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\begin{remark*}
  In proof of previous propositions, we had a
  statement $\Z(\{h_i\}) = \emptyset$, and hence
  by \nameref{null_II}, $1 \in \langle
  \{h_i\} \rangle$, and hence we can write $1 =
  \sum_{i \in I} e_i h_i$ for $I$ a finite index
  set.
\end{remark*}

\newpage
\section{Projective Varieties}

\begin{flashcard}[projective-space-defn]
\begin{definition*}[$\PP^n$]
  \glssymboldefn{P}{$P^n$}{$P^n$}
  \cloze{Let $\KK$ be a field. We define
  \[ \PP^n = (\KK^{n + 1} \setminus \{(0, \ldots,
  0)\}) / \sim \]
  where $(x_0, \ldots, x_n) \sim (\lambda x_0,
  \ldots, \lambda x_n)$ for any $\lambda \in
  \KK^\times \defeq \KK \setminus \{0\}$.

  Alternatively, this is the set of
  one-dimensional sub-vector spaces of $\KK^{n +
  1}$.}
\end{definition*}
\end{flashcard}

\begin{remark*}
  If $\KK = \RR$, then $\PP^n = S^n / \sim$, with
  $x_n \sim -x$ ($S^n \subseteq \RR^{n + 1}$ is
  the unit sphere).
\end{remark*}

\vspace{-1em}
For arbitrary $\KK$: Consider $\P^1$. For $(x_0 :
x_1) \in \P^1$, if $x_1 \neq 0$, then
\[ (x_0 : x_1) \sim \left( \frac{x_0}{x_1} : 1
\right) \in \AA^1 \]
(since there is a unique representative with
seconc coordinate $1$). The missing points are of
the form $(x_0 : 0) \sim (1 : 0)$. Thus we view
$\P^1$ as
\[ \P^1 = \AA^1 \cup \{\ub{(1 : 0)}_{=\infty}\} .\]
This is the Riemann sphere if $\KK = \CC$.

Now $\P^2$: for $(x_0 : x_1 : x_2) \in \P^2$, if
$x_2 \neq 0$, then
\[ (x_0 : x_1 : x_2) \sim \left( \frac{x_0}{x_2} :
\frac{x_1}{x_2} : 1\right) \in \AA^2 .\]
If $x_2 = 0$, we get a point $(x_0 : x_1 : 0) \in
\P^1$. Thus
\[ \P^2 = \AA^2 \cup \P^1 \]
where $\P^1$ can be viewed as the `line at
infinity'.

Algebraic subsets of $\P^n$? When does $f(x_0,
\ldots, x_n) = 0$ make sense?

\begin{flashcard}[homogeneous-poly-defn]
\begin{definition*}[Homogeneous]
  \glsadjdefn{homog}{homogeneous}{polynomial}
  \glspropdefn{deg}{degree}{homogeneous polynomial}
  \cloze{$f \in S = \KK[x_0, \ldots, x_n]$ is
  \emph{homogeneous} if every term of $f$ is of
  the same degree, or equivalently,
  \[ f(\lambda x_0, \ldots, \lambda x_n) =
  \lambda^d f(x_0, \ldots, x_n) \]
  for some $d \ge 0$, where $d$ is the
  \emph{degree} of $d$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  $x_0^3 + x_1 x_2^2$ is \gls{homog} of \gls{deg}
  $3$. $x_0^3 + x_1^2$ is not \gls{homog}.
\end{example*}

\begin{flashcard}[zeroes-in-Pn]
\begin{definition*}[Zero set of $f$ in $\P^n$]
  \glssymboldefn{PZ}{$Z(T)$}{$Z(T)$}
  \cloze{If $T \subseteq S$ is a set of \gls{homog}
  polynomials, define
  \[ Z(T) \defeq \{(a_0, \ldots, a_n) \in \P^n \st
  f(a_0, \ldots, a_n) = 0 ~\forall f \in T\} .\]}
\end{definition*}
\end{flashcard}

\begin{flashcard}[homog-ideal-defn]
\begin{definition*}[Homogeneous ideal]
  \glsadjdefn{homid}{homogeneous}{ideal}
  \cloze{An ideal $I \subseteq S$ is \gls{homog} if $I$
  is generated by \gls{homog} polynomials.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[zero-set-of-ideal]
\begin{definition*}[Zero set of ideal]
  \glssymboldefn{IZ}{$Z(I)$}{$Z(I)$}
  For $I$ a \gls{homid} ideal, we define
  \[ Z(I) = \{(a_0, \ldots, a_n) \in \P^n \st
  f(a_0, \ldots, a_n) = 0 ~\forall f \in I
  \text{ \gls{homog}}\} .\]
\end{definition*}
\end{flashcard}

\begin{flashcard}[algebraic-subset-of-Pn]
\begin{definition*}[Algebraic subset of $\P^n$]
  \glsadjdefn{Palgset}{algebraic}{subset}
  \cloze{A subset of $\P^n$ is \emph{algebraic} if it is
  of the form $\PZ(T)$ for some $T$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  $\PZ(a_0 x_0 + a_1 x_1 + a_2 x_2) \subseteq
  \P^2$, $a_0, a_1, a_2 \in \KK$. In the $\AA^2
  \subseteq \PP^2$ where $x_2 = 1$, we get the
  equation $a_0 x_0 + a_1 x_1 + a_2 = 0$. If $x_2
  = 0$, we get the equation $a_0 x_0 + a_1 x_1 =
  0$, which has the solution $(a_1 : -a_0) \in
  \P^1$ (assuming not both $a_0 = 0$, $a_1 = 0$,
  since otherwise just have $x_2 = 0$, the line at
  $\infty$)
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/cbf651fda45e4b83.png}
  \end{center}
\end{example*}

\vspace{-1em}
\glsadjdefn{pclosed}{closed}{sets}
\glsadjdefn{popen}{open}{sets}
\textbf{Exercise:} Check the \gls{Palgset} sets in
$\P^n$ form the closed sets of a topology on
$\P^n$. This is the \emph{Zariski topology} on
$\P^n$.

\begin{flashcard}[projective-variety-defn]
\begin{definition*}[Projective variety]
  \glsnoundefn{projvty}{projective
  variety}{projective varieties}
  \cloze{A \emph{projective variety} is an \gls{irredsub}
  closed subset of $\P^n$.}
\end{definition*}
\end{flashcard}

\subsubsection*{The standard open affine cover of
$\PP^n$}

Define $U_i \subseteq \P^n$ by
\[ U_i = \P^n \setminus \Z(x_i) ,\]
an open subset of $\P^n$. Note $\bigcup_{i = 1}^n
U_i = \P^n$. We have a bijection $\varphi_i : U_i
\to \AA^n$, given by
\[ \varphi_1(x_0 : \ldots : x_n) = \left(
\frac{x_0}{x_i}, \ldots, \widehat{\frac{x_i}{x_1}},
\ldots, \frac{x_n}{x_i} \right) \]
(hat means this is omitted).

\begin{proposition*}
  With $U_i$ carrying the topology induced from
  $\P^n$, and $\AA^n$ the \glsref[zopen]{Zariski
  topology},
  $\varphi_i$ is a homeomorphism.
\end{proposition*}

\begin{proof}
  Since $\varphi_i$ is a bijection, enough to show
  $\varphi_i$ identifies closed sets of $U_i$ with
  closed sets of $\AA^n$. Can take $c = 0$,
  $\varphi = \varphi_0$, $U = U_0$. Let $S =
  \KK[X_0, \ldots, X_n]$, $S^h$ be the set of
  \gls{homog} polynomials in $S$. Let $A =
  \KK[Y_1, \ldots, Y_n]$. Define maps $\alpha :
  S^n \to A$, $\beta : A \to S^n$ by
  $\alpha(f(x_0, \ldots, x_n)) = f(1, y_1, \ldots,
  y_n)$. If $g \in A$ of degree $e$ (highest
  degree term is degree $e$), then define
  \[ \beta(g) = x_0^e g \left( \frac{x_1}{x_0},
  \ldots, \frac{x_n}{x_0} \right) \]
  \textbf{Remark:} This is a process known as
  \emph{homogenisation}. For example, $y_2^2 -
  y_1^3 - y_1 + y_1 y_2$ becomes
  \[ x_0^3 \left( \frac{x_2^2}{x_0^2} -
  \frac{x_1^3}{x_0^3} - \frac{x_1}{x_0} +
  \frac{x_1 x_2}{x_0^2} \right) =
  x_0 x_2^2 - x_1^3 - x_0^2 x_1 + x_0 x_1 x_2 .\]
  under $\beta$.

  If $Y \subseteq U$ is closed, then $Y$ is the
  intersection $\ol{Y} \cap U$ where $\ol{Y}
  \subseteq \P^n$ is a closed subset, which we can
  take to be the closure of $Y$. $\ol{Y} = \IZ(T)$
  for some $T \subseteq S^h$. Let $T' =
  \alpha(T)$. Then $\varphi(Y) = \Z(\alpha(T))$.

  Check:
  \begin{align*}
    f(a_0 : \ldots : a_n) = 0, (a_0 \neq 0)
    &\iff f \left( f, \frac{a_1}{a_0}, \ldots,
    \frac{a_n}{a_0} \right) = 0 \\
    &\iff (\alpha(f)) \left( \frac{a_1}{a_0},
    \ldots, \frac{a_n}{a_0} \right) \\
    &\iff \alpha(f)(\varphi(a_0, \ldots, a_n)) = 0
  \end{align*}