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  Still to prove that if $W \subseteq \AA^n$ is
  closed, then $\varphi^{-1}(W) \subseteq U = U_0$
  is \gls{pclosed}. We have $W = \Z(T')$ for some
  set $T' \subseteq \polyA = \KK[Y_1, \ldots,
  Y_n]$. Then
  \[ \varphi^{-1}(W) = \Z(\beta(T')) \cap U \]
  ($\beta$ is homogenisation as mentioned
  earlier). Indeed, if $g \in T'$,
  \begin{align*}
    g(b_1, \ldots, b_n) = 0
    &\iff \beta(g)(1, b_1, \ldots, b_n) = 0 \\
    &\iff \beta(g)(\varphi^{-1}(b_1, \ldots, b_n))
    = 0 \qedhere
  \end{align*}
\end{proof}

\begin{example*}
  $f : \P^1 \to \P^3$.
  \[ f(u : t)= (u^3, u^2 t, ut^2, t^3)  \]
  which is well-defined. The image of this map is
  called the \emph{twisted cubic} (recall \es{1}).

  \textbf{Claim:} This is a projective variety.

  Proof: Consider the homomorphism
  \begin{align*}
    \phi : \KK[X_0, \ldots, X_3]
    &\to \KK[u, t] \\
    X_0
    &\mapsto u^3 \\
    X_1
    &\mapsto u^2 t \\
    X_2
    &\mapsto u t^2 \\
    X_3
    &\mapsto t^3
  \end{align*}
  Let $I = \Ker \varphi$. If $g \in I$, then $g$
  vanishes on the image of the map $f$. Thus
  $\Im(p) \le \IZ(I)$.

  Conversely, note that
  \[ X_0 X_3 - X_1 X_2, X_1^2 - X_0 X_2, X_2^2 -
  X_1 X_3 \in I .\]
  Let $p = (a_0 : a_1 : a_2 : a_3) \in \IZ(I)$. 4
  cases:
  \begin{itemize}
    \item $a_0 \neq 0$. So take $a_0 = 1$.
      \[ a_3 - a_1 a_2 = 0, \qquad a_1^2 - a_2 =
      0, \qquad a_2^2 - a_1 a_3 = 0 .\]
      Then $p = (1, a_1, a_2^2, a_1^3) = f(1 :
      a_1)$. Thus $p \in \Im(f)$.
  \end{itemize}
  Similarly check cases $a_1 \neq 0$, $a_2 \neq 0$
  and $a_3 \neq 0$. The conclusion is $p \in
  \Im(f)$ in all $4$ cases, so $\Im f \supseteq
  \IZ(I)$. Therefore $\IZ(I) = \Im f$. Thus the
  twisted cubic is an \gls{Palgset} set.
\end{example*}

\vspace{-1em}
\textbf{Exercise:} Given $X \subseteq \P^n$ an
\gls{Palgset}, define its \emph{ideal} $I(X)$ to
be the ideal in $S = \KK[X_1, \ldots, X_n]$
generated by \gls{homog} polynomials vanishing on
$X$. Then show that $X$ is \gls{irredsub} if and
only if $I(X)$ is prime.

For the twisted cubic, $X = \Im(F)$, $I(X) = I =
\Ker(\varphi)$. But
\[ \frac{\KK[X_0, \ldots, X_3]}{\Ker \varphi} \]
is a subring of the integral domain $\KK[u, t]$.
Hence it is an integral domain, hence $\Ker
\varphi$ is prime. Therefore $X$ is a
\gls{projvty}.

\begin{flashcard}[proj-regf-defn]
\begin{definition*}[Projective regular function]
  \glsadjdefn{pregf}{regular}{function}
  \cloze{Let $X \subseteq \P^n$ be a \gls{projvty}. A
  \emph{regular function} on $U \subseteq X$
  \gls{popen} is a function $f : U \to \KK$ such
  that for every $p \in U$, there exists an
  \gls{popen} neighbourhood $V \subseteq U$ of $p$
  and $g, h \in S$ \gls{homog} of the same degree
  with $h$ nowhere vanishing on $V$, and with
  $f|_V = \frac{g}{h}$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[quasi-variety-defn]
\begin{definition*}[Quasi-variety]
  \glsnoundefn{qaffvty}{quasi-affine
  variety}{quasi-affine varieties}
  \glsnoundefn{qprojvty}{quasi-projective
  variety}{quasi-projective varieties}
  \cloze{A \emph{quasi-affine variety} is an open subset
  of an \gls{affvty}.

  A \emph{quasi-projective variety} is an open
  subset of a \gls{projvty}.}
\end{definition*}
\end{flashcard}

\vspace{-1em}
\glsadjdefn{gregf}{regular}{function}
These types of varieties also have (the same)
notion of \gls{regf} functions.

\glsnoundefn{vty}{variety}{varieties}
A \emph{variety} means an \glsref[affvty]{affine},
\glsref[qaffvty]{quasi-affine},
\glsref[projvty]{projective} or
\glsref[qprojvty]{quasi-projective} variety.

\begin{flashcard}[morphism-varieties-defn]
\begin{definition*}[Morphism between varieties]
  \glsnoundefn{gmorph}{morphism}{morphisms}
  \cloze{A \emph{morphism} $\varphi : X \to Y$ between
  \glspl{vty} is a continuous function $\varphi$
  such that $\forall V \subseteq Y$ open, $f : V
  \to \KK$ \gls{gregf}, $f \circ \varphi :
  \varphi^{-1}(U) \to \KK$ is \gls{gregf}.}
\end{definition*}
\end{flashcard}

\begin{remark*}
  If $X$ is projective, then in fact
  \[ \OX_X(X) = \{X \to \KK \text{ \gls{gregf}}\}
  \]
  is $\KK$. Thus finding \glspl{gmorph} from a
  \gls{projvty} becomes much harder, and this is a
  lot of what Algebraic Geometry is about.
\end{remark*}

\subsubsection*{Example}

Let $Q \subseteq \P^3$ be given by $\Z(xy -
zw)$. This is a quadric surface
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/347df02ba61f4aba.png}
\end{center}
\textbf{Important feature:} For $(a : b) \in
\P^1$, $Q$ contains the line
\[ ax = bz, by = aw \]
(if $a \neq 0$, can take $a = 1$, $(bz) y -
z(by) = 0$, if $a = 0$, $z = 0$, $y = 0$, so $xy
- zw = 0$). This gives a family of lines in $Q$
parametrized by $(a : b) \in \P^1$> We also have
$ax = bw$, $by = az$ for $(a : b) \in \P^1$
contained in $Q$.

If we take a line from one family and a line
from the other, they meet at one point:
\[ ax = bz, by = aw \]
\[ cx = dw, dy = cz \]
has a unique solutoin up to scaling: $(bd, ac,
ad, bc)$.