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\textbf{Exercise:} Check that this gives a $1-1$
correspondence. We checked
\[ \sh f \mapsto f \mapsto \sh f ,\]
so it remains to check that
\[ f \mapsto \sh f \mapsto f .\]

\begin{moral*}
  A \gls{morph} $f : \AA^n \supseteq X \to Y
  \subseteq \AA^m$ is given by choosing polynomial
  functions $f_1, \ldots, f_m \in \KK[X_1, \ldots,
  X_n]$ and defining $f$ by
  \[ f(p) = (f_1(p), \ldots, f_m(p)) .\]
\end{moral*}

\begin{example*}
  \begin{align*}
    f : \AA^1
    &\to \AA^2 \\
    t
    &\mapsto (t, t^2)
  \end{align*}
  The image of this map is $Y = \Z(X^2 - Y)$. This
  defines a \gls{morph} $f : \AA^1 \to Y$. Then
  \begin{align*}
    \sh f : \frac{\KK[X, Y]}{(Y - X^2)}
    &\to \KK[t] \\
    X
    &\mapsto t \\
    Y
    &\mapsto t^2
  \end{align*}
  This is an isomorphism!
\end{example*}

\begin{flashcard}[isomorphic-affvty]
\begin{definition*}[Isomorphism of affine varieties]
  \glsadjdefn{affvtyisic}{isomorphic}{\gls{affvty}}
  \glsnoundefn{isism}{isomorphism}{isomorphisms}
  \cloze{Two \glspl{affvty} are \emph{isomorphic} if
  there exist \glspl{morph} $f : X \to Y$, $g : Y
  \to X$ such that $g \circ f = \id_X$, $f \circ g
  = \id_Y$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[affvty-isomorphic-iff-thm]
\begin{theorem*}
  If $X, Y$ are \glspl{affvty}, then $X$ is
  \gls{affvtyisic} to $Y$ if and only if \cloze{$\cring
  A(X) \cong \cring A(Y)$ as \Kalgebra{\KK}.}
\end{theorem*}
\end{flashcard}

\begin{example*}
  $\AA^1 \cong \Z(X^2 - Y) \subseteq \AA^2$.
\end{example*}

\begin{remark*}
  A \Kalgebra{\KK} $\AA$ is \emph{finitely
  generated} if there exists a surjective
  \Kalgebra{\KK} homomorphism
  \begin{align*}
    \KK[X_1, \ldots, X_n]
    &\to \AA \\
    X_i
    &\mapsto a_i
  \end{align*}
  i.e. every element of $\polyA$ can be written as a
  polynomial in $a_1, \ldots, a_n$ with
  coefficients in $\KK$. If $I$ is the kernel of
  this map then
  \[ \polyA \cong \KK[X_1, \ldots, X_n] / I .\]
  SUppose further that $\polyA$ is an integral
  domain. Then $I$ is a prime ideal of $\KK[X_1,
  \ldots, X_n]$, so
  \[ A = \cring A(X) \]
  where $X = \Z(I)$.
\end{remark*}

\newpage
\section{The proof of Hilbert's Nullstellensatz}

\textbf{Goal:} We want to prove
\nameref{hilbert_null}. That is, if $\KK$ is algebraically closed,
we want to show $\I(\Z(I)) = \sqrtideal{I}$.

\begin{flashcard}[transcendental-over-K-defn]
\begin{definition*}[Transcendental]
  \glsadjdefn{transc}{transcendental}{element}
  \cloze{Let $F / \KK$ be a field extension. We say an
  element $z \in F$ is \emph{transcendental} over
  $\KK$ if it is not algebraic, i.e. $\nexists f
  \in \KK[X]$ with $f \neq 0$, $f(z) = 0$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[alg-indep-defn]
\begin{definition*}[Algebraically independent
  elements]
  \glsadjdefn{algindep}{algebraically
  independent}{elements}
  \cloze{We say $z_1, \ldots, z_d \in F$ are
  \emph{algebraically independent over $\KK$} if
  $\nexists f \in \KK[X_1, \ldots, X_d]$ with $f
  \neq 0$, $f(z_1, \ldots, z_d) = 0$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[transcendence-basis-defn]
\begin{definition*}[Transcendence basis]
  \glsnoundefn{transc_basis}{transcendence
  basis}{transcendence bases}
  \cloze{A \emph{transcendence basis} for $F / \KK$ is a
  set $z_1, \ldots, z_d \in F$ \gls{algindep} over
  $\KK$ and such that $F$ is algebraic over
  $\KK(z_1, \ldots, z_d)$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  If $X$ is a \gls{aavty}, $\K(X)$ is a field
  extension of $\KK$, and it usually has lots of
  \glsref[transc]{transcendentals}.
  \begin{align*}
    \K(\AA^n)
    &= \left\{ \frac{f}{g} \st f, g \in \KK[X_1,
    \ldots, X_n], g \neq 0 \right\} / \sim \\
    &= \KK(X_1, \ldots, X_n) \\
    &= \text{field of rational functions in $X_1,
    \ldots, X_n$}
  \end{align*}
  $X_1, \ldots, X_n$ form a \gls{transc_basis}.
\end{example*}

\begin{flashcard}[finitely-generated-field-extension]
\begin{definition*}[Finitely generated field
  extension]
  \glsadjdefn{fgfe}{finitely generated}{field
  extension}
  \cloze{If $F / \KK$ is a field extension, we say $F$ is
  \emph{finitely generated} over $\KK$ if $F =
  \KK(z_1, \ldots, z_n)$ for some $z_1, \ldots,
  z_n \in F$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  $\K(X) / \KK$ is finitely generated. If $X
  \subseteq \AA^n$, then $\K(X)$ is the fraction
  field of $\cring A(X) = \KK[X_1, \ldots, X_n] /
  I$ and hence $K(X)$ is generated by the images
  of $X_1, \ldots, X_n$.
\end{example*}

\begin{flashcard}[fgfe-has-transc-basis-prop]
\begin{proposition*}
  Every \gls{fgfe} field extension $F / \KK$ has a
  \gls{transc_basis}, and any two
  \glspl{transc_basis} have the same number of
  elements.

  Further, if $F = \KK(z_1, \ldots, z_N)$, then
  there is a \gls{transc_basis} $\{Y_1, \ldots,
  Y_n\} \subseteq \{z_1, \ldots, z_N\}$.
\end{proposition*}

\begin{proof}
  \cloze{Write $F = \KK(z_1, \ldots, z_N)$. If $z_1,
  \ldots, z_n$ are \gls{algindep}, then $z_1,
  \ldots, z_n$ is a \gls{transc_basis}. If $z_1,
  \ldots, z_N$ are algebraic over $\KK$, then the
  \gls{transc_basis} can be taken to be
  empty. Otherwise, assume $\{z_1, \ldots, z_d\}$
  is a maximal subset of algebraically independent
  elements of $\{z_1, \ldots, z_n\}$. I claim
  $z_1, \ldots, z_d$ is a \gls{transc_basis},
  i.e. $F$ is algebraic over $\KK(z_1, \ldots,
  z_d)$. It is neough to show $z_j$ is algebraic
  of $\KK(z_1, \ldots, z_d)$ for any $j > d$.

  By assumption, $z_1, \ldots, z_d, z_j$ are
  \emph{not} algebraically independent, i.e.
  there exists $f_j \in \KK[X_1, \ldots, X_d, X_j]$
  such that $f_j(z_1, \ldots, z_d, z_j) = 0$.

  Write $f_j = \sum_I f_{ji}(X_1, \ldots, X_d)
  X_j^i$. Then
  \[ 0 \neq f_j(z_1, \ldots, z_d, X) = \sum_i f_{ji}(z_1,
  \ldots,z_d) X^i \in \KK(z_1, \ldots,
  z_d)[X] \]
  (the polynomial is non-zero since $z_1, \ldots,
  z_d$ are \gls{algindep}). Then since $f_j(z_1,
  \ldots, z_d, z_j) = 0$, we have $z_j$ algebraic
  over $\KK(z_1, \ldots, z_d)$. Thus $f$ is
  algebraic over $\KK(z_1, \ldots, z_d)$, so $z_1,
  \ldots, z_d$ is a \gls{transc_basis}. Now
  suppose $z_1, \ldots, z_d$ and $w_1, \ldots,
  w_e$ are both \glspl{transc_basis}. Suppose $d
  \le e$. First $w_1$ is algebraic over $\KK(z_1,
  \ldots, z_d)$ since $w_1 \in F$. Then there
  exists a polynomial $f \in \KK[X_1, \ldots, X_d,
  D_{d + i}]$ such that $f(z_1, \ldots, z_d, w_1)
  = 0$. This is obtained by clearing denominators
  of a polynomial $g \in \KK(z_1, \ldots,
  z_d)[X_{d + 1}]$ with $g(w_1) = 0$. Since $w_1$
  is not algebraic over $\KK$, $f$ must involve at
  least one of $X_1, \ldots, X_d$, say $X_1$. Thus
  $z_1$ is algebraic. So $z_1$ is algebraic over
  $\KK(w_1, z_2, \ldots, z_d)$ (as witnessed by
  $f$). So $F$ is algebraic over $\KK(w_1, z_2,
  \ldots, z_d)$. Repeat: $w_2$ is algebraic over
  $\KK(w_1, \ldots, z_2, \ldots, z_d)$ and not
  algebraic over $\KK(w_1)$. So one can find $0
  neq g \in \KK[X_1, \ldots, X_{d + 1}]$ such that
  $g(w_1, z_2, \ldots, z_d, w_2) = 0$ and further
  $g$ involves one of $X_2, \ldots, X_d$: say it
  involves $X_2$. Thus $z_2$ is algebraic over
  $\KK(w_1, w_2, z_3, \ldots, z_d)$. Continuing,
  eventually we find $F$ is algebraic over
  $\KK(w_1, \ldots, w_d)$. But if $e > d$, then
  $w_e$ is algebraic over $\KK(w_1, \ldots, w_d)$,
  contradicting $w_1, \ldots, w_e$ being a
  \gls{transc_basis}.}
\end{proof}
\end{flashcard}