\documentclass{lecture}

\begin{document}

\chapter{Algebraic sets}

\section{Polynomial equations}

Let $k$ be a field.

\begin{definition}
    The \emph{affine space of dimension $n$} is the set $k^{n}$.
\end{definition}

\begin{definition}
    An \emph{algebraic subset} of $k^{n}$ is a subset $V \subseteq k^{n}$ for
    which there exists a subset $A \subseteq k[x_1, \ldots, x_n]$ such
    that
    \begin{salign*}
    V = \{ x \in k^{n}  \mid \forall P \in A\colon P(x) = 0 \}
    .\end{salign*}
    Notation: $V = \mathcal{V}_{k^{n}}(A)$.
\end{definition}

\begin{figure}
    \centering
    \begin{tikzpicture}
        \begin{axis}
        \algebraiccurve[red][$y=x^2-4x+5$]{x^2 - 4*x + 5 - y}
        \end{axis}
    \end{tikzpicture}
    \caption{parabola}
\end{figure}

\begin{figure}
    \centering
    \begin{tikzpicture}
        \begin{axis}
        \algebraiccurve[red][$y^2 = (x+1)x^2$][-1:1][-1:1]{y^2 - (x+1)*x^2}
        \end{axis}
    \end{tikzpicture}
    \caption{nodal cubic}
\end{figure}

\begin{bem}
    If $A \subseteq k[x_1, \ldots, x_n]$ is a subset and $I$ is the ideal generated
    by $A$, then
    \[
    \mathcal{V}(A) = \mathcal{V}(I)
    .\]
\end{bem}

\begin{definition}
    Let $Z \subseteq k^{n}$ be a subset. Define the ideal in $k[x_1, \ldots, x_n]$
    \[
    \mathcal{I}(Z) \coloneqq \{ P \in k[x_1, \ldots, x_n]  \mid \forall x \in Z\colon P(x) = 0\} 
    .\] 
\end{definition}

\begin{bem}
    Since $k[x_1, \ldots, x_n]$ is a Noetherian ring, all ideals are
    finitely generated. For $I \subseteq k[x_1, \ldots, x_n]$ there
    exist polynomials $P_1, \ldots, P_m \in k[x_1, \ldots, x_n]$ such that
    $I = (P_1, \ldots, P_m)$ and
    \[
    \mathcal{V}(I) = \mathcal{V}(P_1, \ldots, P_m) = \mathcal{V}(P_1) \cap \ldots \cap \mathcal{V}(P_m)
    .\] Thus all algebraic subsets of $k^{n}$ are intersections of hypersurfaces.
\end{bem}

\begin{satz}[]
    The maps
    \[
    \mathcal{I}\colon \{ \text{subsets of } k^{n}\} 
    \longrightarrow
    \{\text{ideals in } k[x_1, \ldots, x_n]\}
    \] and
    \[
    \mathcal{V}\colon \{ \text{ideals in } k[x_1, \ldots, x_n] \}
    \longrightarrow
    \{ \text{subsets of } k^{n}\}
    \] satisfy the following properties
    \begin{enumerate}[(i)]
        \item $Z_1 \subseteq Z_2 \implies \mathcal{I}(Z_1) \supseteq \mathcal{I}(Z_2)$
        \item $I_1 \subseteq I_2 \implies \mathcal{V}(I_1) \supseteq \mathcal{V}(I_2)$
        \item $\mathcal{I}(Z_1 \cup Z_2) = \mathcal{I}(Z_1) \cap \mathcal{I}(Z_2)$
        \item $\mathcal{I}(\mathcal{V}(I)) \supseteq I$
        \item $\mathcal{V}(\mathcal{I}(Z)) \supseteq Z$ with equality
            if and only if $Z$ is an algebraic set.
    \end{enumerate}
\end{satz}

\begin{proof}
    Calculation.
\end{proof}

\begin{lemma}
    Let $I, J \subseteq k[x_1, \ldots, x_n]$ be ideals. Then
    \[
    \mathcal{V}(I) \cup \mathcal{V}(J) = \mathcal{V}(IJ)
    \]
    where $IJ$ is the ideal generated by the products $PQ$, where $P \in I$ and $Q \in J$.
    \label{lemma:union-of-alg-sets}
\end{lemma}

\begin{lemma}
    Let $I_j \in k[x_1, \ldots, x_n]$ be ideals. Then
    \[
    \bigcap_{j \in J} \mathcal{V}(I_j) = \mathcal{V}\left( \bigcup_{j \in J} I_j \right) 
    .\] 
    \label{lemma:intersection-of-alg-sets}
\end{lemma}

\section{The Zariski topology}

The algebraic subsets of $k^{n}$ can be used to define a topology on $k^{n}$.

\begin{satz}
    The algebraic subsets of $k^{n}$ are exactly the closed sets of a topology
    on $k^{n}$.
\end{satz}

\begin{proof}
    $\emptyset = \mathcal{V}(1)$ and $k^{n} = \mathcal{V}(0)$. The rest follows from
        \ref{lemma:union-of-alg-sets}
        and \ref{lemma:intersection-of-alg-sets}.
\end{proof}

\begin{definition}
    The topology on $k^{n}$ where the closed sets are exactly the
    algebraic subsets of $k^{n}$, is called the \emph{Zariski topology}.
\end{definition}

\begin{lemma}
    \begin{enumerate}[(i)]
        \item Let $Z \subseteq k^{n}$ be a subset. Then
            \[
            \overline{Z} = \mathcal{V}(\mathcal{I}(Z))
            .\] 
        \item Let $Z \subseteq k^{n}$ be a subset. Then
            \[
            \sqrt{\mathcal{I}(Z)} = \mathcal{I}(Z)
            .\] 
        \item Let $I \subseteq k[x_1, \ldots, x_n]$ be an ideal. Then
            \[
            \mathcal{V}(I) = \mathcal{V}(\sqrt{I})
            .\] 
    \end{enumerate}
\end{lemma}

\begin{proof}
    \begin{enumerate}[(i)]
        \item Let $V = \mathcal{V}(I)$ be a Zariski-closed set such that
            $Z \subseteq V$. Then $\mathcal{I}(Z) \supseteq \mathcal{I}(V)$.
            But $\mathcal{I}(V) = \mathcal{I}(\mathcal{V}(I)) \supseteq I$,
            so $\mathcal{V}(\mathcal{I}(Z)) \subseteq \mathcal{V}(I) = V$.
            Thus
            \[
            \mathcal{V}(\mathcal{I}(Z)) \subseteq \bigcap_{V \text{ closed}, Z \subseteq V} V
            = \overline{Z}
            .\] Since $\mathcal{V}(I(Z))$ is closed, the claim follows.
    \end{enumerate}
\end{proof}

\begin{korollar}
    For ideals $I, J \subseteq k[x_1, \ldots, x_n]$ we have
    \[
    \mathcal{V}(I) \cup \mathcal{V}(J) = \mathcal{V}(IJ) = \mathcal{V}(I \cap J)
    .\] 
\end{korollar}

\begin{proof}
    $\sqrt{I \cap J} = \sqrt{IJ}$
\end{proof}

\begin{satz}
    The Zariski topology turns $k^{n}$ into a Noetherian topological space: If
    $(F_n)_{n \in N}$ is a decreasing sequence of closed sets, then
    $(F_n)_{n \in \N}$ is stationary.
\end{satz}

\begin{proof}
    Let $V_1 \supseteq V_2 \supseteq \ldots$ be a decreasing sequence of closed sets.
    Then $\mathcal{I}(V_1) \subseteq \mathcal{I}(V_2) \subseteq \ldots$
    is an increasing sequence of ideals in $k[x_1, \ldots, x_n]$. As
    $k[x_1, \ldots, x_n]$ is Noetherian, this sequence is stationary. Thus
    there exists $n_0 \in \N$ such that $\forall n \ge n_0$,
    $\mathcal{I}(V_n) = \mathcal{I}(V_{n_0})$. Therefore,
    \[
    V_n = \mathcal{V}(\mathcal{I}(V_n)) = \mathcal{V}(\mathcal{I}(V_{n_0})) = V_{n_0}
    \] for $n \ge n_0$.
\end{proof}

\begin{definition}
    Let $P \in k[x_1, \ldots, x_n]$. The subset
    \[
    D_{k^{n}}(P) \coloneqq k^{n} \setminus \mathcal{V}(P)
    \] is called a \emph{standard} or \emph{principal open set} of $k^{n}$.
\end{definition}

\begin{bem}[]
    Since a Zariski-closed subset of $k^{n}$ is an intersection of finitely many
    $\mathcal{V}(P_i)$, a Zariski-open subset of $k^{n}$ is a union of finitely many
    standard open sets. Thus the standard open sets form a basis for the Zariski topology
    of $k^{n}$.
\end{bem}

\begin{satz}[]
    The affine space $k^{n}$ is quasi-compact in the Zariski topology.
\end{satz}

\begin{proof}
    Let $k^{n} = \bigcup_{i \in J} U_i$ where $U_i$ is open. Since the standard opens form a basis
    of the Zariski topology, we can assume $U_i = D(P_i)$ with $P_i \in k[x_1, \ldots, x_n]$.
    Then
    $\mathcal{V}((P_i)_{i \in I}) = \bigcap_{i \in J} \mathcal{V}(P_i) = \emptyset$. Since
    $k[x_1, \ldots, x_n]$ is Noetherian, we
    can choose finitely many generators $P_{i_1}, \ldots, P_{i_m}$ such that
    $((P_i)_{i \in J}) = (P_{i_1}, \ldots, P_{i_m})$. Thus
    \[
    \bigcap_{j=1}^{m} \mathcal{V}(P_{i_j}) = \mathcal{V}(P_{i_1}, \ldots, P_{i_m})
    = \mathcal{V}((P_i)_{i \in J}) = \emptyset
    .\] By passing to complements in $k^{n}$, we get
    \[
    \bigcup_{j=1}^{m} D(P_{i_j}) = k^{n}
    .\] 
\end{proof}

\begin{satz}[]
    Let $P \in k[x_1, \ldots, x_n]$ and let $f_P \colon k^{n} \to k$ be the associated
    function on $k^{n}$. Then $f_p$ is continuous with respect to the Zariski topology on $k^{n}$
    and $k$.
\end{satz}

\begin{proof}
    The closed proper subsets of $k$ are finite subsets $F = \{t_1, \ldots, t_s\} \subseteq k$.
    The pre-image of a singleton
    $\{t\} \subseteq k$ is
    \[
    f_P^{-1}(\{t\}) = \{x \in k^{n}  \mid P(x) - t = 0\}
     = \mathcal{V}(P - t)
    \]
    which is a closed subset of $k^{n}$. Thus
    \[
    f_P^{-1}(F) = \bigcup_{i=1}^{s} \mathcal{V}(P - t_i)
    \] is closed.
\end{proof}

\begin{satz}
    If $k$ is infinite, $\mathcal{I}(k^{n}) = \{0\} $.
    \label{satz:k-infinite-everywhere-vanish}
\end{satz}

\begin{proof}
    By induction: for $n = 1$, this follows because a non-zero polynomial only has a finite number
    of roots. Let $n \ge 1$ and $P \in \mathcal{I}(k^{n})$. Thus $P(x) = 0$ $\forall x \in k^{n}$.
    Let
    \[
    P = \sum_{i=0}^{m} P_i(X_1, \ldots, X_{n-1}) X_n^{i}
    \] for $P_i \in k[X_1, \ldots, X_{n-1}]$.
    Fix some $x_1, \ldots, x_{n-1} \in k$. Then $P(x_1, \ldots, x_{n-1}, y) \in k[y]$ has an
    infinite number
    of roots. Thus $P(x_1, \ldots, x_{n-1}, y) = 0$ for all $x_2, \ldots, x_n$ by the case $n=1$,
    implying that $P_i(x_1, \ldots, x_{n-1}) = 0$ for all $i$.
    Since this holds for all $(x_1, \ldots, x_{n-1}) \in k^{n-1}$, $P_i = 0$ by induction
    for all $i$.
\end{proof}

\end{document}