) = 0$. So $\dim
  \LD(0) = 1$.}
\end{proof}
\end{flashcard}

\begin{remark*}
  $\LD(0) = \{f : C \to \KK \text{
  \gls{gregf}}\}$, and hence \gls{gregf} functions
  on $C$ are constant.
\end{remark*}

\begin{flashcard}[complete-linear-system-defn]
\begin{definition*}[Complete linear system]
  \glsnoundefn{clsa}{complete linear system}{N/A}
  \glssymboldefn{Dbar}{$|D|$}{$|D|$}
  \cloze{Given a \gls{div} $D$, we define the
  \emph{complete linear system associated} to $D$
  to be
  \begin{align*}
    |D|
    &= \{D' \in \Div C \st \text{$D'$ \gls{eff},
    $D' \sim D$}\} \\
    &= \frac{\LD(D) \setminus \{0\}}{\sim}
    &&(f \sim \lambda f) \\
    &= \P(\LD(D))
  \end{align*}
  a projective space.}
\end{definition*}
\end{flashcard}

\subsubsection*{Morphisms to projective space}

Let $D$ be a \gls{div}, $f_0, \ldots, f_n \in
\LD(D)$ with not all $f_i$ being $0$. This gives a
\gls{gmorph} $f : C \to \P^n$, $p \mapsto (f_0(p)
: \cdots : f_n(p))$.

\begin{flashcard}[fstar-hyperplane-defn]
\begin{definition*}[$f^* H$]
  \glssymboldefn{fstar}{$f^* H$}{$f^* H$}
  \cloze{Let $f : C \to \P^n$ be a \gls{gmorph}. Let $H
  \subseteq \P^n$ be a hyperplane with $f(C)
  \not\subseteq H$. We define $f^* H \in \Div X$ as
  follows. Let $H = \Z(\varphi)$ with $\varphi$ a
  linear homogeneous polynomial and choose $\psi$
  linear homogeneous so that $H' = \Z(\psi)$
  satisfies $f^{-1}(H) \cap f^{-1}(H') =
  \emptyset$. Define
  \[ f^* H = \sum_{p \in f^{-1}(H)} \nuf_p \left(
  \frac{\varphi}{\psi} \circ f \right) p \]}
\end{definition*}
\end{flashcard}

\begin{center}
  \includegraphics[width=0.6\linewidth]{images/8a6da034ac4045ed.png}
\end{center}

\begin{remark*}
  This is independent of the choice of $\psi$. For
  example
  \[ \frac{\varphi}{\psi'} = \frac{\varphi}{\psi}
  \frac{\psi}{\psi'} .\]
\end{remark*}

\subsubsection*{Relations to morphisms}

Let $f_0, \ldots, f_n \in \LD(D)$ to have the
properties:
\begin{enumerate}[(1)]
  \item The $f_i$ aren't all $0$.
  \item $\forall p \in C$, $\exists a_0, \ldots,
    a_n \in \KK$ such that the coefficient of $p$
    in $D + \divzp(\sum_i a_i f_i)$ is $0$.
\end{enumerate}
As above, we get a \gls{gmorph} $f : C \to \P^n$.
Let $H \subseteq \P^n$ be given by an equation
$\sum_i a_i x_i = 0$.
