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\begin{enumerate}[(1)]
  \item[] \begin{proof}
      A line through $0$ can be parametrised by $l
      : \AA^1 \to \AA^n$,
      \[ l(t) = (a_1t, \ldots, a_nt) \]
      for some $a_1, \ldots, a_n$ not all $0$. For
      $t \neq 0$,
      \begin{align*}
        \varphi^{-1}(a_1 t, \ldots, a_n t)
        &= ((a_1 t,\ldots, a_n t), (a_1 t : \cdots
        : a_n t)) \\
        &= ((a_1 t, \ldots, a_n t), (a_1 : \cdots
        : a_n))
      \end{align*}
      Thus the lift of $L \setminus \{0\}$ is
      given parametrically by $t \mapsto ((a_1 t,
      \ldots, a_n t), (a_1 : \cdots : a_n))$,
      $\AA^1 \setminus \{0\} \to
      \varphi^{-1}(\AA^n \setminus \{0\})
      \subseteq X$. This extends to all of $\AA^1$
      and also $\ol{\varphi^{-1}(L \setminus
      \{0\})}$ is the image of this
      parametrisation.
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/d3d0264dbb604fcf.png}
      \end{center}
    \end{proof}
    \setcounter{enumi}{3}
  \item $X$ is \gls{irredsub}.
    \begin{proof}
      $X = (X \setminus \varphi^{-1}(0)) \cup
      \varphi^{-1}(0)$. The first set being
      homeomorphic to $\AA^n \setminus \{0\}$, and
      hence is irreducible. (An open subset of an
      irreducible space is irreducible). But every
      point of $\varphi^{-1}(0)$ is in the closure
      of $X \setminus \varphi^{-1}(0)$, by the
      proof of (3), so $X \setminus
      \varphi^{-1}(0)$ is dense in $X$.

      \textbf{Claim}: If $U \subseteq X$ is a dense
      open set and $U$ is \gls{irredsub}, then $X$ is
      \gls{irredsub}.

      Proof: If $X = Z_1 \cup Z_2$, $Z_1, Z_2$
      closed, then $U = (Z_1 \cap U) \cup (Z_2
      \cap U)$, so $U = Z_1 \cap U$ say. So $U
      \subseteq Z_1$, so $\ol{U} \subseteq Z_1$.
      But $\ol{U} \subseteq Z_1$. But $\ol{U} = X$
      by density of $U$. SO $Z_1 = X$.

      Thus the blowup $X$ is \gls{irredsub}.
    \end{proof}
\end{enumerate}

\begin{flashcard}[blowing-up-defn]
\begin{definition*}[Blowing up]
  \glsnoundefn{blowup}{blow up}{N/A}
  \cloze{If $Y \subseteq \AA^n$ is a closed subvariety
  with $0 \in Y$, we define the \emph{blowing up}
  of $Y$ at $0$ to be
  \[ \tilde{Y} \defeq \ol{\varphi^{-1}(Y \setminus
  \{0\})} \subseteq X ,\]
  where $X = \IZ(\{x_i y_j - x_j y_i \st 1 \le i
  < j \le n\}) \subseteq \AA^n \times \P^{n - 1}$,
  $\varphi : X \to \AA^n$ is given by projection
  of the first $n$ coordinates.}
\end{definition*}
\end{flashcard}

\subsubsection*{Example}

Let $Y \subseteq \AA^2$ be given by
\[ Y = \Z(\ub{x_2^2 - (x_1^3 + x_1^2)}_{x_2^2 -
x_1^2 (x_1 + 1)}) \]
$X \subseteq \AA^2 \times \P^1$, $x_1 y_2 - x_2
y_1 = 0$. Work in two coordinate patches:
\[ U_1 = \{y_1 \neq 0\}, \qquad U_2 = \{y_2 \neq
0\} \]
In $U_2$, we set $y_2 = 1$, and the equation for
$X$ becomes $x_1 = x_2 y_1$. Then
\[ \varphi^{-1}(Y) \cap U_2 = \Z(x_2^2 - (x_1^3
+ x_1^2), x_1 - x_2 y_1) \subseteq \AA^2 \times
\AA^1 = \AA^3 .\]
This is isomorphic to
\[ \Z(x_2^2 - (x_2^3 y-1^3 + x_2^2 y_1^2))
\subseteq \AA^2 .\]
In terms of coordinate rings,
\[ \frac{\KK[x_1, x_2, y_1]}{\langle x_2^2 -
(x_1^3 + x_1^2), x_1 - y_1 x_2 \rangle} \cong
\frac{\KK[x_2, y_1]}{\langle x_2^2 - (x_2^3
y_1^2 + x_2^2 y_1^2) \rangle} \]
Note
\[ x_2^2 - (x_2^2 y_1^2 + x_2^2 y_1^2) = x_2^2
(1 - x_2 y_1^2 - y_1^2) \]
\begin{center}
  \includegraphics[width=0.3\linewidth]{images/fc7e7722f4e94e4a.png}
\end{center}
Noet $\varphi^{-1}(0) \cap U_2 = \Z(x_2)$. The
blowup $\tilde{Y} \cap U_2 = \ol{\varphi^{-1}(Y
\setminus \{0\})} \cap U_2$ is now given by the
equation $1 - x_2 y_1^2 - y_1^2$ in $\AA^2$
($x_2, y_1$).
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/18387955ede4437b.png}
\end{center}
For thoroughness, we will also consider $\tilde{Y}
\cap U_1$, where $y_1 = 1$, $x_2 = x_1 y_2$, so
can eliminate $x_2$ from equation to get
\[ x_1^2 y_2^2 - (x_1^3 + x_1^2) = x_1^2(y_2^2 -
x_1 - 1) \]
$\tilde{Y} \cap U_1$ has equation $y_2^2 - x_1 - 1
= 0$.

\subsubsection*{Rational maps}

\begin{flashcard}[rational-map]
\begin{definition*}[Rational map]
  \glssymboldefn{ratmap}{$\dashrightarrow$}{$\dashrightarrow$}
  \glsnoundefn{ratmap}{rational map}{rational maps}
  \glsadjdefn{rat}{rational}{map}
  \cloze{Let $X, Y$ be \glspl{vty}. Consider the
  equivalence relation on pairs $(U, f)$ where $U
  \subseteq X$ open, and $f : U \to Y$ a
  \gls{gmorph}, with $(U, f) \sim (V, g)$ if $f|_{U
  \cap V} = g|_{U \cap V}$.

  \fcscrap{\textbf{Exercise:} Check that this is an
  equivalence relation.}

  A \emph{rational map} $f : X \dashrightarrow Y$
  is an equivalence class of a pair.}
\end{definition*}
\end{flashcard}

\begin{example*}
  If $X$ is affine, $\varphi = \frac{f}{g} \in
  \K(X)$, then we have a morphism $\varphi : X
  \setminus \Z(g) \to \AA^1$. This defines a
  \glsref[ratmap]{rational} \gls{gmorph} to $\AA^1$.
\end{example*}
