diff --git a/.gitignore b/.gitignore index 5db570e..def21a5 100644 --- a/.gitignore +++ b/.gitignore @@ -17,3 +17,5 @@ *.snm *.bbl *.blg +*.table +*.gnuplot diff --git a/ws2022/rav/lecture/lecture.cls b/ws2022/rav/lecture/lecture.cls index d9f8071..7be03a9 100644 --- a/ws2022/rav/lecture/lecture.cls +++ b/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} diff --git a/ws2022/rav/lecture/rav.pdf b/ws2022/rav/lecture/rav.pdf index 4f9f7d2..64af460 100644 Binary files a/ws2022/rav/lecture/rav.pdf and b/ws2022/rav/lecture/rav.pdf differ diff --git a/ws2022/rav/lecture/rav.tex b/ws2022/rav/lecture/rav.tex index ce49d5f..509fdcb 100644 --- a/ws2022/rav/lecture/rav.tex +++ b/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} diff --git a/ws2022/rav/lecture/rav1.pdf b/ws2022/rav/lecture/rav1.pdf new file mode 100644 index 0000000..5085eef Binary files /dev/null and b/ws2022/rav/lecture/rav1.pdf differ diff --git a/ws2022/rav/lecture/rav1.tex b/ws2022/rav/lecture/rav1.tex new file mode 100644 index 0000000..bbb57e1 --- /dev/null +++ b/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} diff --git a/ws2022/rav/lecture/rav2.pdf b/ws2022/rav/lecture/rav2.pdf new file mode 100644 index 0000000..3da00c5 Binary files /dev/null and b/ws2022/rav/lecture/rav2.pdf differ diff --git a/ws2022/rav/lecture/rav2.tex b/ws2022/rav/lecture/rav2.tex new file mode 100644 index 0000000..2804ec1 --- /dev/null +++ b/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}