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\begin{flashcard}[irred-iff-prime-prop]
\begin{proposition*}
  Let $\KK$ be algebraically closed. Then an
  \gls{algset} set $X \subseteq \AA^n$ is
  \gls{irredsub} if and only if $\I(X)$ is prime.
\end{proposition*}

\begin{proof}
  \phantom{}
  \begin{enumerate}
    \item[$\Rightarrow$] \cloze{If $f \cdot g \in \I(X)$,
      then $X \subseteq \Z(f \cdot g) = \Z(f) \cup
      \Z(g)$. Thus
      \[ X = (X \cap \Z(f)) \cup (X \cap \Z(f)) \]
      By \glsref[irredsub]{irreducibility} of $X$
      we can assume $X = X \cap \Z(f)$, so $X
      \subseteq \Z(f)$, so $f \in \I(X)$.}
    \item[$\Leftarrow$] \cloze{If $P \subseteq \polyA =
      \KK[X_1, \ldots, X_n]$ is prime, suppose
      $\Z(P) = X_1 \cup X_2$ with $X_1, X_2$
      closed. Then
      \[ \I(X_1) \cap \I(X_2) = \I(X_1 \cup X_2) =
      \I(\Z(P)) = \sqrtideal{P} .\]
      The last equality is by
      \nameref{hilbert_null}. But $\sqrtideal{P} =
      P$: if $f^n \in P$ then $f \in P$ by
      primality of $P$. Thus $\I(X_1) \cap \I(X_2)
      = P$, so by the lemma, $P = \I(X_1)$ or $P =
      \I(X_2)$. Thus $\Z(P) = X_1$ or $\Z(P)
      = X_2$. Thus $\Z(P)$ is \gls{irredsub} and
      $\I(\Z(P)) = P$. \qedhere}
  \end{enumerate}
\end{proof}
\end{flashcard}

We now have a $1-1$ correspondence (if $\KK$ is
algebraically closed):
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/680df3b05ae442d5.png}
\end{center}

\begin{flashcard}[algset-union-of-finite-varieties]
\begin{proposition*}
  Any \gls{algset} set in $\AA^n$ can be written as a
  finite union of \glspl{aavty}.
\end{proposition*}

\begin{proof}
  \cloze{Let $\mathcal{R}$ be the set of all
  \gls{algset} sets in $\AA^n$ which can't be written
  as a finite union of \glspl{aavty}. If
  $\mathcal{R} \neq \emptyset$, I claim it has a
  minimal element. Otherwise there exists $X_1,
  X_2, X_3, \ldots \in \mathcal{R}$ with
  \[ X_1 \supsetneq X_2 \supsetneq X_3 \supsetneq
  \cdots \]
  so
  \[ \I(X_1) \subsetneq \I(X_2) \subsetneq I(X_3)
  \subsetneq \cdots \subseteq \polyA = \KK[X_1,
  \ldots, X_n] \]
  This contradicts $\polyA$ being Noetherian
  (\courseref{GRM}).

  Let $X \in \mathcal{R}$ be minimal. $X$ can't be
  \gls{irredsub}, since then $X$ is itself a
  \gls{aavty}. Otherwise, we can write $X = X_1
  \cup X_2$ with $X_1 \subsetneq X$, $X_2
  \subsetneq X$ with $X_1, X_2$ \gls{algset}. Then
  $X_1, X_2 \notin \mathcal{R}$, hence can be
  written as a union of \gls{irredsub} sets. So $X$
  can also be written as a finite union of
  \glsref[irredsub]{irreducibles}, so $X \notin
  \mathcal{R}$, contradiction.}
\end{proof}
\end{flashcard}

\begin{flashcard}[irred-comps-defn]
\begin{definition*}[Irreducible components]
  \glsnoundefn{irredcomp}{irreducible
  components}{irreducible components}
  \cloze{If $X = X_1 \cup \cdots \cup X_n$ with $X$,
  $X_i$ \gls{algset}, $X_i$ \gls{irredsub} and
  $X_i \not\subseteq X_j$ for any $i \neq j$, then
  we say $X_1, \ldots, X_n$ are the
  \emph{irreducible components} of $X$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item In $\AA^2$, $\polyA = \KK[X_1, X_2]$, $X
      = \ZZ(X_1 \cdot X_2) = \Z(X_1) \cup
      \Z(X_2)$.
    \item More generally, $\polyA = \KK[X_1,
      \ldots, X_n]$ is a UFD. If $0 \neq f \in
      \polyA$, we can write $f = \prod f_i^{a_i}$
      with $f_i$ \gls{irredsub}. Since $\polyA$ is
      a UFD, $\langle f_i \rangle$ is prime. Thus
      $\Z(f_i)$ is \gls{irredsub} (assuming $\KK$
      is algebraically closed). Thus $\Z(f) =
      \Z(f_1) \cup \cdots \cup \Z(f_s)$ is the
      \glsref[irredcomp]{irreducible
      decomposition} of $\Z(f)$.
    \item $\Z(X_2^2 - X_1^3 + X_1)$ is
      \gls{irredsub}.
  \end{enumerate}
\end{example*}

\subsection{Regular and rational functions}

In Algebraic Geometry, polynomial functions are
natural. Let $X \subseteq \AA^n$ be an
\gls{algset} set. $f \in \polyA = \KK[X_1, \ldots,
X_n]$. This gives a function $f : \AA^n \to \KK$,
$(a_1, \ldots, a_n) \mapsto f(a_1, \ldots, a_n)
\in \KK$. Then get $f|_X : X \to \KK$.

If $f, g \in \polyA$, and $f|_X = g|_X$, then $f
\cdot $ vanishes on $X$. So $f \cdot g \in \I(X)$.

So it is natural to think of $\polyA / \I(X)$ as
being the set of polynomial functions on $X$.

\begin{flashcard}[coordinate-ring-defn]
\begin{definition*}[Coordinate ring]
  \glsnoundefn{coordring}{coordinate
  ring}{coordinate rings}
  \glssymboldefn{cring}{$A(X)$}{$A(X)$}\ 
  \cloze{Let $X \subseteq \AA^n$ be an \gls{algset} set.
  The \emph{coordinate ring} of $X$ is
  \[ \polyA(X) \defeq \polyA / \I(X) \]
  (sometimes written $\KK[X]$).}
\end{definition*}
\end{flashcard}

\begin{flashcard}[regular-func-defn]
\begin{definition*}[Regular function]
  \glsadjdefn{regf}{regular}{function}%
  \cloze{Let $X \subseteq \AA^n$ be an \gls{algset} set,
  $U \subseteq X$ an \gls{zzopen} subset. A
  function $f : U \to \KK$ is \emph{regular} if
  $\forall p \in U$, there exists an \gls{zzopen}
  neighbourhood $V \subseteq U$ of $p$ and
  functions $g, h \in \cring A(X)$ with $h(q) \neq
  0$ for any $q \in V$ and $f = \frac{g}{h}$ on
  $V$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  Any $f \in \cring A(X)$ defines a \gls{regf} on
  $X$.
\end{example*}

\begin{flashcard}[Ox-notation]
\begin{notation*}
  \glssymboldefn{OX}{$O_X(U)$}{$O_X(U)$}
  We write
  \[ \mathcal{O}_X(U) \defeq \cloze{\{f : U \to
  \KK \st \text{$f$ is \gls{regf}}\}} .\]
\end{notation*}
\end{flashcard}

\begin{note*}
  $\OX_X(U)$ is a ring if $f, g \in \OX_X(U)$,
  then $f \pm g, f \cdot g \in \OX_X(U)$. This is
  also a \Kalgebra{\KK}.

  \begin{flashcard}[algebra-defn]
  \begin{definition*}[Algebra]
    \glsnoundefn{algebra}{algebra}{algebras}
    \cloze{If $A$, $B$ are rings, then an $A$-algebra
    structure on $B$ is the data of a ring
    homomorphism $\varphi : A \to B$. This turns
    $B$ into an $A$-module via
    \[ a \cdot b \defeq \varphi(a) \cdot b \]
    }
  \end{definition*}
  \end{flashcard}

  So $\KK \to \OX_X(U)$ is given by $a \in \KK$
  being mapped to the constant function with value
  $a$.
\end{note*}