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\begin{flashcard}[null-I-thm]
\begin{theorem*}[Nullstellensatz I]
  \label{null_I}
  \cloze{Let $\KK$ be algebraically closed. Then any
  maximal ideal $M \subset \KK[X_1, \ldots, X_n]$
  is of the form
  \[ M = \langle X_1 - a_1, \ldots, X_n - a_n
  \rangle \]
  for some $a_1, \ldots, a_n \in \KK$.}
\end{theorem*}

\begin{proof}
  \cloze{Note we have an isomorphism
  \begin{align*}
    \frac{\KK[X_1, \ldots, X_n]}{\langle X_1 -
    a_1, \ldots, X_n - a_n \rangle}
    &\stackrel{\cong}{\longrightarrow} \KK \\
    X_i
    &\longmapsto a_i
  \end{align*}
  Recall $M \subseteq \polyA$ is a maximal ideal
  if and only if $\polyA / M$ is a field. Thus
  $\langle X_1 - a_1, \ldots, X_n - a_n \rangle$
  is a maximal ideal.

  Conversely, let $M \subseteq \KK[X_1, \ldots,
  X_n]$ be a maximal ideal. Then
  \[ \frac{\KK[X_1, \ldots, X_n]}{M} \cong F \]
  for some field $F$ which is finitely generated
  as a \Kalgebra{\KK} by $X_1, \ldots, X_n$. Thus
  $F = \KK$ by \nameref{hn_lemma4}. We thus have
  an isomorphism
  \[ \varphi : \frac{\KK[X_1, \ldots, X_n]}{M}
  \stackrel{\cong}{\longrightarrow} \KK .\]
  Let $a_i = \varphi(X_i)$. Then
  \[ \varphi(X_i - a_i) = \varphi(X_i) - a_i = a_i
  - a_i = 0 .\]
  Thus $X_i - a_i \in M$ for each $i$. So
  \[ \langle X_1 - a_1, \ldots, X_n - a_n \rangle
  \subseteq M .\]
  But we have already seen that $\langle X_1 -
  a_1, \ldots, X_n - a_n \rangle$ is maximal, so
  we must in fact have equality.}
\end{proof}
\end{flashcard}

\begin{example*}
  $\langle X^2 + 1 \rangle \le \RR[X]$ is a
  maximal ideal, but $\langle X^2 + 1 \rangle \neq
  \langle X - a \rangle$ for any $a \in \RR$.
\end{example*}

\begin{flashcard}[null-II-thm]
\begin{theorem*}[Nullstellensatz II]
  \label{null_II}
  \cloze{Let $\KK$ be algebraically closed, and $I =
  \langle f_1, \ldots, f_r \rangle \subseteq
  \KK[X_1, \ldots, X_n]$. Then either:
  \begin{enumerate}[(1)]
    \item $I = \KK[X_1, \ldots, X_n]$, or
    \item $\Z(I) \neq \emptyset$.
  \end{enumerate}}
\end{theorem*}

\begin{proof}
  \cloze{Suppose $1 \notin I$, i.e. not in case (1). Then
  there exists a maximal ideal $M \subseteq
  \KK[X_1, \ldots, X_n]$ with $I \subseteq M$.
  Thus $\Z(M) \subseteq \Z(I)$. Then by
  \nameref{null_I}, $M = \langle
  X_1 - a_1, \ldots, X_n - a_n \rangle$, and hence
  $\Z(M) = \{(a_1, \ldots, a_n)\}$. So $\Z(M) \neq
  \emptyset$, so $\Z(I) \neq \emptyset$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[null-III-thm]
\begin{theorem*}[Nullstellensatz III]
  \label{null_III}
  \cloze{Let $\KK$ be algebraically closed, $I \subseteq
  \KK[X_1, \ldots, X_n]$ an ideal. Then
  \[ \I(\Z(I)) = \sqrtideal{I} .\]}
\end{theorem*}

\begin{proof}
  \cloze{$\sqrtideal{I} \subseteq \I(\Z(I))$ in any
  event.

  Let $g \in \KK[X_1, \ldots, X_n]$. Define
  \[ V_g = \Z(Z g(X_1, \ldots, X_n) - 1) \subseteq
  \AA^{n + 1} \]
  with coordinates $X_1, \ldots, X_n, Z$.
  Projecting $V_g$ via $(X_1, \ldots, X_n, Z)
  \mapsto (X_1, \ldots, X_n)$ gives the set
  \[ D(g) \defeq \AA^n \setminus \Z(g) .\]
  Now suppose $g \in \I(\Z(I))$. Then $D(g) \cap
  \Z(I) = \emptyset$. If $I = \langle f_1, \ldots,
  f_r \rangle$, consider $J = \langle f_1, \ldots,
  f_r, Zg - 1 \rangle \subseteq \KK[X_1, \ldots,
  X_n, Z]$. Then $\Z(J) = \emptyset$, so $J =
  \KK[X_1, \ldots, X_n, Z]$ by \nameref{null_II}.

  Thus we can write
  \[ 1 = \sum_i h_i(X_1, \ldots, X_n, Z) f_i(X_1, \ldots, X_n)
  + h(X_1, \ldots, X_n, Z) (g(X_1, \ldots, X_n) Z - 1) \]
  with $h_i, h \in \KK[X_1, \ldots, X_n, Z]$.
  Substitute $Z = \frac{1}{g}$. We get
  \[ 1 = \sum_i h_i\left(X_1, \ldots, X_n,
  \frac{1}{g}\right) f_i(X_1, \ldots, X_n) .\]
  Multiplying by a high power of $g$ clears
  denominators, giving:
  \[ g^N = h_i'(X_1, \ldots, X_n) f_i \in I ,\]
  for some $h_i'$. Thus $g^n \in I$, so $g =
  \sqrtideal{I}$.}
\end{proof}
\end{flashcard}

Recall we need the proof of:

\begin{flashcard}[OX-equals-AX-prop]
\begin{proposition*}
  If $X \subseteq \AA^n$ is an \gls{affvty}, then
  $\OX_X(X) = \cring A(X)$.
\end{proposition*}

\fcscrap{
\begin{lemma*}
  Let $f, g : X \to \KK$ be \gls{regf} functions
  on $X$ an \gls{affvty}, and suppose there exists
  \gls{zzopen} $U \subseteq X$ non-empty with $f|_U = g|_U$.
  Then $f = g$.
\end{lemma*}

\begin{proof}
  Consider the map $\varphi = (f, g) : X \to
  \AA^2$. This is a morphism (exercise: check
  this!). Let $\Delta = \{(a, a) \in \AA^2 \st a
  \in \KK\}$, $\Delta = \Z(X - Y)$. Since
  $\varphi$ is continuous, $\varphi^{-1}(\Delta)$
  is closed. But $U \subseteq
  \varphi^{-1}(\Delta)$, and $U$ is a dense subset
  of $X$ (otherwise $X = \ol{U} \cup X \setminus
  U$ is a union of two proper closed subsets,
  violating irreducibility of $X$). Thus $U
  \subseteq \ol{U} = X \subseteq
  \varphi^{-1}(\Delta)$, so $\varphi^{-1}(\Delta)
  = X$.
\end{proof}
}

\begin{proof}[Proof of Proposition]
  \cloze{We know $\cring A(X) \subseteq \OX_X(X)$. So let $f :
  X \to \KK$ be a \gls{regf} function. So there
  exists an open cover $\{U_i\}$ of $X$ with $f$
  is given on $U_i$ by $f|_{U_i} =
  \frac{g_i}{h_i}$, with $g_i, h_i \in \cring A(X)$ and
  $h_i$ nowhere vanishing on $U_i$. Then
  \[ \Z(\{h_i\}) = \bigcap_i \Z(h_i)
  \le \bigcap_i X \setminus U_i = X \setminus
  \bigcup_i U_i = \emptyset .\]
  Thus $\Z(\{h_i\}) = \emptyset$. Thus we can find
  $e_i \in \cring A(X)$ (Remark: Pull back to
  $\KK[X_1, \ldots, X_n]$ and \nameref{null_II} to
  see this) such that $1 = \sum_i e_i h_i$. Note
  on $U_i \cap U_j$, $\frac{g_i}{h_i} =
  \frac{g_j}{h_j}$, so $g_i h_j = g_j h_i$ on $U_i
  \cap U_j$, so by the Lemma, $g_i h_j = g_j h_i$
  on $X$. Thus $\frac{g_i}{h_i} = \frac{g_j}{h_j}$
  in $\K(X)$. Thus we have the equality in $\K(X)$
  \[ f = \sum_i (e_i h_i) \left( \frac{g_i}{h_i}
  \right) = \sum_i g_i \in \cring A(X) \qedhere \]
  }
\end{proof}
\end{flashcard}
