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\begin{flashcard}[birational-map-defn]
\begin{definition*}[Birational map]
  \glsnoundefn{bimap}{birational map}{birational
  maps}
  \glsadjdefn{birat}{birational}{map}
  \cloze{A \emph{birational map} is a \gls{ratmap} $f : X
  \ratmap Y$ with a \gls{rat} inverse $g
  : Y \ratmap X$ such that $f \circ g = \id_Y$ and
  $g \circ f = \id_X$ as \glspl{ratmap}.}
\end{definition*}
\end{flashcard}

\begin{remark*}
  We can't always compose \glspl{ratmap}. Suppose
  given $f : X \ratmap Y$, $g : Y \ratmap Z$, $f :
  U \to Y$, $g : V \to Z$. If $f(U) \subseteq Y
  \setminus V$, we can't compose. If this is not
  the case, then $f^{-1}(Y \setminus V) \subsetneq
  U$ is a proper subset of $U$, and then $g \circ
  f : U \setminus f^{-1}(Y \setminus V) \to Z$
  defines a \gls{ratmap} $g \circ f : X \ratmap
  Z$. Note the ability to compose may depend on
  the representative for $f, g$.
\end{remark*}

\begin{remark*}
  One can show that if $f : X \ratmap Y$ is a
  \gls{bimap}, then $\exists U \subseteq X$, $V
  \subseteq Y$ such that $f$ is defined on $U$,
  $f(U) \subseteq V$ and $f : U \to V$ is an
  \gls{isism}.
\end{remark*}

\begin{flashcard}[birationally-equivalent-defn]
\begin{definition*}[Birationally equivalent]
  \glsadjdefn{biequiv}{birationally
  equivalent}{\glspl{vty}}
  \cloze{We say \glspl{vty} $X, Y$ are \emph{birationally
  equivalent} if there exists a \gls{bimap} $f : X
  \ratmap Y$. Equivalently, $\exists U \subseteq
  X$, $V \subseteq Y$ open subsets with $U \cong
  V$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  $\varphi : X \to \AA^n$, the \gls{blowup} of
  $\AA^n$ at $0 \in \AA^n$. This is a \gls{bimap}
  (\gls{gmorph}) since it induces an \gls{isism}
  $\varphi : \varphi^{-1}(\AA^n \setminus \{0\})
  \to \AA^n \setminus \{0\}$. $\varphi^{-1} :
  \AA^n \ratmap X$ is not a \gls{gmorph}, only
  defined on $\AA^n \setminus \{0\}$.
\end{example*}

\begin{flashcard}[dominant-defn]
\begin{definition*}[Dominant]
  \glsadjdefn{domf}{dominant}{function}
  \cloze{We say that $f : X \ratmap Y$ is a
  \emph{dominant} \gls{ratmap} if whenever
  $\tilde{f} : U \to Y$ is a representative for
  $f$, then $f(U)$ is dense in $Y$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[function-field-vty-defn]
\begin{definition*}[Function field of a variety]
  \glssymboldefn{gKX}{$K(X)$}{$K(X)$}
  \glssymboldefn{gsh}{$f^\#$}{$f^\#$}
  \cloze{The \emph{function field} of a \gls{vty} $X$ is
  \[ K(X) = \{(U, f) \st f : U \to \KK \text{ is a
  \gls{gregf} function}\} / \sim ,\]
  where $(U, f) \sim (V, g)$ if $f|_{U \cap V} =
  g|_{U \cap V}$. In particular, if $X$ is
  \gls{affvty}, then this is the field of
  fractions of $\cring A(X)$. If $f$ is
  \gls{domf}, we obtain
  \begin{align*}
    f^\# : K(Y)
    &\to K(X) \\
    (V, \varphi)
    &\mapsto (f^{-1}(V) \cap U, \varphi \circ f)
  \end{align*}
  Note $f^{-1}(V) \cap U$ is non-empty since $V
  \cap f(U) \neq \emptyset$ by density of $f(U)$.}
\end{definition*}
\end{flashcard}

\begin{note*}
  If $f : X \ratmap Y$ is a \gls{bimap}, with
  \gls{birat} inverse $g : Y \ratmap X$, each are
  \gls{domf} since they induce \glspl{isism}
  between open subsets. Thus we get
  \[ \gsh f : \gK(Y) \to \gK(X), \qquad \gsh g :
  \gK(X) \to \gK(Y) \]
  inverse maps, so $\gK(X) \cong \gK(Y)$.
\end{note*}

\vspace{-1em}
\textbf{Fact:} If $\gK(X) \cong \gK(Y)$, then $X$
and $Y$ are \gls{birat} to each other, i.e.
$\exists f : X \ratmap Y$ \gls{birat}.

\begin{example*}
  $0 \in Y \subseteq \AA^n$, $\tilde{Y} \to Y$ the
  \gls{blowup} of $Y$ at $0$ is a \gls{birat}
  \gls{gmorph}:
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/2105f4a31c6445d6.png}
  \end{center}
\end{example*}

\newpage
\section{Tangent spaces, singularities and
dimension}

\textbf{Recall:} Gvien an equation $f(X_1, \ldots,
X_n) = 0$ in $\RR^n$, $X$ the solution set, $p \in
X$, the tangent space to $X$ is the orthogonal
complement to $(\nabla f)(p)$, i.e. the tangent
space to $X$ at $p$ is
\[ T_p X \defeq \left\{ (v_1, \ldots, v_n) \in
\RR^n ~\Bigg|~ \sum_{i = 1}^n v_i
\pfrac{f}{x_i}(p) = 0 \right\} .\]
This is a vector subspace of $\RR^n$.

\begin{flashcard}[tangent-space-defn]
\begin{definition*}[Tangent space]
  \glsnoundefn{tspace}{tangent space}{tangent
  spaces}
  \glssymboldefn{tspace}{$T_p X$}{$T_p X$}
  \cloze{If $X \subseteq \AA^n$ is an \gls{affvty} with
  $I = \I(X) = \langle f_1, \ldots, f_r \rangle$,
  $f_1, \ldots, f_r \in \KK[X_1, \ldots, X_n]$,
  then we define, for $p \in X$ the tangent space
  to $X$ at $p$ by
  \[ T_p X = \left\{ (v_1, \ldots, v_n) \in \KK^n
  ~\Bigg|~ \sum_{i = 1}^n v_i \pfrac{f_j}{x_i}(f)
  = 0, 1 \le j \le r \right\} .\]
  The derivative is defined using the standard
  differentiation rules for polynomials.}
\end{definition*}
\end{flashcard}

\begin{example*}
  $I = \langle x_2^2 - x_1^3 \rangle \subseteq
  \KK[x_1, x_2]$, $X = \Z(I)$, $p = (a_1, a_2)$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/72468e1e92e4441a.png}
  \end{center}
  \[ \tspace_p X = \{(v_1, v_2) \in \KK^2 \st
  v_1(-3a_1^2) + v_2(2a_2) = 0\} \]
  \[ \dim_{\KK} \tspace_p X = \begin{cases}
    1 & p \neq (0, 0) \\
    2 & p = (0, 0)
  \end{cases} \]
  (assuming $\characteristic \KK \neq 2, 3$).
\end{example*}

\begin{flashcard}[dimension-of-vty-defn]
\begin{definition*}[Dimension of an affine variety]
  \glsnoundefn{avdim}{dimension}{\gls{affvty}}
  \glssymboldefn{avdim}{dim $X$}{dim $X$}
  \cloze{Let $X \subseteq \AA^n$ be an \gls{affvty}. Then
  the \emph{dimension} of $X$ is
  \[ \dim X = \min \{\dim_{\KK} \tspace_p X \st p
  \in X\} .\]
  \glsadjdefn{sing}{singular}{at a point}
  We say $X$ is \emph{singular} at $p$ if
  $\dim_{\KK} \tspace_p X > \dim X$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[dim-tspace-ineq-closed-lemma]
\begin{lemma*}
  $\{p \in X \st \dim_{\KK} \tspace_p X \ge K\}$
  is a closed subset of $X$.
\end{lemma*}
  
\begin{proof}
  \cloze{\[ \tspace_p X = \Ker \ub{\begin{pmatrix}
    \pfrac{f_1}{x_1} & \cdots & \pfrac{f_1}{x_n}
    \\
    \vdots & \ddots & \vdots \\
    \pfrac{f_r}{x_1} & \cdots & \pfrac{f_r}{x_n}
  \end{pmatrix}}_{\KK^n \to \KK^r} \]
  where $\I(X) = \langle f_1, \ldots, f_r \rangle$.
  But $\dim \Ker M + \rank M = n$ (rank-nullity).
  So
  \begin{align*}
    \dim \tspace_p X \ge K
    &\iff n - \rank \ge K \\
    &\iff \rank \le n - K
  \end{align*}
  If $A$ is an $r \times n$ matrix, then $\rank(A)
  \ge k + 1$ if and only if there is a $(k + 1)
  \times (k + 1)$ submatrix of $A$ whose
  determinant is non-zero. So $\rank J \le n - k$
  if and only if all $(n - k + 1) \times (n - k +
  1)$ minors (determinants of $(n - k + 1) \times
  (n - k + 1)$ matrices) vanish. Thus the set:
  \[ \{p \in X \st \dim \tspace_p X \ge k\} =
  \Z(f_1, \ldots, f_r, \text{ all $(n - k + 1)
  \times (n - k + 1)$ minors of $J$}) .\]
  Hence this set is closed.}
\end{proof}
\end{flashcard}