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.gitignore
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*.snm
*.bbl
*.blg
*.table
*.gnuplot
ws2022/rav/lecture/lecture.cls
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\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}
},
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}
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ws2022/rav/lecture/rav.tex
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\tableofcontents
\input{rav1.tex}
\input{rav2.tex}
\input{rav5.tex}
\input{rav6.tex}
\input{rav7.tex}
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\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}
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%& -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}