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\begin{flashcard}[ideal-of-X-defn]
\begin{definition*}[Ideal of a set]
  \glssymboldefn{polyideal}{$I(X)$}{$I(X)$}
  \cloze{
  Let $X \subseteq \AA^n$ be a subset. Define
  \[ I(X) = \{f \in A = \KK[X_1, \ldots, X_n] \st
  f(p) = 0 ~\forall p \in X\} .\]}
\end{definition*}
\end{flashcard}

\begin{remark*}
  $I(X)$ is an ideal: if $f, g \in I(X)$, then $f
  + g \in I(X)$. If $f \in A$, $g \in I(X)$ then
  $f \cdot g \in I(X)$.
\end{remark*}

\begin{remark*}
  If $S_1 \subseteq S_2 \subseteq \polyA$, then
  $\Z(S_2) \subseteq \Z(S_1)$. If $X_1 \subseteq
  X_2 \subseteq \AA^n$, then $I(X_2) \subseteq
  I(X_1)$.
\end{remark*}

\vspace{-1em}
\textbf{Question:} Given an ideal $I$, what is the
relationship between $I$ and $\I(\Z(I))$?

\begin{example*}
  $I = \langle x^2 \rangle \le \KK[X]$.
  \[ \Z(I) = \{0\} \subseteq \AA^1, \qquad
  \I(\Z(I)) = \I(\{0\}) = \langle X \rangle \neq I
  .\]
\end{example*}

\begin{flashcard}[radical-of-ideal]
\begin{definition*}[Radical of an ideal]
  \glssymboldefn{radicalideal}{$sqrtI$}{$sqrtI$}
  \cloze{Let $I \subseteq A$ be an ideal in a commutative
  ring $A$. The \emph{radical} of $I$ is
  \[ \sqrt{I} \defeq \{f \in A \st \text{$f^n \in
  I$ for some $n > 0$}\} .\]}
\end{definition*}
\end{flashcard}

\begin{lemma*}
  $\sqrtideal{I}$ is an ideal.
\end{lemma*}

\begin{proof}
  Suppose $f, g \in \sqrtideal{I}$, say $f^{n_1},
  g^{n_2} \in I$. Then
  \[ (f + g)^{n_1 + n_2 + 1} = \sum_{i = 0}^{n_1 +
  n_2 + 1} {n_1 + n_2 + 1 \choose i} f^i g^{n_1 +
  n_2 + 1 - i} \]
  For each $i$, either $i \ge n_1$ or $(n_1 + n_2
  + 1) - i \ge n_2$. So each term lies in $I$,
  hence $(f + g)^{n_1 + n_2 + 1} \in I$. Thus $f +
  g \in \sqrtideal{I}$. If $f \in \sqrtideal{I}$,
  $g \in A$, then $f^n \in I$ for some $n$, so
  $(fg)^n \in I$ so $fg \in \sqrtideal{I}$.
\end{proof}

\begin{flashcard}[algeset-IZI-is-I-prop]
\begin{proposition*}
  \phantom{}
  \begin{enumerate}[(a)]
    \item If $X \subseteq \AA^n$ is \gls{algset},
      then
      \[ \Z(\I(X)) = X .\]
    \item If $I \subseteq A$ is an ideal, then
      \[ \sqrtideal{I} \subseteq \I(\Z(I)) .\]
  \end{enumerate}
\end{proposition*}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(a)]
    \item \cloze{Since $X$ is \gls{algset}, $X = \Z(I)$
      for some $I$. Certainly, $I \subseteq
      \I(X)$ by definition of $Z$ and $\I(X)$.
      Thus $\Z(\I(X)) \subseteq \Z(I) = X$. But $X
      \subseteq \Z(\I(X))$ is obvious.}
    \item \cloze{Let $f^n \in I$. Then $f^n$ vanishes on
      $\Z(I)$, and hence $f$ vanishes on $\Z(I)$
      also. So $f \in \I(\Z(I))$, hence
      $\sqrtideal{I} \subseteq \I(\Z(I))$.}
      \qedhere
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{flashcard}[hilbert-null-thm]
\begin{theorem*}[Hilbert's Nullstellensatz]
  \label{hilbert_null}
  \cloze{Let $\KK$ be an algebraically closed field. Then
  \[ \sqrtideal{I} = \I(\Z(I)) .\]}
\end{theorem*}
\end{flashcard}

\begin{proof}
  Later.
\end{proof}

\begin{example*}
  $\KK = \RR$. $I = \langle X^2 + Y^2 + 1 \rangle
  \subseteq \RR[X, Y]$. Then $\Z(I) = \emptyset$,
  $\I(\Z(I)) = \RR[X, Y] \neq \sqrtideal{I}$.
\end{example*}

\subsection{Irreducible Subsets}

\begin{flashcard}[irreducible-subset-of-topology-defn]
\begin{definition*}[Irreducible subset]
  \glsadjdefn{irredsub}{irreducible}{subset of a
  topological space}
  \cloze{Let $X$ be a topological space, and $Z \subseteq
  X$ a closed subset. We say $Z$ is
  \emph{irreducible}, if $Z$ is non-empty, and
  whenever $Z = Z_1 \cup
  Z_2$ with $Z_1, Z_2$ closed in $X$, then
  either $Z = Z_1$ or $Z = Z_2$.}
\end{definition*}
\end{flashcard}

\begin{remark*}
  Bad notion in the Euclidean topology in $\CC^n$.
  Only \gls{irredsub} sets are points.
\end{remark*}

\begin{example*}
  $\AA^1$ is \gls{irredsub} as long as $\KK$ is
  infinite.
\end{example*}

\begin{flashcard}[affine-algebraic-variety-defn]
\begin{definition*}[Affine algebraic variety]
  \glsnoundefn{aavty}{variety}{varieties}%
  \glsnoundefn{affvty}{affine variety}{affine
  varieties}
  \cloze{An \emph{(affine algebraic) variety} in $\AA^n$
  is an \gls{irredsub} \gls{algset} set.}
\end{definition*}
\end{flashcard}

\vspace{-1em}
How do we recognize \gls{irredsub} \gls{algset}
sets algebraically?

\begin{flashcard}[ideal-of-union-prop]
\begin{proposition*}
  If $X_1, X_2 \subseteq \AA^n$, then
  \[ \I(X_1 \cloze{\cup} X_2) = \I(X_1)
  \cloze{\cap} \I(X_2) \]
\end{proposition*}

\begin{proof}
  \cloze{Since $X_1, X_2 \subseteq X_1 \cup X_2$, we have
  $\I(X_1 \cup X_2) \subseteq \I(X_1), \I(X_2)$,
  so $\I(X_1 \cup X_2) \subseteq \I(X_1) \cap
  \I(X_2)$.

  Conversely, if $f \in \I(X_1) \cap \I(X_2)$,
  then $f \in \I(X_1 \cup X_2)$.}
\end{proof}
\end{flashcard}

\textbf{Recall} (from \courseref{GRM})\textbf{:}
An ideal $P \subseteq R$ is \emph{prime} if $P
\neq R$ and whenever $f \cdot g \in P$, either
$f \in P$ or $g \in P$.

\begin{flashcard}[ideal-intersection-in-prime-ideal-lemma]
\begin{lemma*}
  Let $P \subseteq A$ be a prime ideal, and let
  $I_1, \ldots, I_n \subseteq A$ be ideals.
  Suppose \cloze{$P \supseteq \bigcap_i I_i$. Then $p
  \supseteq I_i$ for some $i$. In particular, if
  $p = \bigcap_i I_i$, then $P = I_i$ for some
  $i$.}
\end{lemma*}

\cloze{
\begin{example*}
  $A = \ZZ$, $P = \langle p \rangle$, $p$ a
  prime number. Let $I_i = \langle n_i \rangle$.
  Then
  \[ \bigcap_i I_i = \langle \lcm(n_1, \ldots,
  n_s) \rangle \] 
  Then $P \supseteq \bigcap_i I_i \iff p \mid
  \lcm(n_1, \ldots, n_s)$, and the condition on
  the right implies that $p \mid n_i$ for some
  $i$.
\end{example*}
}

\begin{proof}
  \cloze{Suppose $P \not\supseteq I_i$ for any $i$. Thus
  we can find $x_i \in I_i$, $x_i \notin P$. Then
  \[ \prod_{i = 1}^n x_i \in \cap_{i = 1}^n I_i
  \subseteq P ,\]
  so there exists $i$ with $x_i \in P$, a
  contradiction.

  If $P = \bigcap_i I_i$, $P \subseteq I_i$ for
  each $i$ and since we know $I_i \subseteq P$ for
  some $i$, we have $P = I_i$ for that $i$.}
\end{proof}
\end{flashcard}

\begin{proposition*}
  Let $\KK$ be algebraically closed. Then an
  \gls{algset} set $X \subseteq \AA^n$ is
  \gls{irredsub} if and only if $\I(X) \subseteq
  \polyA = \KK[X_1, \ldots, X_n]$ is prime.
\end{proposition*}
