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\begin{center}
  \includegraphics[width=0.6\linewidth]{images/13553c3e5f4a4fe0.png}
\end{center}
This suggests we define a map $\Sigma : \P^1
\times \P^1 \to \P^3$,
\[ \Sigma((a : b), (c : d)) = (bd : ac : ad : bc) \]
\textbf{Claim:} $\Sigma$ is a bijection with $Q =
\PZ(xy - zw)$.

\begin{proof}
  Note $(bd) \cdot (ac) - (ad) \cdot (bc) = 0$, so
  $\Sigma$ has image in $Q$. Injection: suppose
  $a, c \neq 0$, so
  \[ \Sigma((1 : b), (1 : d)) = (bd : 1 : d : b) \]
  so clearly injective on the set where $a, c \neq
  0$. If $a = 0$,
  \[ \Sigma((0 : b), (c : d)) = (bd : 0 : 0 : bc)
  = (d : 0 : 0 : c) \]
  doesn't coincide with any of the previous points
  and is injective on the locus where $a = 0$. If
  $a = c = 0$,
  \[ \Sigma((0 : 1), (0 : 1)) = (1 : 0 : 0 : 0) \]
  If $a \neq 0$, $c = 0$,
  \[ \Sigma((a : b), (0 : 1)) = (b ; 0 : a : 0) \]
  so $\Sigma$ is injective.

  Surjective: Suppose $(a_0 : a_1 : a_2 : a_3) \in
  Q$, i.e. $a_0 a_1 - a_2 a_3 = 0$. If $a_0 \neq
  0$, can take $a_0 = 1$, so $a_1 = a_2 a_3$. So
  \[ (a_0 : a_1 : a_2 : a_3) = (1 : a_2 a_3 : a_2
  : a_3) = \Sigma((a_2 : 1), (a_3 : 1)) \]
  Similar arguments work in the charts where $a_1
  \neq 0$, $a_2 \neq 0$ or $a_3 \neq 0$.
\end{proof}

\begin{moral*}
  $\P^1 \times \P^1$ is not a \gls{projvty}, but
  can be given a \gls{vty} structure by
  identifying it with $Q$, i.e. closed sets of
  $\P^1 \times \P^1$ are of the form
  $\Sigma^{-1}(Z)$ for $Z \subseteq Q$ closed.

  \textbf{Exercise:} Check this is not the product
  topology on $\P^1 \times \P^1$.
\end{moral*}

\vspace{-1em}
\gls{gregf} functions on $U = \Sigma^{-1}(V)$ for
$V \subseteq Q$ open, are functions on $U$ of the
form $\varphi \circ \Sigma$ with $\varphi : V \to
\KK$ \gls{pregf}.

A generalisation:

\begin{flashcard}[Segre-embedding]
The \emph{Segre embedding} is the map\cloze{
\[ \Sigma : \P^n \times \P^m \to \P^{(n + 1)(m +
1) - 1} \]
given by
\[ \Sigma((x_0 : \cdots : x_n), (y_0 : \cdots :
y_m)) = (x_i y_j)_{\substack{0 \le i \le n \\ 0
\le j \le m}} \]

\begin{theorem*}
  $\Sigma$ is injective and its image is an
  algebraic variety.
\end{theorem*}

\vspace{-1em}
Thus $\P^n \times \P^m$ acquires the structure of
an algebraic variety.

\begin{theorem*}
  If $X \subseteq \P^n$, $Y \subseteq \P^m$ are
  \glspl{projvty}, then $\Sigma(X, Y)$ is a
  \gls{projvty} in $\P^{(n + 1)(m + 1) - 1}$.
\end{theorem*}

\begin{moral*}
  This allows us to thin of $X \times Y$ as a
  \gls{projvty}.
\end{moral*}}
\end{flashcard}

\begin{remark*}
  We can also think of the geometry of $\P^n
  \times \P^m$ by thinking about
  \emph{bihomogeneous}
  polynomials in $\KK[x_0, \ldots, x_n, y_0,
  \ldots, y_m]$, i.e. polynomials $f$ satisfying
  \[ f(\lambda x_0, \ldots, \lambda_n x_n, \mu
  y_0, \ldots, \mu y_m) = \lambda^d \mu^e f(x_0,
  \ldots, x_n, y_0, \ldots, y_m) .\]
  We say $f$ is bidegree $(d, e)$. $f = 0$ makes
  sense as an equation in $\P^n \times \P^m$.
\end{remark*}

\begin{remark*}
  If $X$ and $Y$ are
  \glsref[qprojvty]{quasi-projective}, $X \subseteq
  \ol{X} \subseteq \P^n$, $Y \subseteq \ol{Y}
  \subseteq \P^m$, then $X \times Y \subseteq
  \ol{X} \times \ol{Y}$ defines an open subset of
  $\ol{X} \times \ol{Y}$ (check!). This allows us
  to view $X \times Y$ as a \gls{qprojvty}.
\end{remark*}

\subsubsection*{Example: The blowup of $\AA^n$}

By the Remark, $\AA^n \times \P^{n - 1}$ is a
\gls{qprojvty}.

Let
\[ X = \PZ(\{x_i y_j - x_j y_i \st 1 \le i < j
\le n\}) \subseteq \AA^n \times \P^{n - 1} .\]
Let $\varphi : X \to \AA^n$ be given by
\[ \varphi((x_1, \ldots, x_n), (y_1 : \cdots :
y_n)) = (x_1, \ldots, x_n) ,\]
the projection. This is a \gls{gmorph}.

\textbf{Observations:}
\begin{enumerate}[(1)]
  \item If $p \in \AA^n \setminus \{0\}$, then
    $\varphi^{-1}(p)$ consists of one point.

    \begin{proof}
      Let $p = (a_1, \ldots, a_n)$, say $a_1
      \neq 0$. If
      \[ ((a_1, \ldots, a_n), (b_1 : \cdots :
      b_n)) \in \varphi^{-1}(p) ,\]
      then for $j \neq i$, $a_i b_j - a_j b_i =
      0$, or $b_j = \frac{a_j}{a_i}b_i$. So
      $b_1, \ldots, b_n$ are completely
      determined up to scaling. Taking
      $b_i = a_i$, we see
      \[ \varphi^{-1}(p) = \{((a_1, \ldots,
      a_n), (a_1 : \cdots : a_n))\} . \qedhere \]
    \end{proof}

    Defining $\psi : \AA^n \setminus \{0\} \to
    X$ by $\psi(a_1, \ldots, a_n) = ((a_1,
    \ldots, a_n), (a_1 : \cdots : a_n))$ is an
    inverse to $\varphi|_{X \setminus
    \varphi^{-1}(0)} : X \setminus
    \varphi^{-1}(0) \to \AA^n \setminus \{0\}$.
  \item $\varphi^{-1}(0) = \{0\} \times \P^{n -
    1}$
  \item The points of $\varphi^{-1}(0)$ are in
    $1-1$ correspondence with the lines through
    the origin in $\AA^n$. $n = 2$ picture:
    \begin{center}
      \includegraphics[width=0.6\linewidth]{images/72632e47e5c646e8.png}
    \end{center}
\end{enumerate}
