update rav

.gitignore

@@ -17,3 +17,5 @@
 *.snm
 *.bbl
 *.blg
+*.table
+*.gnuplot

ws2022/rav/lecture/lecture.cls

@@ -1,6 +1,7 @@
 \ProvidesClass{lecture}
 \LoadClass[a4paper]{book}
 
+\RequirePackage{xparse}
 \RequirePackage{stmaryrd}
 \RequirePackage[utf8]{inputenc}
 \RequirePackage[T1]{fontenc}
@@ -33,16 +34,42 @@
 \usetikzlibrary{quotes, angles, math}
 \pgfplotsset{
     compat=1.15,
-    default 2d plot/.style={%
-        grid=both,
-        minor tick num=4,
-        grid style={line width=.1pt, draw=gray!10},
-        major grid style={line width=.2pt,draw=gray!50},
-        axis lines=middle,
-        enlargelimits={abs=0.2}
-    },
+    axis lines = middle,
+    ticks = none,
+    %default 2d plot/.style={%
+    %    ticks=none,
+    %    axis lines = middle,
+    %    grid=both,
+    %    minor tick num=4,
+    %    grid style={line width=.1pt, draw=gray!10},
+    %    major grid style={line width=.2pt,draw=gray!50},
+    %    axis lines=middle,
+    %    enlargelimits={abs=0.2}
 }
 
+\newcounter{curve}
+
+\NewDocumentCommand{\algebraiccurve}{ O{} O{$#5 = 0$} O{-4:4} O{-4:4} m }{
+    \addplot[id=curve\arabic{curve}, raw gnuplot, smooth, #1] function{%
+        f(x,y) = #5;
+        set xrange [#3];
+        set yrange [#4];
+        set view 0,0;
+        set isosample 1000,1000;
+        set size square;
+        set cont base;
+        set cntrparam levels incre 0,0.1,0;
+        unset surface;
+        splot f(x,y)
+    };
+    \addlegendentry{#2}
+    \stepcounter{curve}
+}%
+%\newcommand{\algebraiccurve}[3][][hi]{%
+%    %\addlegendentry{#2}
+%    \stepcounter{curve}
+%}%
+
 \geometry{
     bottom=35mm
 }
@@ -71,6 +98,8 @@
 \newtheorem{bem}[satz]{Remark}
 \newtheorem{aufgabe}[satz]{Exercise}
 
+\counterwithin{figure}{chapter}
+
 % enable aufgaben counting
 %\regtotcounter{aufgabe}
 

ws2022/rav/lecture/rav.tex

@@ -19,6 +19,8 @@ Christian Merten (\href{mailto:cmerten@mathi.uni-heidelberg.de}{cmerten@mathi.un
 
 \tableofcontents
 
+\input{rav1.tex}
+\input{rav2.tex}
 \input{rav5.tex}
 \input{rav6.tex}
 \input{rav7.tex}

ws2022/rav/lecture/rav1.tex

@@ -0,0 +1,268 @@
+\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}

ws2022/rav/lecture/rav2.tex

@@ -0,0 +1,360 @@
+%& -shell-escape -enable-write18
+\documentclass{lecture}
+
+\begin{document}
+
+\section{Regular functions}
+
+\begin{lemma}
+    If $U \subseteq k^{n}$ is a Zariski-open set and
+    $f_P \colon k^{n} \to k$ is a polynomial function such that
+    for $x \in U$, $f_P(x) \neq 0$, then the function $\frac{1}{f_P}$ is continuous on $U$.
+
+    \label{lemma:1overP-is-cont}
+\end{lemma}
+
+\begin{proof}
+    For all $t \in k$,
+    \begin{salign*}
+        \left(\frac{1}{f_P}\right)^{-1}(\{t\})
+        &= \left\{ x \in U  \mid \frac{1}{f_P(x)} = t \right\} \\
+        &= \left\{ x \in U  \mid t f_P(x) -1 = 0 \right\} \\
+        &= \mathcal{V}(tf_P -1) \cap U
+    \end{salign*}
+    is closed in $U$.
+\end{proof}
+
+\begin{bem}
+    There can be many continous functions with respect to the Zariski topology. For instance,
+    all bijective maps $f\colon k \to k$ are Zariski-continuous. In algebraic geometry, we will
+    consider only functions which are locally defined by a rational function. We will define
+    them on open subsets of algebrai sets $V \subseteq k^{n}$, endowed with the topology
+    induced by the Zariski topology of $k^{n}$.
+\end{bem}
+
+\begin{bem}[]
+    The open subsets of algebraic sets $V \subseteq k^{n}$ are exactly the
+    \emph{locally closed subsets} of $k^{n}$.
+\end{bem}
+
+\begin{definition}[]
+    Let $X \subseteq k^{n}$ be a locally closed subset of $k^{n}$. A function
+    $f \colon X \to k$ is called \emph{regular at $x \in X$}, if
+    there exist an open subset $x \in U \subseteq X$ and two polynomial functions
+    $P_U, Q_U \colon U \to k$ such that for all $y \in U$, $Q_U(y) \neq 0$ and
+    \[
+    f(y) = \frac{P_U(y)}{Q_U(y)}
+    .\] The function $f\colon X \to k$ is called \emph{regular on $X$} if, for all $x \in X$,
+    $f$ is regular at $x$.
+\end{definition}
+
+\begin{bsp}[]
+    A rational fraction $\frac{P}{Q} \in k(T_1, \ldots, T_n)$ defines a regular
+    function on the standard open set $D(Q)$.
+\end{bsp}
+
+\begin{satz}[]
+    Let $X \subseteq k^{n}$ be a locally closed subset. If $f\colon X \to k$ is regular,
+    then $f$ is continous.
+\end{satz}
+
+\begin{proof}
+    Since continuity is a local property, we may assume $X = \Omega \subseteq k^{n}$ open
+    and $f = \frac{P}{Q}$ for polynomial functions $P, Q\colon \Omega \to k$ such that
+    $Q(y) \neq 0$. By \ref{lemma:1overP-is-cont} it suffices to prove that
+    if $P, R\colon \Omega \to k$ are continuous, then $PR\colon \Omega \to k$,
+    $z \mapsto P(z)R(z)$ is continuous. Let $t \in k$. Then
+    \begin{salign*}
+        (PR)^{-1}(\{t\}) &= \{ z \in \Omega  \mid P(z) R(z) - t = 0\} \\
+                         &= \mathcal{V}(PR - t) \cap \Omega
+    \end{salign*}
+    is closed in $\Omega$.
+\end{proof}
+
+\begin{bem}
+    Being a regular function is a local property.
+\end{bem}
+
+\begin{satz}
+    Let $X \subseteq k^{n}$ be a locally closed subset of $k^{n}$, endowed
+    with the induced topology. The map
+    \begin{salign*}
+        \mathcal{O}_X\colon \{\text{open sets of }X\} &\longrightarrow k-\text{algebras}\\
+        U &\longmapsto \{ \text{regular functions on }U\} 
+    \end{salign*}
+    defines a sheaf of sheaf of $k$-algebras on $X$, which is a subsheaf of the sheaf of functions.
+\end{satz}
+
+\begin{proof}
+    Constants, sums and products of regular functions are regular, thus
+    $\mathcal{O}_X(U)$ is a sub algebra of the $k$-algebra of functions
+    $U \to k$.
+    Since restricting a function preserves regularity, $\mathcal{O}_X$ is a presheaf. Since
+    being regular is a local property and the presheaf of functions is a sheaf,
+    $\mathcal{O}_X$ is also a sheaf.
+\end{proof}
+
+\section{Irreducibility}
+
+\begin{definition}
+    Let $X$ be a topological space. $X$ is
+    \begin{enumerate}[(i)]
+        \item \emph{irreducible} if $X \neq \emptyset$ and $X$ is not the union
+            of two proper closed subets, i.e. for $X = F_1 \cup F_2$ with $F_1, F_2 \subseteq X$ closed,
+            we have $X = F_1$ or $X = F_2$.
+        \item \emph{connected} if $X$ is not the union of two disjoint proper closed subets, i.e.
+            for $X = F_1 \cup F_2$ with $F_1, F_2 \subseteq X$ closed and $F_1 \cap F_2 = \emptyset$,
+            we have $X = F_1$ or $X = F_2$.
+    \end{enumerate}
+    A space $X$ which is not irreducible, is called \emph{reducible}.
+\end{definition}
+
+\begin{lemma}
+    If $k$ is infinite, $k$ is irreducible in the Zariski topology.
+\end{lemma}
+
+\begin{proof}
+    Closed subsets of $k$ are $k$ and finite subsets of $k$.
+\end{proof}
+
+\begin{bem}
+    If $k$ is finite, $k^{n}$ is the finite union of its points, which are closed, so
+    $k^{n}$ is reducible.
+\end{bem}
+
+\begin{bem}
+    $X$ irreducible $\implies$ $X$ connected, but the converse is false: Let $k$ be infinite and
+    consider $X = \mathcal{V}_{k^2}(x^2 - y^2)$ (see figure \ref{fig:reducible-alg-set}).
+    Since $x^2 - y^2 = (x-y)(x+y) = 0$ in $k$
+    if and only if $x = -y$ or $x = y$, we have
+    $X = \mathcal{V}_{k^2}(x - y) \cup \mathcal{V}_{k^2}(x+y)$. Thus $X$ is reducible. But
+    $\mathcal{V}_{k^2}(x-y)$ and $\mathcal{V}_{k^2}(x+y)$ are homeomorphic to $k$, in particular
+    irreducible and thus connected. Since $\mathcal{V}_{k^2}{x-y} \cap \mathcal{V}_{k^2}(x+y) \neq \emptyset$,
+    $X$ is connected.
+\end{bem}
+
+\begin{figure}
+    \centering
+    \begin{tikzpicture}
+        \begin{axis}
+        \algebraiccurve[red]{x^2 - y^2}
+        \end{axis}
+    \end{tikzpicture}
+    \caption{Reducible connected algebraic set}
+    \label{fig:reducible-alg-set}
+\end{figure}
+
+\begin{satz}
+    Let $X$ be a non-empty topological space. The following conditions are equivalent:
+    \begin{enumerate}[(i)]
+        \item $X$ is irreducible
+        \item If $U_1 \cap U_2 = \emptyset$ with $U_1, U_2$ open subsets of $X$, then
+            $U_1 = \emptyset$ or $U_2 = \emptyset$.
+        \item If $U \subseteq X$ is open and non-empty, then $U$ is dense in $X$.
+    \end{enumerate}
+    \label{satz:equiv-irred}
+\end{satz}
+
+\begin{proof}
+    Left as an exercise to the reader.
+\end{proof}
+
+\begin{satz}
+    Let $X$ be a topological space and $V \subseteq X$. Then
+    $V$ is irreducible if and only if $\overline{V}$ is irreducible.
+    \label{satz:closure-irred}
+\end{satz}
+
+\begin{proof}
+    Since $\emptyset$ is closed in $X$, we have
+    $ V = \emptyset \iff \overline{V} = \emptyset$.
+
+    ($\Rightarrow$)
+    Let $\overline{V} \subseteq Z_1 \cup Z_2$ with $Z_1, Z_2 \subseteq X$ closed.
+    Then $V \subseteq Z_1 \cup Z_2$ and by irreducibility of $V$ we may assume $V \subseteq Z_1$.
+    Since $Z_1$ is closed, it follows $\overline{V} \subseteq Z_1$.
+
+    ($\Leftarrow$) Let $V \subseteq Z_1 \cup Z_2$ with $Z_1, Z_2 \subseteq X$ closed.
+    Since $Z_1 \cup Z_2$ is closed, we get $\overline{V} \subseteq Z_1 \cup Z_2$. By
+    irreducibility of $\overline{V}$ we may assume $\overline{V} \subseteq Z_1$, thus
+    $V \subseteq Z_1$.
+\end{proof}
+
+\begin{korollar}
+    Let $X$ be an irreducible topological space. Then every non-empty open
+    subset $U \subseteq X$ is irreducible.
+    \label{kor:non-empty-open-of-irred}
+\end{korollar}
+
+\begin{proof}
+    By \ref{satz:equiv-irred}, $\overline{U} = X$ and thus irreducible. The claim
+    follows now from \ref{satz:closure-irred}.
+\end{proof}
+
+\begin{lemma}[prime avoidance]
+    Let $\mathfrak{p}$ be a prime ideal in a commutative ring $A$. If $I, J \subseteq A$ are
+    ideals such that $IJ \subseteq \mathfrak{p}$, then
+    $I \subseteq \mathfrak{p}$ or $J \subseteq \mathfrak{p}$.
+
+    \label{lemma:prime-avoidance}
+\end{lemma}
+
+\begin{proof}
+    Assume that $I \not\subset \mathfrak{p}$ and $J \not\subset \mathfrak{p}$. Then
+    there exist $a \in I$, such that $a \not\in  \mathfrak{p}$ and $b \in J$ such that
+    $b \not\in \mathfrak{p}$. But $ab \in IJ \subseteq \mathfrak{p}$. Since
+    $\mathfrak{p}$ is prime, this implies $a \in \mathfrak{p}$ or
+    $b \in \mathfrak{p}$. Contradiction.
+\end{proof}
+
+\begin{theorem}
+    Let $V \subseteq k^{n}$ be an algebraic set. Then $V$ is irreducible in the Zariski
+    topology if and only if $\mathcal{I}(V)$ is a prime ideal in $k[T_1, \ldots, T_n]$.
+\end{theorem}
+
+\begin{proof}
+    ($\Rightarrow$) Since $V \neq \emptyset$,
+    $\mathcal{I}(V) \subsetneq k[T_1, \ldots, T_n]$.
+    Let $P, Q \in k[T_1, \ldots, T_n]$
+    such that $PQ \in \mathcal{I}(V)$. For $x \in V$, $(PQ)(x) = 0$ in $k$, hence
+    $P(x) = 0$ or $Q(x) = 0$. Thus $x \in \mathcal{V}(P) \cup \mathcal{V}(Q)$. Therefore
+    $V = (\mathcal{V}(P) \cap V) \cup (\mathcal{V}(Q) \cap V)$ is the union of
+    two closed subsets. Since $V$ is irreducible,
+    we may assume $V = \mathcal{V}(P) \cap V \subseteq \mathcal{V}(P)$, hence
+    $P \in \mathcal{I}(V)$ and $\mathcal{I}(V)$ is prime.
+
+    ($\Leftarrow$) $V \neq \emptyset$, since $\mathcal{I}(V)$ is a proper ideal. Let
+    $V = V_1 \cup V_2$ with $V_1, V_2$ closed in $V$. Then
+    \[
+    \mathcal{I}(V) = \mathcal{I}(V_1 \cup V_2) = \mathcal{I}(V_1) \cap \mathcal{I}(V_2)
+    \supseteq \mathcal{I}(V_1) \mathcal{I}(V_2)
+    .\] By \ref{lemma:prime-avoidance}, we may assume
+    $\mathcal{I}(V_1) \subseteq \mathcal{I}(V)$. But then
+    \[
+    V_1 = \mathcal{V}(\mathcal{I}(V_1)) \supseteq \mathcal{V}(\mathcal{I}(V)) = V
+    \] since $V_1$ and $V$ are closed. Therefore $V = V_1$ and $V$ is irreducible.
+\end{proof}
+
+\begin{korollar}
+    If $k$ is infinite, the affine space $k^{n}$ is irreducible with respect to the Zariski
+    topology.
+\end{korollar}
+
+\begin{proof}
+    Since $k$ is infinite, $\mathcal{I}(k^{n}) = (0)$ by \ref{satz:k-infinite-everywhere-vanish}
+    which is a prime ideal in the integral domain $k[T_1, \ldots, T_n]$.
+\end{proof}
+
+\begin{theorem}
+    Let $V \subseteq k^{n}$ be an algebraic set. Then there exists a decomposition
+    \[
+    V = V_1 \cup \ldots \cup V_r
+    \] such that
+    \begin{enumerate}[(i)]
+        \item $V_i$ is a closed irreducible subset of $k^{n}$ for all $i$.
+        \item $V_{i} \not\subset V_j$ for all $i \neq j$.
+    \end{enumerate}
+    This decomposition is unique up to permutations.
+    \label{thm:decomp-irred}
+\end{theorem}
+
+\begin{definition}[]
+    For an algebraic set $V \subseteq k^{n}$, the $V_i$'s in the decomposition
+    in \ref{thm:decomp-irred} are called the \emph{irreducible components} of $V$.
+\end{definition}
+
+\begin{proof}[Proof of \ref{thm:decomp-irred}]
+    Existence: Let $A$ be the set of algebraic sets $V \subseteq k^{n}$ that
+    admit no finite decomposition into a union of closed irreducible subsets. Assume
+    $A \neq \emptyset$. By noetherianity of $k^{n}$,
+    there exists a minimal element $V \in A$. In particular
+    $V$ is not irreducible, so $V = V_1 \cup V_2$ with $V_1, V_2 \subsetneq V$. By
+    minimality of $V$, $V_1, V_2 \not\in A$, thus they admit
+    a finite decomposition into a union of closed irreducible subsets. Since
+    $V = V_1 \cup V_2$, the same holds for $V$. Contradiction. Removing the
+    $V_i's$ for which $V_i \subseteq V_j$ for some $j$, we may assume that
+    $V_i \not\subset V_j$ for $i \neq j$.
+
+    Uniqueness: Assume that $V = V_1 \cup \ldots \cup V_r$
+    and $V = W_1 \cup \ldots \cup W_s$
+    are decompositions that satisfiy (i) and (ii). Then
+    \[
+    W_1 = W_1 \cap V = (W_1 \cap V_1) \cup \ldots \cup (W_1 \cap V_r)
+    .\] Since $W_1$ is irreducible and $W_1 \cap V_i$ is closed in $W_1$,
+    there exists $j$ such that $W_1 = W_1 \cap V_j \subseteq V_j$. Likewise,
+    there exists $k$ such that $V_j \subseteq W_k$. Hence $W_1 \subseteq W_k$,
+    which forces $k = 1$ (because for $k \neq 1$, we have $W_1 \subsetneq W_k)$. Thus
+    $W_1 = V_j$ and we can repeat the procedure
+    with $W_2 \cup \ldots \cup W_s = \bigcup_{i \neq j} V_i$.
+\end{proof}
+
+\begin{korollar}[]
+    Let $V \subseteq k^{n}$ be an algebraic set and
+    denote by $V_1, \ldots, V_r$ the irreducible components of $V$. Let $W \subseteq V$
+    be an irreducible subset. Then $W \subseteq V_i$ for some $i$.
+    \label{cor:irred-sub-of-alg-set}
+\end{korollar}
+
+\begin{proof}
+    We have
+    \[
+    W = W \cap V = \bigcup_{i=1}^{r} \underbrace{W \cap V_i}_{\text{closed in }W}
+    .\] 
+    Since $W$ is irreducible, there exists an $i$ such that
+    $W = W \cap V_i \subseteq V_i$.
+\end{proof}
+
+\begin{bem}
+    \begin{enumerate}[(i)]
+        \item The $i$ in \ref{cor:irred-sub-of-alg-set} is not unique in general. Consider
+        \[
+        V = \{ x^2 - y^2 = 0\} = \{ x - y = 0\} \cup \{ x + y = 0\}
+        .\] The closed irreducible subset $\{(0, 0)\} $ lies in the intersection of
+        the irreducible components of $V$.
+    \item In view of the corollary \ref{cor:irred-sub-of-alg-set},
+        theorem \ref{thm:decomp-irred} implies that an algebraic
+        set $V \subseteq k^{n}$ has a unique minimal decomposition into a union of closed irreducible
+        subsets.
+    \end{enumerate}
+\end{bem}
+
+\begin{korollar}
+    Let $V \subseteq k^{n}$ be an algebraic set. The irreducible components of $V$
+    are exactly the maximal closed irreducible subsets of $V$. In terms
+    of ideals in $k[T_1, \ldots, T_n]$, a closed subset $W \subseteq V$
+    is an irreducible component of $V$, if and only if the ideal
+    $\mathcal{I}(W)$ is a prime ideal which is minimal among those containing
+    $\mathcal{I}(V)$.
+\end{korollar}
+
+\begin{proof}
+    A closed irreducible subset $W \subseteq V$ is
+    contained in an irreducible component $V_j \subseteq V$
+    by \ref{cor:irred-sub-of-alg-set}. If $W$ is maximal, then $W = V_j$.
+
+    Conversely, if $V_j$ is an irreducible component of $V$ and
+    $V_j \subseteq W$ for some irreducible and closed subset $W \subseteq V$, again
+    by \ref{cor:irred-sub-of-alg-set} we have $W \subseteq V_i$ for some $i$, therefore
+    $V_j \subseteq V_i$ which implies $i = j$ and $V_j = W$.
+\end{proof}
+
+\begin{satz}[Identity theorem for regular functions]
+    Let $X \subseteq k^{n}$ be an irreducible algebraic set and let $U \subseteq X$
+    be open. Let $f, g \in \mathcal{O}_X(U)$ be regular functions on $U$. If
+    there is a non-empty open set $U' \subseteq U$ such that
+    $f|_{U'} = g|_{U'}$, then $f = g$ on $U$.
+\end{satz}
+
+\begin{proof}
+    The set $Y = \mathcal{V}_U(f-g)$ is closed in $U$ and
+    contains $U'$. Thus the closure $\overline{U'}^{(U)}$ of $U'$ in $U$
+    is also contained in $Y$. By \ref{kor:non-empty-open-of-irred}
+    $U$ is irreducible, so $U'$ is dense in $U$. Therefore $Y = U$.
+\end{proof}
+
+\begin{bsp}
+    If $k$ is infinite and $P \in k[T_1, \ldots, T_n]$ is zero
+    outside an algebraic set $V \subseteq k^{n}$, then $P = 0$ on $k^{n}$.
+\end{bsp}
+
+\end{document}