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\begin{flashcard}[Tp-mp-mp-mod-thm]
\begin{theorem*}
  If $X \subseteq \AA^n$ is an \gls{affvty} then
  $\tspace_p X \cong (m_p / m_p^2)^*$ where $V^*$
  is the dual of the $\KK$-vector space $V$.
\end{theorem*}

\begin{proof}
  \cloze{Note that there is an isomorphism
  \begin{align*}
    \OXp_{X, p} / m_p
    &\to \KK \\
    f
    &\mapsto f(p)
  \end{align*}
  This map is surjective since constants are
  regular functions, and injective by definition
  of $m_p$. Thus we can define the $\KK$-vector
  space on $m_p / m_p^2$ by identifying $\KK$ with
  $\OXp_{X, p} / m_p$, and
  \[ (f + m_p) \cdot (g + m_p^2) = (f \cdot g +
  m_p^2) .\]
  We will show $\Der(\cring A(X), p) \cong (m_p /
  m_p^2)^*$. Given $D \in \Der(\cring A(X), p)$,
  we define
  \[ \varphi_D : m_p / m_p^2 \to \KK \]
  defined as follows: for $f, g \in \cring A(X)$,
  $g(p) \neq 0$, $f(p) = 0$, $(X \setminus \Z(g),
  \frac{f}{g}) \in m_p \OXp_{X, p}$. Set
  \begin{align*}
    \varphi_D \left( \frac{f}{g} \right)
    &= \text{``} D \left( \frac{f}{g} \right)
    \text{''} \\
    &= \frac{g(p) D(f) - f(p) D(g)}{g(p)^2} \\
    &= \frac{D(f)}{g(p)}
  \end{align*}
  since $f(p) = 0$. Note if $\frac{f_1}{g_1},
  \frac{f_2}{g_2} \in m_p$, then
  \[ \varphi_D \left( \frac{f_1 f_2}{g_1 g_2}
  \right) = \frac{f_2(p)}{g_2(p)} \cdot \varphi_D
  \left( \frac{f_1}{g_1} \right) +
  \frac{f_1(p)}{g_1(p)} \varphi_D \left(
  \frac{f_2}{g_2} \right) = 0 .\]
  Thus $\varphi_D(m_p^2) = 0$, so we obtain a
  well-defined map $\varphi_D : m_p / m_p^2 \to
  \KK$.

  Conversely, if given $\varphi : m_p / m_p^2 \to
  \KK$, $p = (a_1, \ldots, a_n) \in X \subseteq
  \AA^n$. Note $x_i - a_i \in m_p$ for all $i$,
  and we define
  \[ D_\varphi(x_i - a_i) = \varphi(x_i - a_i) .\]
  This is sufficient to determine $D_\varphi$ as
  before.}
\end{proof}
\end{flashcard}

\begin{example*}
  Suppose $X = \AA^n$, $p = 0$. Then
  \[ m_p / m_p^2 = \ub{(x_1, \ldots,
  x_n)}_{\subseteq \KK[x_1, \ldots, x_n]} / (x_1,
  \ldots, x_n)^2 \]
  (exercise).
\end{example*}

\begin{flashcard}[zariski-tangent-space]
\begin{definition*}[Zariski tangent space]
  \glssymboldefn{zspace}{$T_p X$}{$T_p X$}
  \cloze{If $X$ is any \gls{vty}, and $p \in X$, then the
  \emph{Zariski tangent space} to $X$ at $p$ is
  \[ T_p X = (m_p / m_p^2)^* ,\]
  where $m_P \subseteq \OXp_{X, p}$ is the maximal
  ideal.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[open-cover-aff-vty-thm]
\begin{theorem*}
  Any \gls{vty} has an open cover by
  \glspl{affvty} (i.e. open subsets isomorphic to
  \glspl{affvty}).
\end{theorem*}

\fcscrap{
\begin{note*}
  If $X \subseteq \P^n$ is projective, $\{U_i \cap
  X \st 0 \le i \le n\}$ ($U_i = \P^n \setminus
  \Z(x_i)$) is a cover of $X$ by affines.
\end{note*}
}

\begin{proof}
  \cloze{Consider the most general case where $X$ is a
  \gls{qprojvty}, $X \subseteq \P^n$. Each $U_i
  \cap X$ is a \gls{qaffvty}. So enough to show
  each \gls{qprojvty} is covered by
  \glspl{affvty}. Let $p \in X \subseteq \AA^n$.
  Will find an affine neighbourhood of $p$ in $X$.
  Then $\ol{X} \subseteq \AA^n$, the closure, is
  an \gls{affvty}, and $Z = \ol{X} \setminus X$ is
  closed in $\ol{X}$. Choose $f \in \I(Z)$ with
  $f(p) \neq 0$. Then $\langle f \rangle \subseteq
  \I(X)$, so $\Z(f) \supseteq \Z(\I(Z)) = Z$, so
  $p \in \ol{X} \setminus \Z(f) \subseteq \ol{X}
  \setminus Z = X$. But $\ol{X} \setminus \Z(f)$
  can be identified with the closed subset of
  $\AA^{n + 1}$ given by $\Z(\I(\ol{X}), yf - 1)$
  as in \es{1}.}
\end{proof}
\end{flashcard}

\begin{remark*}
  The definition of dimension of singular points
  goes through unchanged with the Zariski tangent
  space.
  \[ \dimX X = \inf \{\dim \zspace_p X \st p \in
  X\} .\]
  $p \in X$ is \gls{sing} if $\dimX X < \dim
  \zspace_p X$. By applying the above theorem, in
  fact the set of \gls{sing} points of an
  arbitrary \gls{vty} $X$ is closed in $X$. This
  also shows dimension and singularity are
  intrinsic to $X$.
\end{remark*}

\subsubsection*{Alternative definitions of
dimension (we won't prove stuff here)}

\begin{flashcard}[transcendence-deg-defn]
\begin{definition*}[Transcendence degree]
  \cloze{If $F / \KK$ is a finitely generated field
  extension, then the \emph{transcendence degree}
  of $F / \KK$, written $\Trdeg_{\KK} F$ is the
  cardinality of any \gls{transc_basis}.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[krull-dimension-defn]
\begin{definition*}[Krull dimension of a ring]
  \glsadjdefn{kdimr}{Krull dimension}{ring}
  \cloze{If $A$ is a ring, the \emph{Krull dimension} of
  $A$ is the largest $n$ such that there exists a
  chain of prime ideals
  \[ P_0 \subsetneq P_1 \subsetneq \cdots
  \subsetneq P_n \subseteq A .\]}
\end{definition*}
\end{flashcard}

\begin{flashcard}[krull-dimension-top-defn]
\begin{definition*}[Krull dimension of a
  topological space]
  \glsadjdefn{kdimt}{Krull dimension}{topological
  space}
  \cloze{If $X$ is a topological space, the \emph{Krull
  dimension of $X$} is the largest $n$ such that
  there exists a chain of irreducible subsets
  \[ Z_0 \subsetneq Z_1 \subsetneq \cdots
  \subsetneq Z_n \subseteq X .\]}
\end{definition*}
\end{flashcard}