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\begin{example*}
  $\rOmega_{\KK[x] / \KK}$ For $f \in \KK[x]$, $
  \dd f
  = f'(x) \dd x$. Thus $\rOmega_{\KK[x] / \KK}$ is
  the free $\KK[x]$-module with one generator $\dd
  x$.

  Similarly $\rOmega_{\KK(x) / \KK}$, $f \in
  \KK(x)$, $\dd f = f'(x) \dd x$. Thus
  $\rOmega_{\KK(x) / \KK}$ is the 1-dimensional
  vector space over $\KK(x)$ with basis $\dd x$.
\end{example*}

\begin{flashcard}[Omega-L-K-separable-zero-prop]
\begin{proposition*}
  If $L / K$ is a separable algebraic field
  extension, then $\rOmega_{L / K} = 0$.
\end{proposition*}

\fcscrap{
\vspace{-1em}
A field extension $L / K$ is separable algebraic
if everything in $L$ is a solution to some
irreducible polynomial equation $f(x) = 0$ with
$f(\alpha) \in K[X]$, and $f'(\alpha) \neq 0$,
i.e. $\alpha$ is not a multiple root.
}

\begin{proof}
  \cloze{Given $\alpha \in L$, $f(x) \in K[x]$ with
  $f(\alpha) = 0$, $f'(\alpha) \neq 0$, then $0 =
  f(\alpha)$ implies $0 = \dd(f(\alpha)) =
  f'(\alpha) \dd \alpha$, so $\dd \alpha = 0$
  since $f'(\alpha) \neq 0$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[Omega-KC-K-dt-lemma]
\begin{lemma*}
  Let $C$ be a \gls{curve}, $p \in C$, and $t$ a
  local parameter for $C$ at $p$. Then
  \[ \rOmega_{\K(C) / \KK} = \K(C) \dd t .\]
\end{lemma*}

\begin{proof}
  \cloze{$t$ local parameter implies $t$ is not a
  constant function, and hence defines a
  non-constant map $t : C \to \P^1$, inducing a
  finite field extension $\K(\P^1) = \KK(t) \to
  \K(C)$. This extension is separable (proof
  omitted, not required if $\characteristic \KK = 0$. The
  idea is that if the extension is not separable,
  then $\characteristic \KK \mid e_Q$ for all $Q
  \in C$. However, since $t$ is a local parameter
  at $p$, $e_p = 1$). If $\alpha \in \K(C)$, then
  there exists $f \in \KK(t)[x]$ such that
  $f(\alpha) = 0$, $f'(\alpha) = 0$. Write
  \[ f(x) = \sum_{i \ge 0} f_i(t) x^i \]
  for some $f_(t) \in \KK(t)$. Then
  \begin{align*}
    0 = \dd (f(\alpha))
    &= \dd \left( \sum_{i \ge 0} f_i(t) \alpha^i
    \right) \\
    &= \left( \sum_{i \ge 0} f_i'(t) \alpha^i
    \right) \dd f
    + \ub{\left( \sum_{i \ge } i f_i(t) \alpha^{i - 1}
    \right)}_{=f'(\alpha) \neq 0} \dd \alpha
  \end{align*}
  Thus we can solve for $\dd \alpha$, getting $\dd
  \alpha = g \dd t  \in \K(C) \dd t$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[order-of-pole-zero-defn]
\begin{definition*}[$\nu_p(\omega)$]
  \glssymboldefn{rdiv}{div$(\omega)$}{div$(\omega)$}
  \cloze{Let $C$ be a projective non-\gls{sing}
  \gls{curve}, $\omega \in \rOmega_{\K(C) / \KK}$,
  $p \in C$. We define $\nu_p(\omega)$ as follows.
  Let $t \in \OXp_{C, p}$ a local parameter and
  write $w = f \dd t$ for $f \in \K(C)$. Define
  \[ \nuf_p(\omega) = \nu_p(f) .\]
  We define $\rdiv(\omega) = \sum_{p \in C}
  \nu_p(\omega) p \in \Div C$. We say $\omega$ is
  regular at $p$ if $\nu_p(\omega) \ge 0$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[order-pole-well-def-lemma]
\begin{lemma*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item $f \in \OXp_{C, p} \implies \nu_p(\dd f)
      \ge 0$.
    \item If $t'$ is another local parameter at
      $p$, then $\nu_p(\dd t') = 0$ and $\nu_p(f
      \dd t') = \nu_p(f) + \nu_p(\dd t')$ is
      independent of $t$.
    \item If $f \in \K(C)$ and $\nu_p(f) \neq 0$
      in $\KK$ (i.e., $\characteristic \KK \mid
      \nuf_p(f)$) then $\nu_p(\dd f) = \nuf_p(f) -
      1$.
  \end{enumerate}
\end{lemma*}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(1)]
    \item \cloze{Let $p \in C \subseteq \P^n$, $p \in C
      \cap U_i$, where $U_i = \P^n \setminus
      \Z(x_i)$. Work on $U_i \cap C$, where
      rational functions are just ratios of
      polynomials. If $f = g / h$, $h(p) \neq 0$,
      we have
      \[ \dd f = \frac{h \dd g - g \dd h}{h^2} =
      \sum_i \gamma_i \dd x_i \]
      with $\gamma_i \in \OXp_{C, p}$. So
      \[ \nu_p(\dd f) \ge
      \min\{\nu_p(\gamma_i \dd x_i) \mit 1 \le i
      \le n\} \ge \min\{\nu_p(\dd x_i) \mid 1 \le i
      \le n\} .\]
      Thus $\nu_p(\dd f)$ is bounded below
      independently of $f$. Choose $f \in \OXp_{C,
      p}$ such that $\nu_p(\dd f)$ is minimal, $t$
      a local parameter at $p \in C$. Then
      $\nu_p(f - f(p)) \ge 1$, so can write $f -
      f(p) = t f_1$, for some $f_1 \in \OXp_{C, p}$.
      So
      \begin{align*}
        \dd f
        &= \dd (f - f(p)) \\
        &= \dd (f t_1) \\
        &= f_1 \dd t + t \dd f_1 \tag{$*$}
        \label{lec22_l138_eq}
      \end{align*}
      If $\nu_p(\dd f) < 0$, note $\nu_p(f_1 \dd
      t) \ge 0$, and hence \eqref{lec22_l138_eq}
      implies
      \[ \nu_p(\dd f) = \nu_p(t \dd f_i) =
      \nu_p(t) + \nu_p(\dd f_1) = 1 + \nu_p(\dd
      f_1) .\]
      So $\nu_p(\dd f_1) < \nu_p(\dd f)$. This
      contradicts the minimality of $\nu_p(\dd
      f)$. Thus $\nu_p(\dd f) \ge 0$.}
    \item \cloze{We may write $t' = u \cdot t$ for $u$ a
      unit, $u \in \OXp_{C, p}^\times$ (the group
      of units). Then $\dd t' = u \dd t + t \dd
      u$. $\dd u = g \cdot \dd t$ for some $g$
      with $\nuf_p(g) \ge 0$ by (1). So
      \[ \dd t' = \ub{(u + tg)}_{\nu_p = 0} \dd t
      ,\]
      so $\nu_p(\dd t') = 0$ by definition. If $f
      \dd t = h \dd t' = h(u + tg) \dd t$, then
      \[ \nuf_p(h(u + tg)) = \nuf_p(h) + \nuf_p(u +
      tg) = \nuf_p(h) .\]
      Hence $\nu_p$ is independent of choice of
      $t$.}
    \item \cloze{Suppose $f = t^n u$ where $n =
      \nu_p(f)$, $u \in \OXp_{C, p}^\times$. Then
      $\dd f = nt^{n - 1} u \dd t + t^n \dd u$. If
      $\characteristic \KK \nmid n$, then
      \[ \nuf_p(f) \ge \min\{\nuf_p(nt^{n - 1} u
      \dd t), t^n \dd u\} = \min\{n - 1, n\} = n -
      1 \]
      and equality holds since $n \neq n - 1$.
      Thus $\nu_p(\dd f) = \nuf_p(f) - 1$.}
      \qedhere
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{proposition*}
  If $\omega \in \rOmega_{\K(C) / \KK}$, then
  $\nu_p(\omega) = 0$ for all but a finite number
  of $p$.
\end{proposition*}

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

Thus $\rdiv(\omega) \in \Div(C)$.