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\begin{flashcard}[fstar-H-formula-thm]
\begin{theorem*}
  $\fstar f H = \cloze{D + \divzp (\sum_i a_i
  f_i)}$.
\end{theorem*}
  
\begin{proof}
  \cloze{Let $p \in f^{-1}(H)$. Suppose the coefficient
  of $p$ in $D$ is $0$. Let $\varphi = \sum_i a_i
  x_i$. Let $b_0, \ldots, b_n$ be such that $p
  \notin \Z(\sum_i b_i x_i)$. Let $\psi = \sum_i
  b_i x_i$. Then the coefficient of $p$ in $\fstar
  f H$ is
  \[ \nuf_p \left( \frac{\varphi}{\psi} \circ f \right) \]
  Necessarily, $f_0, \ldots, f_n$ do not have a
  pole at $p$, since otherwise $D + \divzp(f_i)$
  has a negative coefficient for $p$. Thus, $f_0,
  \ldots, f_n$ are \gls{gregf} on a neighbourhood
  of $p$, so we can write $f = (f_0 : \ldots : f_n)$
  in this neighbourhood. Now
  \[ \nuf_p \left( \frac{\varphi}{\psi} \circ f
  \right) = \nuf_p \left( \frac{\sum_i a_i
  f_i}{\sum_i b_i f_i} \right) = \nuf_p \left(
  \sum_i a_i f_i \right) \]
  since $\sum_i b_i f_i$ is non-vanishing and
  \gls{gregf} at $p$. But $\nuf_p \left( \sum_i
  a_i f_i \right)$ is the coefficient of $p$ in $D
  + \left( \sum_i a_i f_i \right)$. If $p$
  appears in $D$ with coefficient $m$< then
  \[ \nuf_p \left( \sum_i b_i f_i \right) \ge -m \]
  for any $b_0, \ldots, b_n \in \KK$. There is
  also some choice of $b_0, \ldots, b_n$ with
  $\nuf_p \left( \sum_i b_i f_i \right) = -m$ by
  assumption (2). In a neighbourhood of $p$, the
  \gls{gmorph} $f$ is given by
  \[ f = (t^m f_0 : \cdots : t^m f_n) \]
  where $t$ is a local parameter at $p$. The
  coefficient of $p$ in $\fstar f H$ is
  \[ \nuf_p \left( \frac{\sum_i a_i t^m
  f_i}{\ub{\sum_i b_i t^m f_i}_{\nuf_p = 0}} \right)
  = \nuf_p \left( \sum_i a_i t^m f_i \right) = m +
  \nuf_p \left( \sum_i a_i f_i \right) ,\]
  which is the coefficient of $p$ in $D + \divzp (\sum_i
  a_i f_i)$. Thus $\fstar f H = D + \divzp(\sum_i
  a_i f_i)$.}
\end{proof}
\end{flashcard}

\vspace{-1em}
Picture so far: $f_0, \ldots, f_n$ span a subspace
$V \subseteq \LD(D)$. This gives a linear subspace
\[ \mathcal{D} = \frac{V \setminus \{0\}}{\KK^1} =
\P(V) \subseteq \dbar|D| = \P(\LD(D)) .\]
We call $\mathcal{D}$ the \emph{linear system}.

\begin{flashcard}[div-support-defn]
\begin{definition*}[Support of a divisor]
  \glsnoundefn{supp}{support}{N/A}
  \glssymboldefn{supp}{Supp$(D)$}{Supp$(D)$}
  \cloze{For a \gls{div} $D = \sum_{i = 1}^n a_i p_i$ with
  $a_i \neq 0$, we define the \emph{support} of $D$
  to be $\Supp(D) = \{p_1, \ldots, p_n\}$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[bp-free-defn]
\begin{definition*}[Base-point free]
  \glsadjdefn{bpfree}{base-point free}{P(V)}
  \cloze{We say $\mathcal{D} = \P(V)$ is \emph{base-point
  free} if $\forall p \in C$, $\exists D' \in
  \mathcal{D}$ (where we identify $[f] \in D$ with
  $D + \divzp(f)$) with $p \notin \Supp D'$.}
\end{definition*}
\end{flashcard}

\vspace{-1em}
(This is assumption (2): $\forall p \in C$, there
exists $b_0, \ldots, b_n$ such that $p \notin
\Supp(D + \divzp(\sum_i b_i f_i))$).

In this case, the theorem applies, and we obtain
$f : C \to \P^n$ with the property that
\[ \mathcal{D} = \{\fstar f H \st H \subseteq \P^n
\text{ hyperplane}\} .\]

\textbf{Converse:} Suppose $f : L \to \P^n$ is a
\gls{gmorph}. Set $D = \fstar f \Z(x_0)$. (Assume
$f(C) \subseteq \Z(x_0)$). Let $f_i \in \K(C)$ be
given by
\[ r_i = \frac{x_1}{x_0} \circ f ,\]
a rational function on $C$ which is \gls{gregf} on
$C \setminus f^{-1}(\Z(x_0))$. Then $f = (f_0 :
f_1 : \cdots : f_n)$ on $C \setminus
f^{-1}(\Z(x_0))$ and hence $f$ is induced by the
linear system $\mathcal{D} \subseteq \dbar|D|$,
$\mathcal{D} = \P(V)$ with $V$ spanned by $f_0,
\ldots, f_n \in \LD(D)$.

By the previous theorem, $\fstar f \Z(\sum_i a_i
x_i) = D + \divzp(\sum_i a_i f_i) \in
\mathcal{D}$. Note $\mathcal{D}$ is \gls{bpfree},
since given $p \in C$, can find a hyperplane $H
\subseteq \P^n$ with $f(p) \notin H$, so $p \notin
\Supp \fstar f H$, while $\fstar f H \in
\mathcal{D}$.

\begin{remark*}
  If $f : C \hookrightarrow \P^n$ is an embedding,
  then $\fstar f H$ can be viewed as ``$H \cap C$
  iwth multiplicaion'', and
  \[ D = \{H \cap C \st H \subseteq \P^n \text{
  hyperplane}\} .\]
\end{remark*}

\begin{remark*}
  Can also pull-back hypersurfaces $H \subseteq
  \P^n$, with $H = \Z(\varphi)$, $\varphi$ a
  homogeneous polynomial of degree $d$, as
  follows. For $p \in f^{-1}(H)$, choose a
  homogeneous polynomial $\psi$ which doesn't
  vanish at $f(p)$ and take the coefficient of $p$
  in $\fstar f H$ to be
  \[ \nuf_p \left( \frac{\varphi}{\psi} \circ f
  \right) .\]
\end{remark*}

\begin{flashcard}[degree-of-f-morphism-defn]
\begin{definition*}[Degree of a curve morphism]
  \glspropdefn{degcm}{degree}{\gls{gmorph}}
  \cloze{Let $f : C \to \P^n$ be a \gls{gmorph}, $L
  \subseteq \P^n$ a hyperplane, $f(C)
  \not\subseteq L$. The \emph{degree of $f$} is
  the degree of the \gls{div} $\fstar f L$. This
  is well-defined since $\fstar f L$, $\fstar f
  L'$ are \gls{linequiv} and
  \gls{linequiv} \glspl{div} have the same
  defgree.}
\end{definition*}
\end{flashcard}

\begin{example*}
  Let $f : C \hookrightarrow \P^2$ identify $C$
  with $\Z(\varphi)$ where $\varphi$ has degree
  $d$. In this case, the \gls{degcm} of $f$ is
  $d$. (Check this: need to compare coefficients
  in $\fstar f L$ with the multiplicativity of
  zeroes of $\varphi|_L$).
\end{example*}

\begin{flashcard}[deg-fstar-H-formula-thm]
\begin{theorem*}
  Let $f : C \to \P^n$ be a \gls{gmorph}. $H
  \subseteq \P^n$ a hypersurface with $f(C)
  \not\subseteq H$. $H = \Z(\varphi)$. $\deg
  \varphi = e$. Then $\deg \fstar f H = (\deg f)
  \cdot e$.
\end{theorem*}
  
\begin{proof}
  \cloze{Choose some $x_i$ such that $f(C) \not\subseteq
  \Z(x_i)$. Then $\frac{\varphi}{x_i^e}$ is a
  rational function in $\P^n$ and
  $\frac{\varphi}{x_i^e} \circ f$ is a rational
  function on $C$. Assume $H \cap L \cap f(C) =
  \emptyset$. Then
  \begin{align*}
    \left( \frac{\varphi}{x_i^e} \circ f \right)
    &= \sum_{p \in f^{-1}(H)} \nuf_p \left(
    \frac{\varphi}{x_i^e} \circ f \right) p -
    \sum_{p \in f^{-1}(L)} \left(
    \frac{x_i^e}{\varphi} \circ f \right) \\
    &= \fstar f H - e \fstar f L
  \end{align*}
  Since the degree of a principal \gls{div} is
  $0$, we get $\deg \fstar f H = e \cdot \deg
  \fstar f L$.}
  % Thus $\fstar f H$ is \gls{linequiv} to $e \fstar
  % f L$, so $\deg \fstar f H = e \cdot \deg \fstar
  % f L = e \cdot \deg f$.}
\end{proof}
\end{flashcard}