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\begin{flashcard}[nonsing-proj-curve-mp-prin-ideal]
\begin{corollary*}
  Let $C \subseteq \P^n$ be a non-\gls{sing}
  projective \gls{curve}. Then $m_p \subseteq
  \OXp_{C, p}$ \cloze{is a principal ideal.}
\end{corollary*}

\begin{proof}
  \cloze{We begin by proving $\OXp_{C, p}$ is Noetherian.
  Replace $C$ by an open affine neighbourhood of
  $p \in C$, $C'$. This does not change $\OXp_{C,
  p}$, i.e. $\OXp_{C, p} = \OXp_{C', p}$. Then
  \[ \OXp_{C', p} = \left\{ \frac{f}{g} ~\bigg|~
  f, g \in \cring A(C') = \frac{\KK[x_1, \ldots,
  x_n]}{\I(C')} \right\} \subseteq \K(C') .\]
  If $J \subseteq \OXp_{C', p}$, then
  \[ J = \left\{ \frac{f}{g} ~\bigg|~ f \in \cring
  A(C') \cap J, g \in \cring A(C'), g(p) \neq 0)
  \right\} \subseteq \OXp_{C', p} .\]
  Prove $\subseteq$: if $f / g \in J$, then $g
  \cdot \left( \frac{f}{g} \right) = f \in J$, so
  $f \in \cring A(C') \cap J$. Prove $\supseteq$:
  if $f \in \cring A(C') \cap J$, then
  $\frac{f}{g} = \frac{1}{g} \cdot f \in J$ (if
  $g(p) \neq = 0$).
  
  Now $\KK[x_1, \ldots, x_n]$ is Noetherian by
  Hilbert's basis theorem. Hence $\cring A(C') =
  \KK[x_1, \ldots, x_n] / \I(C')$ is Noetherian.
  Hence $\cring A(C') \cap J$ is finitely
  generated, and by the equation for $J$, the set
  of generators of $\cring A(C')$ generate $J$ as
  an ideal in $\OXp_{C', p}$. Since $C$ is
  non-\gls{sing} of \gls{avdim} $1$,
  \[ 1 = \dim \tspace_p C = \dim(m_p / m_p^2)^* .\]
  Also, $\OXp_{C, p} / m_p \stackrel{\cong}{\to}
  \KK$, $f + m_p \mapsto f(p)$. Thus $m_p / m_p^2$
  is a 1-dimensional vector space over $\OXp_{C,
  p} / m_p$, hence by the previous Corollorary to
  \nameref{nakayama}, $m_p$ is generated by the lift
  of a 1 element basis of $m_p / m_p^2$. Thus
  $m_p$ is principal (we need $m_p$ finitely
  generated here!).}
\end{proof}
\end{flashcard}

\begin{flashcard}[nu-defn]
\begin{remark*}
  \glssymboldefn{nu}{$\nu$}{$\nu$}
  \cloze{Let $t \in m_p$ be a generator. We get a chain
  of ideals
  \[ \cdots \subseteq (t^3) \subseteq (t^2)
  \subseteq (t) = m_p \subset \OXp_{C, p} .\]
  Notice if $(t^{k + 1}) = (t^k)$, then $m_p \cdot
  (t^k) = (t^k)$. But then \nameref{nakayama}
  tells us that $(t^k) = 0$. But $t^k \neq 0$
  since $\OXp_{C, p}$ is an integral domain.

  Also, consider $I = \bigcap_{k = 1}^\infty
  (t^k)$. Clearly $t \cdot I = I$, so $m_p \cdot I
  = I$, so $I = 0$.

  \textbf{Consequence:} If $f \in \OXp_{C, p}
  \setminus \{0\}$, then there exists a unique
  $\nu \ge 0$ such that $f \in (t^\nu)$ but $f \notin
  (t^{\nu + 1})$.} Define $\nu : \cloze{\OXp_{C, p}
  \setminus \{0\} \to \ZZ}$ \cloze{by $\nu(f) = \nu$ as
  above.}
\end{remark*}
\end{flashcard}

\begin{remark*}
  \phantom{}
  \begin{itemize}
    \item $\nuf(f \cdot g) = \nuf(f) + \nuf(g)$.
    \item $\nuf(f + g) \ge \in\{\nuf(f),
      \nuf(g)\}$ with equality if $\nuf(f) \neq
      \nuf(g)$.
  \end{itemize}
  Can extend $\nuf$ to a map
  \[ \nuf : \K(C) \setminus \{0\} \eqdef
  \K(C)^\times \to \ZZ \]
  by
  \[ \nuf \left( \frac{f}{g} \right) = \nuf(f) -
  \nuf(g) .\]
  $\nuf$ is an example of a \emph{discrete
  valuation}.
\end{remark*}

\begin{flashcard}[discrete-valuation-defn]
\begin{definition*}[Discrete valuation]
  \glsnoundefn{discv}{discrete valuation}{discrete
  valuations}
  \cloze{Let $K$ be a field. A \emph{discrete valuation}
  on $K$ is a function $\nu : K^\times \to \ZZ$
  such that
  \begin{enumerate}[(1)]
    \item $\nu(f \cdot g) = \nu(f) + \nu(g)$.
    \item $\nu(f + g) \ge \min\{\nu(f), \nu(g)\}$
      with equality if $\nu(f) \neq \nu(g)$.
  \end{enumerate}
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[DVR-defn]
\begin{definition*}[Discrete valuation ring]
  \glsnoundefn{dvr}{discrete valuation
  ring}{discrete valuation rings}
  \cloze{Given a \gls{discv}, we define the corresponding
  \emph{discrete valuation ring} (DVR) by
  \[ R = \{f \in K^\times \mid \nu(f) \ge 0\} \cup
  \{0\} \]
  which is a subring of $K$. We also define
  \[ m = \{f \in K^\times \st \nu(f) \ge 1\} \cup
  \{0\} .\]
  Note $m$ is the unique maximal ideal of $R$:
  if $f \in R \setminus m$, then $\nu(f) = 0$, so
  $\nu(f^{-1}) = 0$, so $f^{-1} \in R$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item $R = \OXp_{C, p} \subseteq K = \K(C)$.
      $\nuf$ the \gls{discv} we defined.
    \item Let $p \in \ZZ$ be prime, $K = \QQ$. Any
      rational number can be written as
      $\frac{a}{b} p^\nu$ with $(a, p) = 1$, $(b,
      p) = 1$. Then define
      \[ \nu_p \left( \frac{a}{b} p^\nu \right) =
      \nu .\]
      This is a \gls{discv}, with \gls{dvr}
      \[ \ZZ_{(p)} = \left\{ \frac{a}{b} \in \QQ
      \st p \nmid b \right\} .\]
    \item $K = \KK(x)$, $a \in \KK$. Define
      \[ \nu\left((x - a)^\nu \frac{f}{g}\right) = \nu \]
      where $f, g$ are relatively prime to $x -
      a$. Here the \gls{dvr} is $\OXp_{\AA^1, a}$.
    \item Let $K = \KK(x)$,
      \[ \nu \left( \frac{f}{g} \right) = \deg g -
      \deg f .\]
      This is the ``order of $0$ at $\infty$''.
  \end{enumerate}
\end{example*}

\vspace{-1em}
\textbf{Setup:} $C \subseteq \P^n$ a projective
non-\gls{sing} \gls{curve}. Each point $p \in C$
gives a \gls{discv} $\nuf_p : \K(C)^\times \to
\ZZ$ with \gls{dvr} $\OXp_{C, p}$. For $f \in
\K(C)^\times$, we define the \gls{div} of zeroes and
poles of $f$ to be
\[ (f) \defeq \sum_{p \in C} \nuf_p(f) p \]

Next time: need to check that this is a finite
sum!