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\begin{remark*}
  If $\KK$ is algebraically closed, then $\dim
  \KK[x_1, \ldots, x_n]$ agrees with the
  \gls{kdimt} of $\AA^n$. If $X \subseteq \AA^n$
  is an \gls{affvty}, then $\dim \cring A(X)$ equals the
  \gls{kdimt} of $X$ (check: there exists a $1-1$
  correspondence between prime ideals of $\cring
  A(X)$ and irreducible subsets of $X$).
\end{remark*}

\begin{flashcard}[dimension-equalities-thm]
\begin{theorem*}
  If $X$ is a \gls{vty}, then
  \[ \dimX X = \cloze{\Trdeg_{\KK} \K(X)} =
  \cloze{\text{\gls{kdimt}
  of $X$}} = \cloze{\text{\gls{kdimr} of $\OXp_{X,
  p}$}} \]
  for $p \in X$.
\end{theorem*}

\begin{proof}
  ``Dimension theory'' -- non-examinable proof.
\end{proof}
\end{flashcard}

\begin{example*}
  In \es{1}, we showed that if $X = \Z(f)
  \subseteq \AA^2$, then the clsoed subsets of $X$
  are $X$ and finite subsets of $X$. Thus the
  \gls{kdimt} of $X$ is $1$.
\end{example*}

\newpage
\section{Curves}

\begin{flashcard}[algebraic-curve-defn]
\begin{definition*}[Algebraic curve]
  \glsnoundefn{curve}{curve}{curves}
  \cloze{An \emph{(algebraic) curve} is a \gls{vty} $C$
  with $\dimX C = 1$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[curve-Dn-defn]
\begin{definition*}
  \glssymboldefn{Div}{Div$C$}{Div$C$}
  \glsnoundefn{div}{divisor}{divisors}
  Let $C \subseteq \P^n$ be a \cloze{projective
  non-\gls{sing} \gls{curve}}. We define $\Div C$
  to be the \cloze{free abelian group generated by the
  points of $C$. This is called the group of
  \emph{divisors} of $C$.

  An element of $\Div C$ is of the form $\sum_{i =
  1}^n a_i p_i$, $a_i \in \ZZ$, $p_i \in C$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  Consider $C = \P^1$. An element of $\K(C)$ is
  the raito $\frac{f(x_0, x_1)}{g(x_0, x_1)}$
  where $f, g$ are \gls{homog} polynomials of the
  same degree. we can write
  \[ \frac{f}{g} = \frac{\prod_i (b_i x_0 - a_i
  x_1)^{m_i} }{\prod_j (d_j x_0 - c_j
  x_1)^{n_j}} \]
  $\sum m_i = d = \sum n_j$. Let $P_i = (a_i :
  b_i)$, $Q_j = (c_j : d_j)$. $\frac{f}{g}$ has a
  zero of order $m_i$ at $P_i$ and a pole of order
  $n_j$ at $Q_j$. The divisors of zeroes and poles
  of $\frac{f}{g}$ is
  \[ \left( \frac{f}{g} \right) = \sum_i m_i P_i -
  \sum_j n_j Q_j .\]
\end{example*}

\begin{flashcard}[principal-divisor-defn]
\begin{definition*}[Principal divisor]
  \glssymboldefn{Prin}{Prin$C$}{Prin$C$}
  \glssymboldefn{Cl}{Cl$C$}{Cl$C$}
  \glsadjdefn{prin}{principal}{divisor}
  \cloze{We call a \gls{div} $D \in \Div C$
  \emph{principal} if it is of the form $\left(
  \frac{f}{g} \right)$. Let $\Prin C \subseteq
  \Div C$ be the subgroup of principal divisors
  and define the \emph{class group} of $C$ to be
  \[ \Cl C = \frac{\Div C}{\Prin C} .\]}
\end{definition*}
\end{flashcard}

\begin{example*}
  We see $\Cl \P^1 = \ZZ$.
\end{example*}

\vspace{-1em}
\textbf{Goal:} Given any non-\gls{sing}
\gls{curve}, $f \in \K(X)$, want to define the
order of $0$ or pole at $p \in X$.

\begin{flashcard}[module-annihilation-lemma]
\begin{lemma*}
  Let $A$ be a ring, $M$ a finitely generated
  $A$-module and $I \subsetneq A$ an ideal such
  that $I \cdot M = M$. Then there exists $x \in
  A$ such that $x \equiv 1 \pmod{I}$ and $x \cdot
  M = 0$.
\end{lemma*}

\begin{proof}
  \cloze{Recall if we have $\phi : M \to M$ an $A$-module
  homomorphism with $\phi(M) \subseteq IM$, then
  there exists $a_1, \ldots, a_n \in I$ such that
  \[ \phi^n + a_1 \phi^{n - 1} + \cdots + a_n = 0
  .\]
  Take $\varphi$ to be the identity map. So this
  means multiplication by
  \[ 1 + a_1 + a_2 + \cdots + a_n \]
  is the zero homomorphism of $M$. Then taking
  this to be $x$, $x \equiv 1 \pmod{I}$ and $xM =
  0$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[Nakayamas-lemma]
\begin{theorem*}[Nakayama's lemma]
  \label{nakayama}
  \cloze{Let $A$ be a \gls{locring} with maximal ideal
  $m$. Let $I \subseteq m$ be an ideal.
  Let $M$ be a finitely generated $A$-module.
  Then $I \cdot M = M$ implies $M = 0$.}
\end{theorem*}

\begin{proof}
  \cloze{There exists $x \in A$ with $x \cdot M = 0$ and
  $x \equiv 1 \pmod{I}$, so $x \equiv 1 \pmod{m}$.
  So $x \notin m$. But this implies $x$ is
  invertible: otherwise, $\langle x \rangle \neq
  A$ and hence $\langle x \rangle \subseteq m$.
  Then $M = x^{-1} \cdot (xM) = 0$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[nakayama-lemma-coro]
\begin{corollary*}
  Let $A$ be a \gls{locring} with maximal ideal
  $m$, $M$ a finitely-generated $A$-module, $I
  \subseteq m$ an ideal. Then if \cloze{$M = IM + N$ for
  a submodule $N \subseteq M$, we have $M = N$.}
\end{corollary*}

\begin{proof}
  \cloze{Note $M / N$ satisfies
  \[ I \left( \frac{M}{N} \right) = \frac{IM +
  N}{N} .\]
  If $M = IM + N$, we get $I \left( \frac{M}{N}
  \right) = \frac{M}{N}$, so $\frac{M}{N} = 0$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[nakayama-lemma-coro-2]
\begin{corollary*}
  Let $A$ be a \gls{locring} with $m$ its maximal
  ideal. Let $x_1, \ldots, x_n \in M$ be a set of
  elements of a finitely generated module $M$ such
  that the images $\ol{x}_1, \ldots, \ol{x}_n \in
  M / mM$ form a basis for $M / mM$ as an $A /
  m$-vector space. Then $x_1, \ldots, x_n$
  generate $M$ as an $A$-module.
\end{corollary*}

\fcscrap{
\begin{remark*}
  $A / m$ is a field since $m$ is maximal.
  Further, $M / mM$ is a vector space over $A / m$
  via
  \[ (a + m) (\alpha + mM) = a \alpha + mM ,\]
  which is well-defined.
\end{remark*}
}

\begin{proof}
  \cloze{Let $N \subseteq M$ be the submodule of $M$
  generated by $x_1, \ldots, x_n$. Then the
  composition
  \[ N \hookrightarrow M \to M / mM \]
  is surjective. Thus $M = N + mM$. So by the
  previous Corollary, $M = N$.}
\end{proof}
\end{flashcard}
