%! TEX root = AG.tex
% vim: tw=50
% 19/01/2024 12PM

% Prerequisites:
% - GRM:
%   - commutative rings,
%   - ideals
%   - prime ideals, maximal ideals
%   - fields
% - Basic topology. Need to know what a
%   topological space is and what continuous maps
%   are.

\subsubsection*{What is Algebraic Geometry?}

Study of solution sets to systems of polynomial
equations.

For example:
\begin{itemize}
  \item $\{(x, y) \in \RR^2 \st x^2 + y^2 = 1\}$
    \begin{center}
      \includegraphics[width=0.6\linewidth]{images/42e6552d5beb4585.png}
    \end{center}
  \item $\{(x, y) \in \RR^2 \st y^2 = x^3 - x\}$
    \begin{center}
    \begin{asy}
      import graph;
      size(5cm);
      real f(real x) {
        return sqrt(x^3 - x);
      }
      real g(real x) {
        return -sqrt(x^3 - x);
      }
      draw(graph(f,-1,0));
      draw(graph(f,1,3));
      draw(graph(g,-1,0));
      draw(graph(g,1,3));
    \end{asy}
    \end{center}
    % \begin{center}
    %   \includegraphics[width=0.6\linewidth]{images/854adfa01e704a81.png}
    % \end{center}
\end{itemize}

We could look for solutions over $\CC$:
\[ \{(x, y) \in \CC^2 \st y^2 = x^3 - x\} \]
`Looks like' a torus with one point removed.
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/5fe4b95a19b8406a.png}
\end{center}
$\RR^3$: $x^2 + y^2 + z^2 = 1$
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/b1110b9b7749439d.png}
\end{center}
$\{(x, y, z) \in \CC^3 \st x^3 + y^3 + z^3 = 1\} =
X$. $X$ contains $27$ lines:
\begin{align*}
  \text{9 lines}
  &: \begin{cases}
    x = -\xi^j y \\
    z = \xi^k
  \end{cases} \\
  \text{9 lines}
  &: \begin{cases}
    x = -\xi^j z \\
    y = \xi^k
  \end{cases} \\
  \text{9 lines}
  &: \begin{cases}
    y = -\xi^j z \\
    x = \xi^k
  \end{cases}
\end{align*}
(where $\xi = e^{2\pi i / 4}$).

\newpage
\section{Affine Varieties}

\subsection{Basic setup}

\glssymboldefn{K}{K}{K}%
Fix a field $\KK$.

\begin{flashcard}[affine-n-space-defn]
\begin{definition*}[Affine $n$-space]
  \glssymboldefn{A}{A}{A}
  \cloze{Affine $n$-space over $\KK$ is $\AA^n =
  \KK^n$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[zero-set-defn]
\begin{definition*}[Zero set]
  \glsnoundefn{zero_set}{zero set}{zero sets}%
  \glssymboldefn{Z}{$Z(S)$}{$Z(S)$}
  \glssymboldefn{polyA}{$A$}{$A$}
  \cloze{
  Let $\polyA \defeq \KK[X_1, \ldots, X_n]$, $S
  \subset \polyA$ a
  subset. Define
  \begin{align*}
    Z(S)
    &\defeq \text{``zero set of $S$''} \\
    &= \{(a_1, \ldots, a_n) \in \AA^n \st f(a_1,
    \ldots, a_n) = 0 ~\forall f \in S\}
  \end{align*}}
\end{definition*}
\end{flashcard}

\begin{flashcard}[zero-set-topology-properties]
\begin{proposition*}
  \glssymboldefn{cdotset}{$\cdot$}{$\cdot$}
  Basic properties of the zero set:
  \begin{enumerate}[(a)]
    \item \cloze{$\Z(\{0\}) = \AA^n$.}
    \item \cloze{$\Z(\polyA) = \emptyset$.}
    \item \cloze{$\Z(S_1 \cdot S_2) = \Z(S_1) \cup
      \Z(S_2)$ where
      \[ S_1 \cdot S_2 = \{f \cdot g \st f \in
      S_1, g \in S_2\} .\]}
    \item \cloze{Let $I$ be an index set and suppose for
      each $i \in I$, we are given $S_i \subset
      \polyA$. Then
      \[ \bigcap_{i \in I} \Z(S_i) = \Z \left(
      \bigcup_{i \in I} S_i \right) .\]}
  \end{enumerate}
\end{proposition*}
\end{flashcard}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(a)]
    \item Obvious
    \item Obvious
    \item If $p \in \Z(S_1) \cup \Z(S_2)$, then
      either $f(p) = 0 ~\forall f \in S_1$ or
      $g(p) = 0 ~\forall g \in S_2$. Thus $(f
      \cdot g)(p) = 0$ for all $f \in S_1$, $g \in
      S_2$. So $p \in \Z(S_1 \cdotset S_2)$.

      Conversely, suppose $p \in \Z(S_1 \cdotset
      S_2)$, and suppose $p \notin \Z(S_1)$. So
      there exists $f \in S_1$ with $f(p) \neq 0$.
      But $(f \cdot g)(p) = 0 ~\forall g \in S_2$
      and $(f \cdot g)(p) = f(p) \cdot g(p)$, so
      $g(p) = 0 ~\forall g \in S_2$. Thus $p \in
      \Z(S_2)$. Thus $\Z(S_1 \cdotset S_2) \subseteq
      \Z(S_1) \cup \Z(S_2)$.
    \item If $p \in \Z(S_1) ~\forall i$, then $p
      \in \Z \left( \bigcup_{i \in I} S_i
      \right)$.

      Conversely, if $p \in \Z \left( \bigcup_{i
      \in I} S_i \right)$, then $p \in \Z(S_i)
      ~\forall i$, so $p \in \bigcap_{i \in I}
      \Z(S_i)$. \qedhere
  \end{enumerate}
\end{proof}

\textbf{Moral:} This says that sets of the form
$\Z(S)$ form the closed sets of a topology on
$\AA^n$

\textbf{Moral:} This says that sets of the form
$\Z(S)$ form the closed sets of a topology on
$\AA^n$.

\begin{flashcard}[algebraic-subset-defn]
\begin{definition*}[Algebraic subset]
  \glsadjdefn{zclosed}{Zariski closed}{set}
  \glsadjdefn{zzclosed}{closed}{set}%
  \glsadjdefn{algset}{algebraic}{set}
  \cloze{A subset of $\AA^n$ is \emph{algebraic}
  (or \emph{Zariski closed}) if it is
  of the form $\Z(S)$ for some $S \subset
  \polyA$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[zariski-open-subset-defn]
\begin{definition*}[Zariski open subset]
  \glsadjdefn{zopen}{Zariski open}{set}%
  \glsadjdefn{zzopen}{open}{set}
  A \emph{Zariski open subset} of $\AA^n$ is a
  set of the form $\AA^n \setminus \Z(S)$ for some
  $S \subseteq \polyA$. This defines the Zariski
  topology on $\AA^n$.
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item If $K = \CC$, \gls{zopen} or
      \glsref[zclosed]{closed}
      subsets are also open and closed in the
      ``usual'' topology.
    \item For any $\KK$, consider $\AA^1$, $\polyA
      = \KK[X]$, $S \subseteq \KK[X]$ containing a
      non-zero element. Then $\Z(S)$ is finite. So
      \gls{zclosed} sets are $\AA^1$ and all
      finite sets. \gls{zopen} sets are
      $\emptyset$ and ``co-finite sets''.
  \end{enumerate}
\end{example*}

\vspace{-1em}
\textbf{Recall:} If $A$ is a commutative ring,
$S \subseteq A$ a subset, the \emph{ideal
generated by $S$} is the ideal $\langle S \rangle
\subseteq A$ given by
\begin{align*}
  \langle S \rangle
  &= \left\{ \sum_{i = 1}^q f_i
  g_i \st q \ge 0, f_i \in S, g_i \in A \right\}
  \\
  &= \text{the smallest ideal of $A$ containing
  $S$}
\end{align*}

\begin{flashcard}[zero-set-equals-zero-ideal-lemma]
\begin{lemma*}
  Let $S \subseteq \polyA = \KK[X_1, \ldots,
  X_n]$, $I = \langle S \rangle$. Then
  \[ \Z(S) = \Z(I) .\]
\end{lemma*}

\begin{proof}
  If $p \in \Z(S)$, $f_1, \ldots, f_q \in S$,
  $g_1, \ldots, g_q \in A$, then
  \[ \sum_{i = 1}^q (f_i g_i)(p) = \sum_{i = 1}^q
  \ub{f_i(p)}_{=0} g_i(p) = 0 .\]
  Thus $\Z(S) \subseteq \Z(I)$.

  Conversely, since $S \subseteq I$, $\Z(I)
  \subseteq \Z(S)$.
\end{proof}
\end{flashcard}