rav: removed, moved into submodule

ws2022/rav/lecture/lecture.cls

@@ -1,276 +0,0 @@
-\ProvidesClass{lecture}
-\LoadClass[a4paper]{book}
-
-\RequirePackage{faktor}
-\RequirePackage{xparse}
-\RequirePackage{stmaryrd}
-\RequirePackage[utf8]{inputenc}
-\RequirePackage[T1]{fontenc}
-\RequirePackage{textcomp}
-\RequirePackage{babel}
-\RequirePackage{amsmath, amssymb, amsthm}
-\RequirePackage{mdframed}
-\RequirePackage{tikz-cd}
-\RequirePackage{geometry}
-\RequirePackage{import}
-\RequirePackage{pdfpages}
-\RequirePackage{transparent}
-\RequirePackage{xcolor}
-\RequirePackage{array}
-\RequirePackage[shortlabels]{enumitem}
-\RequirePackage{tikz}
-\RequirePackage{pgfplots}
-\RequirePackage[pagestyles, nobottomtitles]{titlesec}
-\RequirePackage{listings}
-\RequirePackage{mathtools}
-\RequirePackage{forloop}
-\RequirePackage{totcount}
-\RequirePackage[hidelinks, unicode]{hyperref} %[unicode, hidelinks]{hyperref}
-\RequirePackage{bookmark}
-\RequirePackage{wasysym}
-\RequirePackage{environ}
-\RequirePackage{stackrel}
-\RequirePackage{subcaption}
-
-\usetikzlibrary{quotes, angles, math}
-\pgfplotsset{
-    compat=1.15,
-    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
-}
-
-%\DeclareOption*{\PassOptionsToClass{\CurrentOption}{article}}
-\DeclareOption{uebung}{
-    \makeatletter
-    \lhead{\@title}
-    \rhead{\@author}
-    \makeatother
-}
-\ProcessOptions\relax
-
-% PARAGRAPH no indent but skip
-%\setlength{\parskip}{3mm}
-%\setlength{\parindent}{0mm}
-
-\newtheorem{satz}{Proposition}[chapter]
-\newtheorem{theorem}[satz]{Theorem}
-\newtheorem{lemma}[satz]{Lemma}
-\newtheorem{korollar}[satz]{Corollary}
-\theoremstyle{definition}
-\newtheorem{definition}[satz]{Definition}
-
-\newtheorem{bsp}[satz]{Example}
-\newtheorem{bem}[satz]{Remark}
-\newtheorem{aufgabe}[satz]{Exercise}
-
-\counterwithin{figure}{chapter}
-
-% enable aufgaben counting
-%\regtotcounter{aufgabe}
-
-\newcommand{\N}{\mathbb{N}}
-\newcommand{\R}{\mathbb{R}}
-\newcommand{\Z}{\mathbb{Z}}
-\newcommand{\Q}{\mathbb{Q}}
-\newcommand{\C}{\mathbb{C}}
-
-% HEADERS
-
-%\newpagestyle{main}[\small]{
-%  \setheadrule{.55pt}%
-%  \sethead[\thepage]%  even-left
-%          []%                              even-center
-%          [\thechapter~\chaptertitle]%     even-right
-%          {\thesection~\sectiontitle}%     odd-left
-%          {}%                              odd-center
-%          {\thepage}%  odd-right
-%}
-%\pagestyle{main}
-
-\newcommand{\incfig}[1]{%
-    \def\svgwidth{\columnwidth}
-    \import{./figures/}{#1.pdf_tex}
-}
-\pdfsuppresswarningpagegroup=1
-
-% horizontal rule
-\newcommand\hr{
-    \noindent\rule[0.5ex]{\linewidth}{0.5pt}
-}
-
-% code listings, define style
-\lstdefinestyle{mystyle}{
-    commentstyle=\color{gray},
-    keywordstyle=\color{blue},
-    numberstyle=\tiny\color{gray},
-    stringstyle=\color{black},
-    basicstyle=\ttfamily\footnotesize,
-    breakatwhitespace=false,         
-    breaklines=true,                 
-    captionpos=b,                    
-    keepspaces=true,                 
-    numbers=left,                    
-    numbersep=5pt,                  
-    showspaces=false,                
-    showstringspaces=false,
-    showtabs=false,                  
-    tabsize=2
-}
-
-% activate my colour style
-\lstset{style=mystyle}
-
-% better stackrel
-\let\oldstackrel\stackrel
-\renewcommand{\stackrel}[3][]{%
-    \oldstackrel[\mathclap{#1}]{\mathclap{#2}}{#3}
-}%
-
-% integral d sign
-\makeatletter \renewcommand\d[2][]{\ensuremath{%
-		\,\mathrm{d}^{#1}#2\@ifnextchar^{}{\@ifnextchar\d{}{\,}}}}
-\makeatother
-
-% remove page before chapters
-\let\cleardoublepage=\clearpage
-
-%josua
-\newcommand{\norm}[1]{\left\Vert#1\right\Vert}
-
-% contradiction
-\newcommand{\contr}{\text{\Large\lightning}}
-
-% people seem to prefer varepsilon over epsilon
-\renewcommand{\epsilon}{\varepsilon}
-
-\ExplSyntaxOn
-
-% S-tackrelcompatible ALIGN environment
-% some might also call it the S-uper ALIGN environment
-% uses regular expressions to calculate the widest stackrel
-% to put additional padding on both sides of relation symbols
-\NewEnviron{salign}
-{
-    \begin{align}
-        \lec_insert_padding:V \BODY
-    \end{align}
-}
-% starred version that does no equation numbering
-\NewEnviron{salign*}
-{
-    \begin{align*}
-        \lec_insert_padding:V \BODY
-    \end{align*}
-}
-
-% some helper variables
-\tl_new:N \l__lec_text_tl
-\seq_new:N \l_lec_stackrels_seq
-\int_new:N \l_stackrel_count_int
-\int_new:N \l_idx_int
-\box_new:N \l_tmp_box
-\dim_new:N \l_tmp_dim_a
-\dim_new:N \l_tmp_dim_b
-\dim_new:N \l_tmp_dim_c
-\dim_new:N \l_tmp_dim_needed
-
-% function to insert padding according to widest stackrel
-\cs_new_protected:Nn \lec_insert_padding:n
- {
-  \tl_set:Nn \l__lec_text_tl { #1 }
-  % get all stackrels in this align environment
-  \regex_extract_all:nnN { \c{stackrel}(\[.*?\])?{(.*?)}{(.*?)} } { #1 } \l_lec_stackrels_seq
-  % get number of stackrels
-  \int_set:Nn \l_stackrel_count_int { \seq_count:N \l_lec_stackrels_seq }
-  \int_set:Nn \l_idx_int { 1 }
-  \dim_set:Nn \l_tmp_dim_needed { 0pt }
-  % iterate over stackrels
-  \int_while_do:nn { \l_idx_int <= \l_stackrel_count_int }
-  {
-      % calculate width of text
-      \hbox_set:Nn \l_tmp_box {$\seq_item:Nn \l_lec_stackrels_seq { \l_idx_int + 1 }$}
-      \dim_set:Nn \l_tmp_dim_a {\box_wd:N \l_tmp_box}
-      \hbox_set:Nn \l_tmp_box {$\seq_item:Nn \l_lec_stackrels_seq { \l_idx_int + 2 }$}
-      \dim_set:Nn \l_tmp_dim_c {\box_wd:N \l_tmp_box}
-      \dim_set:Nn \l_tmp_dim_a {\dim_max:nn{ \l_tmp_dim_c} {\l_tmp_dim_a}}
-      % calculate width of relation symbol
-      \hbox_set:Nn \l_tmp_box {$\seq_item:Nn \l_lec_stackrels_seq { \l_idx_int + 3 }$}
-      \dim_set:Nn \l_tmp_dim_b {\box_wd:N \l_tmp_box}
-      % check if 0.5*(a-b) > minimum padding, if yes updated minimum padding
-      \dim_compare:nNnTF
-        { 1pt * \dim_ratio:nn { \l_tmp_dim_a - \l_tmp_dim_b } { 2pt } } > { \l_tmp_dim_needed }
-        { \dim_set:Nn \l_tmp_dim_needed { 1pt * \dim_ratio:nn { \l_tmp_dim_a - \l_tmp_dim_b } { 2pt } } }
-        { }
-      % increment list index by three, as every stackrel produces three list entries
-      \int_incr:N \l_idx_int
-      \int_incr:N \l_idx_int
-      \int_incr:N \l_idx_int
-      \int_incr:N \l_idx_int
-  }
-  % replace all relations with align characters (&) and add the needed padding
-  \regex_replace_all:nnN
-  { (\c{simeq}&|&\c{simeq}|\c{leq}&|&\c{leq}|\c{geq}&|&\c{geq}|\c{iff}&|&\c{iff}|\c{impliedby}&|&\c{impliedby}|\c{implies}&|&\c{implies}|\c{approx}&|&\c{approx}|\c{equiv}&|&\c{equiv}|=&|&=|\c{le}&|&\c{le}|\c{ge}&|&\c{ge}|&\c{stackrel}(\[.*?\])?{.*?}{.*?}|\c{stackrel}(\[.*?\])?{.*?}{.*?}&|&\c{neq}|\c{neq}&|>&|&>|<&|&<) }
-      { \c{kern} \u{l_tmp_dim_needed} \1 \c{kern} \u{l_tmp_dim_needed} }
-      \l__lec_text_tl
-  \l__lec_text_tl
- }
-\cs_generate_variant:Nn \lec_insert_padding:n { V }
-
-\NewEnviron{leftright}
-{
-    \lec_replace_parens:V \BODY
-}
-
-% function to replace parens with left right
-\cs_new_protected:Nn \lec_replace_parens:n
- {
-  \tl_set:Nn \l__lec_text_tl { #1 }
-  % replace all parantheses with \left( \right)
-  \regex_replace_all:nnN { \( } { \c{left}( } \l__lec_text_tl
-  \regex_replace_all:nnN { \) } { \c{right}) } \l__lec_text_tl
-  \regex_replace_all:nnN { \[ } { \c{left}[ } \l__lec_text_tl
-  \regex_replace_all:nnN { \] } { \c{right}] } \l__lec_text_tl
-  \l__lec_text_tl
- }
-\cs_generate_variant:Nn \lec_replace_parens:n { V }
-
-\ExplSyntaxOff
-
-% add one equation tag to the current line to otherwise unnumbered environment
-\newcommand{\tageq}{\stepcounter{equation}\tag{\theequation}}

ws2022/rav/lecture/rav.tex

@@ -1,42 +0,0 @@
-\documentclass{lecture}
-
-\usepackage{standalone}
-\usepackage{tikz}
-\usepackage{subcaption}
-
-\title{Real algebraic varieties}
-\author{Florent Schaffhauser\\[5mm]
-Transcript of\\[1mm]
-Christian Merten (\href{mailto:cmerten@mathi.uni-heidelberg.de}{cmerten@mathi.uni-heidelberg.de})\\
-}
-\date{WiSe 2022}
-
-\begin{document}
-
-\newgeometry{right=15mm, left=15mm}
-\maketitle
-\restoregeometry
-
-\tableofcontents
-
-\input{rav1.tex}
-\input{rav2.tex}
-\input{rav3.tex}
-\input{rav4.tex}
-\input{rav11.tex}
-\input{rav5.tex}
-\input{rav6.tex}
-\input{rav7.tex}
-\input{rav8.tex}
-\input{rav9.tex}
-\input{rav10.tex}
-\input{rav15.tex}
-\input{rav16.tex}
-\input{rav17.tex}
-\input{rav18.tex}
-\input{rav21.tex}
-\input{rav19.tex}
-\input{rav20.tex}
-\input{rav22.tex}
-
-\end{document}

ws2022/rav/lecture/rav1.tex

@@ -1,268 +0,0 @@
-\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/rav10.tex

@@ -1,378 +0,0 @@
-\documentclass{lecture}
-
-\begin{document}
-
-\section{Examples of algebraic varieties}
-
-\begin{aufgabe}[]
-    Let $f\colon X \to Y$ be a morphism of algebraic pre-varieties. Assume
-    \begin{enumerate}[(i)]
-        \item $Y$ is a variety.
-        \item There exists an open covering $(Y_i)_{i \in I}$ of $Y$ such that the open subset
-            $f^{-1}(Y_i)$ is a variety.
-    \end{enumerate}
-    Show that $X$ is a variety.
-\end{aufgabe}
-
-\begin{aufgabe}[]
-    Let $X$ be a topological space. Assume that there exists a covering $(X_i)_{i \in I}$ of
-    $X$ by irreducible open subsets such that for all $(i,j)$, $(X_i \cap X_j) \neq \emptyset$.
-    Show that $X$ is irreducible.
-\end{aufgabe}
-
-\subsection{Grassmann varieties}
-
-Let $0 \le p \le n$ be integers. The Grassmannian $\text{Gr}(p, n)$ is the set
-of $p$-dimensional linear subspaces of $k^{n}$. In order to endow this set with a structure
-of algebraic prevariety, there are various possibilities:
-
-\begin{enumerate}[(i)]
-    \item To a $p$-dimensional linear subspace $E \subseteq k^{n}$, we associate the line
-        $\Lambda^{p} E \subseteq \Lambda^{p} k^{n} \simeq k^{\binom{n}{p}}$, which
-        defines a point in the projective space $k\mathbb{P}^{\binom{n}{p}-1}$.
-
-        Claim: The map $\text{Gr}(p, n) \to k\mathbb{P}^{\binom{n}{p} -1}$
-        is an injective map whose image is a Zariski-closed subset of $k\mathbb{P}^{\binom{n}{p} -1}$.
-
-        This identifies $\text{Gr}(n, p)$ canonically to a projective variety. In particular
-        one obtains in this way a structure of \emph{algebraic variety}
-        on $\text{Gr}(p, n)$.
-    \item For the second approach, recall that $\text{GL}(n, k)$ acts transitively on
-        $\text{Gr}(p, n)$. But the identification of $k^{n}$ to $(k^{n})^{*}$
-        via the canonical basis of $k^{n}$ enables one to define, for all $E \in \text{Gr}(p, n)$,
-        a canonical complement $E^{\perp} \in \text{Gr}(n-p, n)$, i.e.
-        an $(n-p)$-dimensional linear subspace such that $E \oplus E^{\perp} = k^{n}$.
-        
-        So the stabiliser of $E \in \text{Gr}(p, n)$ for the action of
-        $\text{GL}(n, k)$ is conjugate to the subgroup
-        \begin{salign*}
-        \text{P}(p, n) \coloneqq
-        \left\{ g \in \text{GL}(n, k)  \middle \vert
-            \begin{array}{l}
-                g = \begin{pmatrix} A & B \\ 0 & C \end{pmatrix} \\
-                \text{with } A \in \text{GL}(p, k), B \in \text{Mat}(p \times (n-p), k),\\
-                \text{and } C \in \text{GL}(n-p, k) 
-            \end{array}
-        \right\} 
-        .\end{salign*}
-        This shows that the Grassmannian $\text{Gr}(p, n)$ is a homogeneous space
-        under $\text{GL}(n, k)$ and that
-        \begin{salign*}
-        \text{Gr}(p, n) \simeq \text{GL}(n, k) / \text{P}(p, n)
-        \end{salign*}
-        which is useful if one knows that, given an affine algebraic group $G$ and
-        a closed subgroup $H$, the homogeneous space $G / H$ is an algebraic variety. We
-        will come back to this later on.
-    \item The third uses the gluing theorem. In particular, it also constructs
-        a standard atlas on $\text{Gr}(p, n)$, like the one we had on
-        $k\mathbb{P}^{n-1} = \text{Gr}(1, n)$.
-        The idea is that, in order to determine a $p$-dimensional subspace of $k^{n}$,
-        it suffices to give a basis of that subspace, which is a family of $p$ vectors
-        in $k^{n}$. Geometrically, this means that the subspace in question is seen
-        as the graph of a linear map $A\colon k^{p} \to k^{n}$.
-
-        Take $E \in \text{Gr}(p, n)$ and let $(v_1, \ldots, v_p)$ be a basis of $E$ over $k$.
-        Let $M$ be the $(n \times p)$-matrix representing the coordinates
-        of $(v_1, \ldots, v_p)$ in the canonical basis of $k^{n}$. Since $M$ has rank $p$,
-        there exists a $(p \times p)$-submatrix of $M$ with non-zero determinant: We set
-        \begin{salign*}
-            J &\coloneqq \{ \text{indices } j_1 < \ldots < j_p \text{ of the rows of that submatrix}\} \\
-            M_J &\coloneqq \text{the submatrix in question}
-        .\end{salign*}
-        Note that if $M' \in \text{Mat}(n \times p, k)$ corresponds to a basis
-        $(v_1', \ldots, v_p')$, there exists a matrix $g \in \text{GL}(p, k)$ such that
-        $M' = Mg$. But then $(M')_J = (Mg)_J = M_J g$, so
-        \[
-        \text{det }(M')_J = \text{det } (M_J g) = \text{det}(M_J) \text{det}(g)
-        ,\]
-        which is non-zero if and only if $\text{det}(M_J)$ is non-zero. As a consequence,
-        given a subset $J \subseteq \{1, \ldots, n\} $ of cardinal $p$, there is a well-defined
-        subset
-        \begin{salign*}
-        G_J \coloneqq \left\{ E \in \text{G}(p, n)  \mid 
-                \exists M \in \text{Mat}(n \times p, k), E = \text{im }M \text{ and }
-                \text{det}(M_J) \neq 0
-        \right\} 
-        .\end{salign*}
-        Moreover, if $M$ satisfies the conditions $E = \text{im }M$ and
-        $\text{det}(M_J) \neq 0$, then
-        $(M M_J^{-1})_J = I_p$ and $\text{im}(MM^{-1}_J) = \text{im }M = E$.
-        In fact, if $E \in G_J$, there is a unique matrix $N \in \text{Mat}(n \times p, k)$,
-        such that $E = \text{im }N$ and $N_J = I_p$, for if $N_1, N_2$ are two
-        such matrices, the columns of $N_2$ are linear combinations of those of $N_1$,
-        thus $\exists g \in \text{GL}(p, k)$ such that $N_2 = N_1g$. But then
-        \[
-        I_p = (N_2)_J = (N_1g)_J = (N_1)_J g = g
-        .\] 
-        So, there is a well-defined map
-        \begin{salign*}
-            \hat{\varphi}_J: G_J &\longrightarrow \operatorname{Hom}(k^{J}, k^{n}) \\
-            E &\longmapsto N \text{ such that } E = \text{im }N \text{ and } N_J = I_p
-        \end{salign*}
-        whose image can be identified to the subspace
-        $\text{Hom}(k^{J}, k^{J^{c}})$, where $J^{c}$ is the complement of $J$ in
-        $\{1, \ldots, n\} $, via the map $N \mapsto N_{J^{c}}$. Conversely, a
-        linear map $A \in \text{Hom}(k^{J}, k^{J^{c}})$ determines a rank $p$ map
-        $N \in \text{Hom}(k^{J}, k^{n})$ such that $N_J = I_p$ via the formula
-        $N(x) = x + Ax$.
-
-        Geometrically, this means that the $p$-dimensional subspace
-        $\text{im }N \subseteq k^{n}$ is equal to the graph of $A$.
-        This also means that we can think of $G_J$ as the set
-        \begin{salign*}
-        \{E \in \text{Gr}(p, n)  \mid E \cap k^{J^{c}} = \{0_{k^{n}}\} \} 
-        .\end{salign*}
-        The point is that $\text{im } \hat{\varphi}_J = \text{Hom}(k^{J}, k^{J^{c}})$
-        can be canonically identified with the affine space $k^{p(n-p)}$ and that we
-        have a bijection
-        \begin{salign*}
-            \varphi_J \colon G_J &\xlongrightarrow{\simeq} \text{Hom}(k^{J}, k^{J^{c}})
-        \simeq k^{p(n-p)} \\
-            E &\longmapsto A  \mid \text{gr}(A) = E \\
-            \text{gr}(A) &\longmapsfrom A
-        .\end{salign*}
-        Note that the matrix $N \in \text{Mat}(n \times p, k)$
-        such that $\text{im }N = E$ and $N_J = I_p$
-        is row-equivalent to $\begin{pmatrix} I_p \\ A \end{pmatrix} $
-        with $A \in \text{Mat}((n-p) \times p, k)$.
-
-        Now, if $E \in G_{J_1} \cap G_{J_2}$, then, for all
-        $M \in \text{Mat}(p \times n, k)$ such that $\text{im } M = E$,
-        $\hat{\varphi}_{J_1}(E) = M M_{J_1}^{-1}$ and
-        $\hat{\varphi}_{J_2}(E) = M M_{J_2}^{-1}$. So
-        \begin{salign*}
-        \text{im } \hat{\varphi}_{J_1}
-        &= \left\{ N \in \text{Hom}(k^{J_1}, k^{n})  \mid N_{J_1} = I_p,
-            \text{im } N_{J_1} = E \text{ and }
-            \text{det}(N_{J_2})  \neq 0
-        \right\}  \\
-        &= \{ N \in \text{im } \hat{\varphi}_{J_1}  \mid \text{det}(N_{J_2})  \neq 0\}
-        \end{salign*}
-        which is open in $\text{im } \hat{\varphi}_{J_1} \simeq \text{im } \varphi_{J_1}$.
-
-        Moreover, for all $N \in \text{im }\hat{\varphi}_{J_1}$,
-        \[
-        \hat{\varphi}_{J_2} \circ \hat{\varphi}_{J_1}^{-1}(N) = N N_{J_2}^{-1}
-        \] and, by Cramer's formulae, this is a regular function
-        on $\text{im }\hat{\varphi}_{J_1}$.
-
-        We have therefore constructed a covering
-        \[
-        \text{Gr}(p, n) = \bigcup_{J \subseteq \{1, \ldots, n\}, \# J = p } G_J
-        \] 
-        of the Grassmannian $\text{Gr}(p, n)$ by subsets $G_J$
-        that can be identified to the affine variety $k^{p(n-p)}$ via bijective
-        maps $\varphi_J\colon G_j \to k^{p(n-p)}$ such that,
-        for all $(J_1, J_2)$, $\varphi_{J_1}(G_{J_1} \cap G_{J_2})$ is open
-        in $k^{p(n-p)}$ and the map
-        $\varphi_{J_2} \circ \varphi_{J_1}^{-1}\colon \varphi_{J_1}(G_{J_1} \cap G_{J_2}) \to \varphi_{J_2}(G_{J_1} \cap G_{J_2})$
-        is a morphism of affine varieties. By the gluing theorem,
-        this endows $\text{Gr}(p, n)$ with a structure of algebraic prevariety.
-\end{enumerate}
-
-\subsection{Vector bundles}
-
-\begin{definition}[]
-    A \emph{vector bundle} is a triple
-    $(E, X, \pi)$ consisting of two algebraic varieties $E$ and $X$, and
-    a morphism $\pi\colon E \to X$ such that
-    \begin{enumerate}[(i)]
-        \item for $x \in X$, $\pi^{-1}(\{x\} )$ is a $k$-vector space.
-        \item for $x \in X$, there exists an open neighbourhood $U$ of $x$
-            and an isomorphism of algebraic varieties
-            \[
-            \Phi\colon \pi^{-1}(U) \xlongrightarrow{\simeq} U \times \pi^{-1}(\{x\} )
-            \] such that
-            \begin{enumerate}[(a)]
-                \item $\text{pr}_1 \circ \Phi = \pi |_{\pi^{-1}(U)}$ and
-                \item for $y \in U$, $\Phi|_{\pi^{-1}(\{y\})}\colon \pi^{-1}(\{y\})
-                    \to \{y\} \times \pi^{-1}(\{x\})$ is
-                    an isomorphism of $k$-vector spaces.
-            \end{enumerate}
-    \end{enumerate}
-    A morphism of vector bundles is a morphism of algebraic varieties $f\colon E_1 \to E_2$
-    such that $\pi_2 \circ f = \pi_1$ and $f$ is $k$-linear in the fibres.
-\end{definition}
-
-\begin{bem}
-    In practice, one often proves that a variety $E$ is a vector bundle over $X$ by
-    finding a morphism $\pi\colon E \to X$ and an open covering
-    \[
-    X = \bigcup_{i \in  I} U_i
-    \] such that $E|_{U_i} \coloneqq \pi^{-1}(U_i)$ is isomorphic to
-    $U_i \times k^{n_i}$ for some integer $n_i$, in such a way that, on $U_i \cap U_j$,
-    the morphism
-    \[
-    \Phi_j \circ \Phi_i^{-1}\Big|_{\Phi_i(\pi^{-1}(U_i \cap U_j))}\colon
-    (U_i \cap U_j) \times k^{n_i} \longrightarrow
-    (U_i \cap U_j) \times k^{n_j}
-    \] is an isomorphism of algebraic varieties such that the following diagram commutes
-    and $\Phi_j \circ \Phi_i^{-1}$ is linear fibrewise:
-    \[
-    \begin{tikzcd}
-        (U_i \cap U_j) \times k^{n_i} \arrow{dr}{\text{pr}_1} \arrow{rr}{\Phi_j \circ \Phi_i^{-1}}
-        & & (U_i \cap U_j) \times k^{n_j} \arrow{dl}{\text{pr}_1}\\
-        & U_i \cap U_j & \\
-    \end{tikzcd}
-    .\] In particular $k^{n_i} \simeq k^{n_j}$ as $k$-vector spaces, so
-    $n_i = n_j$ if $U_i \cap U_j \neq \emptyset$, and
-    $\Phi_j \circ \Phi_i^{-1}$ is necessarily of the form
-    \[
-        (x, v) \longmapsto (x, g_{ji}(x) \cdot v)
-    \] for some morphism of algebraic varieties
-    \[
-    g_{ji}\colon U_i \cap U_j \longrightarrow \text{GL}(n, k)
-    .\]
-    These maps $(g_{ij})_{(i, j) \in I \times I}$ then
-    satisfy for $x \in U_i \cap U_j \cap U_l$
-    \[
-    g_{lj}(x) g_{ji}(x) = g_{li}(x)
-    \] and for $x \in U_i$, $g_{ii}(x) = \text{I}_n$.
-\end{bem}
-
-\begin{satz}
-    If $\pi\colon E \to X$ is a morphism of algebraic varieties and
-    $X$ has an open covering $(U_i)_{i \in I}$ over which $E$ admits
-    local trivialisations
-    \[
-    \Phi_i \colon E|_{U_i} = \pi^{-1}(U_i) \xlongrightarrow{\simeq} U_i \times k^{n}
-    \]
-    with $\text{pr}_1 \circ \Phi_i = \pi|_{\pi^{-1}(U_i)}$
-    such that the isomorphisms
-    \[
-    \Phi_j \circ \Phi_i^{-1} \colon (U_i \cap U_j) \times k^{n}
-    \longrightarrow (U_i \cap U_j) \times k^{n}
-    \] are
-    linear in the fibres, then for all $x \in X$, $\pi^{-1}(\{x\})$ has
-    a well-defined structure of $k$-vector space and the local trivialisations
-    $(\Phi_i)_{i \in I}$ are linear in the fibres. In particular,
-    $E$ is a vector bundle.
-\end{satz}
-
-\begin{proof}
-    For $x \in U_i$ and $a, b \in \pi^{-1}(\{x\})$, let
-    \[
-    a + \lambda b \coloneqq \Phi_i^{-1}(x, \text{pr}_2 (\Phi_i(a)) + \lambda \text{pr}_2 (\Phi_i(b)))
-    .\]
-    By using the linearity in the fibres of $\Phi_j \circ \Phi_i^{-1}$, one verifies
-    that this does not depend on the choice of $i \in I$.
-\end{proof}
-
-\begin{bem}[]
-    Assume given an algebraic prevariety $X$ obtained by gluing affine varieties
-    $(X_i)_{i \in I}$ along isomorphisms $\varphi_{ji}\colon X_{ij} \xrightarrow{\simeq} X_{ji}$
-    defined on open subsets $X_{ij} \subseteq X_i$,
-    such that $X_{ii} = X_i$, $\varphi_{ii} = \text{Id}_{X_i}$
-    %, $\varphi_{ji}(X_{ij})$ is open in $X_{ji}$
-    and
-    $\varphi_{lj} \circ \varphi_{ji} = \varphi_{li}$ on $X_{ij} \cap X_{il} \subseteq X_i$.
-
-    Recall that such an $X$ comes equipped with a canonical
-    map $p \colon \bigsqcup_{i \in I} \to X$ such that
-    $p_i \coloneqq p|_{X_i}\colon X_i \to X$ is an isomorphism onto an affine open subset
-    $U_i \coloneqq p_i(X_i) \subseteq X$ and, if we set $\varphi_i = p_i^{-1}$,
-    we have $\varphi_j \circ \varphi_i^{-1} = \varphi_{ji}$
-    on $\varphi_i(U_i \cap U_j)$.
-
-    Let us now consider the vector bundle $X_i \times k^{n}$ on each of the affine varieties
-    $X_i$ and assume that an isomorphism of algebraic prevarieties of the form
-    \begin{salign*}
-        \Phi_{ji}\colon X_{ij} \times k^{n} &\longrightarrow X_{ji} \times k^{n} \\
-        (x, v) &\longmapsto (\varphi_{ji}(x), h_{ji}(x) \cdot v)
-    \end{salign*}
-    has been given, where $h_{ij}\colon X_{ij} \to \text{GL}(n, k)$
-    is a morphism of algebraic varieties, in such a way that the following compatibility
-    conditions are satisfied:
-    \begin{salign*}
-    \Phi_{ii} = \text{Id}_{X_{ii} \times k^{n}}
-    \end{salign*}
-    and, for all $(i, j, l)$ and all $(x, v) \in (X_{ij} \cap X_{il}) \times k^{n}$
-    \[
-    \Phi_{lj} \circ \Phi_{ji}(x, v) = \Phi_{li}(x, v)
-    .\] 
-    Then there is associated to this gluing data an algebraic vector bundle
-    $\pi\colon E \to X$, endowed with 
-    local trivialisations $\Phi_i \colon E|_{U_i} \xrightarrow{\simeq} U_i \times k^{n}$,
-    where as earlier $U_i = p(X_i) \subseteq X$,
-    in such a way that, for all $(i, j)$ and all $(\xi, v) \in (U_i \cap U_j) \times k^{n}$,
-    \[
-    \Phi_j \circ \Phi_i^{-1}(\xi, v) =
-    (\xi, g_{ji}(\xi) \cdot v)
-    \] where $g_{ji}(x) = h_{ji}(\varphi_i(\xi)) \in \text{GL}(n, k)$, so
-    $g_{ii} = \text{I}_n$ on $U_i$, and, for all $(i, j, l)$ and
-    all $\xi \in U_i \cap U_j \cap U_l$,
-    \begin{salign*}
-        g_{lj}(\xi) g_{ji}(\xi) &= h_{lj}(\varphi_j(\xi)) h_{ji}(\varphi_i(\xi)) \\
-    &= h_{lj}(\varphi_{ji}(\varphi_i(\xi))) h_{ji}(\varphi_i(\xi)) \\
-    &= h_{li}(\varphi_i(\xi)) \\
-    &= g_{li}(\xi)
-    .\end{salign*}
-
-    Indeed, we can simply set
-    \begin{salign*}
-    E \coloneqq \left( \bigsqcup_{i \in I} X_i \times k^{n} \right) / \sim 
-    \end{salign*}
-    where $(x, v) \sim (\varphi_{ji}(x), h_{ji}(x) \cdot v)$, and, by the 
-    gluing theorem, this defines an algebraic prevariety, equipped
-    with a morphism $\pi\colon E \to X$ induced
-    by the first projection $\text{pr}_1\colon \bigsqcup_{i \in I} (X_i \times k^{n})
-    \to \bigsqcup_{i \in I} X_i$.
-    The canonical map $\hat{p}\colon \bigsqcup_{ i \in I} (X_i \times k^{n}) \to E$
-    makes the following diagram commute
-    \[
-    \begin{tikzcd}
-        \bigsqcup_{i \in I} (X_i \times k^{n}) \arrow{d}{\text{pr}_1}
-        \arrow{r}{\hat{p}} & E \arrow{d}{\pi} \\
-        \bigsqcup_{i \in I} X_i \arrow{r}{p} & X \\
-    \end{tikzcd}
-    \]
-    and it induces an isomorphism of prevarieties
-    \[
-    \hat{p}|_{X_i \times k^{n}}\colon X_i \times k^{n}
-    \xrightarrow{\simeq} E|_{p(X_i)}
-    = \pi^{-1}(p(X_i))
-    \]
-    such that $\pi \circ \hat{p}|_{X_i \times k^{n}} = p|_{X_i} \circ \text{pr}_1$.
-    Since $p|_{X_i}$ is an isomorphism between $X_i$ and the open subset
-    $U_i = p(X_i) \subseteq X$ with inverse $\varphi_i$, the
-    isomorphism $\hat{p}|_{X_i \times k^{n}}$
-    induces a local trivialisation
-    \begin{salign*}
-        \Phi_i \colon E|_{U_i} &\longrightarrow U_i \times k^{n} \\
-        w &\longmapsto (\pi(w), v)
-    \end{salign*}
-    where $v$ is defined as above by $\hat{p}(x, v) = w$. Note that $p(x) = \pi(w)$ in this
-    case, and that $\pi^{-1}(\{\pi(w)\}) \simeq k^{n}$
-    via $\Phi|_{\pi^{-1}(\{\pi(w)\})}$. As the isomorphism of algebraic prevarieties
-    \[
-    \Phi_j \circ \Phi_i^{-1}\colon (U_i \cap U_j) \times k^{n}
-    \longrightarrow (U_i \cap U_j) \times k^{n}
-    \] 
-    thus defined is clearly linear fibrewise, we have indeed constructed in this way
-    a vector bundle $\pi\colon E \to X$, at least in the category of algebraic prevarieties.
-
-    Note that if the prevariety $X$ obtained via the gluing of the $X_i$ is
-    a variety, then we can show that $E$ is actually a variety
-    (because the product variety $U_i \times k^{n}$ is separated). The rest of the verifications,
-    in particular the fact that for all $(\xi, v) \in U_i \cap U_j \times k^{n}$
-    \[
-    \Phi_j \circ \Phi_i^{-1}(\xi, v) = (\xi, h_{ji}(\varphi_i(\xi)) \cdot v)
-    \] is left to the reader.
-\end{bem}
-
-\begin{aufgabe}[]
-    Consider the set
-    \[
-    E \coloneqq \{ (\rho, v) \in k \mathbb{P}^{1} \times k\mathbb{P}^{2}  \mid v \in \rho\}
-    \] and the canonical map $\pi\colon E \to k\mathbb{P}^{1}$.
-
-    Show that $E$ is a vector bundle on $k\mathbb{P}^{1}$ and compute
-    its ,,cocycle of transition functions`` $g_{10}$ on the standard atlas
-    $(U_0, U_1)$ of $k\mathbb{P}^{1}$ with
-    \begin{salign*}
-        \varphi_{10}\colon k \setminus \{0\} &\longrightarrow k \setminus \{0\} \\
-        t &\longmapsto \frac{1}{t}
-    .\end{salign*}
-\end{aufgabe}
-
-\end{document}

ws2022/rav/lecture/rav11.tex

@@ -1,350 +0,0 @@
-\documentclass{lecture}
-
-\begin{document}
-
-\section{The tangent cone and the Zariski tangent space}
-
-\subsection{The tangent cone at a point}
-
-Let $X \subseteq k^{n}$ be a non-empty Zariski-closed subset.
-
-Let $P \in k[T_1, \ldots, T_n]$ be a polynomial. For all $x \in k^{n}$, we have a Taylor expansion
-at $x$: For all $h \in k^{n}$:
-\begin{salign*}
-    P(x+h) &= P(x) + P'(x)h + \frac{1}{2} P''(x) (h, h) + \underbrace{\ldots}_{\text{finite number of terms}} \\
-&= \sum_{d=0}^{\infty} \frac{1}{d!} P^{(d)}(x) (\underbrace{h, \ldots, h}_{d \text{ times}})
-.\end{salign*}
-
-\begin{bem}[]
-    The term $\frac{1}{d!} P^{(d)}(x)$ is a homogeneous polynomial of degree $d$
-    in the coordinates of $h = (h_1, \ldots, h_n)$:
-    \begin{salign*}
-    P^{(d)}(x) (h, \ldots, h)
-    &= \sum_{\alpha \in \N_0^{n}, |\alpha| = d} \frac{d!}{\alpha_1! \cdots \alpha_n!}
-    \frac{\partial^{|\alpha|}}{\partial T_1^{\alpha_1} \cdots \partial T_n^{\alpha_n}} P(x)
-    h_1^{\alpha_1} \cdots h_n^{\alpha_n}
-    .\end{salign*}
-
-    Also, when $x = 0_{k^{n}}$ and if we write
-    \[
-    P = P(0) + \sum_{d=1}^{\infty} Q_d
-    \] with $Q_d$ homogeneous of degree $d$, then for all $h = (h_1, \ldots, h_n) \in k^{n}$, we
-    have
-    \[
-    \frac{1}{d!}P^{(d)}(0) \cdot (h, \ldots, h) = Q_d(h_1, \ldots, h_n)
-    .\]
-    For all $P \in \mathcal{I}(X) \setminus \{0\} $, we denote by
-    $P_x^{*}$ the \emph{initial term} in the Taylor expansion of $P$ at $x$, i.e.
-    the term $\frac{1}{d!} P^{(d)}(x) \cdot  (h, \ldots, h)$ for the smallest
-    $d \ge 1$ such that this is not zero. If $P = 0$, we put $P_x^{*} \coloneqq 0$.
-\end{bem}
-
-\begin{definition}
-    We set $\mathcal{I}(X)_x^{*}$ to be the ideal generated
-    by $P_x^{*}$ for all $P \in \mathcal{I}(X)$.
-\end{definition}
-
-%\begin{satz}
-%    The set $\mathcal{I}(X)_x^{*}$ is an ideal of $k[T_1, \ldots, T_n]$.
-%\end{satz}
-%
-%\begin{proof}
-%    By definition, $0 \in \mathcal{I}(X)_x^{*}$. Let $P_x^{*}, Q_x^{*}$ be elements
-%    of $\mathcal{I}(X)_x^{*}$ coming from $P, Q \in \mathcal{I}(X)$. Then
-%    $P_x^{*} - Q_x^{*}$ is of the form $R_x^{*}$ for some $R \in \mathcal{I}(X)$, where
-%    $R = 0$, $R = P$, $R = Q$ or $R = P-Q$. Moreover, for $Q \in k[T_1, \ldots, T_n]$,
-%    we have $P_x^{*} Q = (PQ)_x^{*} \in \mathcal{I}(X)_x^{*}$.
-%\end{proof}
-
-\begin{bem}[]
-    The ideal $\mathcal{I}(X)^{*}$ is finitely generated. However,
-    if $\mathcal{I}(X) = (P_1, \ldots, P_m)$, it is not true in general that
-    $\mathcal{I}(X)_x^{*} = ((P_1)_x^{*}, \ldots, (P_m)_{x}^{*})$. We may need
-    to add the initial terms at $x$ of some other polynomials of the
-    form $\sum_{k=1}^{m} P_k Q_k \in \mathcal{I}(X)$.
-
-    If $\mathcal{I}(X) = (P)$ is principal though, we have $\mathcal{I}(X)_x^{*}
-    = (P_x^{*})$.
-\end{bem}
-
-\begin{definition}
-    The \emph{tangent cone} to $X$ at $x$ is the affine algebraic
-    set
-    \[
-    \mathcal{C}_x^{(X)} \coloneqq x + \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*})
-    = \{ x + h \colon h \in \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*})\} 
-    .\] 
-\end{definition}
-
-\begin{bem}
-    The algebraic set $\mathcal{C}_x(X)$ is a cone at $x$: It contains $x$ and
-    for all $x + h \in \mathcal{C}_x(X)$ for some $h \in \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*})$,
-    we have for all
-    $\lambda \in k^{\times}$,
-    $\lambda h \in \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*}$), i.e. $x + \lambda h \in
-    \mathcal{C}_x(X)$.
-
-    Indeed, $P_x^{*} \in \mathcal{I}(X)_x^{*}$ is either zero or a homogeneous polynomial of
-    degree $r \ge 1$. Thus for $h \in k^{n}$ and $\lambda \in k^{\times}$:
-    $P_x^{*}(\lambda h) = \lambda^{r} P_x^{*}(h)$ which
-    is $0$ if and only if $P_x^{*}(h) = 0$.
-\end{bem}
-
-\begin{bsp}[]
-    Let $k$ be an infinite field and let $P \in k[x,y]$ be an irreducible polynomial
-    such that $X \coloneqq \mathcal{V}(P)$ is infinite. Then we know that
-    $\mathcal{I}(X) = (P)$. Then we can determine $\mathcal{C}_X(X)$ by computing
-    the successive derviatives of $P$ at $x$: In this case
-    $\mathcal{I}(X)_x^{*} = (P_x^{*})$. For convenience wie will mostly consider examples
-    for which $x = 0_{k^2}$.
-    \begin{enumerate}[(i)]
-        \item $P(x,y) = y^2 - x^{3}$. Then $P^{*}_{(0,0)} = y^2$, so the tangent cone
-            at $(0, 0)$ is the algebraic set
-            \[
-            \mathcal{C}_{(0,0)}(X) = \{ (x,y) \in k^2  \mid y^2 = 0\} 
-            .\]
-
-        \begin{figure}[h]
-            \centering
-            \begin{tikzpicture}
-                \begin{axis}[
-                    legend style={at={(0.02, 0.98)}, anchor=north west}
-                    ]
-                \algebraiccurve[red]{y^2 - x^3}
-                \algebraiccurve[green][$y^2 = 0$]{y}
-                \algebraiccurve[blue][$y = \frac{3}{2}x - \frac{1}{2}$]{y-1.5*x + 0.5}
-                \end{axis}
-            \end{tikzpicture}
-            \caption{The green line is the tangent cone at $(0,0)$ and the blue line
-            the tangent cone at $(1,1)$.}
-        \end{figure}
-        Note that $P_{(1,1)}^{*}(h_1, h_2) = 2h_2 - 3 h_1$, so the tangent cone at
-        $(1,1)$ is
-        \begin{salign*}
-            \mathcal{C}_{(1,1)}(X) &= \{ (1 + h_1, 1 + h_2)  \mid 2h_2 - 3h_1 = 0\} \\
-                                   &= \left\{ (x,y) \in k^2  \mid y = \frac{3}{2} x - \frac{1}{2}\right\} 
-        .\end{salign*}
-    \item $P(x,y) = y^2 - x^2(x+1)$. Then $P_{(0,0)}^{*} = y^2 - x^2$ so
-        \[
-        \mathcal{C}_{(0,0)}(X) = \{ y^2 - x^2 = 0\} 
-        \] which
-        is a union of two lines.
-        \begin{figure}[h]
-            \centering
-            \begin{tikzpicture}
-                \begin{axis}[
-                    legend style={at={(0.02, 0.98)}, anchor=north west}
-                    ]
-                \algebraiccurve[red][$y^2 = x^2(x+1) $]{y^2 - x^2*(x+1)}
-                \algebraiccurve[green]{y^2 - x^2}
-                \end{axis}
-            \end{tikzpicture}
-            \caption{The green line is the tangent cone at $(0,0)$.}
-        \end{figure}
-
-        In contrast, $P_{(1,1)}^{*}(h_1, h_2) = 2h_2 - 5h_1$ so
-        \[
-        \mathcal{C}_{(1,1)}(X) = \left\{ (x,y) \in k^2  \mid y = \frac{5}{2} x - \frac{3}{2}\right\}
-        ,\] which is just one line.
-        Evidently this is related to the origin being a ,,node`` of the curve of equation
-        $y^2 - x^2(x+1) = 0$.
-    \end{enumerate}
-\end{bsp}
-
-\begin{bem}
-    \begin{enumerate}[(i)]
-        \item The tangent cone $\mathcal{C}_x(X)$ represents all directions coming out
-            of $x$ along which the initial term $P_x^{*}$
-            vanishes, for all $P \in \mathcal{I}(X)$. In that sense, it is the least complicated
-            approximation to $X$ around $x$, in terms of the degrees of the polynomials involved.
-        \item The notion of tangent cone at a point enables us to define singular points of algebraic
-            sets and even distinguish between the type of singularities:
-            Let $\mathcal{I}(X) = (P)$.
-
-            When $\text{deg}(P_x^{*}) = 1$, the tangent cone to $X \subseteq k^{n}$ at $x$
-            is just an affine hyperplane, namely $x + \text{ker } P'(x)$, since
-            $P_x^{*} = P'(x)$ in this case. The point $x$ is then called \emph{non-singular}.
-
-            When $\text{deg}(P_x^{*}) = 2$, we say that $X$ has a \emph{quadratic singularity}
-            at $x$. If $X \subseteq k^2$, a quadratic singularity is called a \emph{double point}.
-            In that case,
-            $P_x^{*} = \frac{1}{2} P''(x)$ is a quadratic form on $k^2$. If it is non-degenerate,
-            then $x$ is called an \emph{ordinary} double point. For instance,
-            if $X$ is the nodal cubic of equation $y^2 = x^2(x+1)$, then the origin is
-            an ordinary double point (also called a \emph{node}), since
-            $\frac{1}{2}P''(0,0)$ is the quadratic form associated to the symmetric matrix
-            $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} $.
-            But if $X$ is the cuspidal cubic of equation $y^2 = x^{3}$, then
-            the origin is \emph{not} an ordinary double point, since
-            $\frac{1}{2}P''(0,0)$ corresponds to $\begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix} $.
-            Instead, the origin is a \emph{cusp} in the following sense. We can write
-            \[
-            P(x,y) = l(x,y)^2 + Q_3(x,y) + \ldots
-            \] with $l(x,y) = \alpha x + \beta y$ a linear form in $(x,y)$, and the double point
-            $(0,0)$ is called a cusp if $Q_3(\beta, -\alpha) \neq 0$. This means that
-            \[
-            t ^{4}X P(\beta t, - \alpha t)
-            \] in $k[t]$. And this is indeed what happens for $P(x,y) = y^2 - x^{3}$, since
-            $l(x,y) = y$ and $Q_3(x,y) = -x^{3}$.
-    \end{enumerate}
-\end{bem}
-
-\begin{bem}[]
-    One can define the \emph{multiplicity} of a point $(x,y) \in \mathcal{V}_{k^2}(P)$ as
-    the smallest integer $r \ge 1$ such that $P^{(r)}(x,y)\neq 0$.
-    If $P^{(r)}(x,y) \cdot (h, \ldots, h) = 0 \implies h = 0_{k^2}$, the singularity
-    $(x,y)$ is called \emph{ordinary}. If $k$ is algebraically closed and
-    $(x,y) = (0,0)$, we can write
-    $P^{(r)}(0, 0) = \prod_{i=1}^{m} (\alpha_i x + \beta_i y)^{r_i} $,
-    with $r_1 + \ldots + r_m = r$. Then $(0,0)$ is an ordinary singularity of multiplicity $r$
-    iff $r_i = 1$ for all $i$. For instance, $(0,0)$ is an ordinary triple point of the trefoil
-    curve $P(x,y) = (x^2 + y^2)^2 + 3x^2 y - y^{3}$.
-\end{bem}
-
-\subsection{The Zariski tangent space at a point}
-
-Let $X \subseteq k^{n}$ be a Zariski-closed subset and $x \in X$.
-
-The tangent cone is in general not a linear approximation. To remedy this, one can
-consider the Zariski tangent space to $X$ at a point $x \in X$.
-
-\begin{definition}
-    The \emph{Zariski tangent space} to $X$ at $x$ is the affine subspace
-    \[
-    T_xX \coloneqq x + \bigcap_{P \in \mathcal{I}(X)} \text{ker } P'(x)
-    .\]
-\end{definition}
-
-\begin{bem}[]
-    By translation, $T_xX$ can be canonically identified to the vector space
-    $\bigcap_{P \in \mathcal{I}(X)} \text{ker } P'(x) $.
-\end{bem}
-
-\begin{satz}[]
-    View the linear forms
-    \[
-    P'(x) \colon h \mapsto P'(x) \cdot h
-    \] as homogeneous polynomials of degree $1$ in the coordinates of $h \in k^{n}$ and
-    denote by
-    \[
-    \mathcal{I}(X)_x \coloneqq (P'(x) : P \in \mathcal{I}(X))
-    \] the ideal generated by these polynomials. Then
-    \[
-    T_xX = x + \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x)
-    .\]
-\end{satz}
-
-\begin{proof}
-    It suffices to check that
-    \[
-    \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x) = \bigcap_{P \in \mathcal{I}(X)} \text{ker } P'(x)
-    \]
-    which is obvious because the $(P'(x))_{P \in \mathcal{I}(X)}$ generate $\mathcal{I}(X)_x$.
-\end{proof}
-
-\begin{korollar}
-    $T_xX \supseteq \mathcal{C}_x(X)$
-    \label{kor:cone-in-tangent-space}
-\end{korollar}
-
-\begin{proof}
-    Since $\mathcal{I}(X)_x \subseteq \mathcal{I}(X)_x^{*}$, one has
-    $\mathcal{V}_{k^{n}}(\mathcal{I}(X)_x) \supseteq \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*})$.
-\end{proof}
-
-\begin{definition}
-    If $T_xX = \mathcal{C}_x(X)$, the point $x$ is called \emph{non-singular}.
-\end{definition}
-
-\begin{satz}
-    If $\mathcal{I}(X) = (P_1, \ldots, P_m)$, then
-    $\mathcal{I}(X)_x = (P_1'(x), \ldots, P_m'(x))$
-\end{satz}
-
-\begin{proof}
-    By definition,
-    \[
-        (P_1'(x), \ldots, P_m'(x)) \subseteq (P'(x) : P \in \mathcal{I}(X)) = \mathcal{I}(X)_x
-    .\] But for $P \in \mathcal{I}(X)$, there exist $Q_1, \ldots, Q_m \in k[T_1, \ldots, T_n]$ such
-    that $P = \sum_{i=1}^{m} Q_i P_i$, so
-    \begin{salign*}
-        P'(x) &= \sum_{i=1}^{m} (Q_i P_i)'(x) \\
-              &= \sum_{i=1}^{m} (Q_i'(x) \underbrace{P_i(x)}_{= 0} + \overbrace{Q_i(x)}^{\in k}
-              P_i'(x))
-    \end{salign*}
-    since $x \in X$. This proves that $P'(x)$ is in fact a linear combination of the linear
-    forms $(P_i'(x))_{1 \le i \le m}$.
-\end{proof}
-
-\begin{korollar}
-    If $\mathcal{I}(X) = (P_1, \ldots, P_m)$, then
-    $T_xX = x + \bigcap_{i=1}^{m} \operatorname{ker } P_i'(x)$.
-    Moreover, if we write $P = (P_1, \ldots, P_m)$, and view this
-    $P$ as a polynomial map $k^{n} \to k^{m}$, then
-    \[
-    T_xX = x + \text{ker } P'(x)
-    \] with $P'(x)$ the Jacobian of $P$ at $x$, i.e.
-    \[
-        P'(x) = \begin{pmatrix} \frac{\partial P_1}{\partial T_1}(x) & \cdots & \frac{\partial P_1}{\partial T_n}(x) \\
-        \vdots & & \vdots \\
-        \frac{\partial P_m}{\partial T_1}(x) & \cdots & \frac{\partial P_m}{\partial T_n}(x)
-    \end{pmatrix} 
-    .\] In particular, $\text{dim } T_xX = n - \operatorname{rk } P'(x)$.
-    \label{kor:tangent-kernel-jacobian}
-\end{korollar}
-
-\begin{bsp}
-    \begin{enumerate}[(i)]
-        \item $X = \{ y^2 - x^{3} = 0\} \subseteq k^2$. Then $\mathcal{I}(X) = (y^2 - x^{3})$,
-            so,
-            \[
-            T_{(0,0)}X = (0,0) + \text{ker} \begin{pmatrix} 0 & 0 \end{pmatrix} = k^2
-            .\] 
-            which strictly contains the tangent cone $\{y^2 = 0\} $. In particular,
-            the origin is indeed a singular point of the cuspidal cubic. In general,
-            \[
-            T_{(x,y)}X = (x,y) + \text{ker} \begin{pmatrix} -3x^2 & 2y \end{pmatrix} 
-            ,\] 
-            which is an affine line if $(x,y) \neq (0,0)$.
-        \item $X = \{ y^2 - x^2 - x^{3} = 0\} \subseteq k^2$. Then
-            $\mathcal{I}(X) = (y^2 - x^2 - x^{3})$, so
-            \[
-                T_{(0,0)}X = (0,0) + \text{ker} \begin{pmatrix} 0 & 0 \end{pmatrix} = k^2
-            \] which again strictly contains the tangent cone $\{y = \pm x\} $. In general,
-            \[
-                T_{(x,y)}X = (x,y) + \text{ker} \begin{pmatrix} -2x & 2y \end{pmatrix}
-            ,\] which is an affine line if $(x,y) \neq (0,0)$.
-    \end{enumerate}
-\end{bsp}
-
-\begin{bem}
-    The dimension of the Zariski tangent space at $x$ (as an affine subspace of $k^{n}$)
-    may vary with $x$.
-\end{bem}
-
-%\begin{satz}[a Jacobian criterion]
-%    If $(P_1, \ldots, P_m)$ are polynomials such that
-%    $\mathcal{I}(X) = (P_1, \ldots, P_m)$ and $\operatorname{rk } P'(x) = m$, where
-%    $P = (P_1, \ldots, P_m)$, then $x$ is a non-singular point of $X$.
-%\end{satz}
-%
-%\begin{proof}
-%    By \ref{kor:cone-in-tangent-space} and \ref{kor:tangent-kernel-jacobian} it suffices to show that
-%    \[
-%    \mathcal{C}_x(X) \supseteq x + \bigcap_{i=1} ^{m} \text{ker } P_i'(x)
-%    .\] By definition
-%    \[
-%    \mathcal{C}_x(X) = x + \mathcal{V}_{k^{n}}(\mathcal{I}(X)_x^{*})
-%    \] and $\mathcal{I}(X)_x^{*} = \{ Q_x^{*} : Q \in \mathcal{I}(X)\} $. If $Q \in \mathcal{I}(X)$,
-%    there exist polynomials $Q_1, \ldots, Q_m$ such that
-%    $Q = \sum_{i=1}^{m} Q_i P_i$, so $Q_x^{*}$ is a linear combination of the $(P_i)_x^{*}$.
-%    Since $\text{rk }(P_1'(x), \ldots, P_m'(x)) = m$, we have
-%    $P_i'(x) \neq 0$ for all $i$. So $(P_i)_x^{*} = P_i'(x)$ in the Taylor expansion
-%    of $P_i$ at $x$. So $Q_x^{*}$ is a linear combination
-%    of $(P_1'(x), \ldots, P_m'(x))$,
-%    which proves that if $h \in \bigcap_{i=1}^{m} \text{ker } P_i'(x)$, then
-%    $Q_x^{*}(h) = 0$ for all $Q \in \mathcal{I}(X)$, hence
-%    $x + h \in \mathcal{C}_x(X)$.
-%\end{proof}
-
-\end{document}

ws2022/rav/lecture/rav15.tex

@@ -1,311 +0,0 @@
-\documentclass{lecture}
-
-\begin{document}
-
-\usetikzlibrary{shapes.misc}
-\tikzset{cross/.style={cross out, draw=black, minimum size=2*(#1-\pgflinewidth), inner sep=0pt, outer sep=0pt},
-%default radius will be 1pt. 
-cross/.default={1pt}}
-
-\chapter{Hilbert's Nullstellensatz and applications}
-
-\section{Fields of definition}
-
-When $k$ is an algebraically closed field, Hilbert's Nullstellensatz gives us a bijection
-between algebraic subsets of $k^{n}$ and radical ideals in $k[T_1, \ldots, T_n]$.
-
-This correspondence induces an anti-equivalence of categories
-\begin{salign*}
-    \{\text{affine } k\text{-varieties}\} &\longleftrightarrow
-    \{\text{finitely-generated reduced } k \text{-algebras}\} \\
-    (X, \mathcal{O}_X) &\longmapsto \mathcal{O}_X(X) \\
-    \hat{A} = \operatorname{Hom}_{k\mathrm{-alg}}(A, k) &\longmapsfrom A
-.\end{salign*}
-
-\begin{lemma}
-    Let $k$ be algebraically closed and $A$ a finitely-generated $k$-Algebra. Then
-    the map
-    \begin{salign*}
-        \hat{A} = \operatorname{Hom}_{k\text{-alg}}(A, k) &\longrightarrow \operatorname{Spm } A \\
-    \xi &\longmapsto \text{ker } \xi
-    \end{salign*}
-    is a bijection.
-\end{lemma}
-
-\begin{proof}
-    The map
-    admits an inverse
-    \begin{salign*}
-        \operatorname{Spm } A &\longrightarrow \operatorname{Hom}_{k\text{-alg}}(A, k) \\
-        \mathfrak{m} &\longmapsto (A \to A / \mathfrak{m})
-    .\end{salign*}
-    This is well-defined, since $A / \mathfrak{m}$ is a finite extension of the algebraically closed field
-    $k$, so $k \simeq A / \mathfrak{m}$.
-\end{proof}
-
-Since we have defined a product on the left-hand side of the anti-equivalence, this must correspond
-to coproduct on the right-hand side. Since the coproduct in the category of commutative
-$k$-algebras with unit is given by the tensor product, we have
-\[
-\mathcal{O}_{X \times Y} (X \times Y) \simeq \mathcal{O}_X(X) \otimes_k \mathcal{O}_Y(Y)
-.\]
-
-\begin{korollar}
-    Let $k$ be algebraically closed. Then the tensor product of two
-    reduced (resp. integral) finitely-generated $k$-algebras is reduced (resp. integral).
-    \label{kor:k-alg-closed-tensor-of-reduced}
-\end{korollar}
-
-\begin{proof}
-    This follows from the anti-equivalence of categories: Reduced since products of affine
-    $k$-varieties exist and integral since the product of two irreducible affine $k$-varieties is irreducible.
-\end{proof}
-
-\begin{bem}
-    \ref{kor:k-alg-closed-tensor-of-reduced} is false in general if $k = \overline{k}$. For instance
-    $\mathbb{C}$ is an integral $\R$-algebra, but
-    \begin{salign*}
-        \mathbb{C} \otimes_{\R} \mathbb{C}
-        &= \R[x]/(x^2 + 1) \otimes_{\R} \mathbb{C} \\
-        &= \mathbb{C}[x]/(x^2 + 1) \\
-        &= \mathbb{C}[x]/((x-i)(x+i)) \\
-        &\stackrel{(*)}{\simeq} \mathbb{C}[x]/(x-i) \times \mathbb{C}[x]/(x+i) \\
-        &\simeq \mathbb{C} \times \mathbb{C}
-    \end{salign*}
-    is not integral, where $(*)$ follows from the Chinese remainder theorem.
-
-    For a non-reduced example, consider $k = \mathbb{F}_{p}(t)$ and choose a $p$-th root
-    $\alpha = t ^{\frac{1}{p}}$ in $\overline{\mathbb{F}_p(t)}$. Then $\alpha \not\in k$
-    but $\alpha ^{n} \in k$. If we put $L = k(\alpha)$, then
-    $\alpha \otimes 1 - 1 \otimes \alpha \neq 0$ in $L \otimes_k L$ since
-    the elements $(\alpha ^{i} \otimes \alpha ^{j})_{0 \le i, j \le p-1}$ form a basis
-    of $L \otimes_k L$ as a $k$-vector space, but
-    \[
-        (\alpha \otimes 1 - 1 \otimes \alpha)^{p}
-        = \alpha ^{p} \otimes 1 - 1 \otimes \alpha ^{p}
-        = 1 \otimes \alpha ^{p} - 1 \otimes \alpha ^{p} = 0
-    .\] 
-\end{bem}
-
-We now consider more generally finitely generated reduced $k$-algebras when $k$ is not
-necessarily closed.
-
-\begin{bsp}
-    Let $A = \R[X]/(x^2 +1)$. Since $x^2 + 1 $ is irreducible in $\R[x]$, it
-    generates a maximal ideal, thus the finitely-generated $\R$-algebra $A$ is a field and in
-    particular reduced. We can equip the topogical space
-    $X \coloneqq \operatorname{Spm } A = \{ (0)\} $ with a sheaf of regular functions, defined
-    by $\mathcal{O}_X(\{(0)\}) = A$. In other words, $\operatorname{Spm } A$ is just a point,
-    but equipped with the reduced $\R$-algebra $A$. It thus differs from the
-    point $\operatorname{Spm } \R$, which is equipped with the reduced $\R$-algebra $\R$,
-    since $\R[x]/(x^2 + 1) \not\simeq \R$ as $\R$-algebras. Indeed, the $\R$-algebra $\R[x]/(x^2+1)$
-    is $2$ dimensional as a real vector space.
-
-    $A$ possesses a non-trivial $\R$-algebra automorphism induced by the automorphism of $\R$-algebras,
-    $P \mapsto P(-x)$ in $\R[x]$. Indeed, $\R[x]/(x^2+1) \simeq \mathbb{C}$ as $\R$-algebras,
-    with the previous automorphism corresponding to the complex conjugation $z \mapsto \overline{z}$.
-\end{bsp}
-
-\begin{bsp}
-    By analogy with the Zariski topology on maximal spectra of (finitely generated, reduced)
-    $\mathbb{C}$-algebras, we can equip $X = \operatorname{Spm } A$ with
-    a Zariski topology for all (finitely generated reduced) $\R$-algebras $A$: the closed subsets
-    of this topology are given by
-    \[
-    \mathcal{V}_X(I) \coloneqq \{ \mathfrak{m} \in \operatorname{Spm } A  \mid \mathfrak{m} \supset I\} 
-    \] for any ideal $I \subseteq A$.
-    Note that $X = \operatorname{Spm } A$ contains
-    $\hat{A} = \operatorname{Hom}_{k\text{-alg}}(A, k)$ as a subset: the points
-    of $\hat{A}$ correspond to maximal ideals $\mathfrak{m}$ of $A$ with residue field
-    $A / \mathfrak{m} \simeq k$. But when $k \not\simeq \overline{k}$, the set
-    $\operatorname{Spm } A$ is strictly larger than $\hat{A}$: it contains maximal ideals $\mathfrak{m}$
-    such that $A / \mathfrak{m}$ is a non-trivial finite extension of $k$. The induced topology on
-    $\hat{A} \subseteq \operatorname{Spm } A$ is the Zariski topologoy of $\hat{A}$ that was
-    introduced earlier.
-
-    Let $A = \R[x]$. Maximal ideals in the principal ring $\R[x]$ are generated
-    by a single irreducible polynomial $P$, which is either of degree $1$ or of degree $2$ with
-    negative discriminant.
-    
-    In the first case, $P = x-a$ for some $a \in \R$ and the residue field is $\R[x]/(x - a) \simeq \R$,
-    while, in the second case, $P = x^2 + bx + c$ for $b, c \in \R$ and $b^2 - 4c < 0$ and
-    by choosing a root $z_0$ of $P$ in $\mathbb{C}$, the map
-    \begin{salign*}
-        \eta_{z_0} \colon \R[x]/(x^2 + bx + c) &\longrightarrow \mathbb{C} \\
-                             \overline{P} &\longmapsto P(z_0)
-    \end{salign*}
-    is a field-homomorphism. In particular it is injective. Since $\mathbb{C}$ and
-    $\R[x]/(x^2 + bx + c)$ are both degree $2$ extensions of $\R$, we have
-    $\R[x]/(x^2 + bx + c) \simeq \mathbb{C}$.
-    Note that the other root of $x^2 + bx +c $ is $\overline{z_0}$ and that
-    $\eta_{\overline{z_0}} = \sigma \circ \eta_{z_0}$ where $\sigma$ is complex conjugation on $\mathbb{C}$.
-    So we have to ways to identify $\R[x]/(x^2 + bx +c)$ to $\mathbb{C}$ and they are
-    related by the action of $\text{Gal}(\mathbb{C}/ \R)$ on $\mathbb{C}$.
-
-    To sum up, the difference between the two possible types of maximal ideals $\mathfrak{m} \subseteq \R[x]$
-    is the residue field, which is either $\R$ or $\mathbb{C}$. When it is $\R$, we
-    find exactly the points of
-    \begin{salign*}
-        \widehat{\R[x]}  &= \operatorname{Hom}_{\R\text{-alg}}(\R[x], \R) \\
-                     &\simeq \{ \mathfrak{m} \in \operatorname{Spm } \R[x]  \mid \R[x]/\mathfrak{m} \simeq \R\} \\
-                     &\simeq \{ (x-a) \colon a \in \R\} \\
-                     &\simeq \R
-    .\end{salign*}
-    And when the residue field is $\mathbb{C}$, we have $\mathfrak{m} = (x^2 + bx + c)$ with
-    $b, c \in \R$ such that $b^2 - 4c < 0$. If we choose $z_0$ to be the root
-    of $x^2 + bx +c$ with $\text{Im}(z_0) > 0$, we can identify the set of these maximal ideals with
-    the subset
-    \[
-    H \coloneqq \{ z \in \mathbb{C}  \mid \text{Im}(z) > 0\} 
-    .\]
-    In other words, the following pictures emerges, where we identify
-    $\operatorname{Spm } \R[x]$ with
-    \[
-    \hat{H} \coloneqq \{ z \in \mathbb{C} \mid \text{Im}(z) \ge 0\} 
-    \] 
-    via the map
-    \begin{salign*}
-        \operatorname{Spm } \R[x] &\longrightarrow \hat{H} \\
-        \mathfrak{m} &\longmapsto \begin{cases}
-            a \in \R & \mathfrak{m} = (x-a) \\
-            z_0 \in H & \mathfrak{m} = ((x-z_0)(x-\overline{z_0})) \text{ and } \text{Im}(z_0) > 0
-        \end{cases}
-    \end{salign*}
-    which is indeed bijective.
-    %\begin{figure}
-    %    \centering
-    %    \begin{tikzpicture}
-    %        \draw[red] (-2, 0) -- (2,0) node[right] {$\R \simeq \operatorname{Hom}_{\R\text{-alg}}(\R[x], \R)$};
-    %        \draw[->] (0, 0) -- (0,4);
-    %    \end{tikzpicture}
-    %    \caption{$\operatorname{Spm } \R[x] \simeq \hat{H}
-    %        = \left\{ z \in \mathbb{C} : \text{Im}(z) \ge 0 \right\}$}
-    %\end{figure}
-    We see that $\operatorname{Spm } \R[x]$ contains a lot more points
-    that $\R$. One could go further and add the ideal $(0)$: This would give the set
-    \[
-    \mathbb{A}^{1}_{\R} = \operatorname{Spec } \R[x]
-    = \operatorname{Spm } \R[x] \cup \{(0)\}
-    .\] 
-\end{bsp}
-
-\begin{bem}
-    If $A$ is a $k$-algebra and $\overline{k}$ is an algebraic closure of $k$, the
-    group $\text{Aut}_k(\overline{k})$ acts on the $\overline{k}$-algebra
-    $A_{\overline{k}} \coloneqq A \otimes_k \overline{k}$ via
-    $\sigma (a \otimes \lambda) \coloneqq a \otimes \sigma(\lambda)$. Moreover, the map
-    $a \mapsto a \otimes 1$ induces an injective morphism of $k$-algebras
-    $A \xhookrightarrow{} A \otimes_k \overline{k}$ since
-    the tensor product over fields is left-exact.
-    Its image is contained in the $k$-subalgebra
-    $\operatorname{Fix}_{\operatorname{Aut}_k(\overline{k})} A_{\overline{k}} \subseteq A_{\overline{k}}$. When
-    $k$ is a perfect field, this inclusion is an equality.
-\end{bem}
-
-\begin{bsp}
-    If $A = \R[x]$, then $A \otimes_{\R} \mathbb{C} \simeq \mathbb{C}[x]$. The group
-    $\text{Aut}_{\R}(\mathbb{C}) = \text{Gal}(\mathbb{C}/\R) = \langle \sigma \rangle$  with
-    $\sigma\colon z \mapsto \overline{z}$, acts naturally on $\mathbb{C}[x]$. This
-    is an action by $\R$-algebra automorphisms. Clearly,
-    $\text{Fix}_{\langle\sigma\rangle} \mathbb{C}[x] = \R[x]$. There
-    is an induced action on $\operatorname{Spm } \mathbb{C}[x]$,
-    defined by
-    \[
-    \sigma(\mathfrak{m}) = \sigma((x-z)) \coloneqq (x - \sigma(z)) = (x - \overline{z})
-    .\] 
-    When we identify $\operatorname{Spm } \mathbb{C}[x]$ with $\mathbb{C}$
-    via $(x-z) \mapsto z$, this action is just $z \mapsto \overline{z}$. This
-    ,,geometric action'' induces an action of $\text{Gal}(\mathbb{C}/\R)$ on
-    regular functions on $\mathbb{C}$: to $h \in \mathcal{O}_{\mathbb{C}}(U)$, there
-    is associated a regular function $h \in \mathcal{O}_{\mathbb{C}}(\sigma(U))$, defined for
-    all $x \in \sigma(U)$, by
-    \[
-    \sigma(h)(z) \coloneqq \sigma \circ h \circ \sigma ^{-1}(z) = \overline{h(\overline{z})}
-    .\]
-    In particular, if $h = P \in \mathcal{O}_{\mathbb{C}}(\mathbb{C}) = \mathbb{C}[x]$, then
-    $P \mapsto \sigma(P)$ coincides with the natural
-    $\text{Gal}(\mathbb{C} / \R)$ action on $\mathbb{C}[x]$. We will see momentarily that this
-    defines a sheaf of $\R$-algebras on $\operatorname{Spm } \R[x]$. To that end,
-    let us first look more closely at the $\text{Gal}(\mathbb{C}/ \R)$ action
-    on $\operatorname{Spm } \mathbb{C}[x]$. Its fixed-point set
-    is
-    \[
-    \{ \mathfrak{m} \in \operatorname{Spm } \mathbb{C}[x]  \mid \mathfrak{m} = (x-a), a \in \R\}
-    \simeq \R = \operatorname{Fix}_{z \mapsto \overline{z}}(\mathbb{C})
-    .\]
-    Moreover, there is a map
-    \begin{salign*}
-        \operatorname{Spm } \mathbb{C}[x] &\longrightarrow \operatorname{Spm } \R[x] \\
-        \mathfrak{m} &\longmapsto \mathfrak{m} \cap \R[x]
-    \end{salign*}
-    sending $(x-a) \mathbb{C}[x]$ to $(x-a)\R[x]$ if $a \in \R$,
-    and $(x-z)\mathbb{C}[x]$ to $(x-z)(x-\overline{z})\R[x]$ if $z \in \mathbb{C} \setminus \R$.
-    This map is surjective and induces a bijection
-    \[
-        (\operatorname{Spm } \mathbb{C}[x]) / \operatorname{Gal}(\mathbb{C} / \R)
-        \xlongrightarrow{\simeq} \operatorname{Spm } \R[x]
-    .\]
-    Geometrically, the quotient map $\pi\colon \operatorname{Spm } \mathbb{C}[x] \to \operatorname{Spm } \R[x]$
-    is the ,,folding map``
-    \begin{salign*}
-        \mathbb{C} &\longrightarrow \hat{H} \\
-        z = u + iv &\longmapsto u + i |v|
-    .\end{salign*}
-    \begin{figure}
-        \centering
-        \begin{tikzpicture}
-            \draw[red] (-2, 0) -- (2,0) node[right] {$\R$};
-            \fill (1, -1) circle[radius=0.75pt] node[right] {$z_0$};
-            \draw[->] (0,-1.5) -- (0,2);
-            \draw[->] (3.2,0) -- node[above] {$\pi$} (4.2,0);
-            \draw[red] (5, 0) -- (9,0) node[right] {$\R$};
-            \fill (8, 1) circle[radius=0.75pt] node[right] {$\pi(z_0)$};
-            \draw[->] (7, 0) -- (7,2);
-        \end{tikzpicture}
-        \caption{The quotient map
-        $\pi\colon \operatorname{Spm } \mathbb{C}[x] \to \operatorname{Spm } \R[x]$ is geometrically a folding.}
-    \end{figure}
-    In view of this, it is natural to
-    \begin{enumerate}[(i)]
-        \item put the quotient topology on
-            \[
-            \operatorname{Spm } \R[x] = \left( \operatorname{Spm } \mathbb{C}[x] \right)
-            / \operatorname{Gal}(\mathbb{C}/\R)
-            \]
-            where $\operatorname{Spm } \mathbb{C}[x] \simeq \mathbb{C}$ is equipped with its topology
-            of algebraic variety.
-        \item define a sheaf of $\R$-algebras on $\operatorname{Spm } \R[x]$ by pushing-forward
-            the structure sheaf on $\operatorname{Spm } \mathbb{C}[x]$
-            and then taking the $\operatorname{Gal}(\mathbb{C}/ \R)$-invariant subsheaf:
-            \[
-            \mathcal{O}_{\operatorname{Spm } \R[x]}(U)
-            \coloneqq \mathcal{O}_{\operatorname{Spm } \mathbb{C}[x]}
-            (\pi^{-1}(U))^{\operatorname{Gal}(\mathbb{C} / \R)}
-            \] where
-            $\pi\colon \operatorname{Spm } \mathbb{C}[x] \to \operatorname{Spm } \R[x]$,
-            $\mathfrak{m} \mapsto \mathfrak{m} \cap \R[x]$ is the quotient map,
-            and $\operatorname{Gal}(\mathbb{C}/ \R)$ acts on
-            $\mathcal{O}_{\operatorname{Spm } \mathbb{C}[x]}(\pi^{-1}(U))$ via
-            $h \mapsto \sigma(h) = \sigma \circ h \circ \sigma ^{-1}$ (note that the open set
-            $\pi^{-1}(U)$ is $\operatorname{Gal}(\mathbb{C} / \R)$-invariant).
-    \end{enumerate}
-    Observe that
-    \[
-    \mathcal{O}_{\operatorname{Spm } \R[x]}(\operatorname{Spm } \R[x])
-    = \mathbb{C}[x]^{\operatorname{Gal}(\mathbb{C} / \R)} = \R[x]
-    .\]
-    Also, if $h = \frac{f}{g}$ around $x \in U$, then, around
-    $\sigma(x) \in U$, one has $\sigma(h) = \frac{\sigma(f)}{\sigma(g)}$ and,
-    for all $\lambda \in \mathbb{C}$, $\sigma(\lambda h) = \overline{\lambda} \sigma(h)$.
-
-    Remarkably, we will see that we can reconstruct the algebraic $\mathbb{C}$-variety
-    \[
-        (X_{\mathbb{C}}, \mathcal{O}_{X_{\mathbb{C}}})
-        \coloneqq (\operatorname{Spm } \mathbb{C}[x], \mathcal{O}_{\operatorname{Spm } \mathbb{C}[x]})
-    \] from the ringed space
-    \[
-        (X, \mathcal{O}_X) \coloneqq (\operatorname{Spm } \R[x], \mathcal{O}_{\operatorname{Spm } \R[x]}
-    \] that we have just constructed.
-\end{bsp}
-
-\end{document}

ws2022/rav/lecture/rav16.tex

@@ -1,284 +0,0 @@
-\documentclass{lecture}
-
-\begin{document}
-
-\chapter{Real algebra}
-
-\section{Ordered fields and real fields}
-
-\begin{definition}[]
-    An \emph{ordered field} is a pair $(k, \le)$ consisting of a field $k$ and
-    an order relation $\le$ such that
-    \begin{enumerate}[(i)]
-        \item $\le $ is a total order: if $x, y \in k$, then $x \le y$ or $y \le x$.
-        \item $\le $ is compatible with addition in $k$:
-            if $x, y, z \in k$, then $x \le y$ implies $x + z \le y + z$.
-        \item $\le $ is compatible with multiplication in $k$:
-            if $x, y\in k$, then $0 \le x$ and $0 \le y$ implies $0 \le xy$.
-    \end{enumerate}
-    A morphism between two ordered fields $(k, \le)$ and $(L, \le)$ is a field homomorphism
-    $\varphi\colon k \to L$ such that $x \le y$ in $k$ implies $\varphi(x) \le \varphi(y)$ in $L$.
-\end{definition}
-
-\begin{bsp}[]
-    \begin{enumerate}[(1)]
-        \item The fields $\Q$ and $\R$, equipped with their usual orderings, are
-            ordered fields.
-        \item The field $\mathbb{C}$ can be equipped with a total ordering
-            (the ,,lexicographic order``) but not with a structure of
-            ordered field.
-        \item The field $\R(t)$ of rational fractions with coefficients in $\R$, can
-            be equipped with a structure of ordered field in multiple ways:
-
-            Fix an $x \in \R$ and, for all polynomial $P \in \R[t]$, use
-            Taylor expansion at $x$ to write
-            \[
-            P(t) = a_p (t - x)^{p} + \text{higher order terms}
-            .\] 
-            with $a_p \neq 0$, then define
-            $P(t) >_{x^{+}} 0$ if $a_p > 0$, i.e. if the function
-            $t \mapsto P(t)$ is positive on a small interval $(x, x + \epsilon)$. Set also
-            $\frac{P(t)}{Q(t)} >_{x^{+}} 0$ if $P(t)Q(t) >_{x^{+}} 0$,
-            and define $f \le_{x^{+}} g$ in $\R(t)$ if either $f = g$ or
-            $g - f >_{x^{+}} 0$.
-            Equivalently $f \le_{x^{+}} g$ in $\R(t)$ if
-            either $f = g$ or $g - f$ is positively-valued on $(x, x + \epsilon)$ for $\epsilon > 0$ small
-            enough.
-
-            It is clear that this is a total ordering on $\R(t)$, and that this ordering is compatible
-            with addition and multiplication in the sense of the definition of an ordered field.
-            Moreover, the substitution homomorphism
-            $h(t) \mapsto h(t - x)$ induces an isomorphism of ordered fields
-            $(\R(t), \le_{0^{+}}) \xlongrightarrow{\simeq} (\R(t), \le_{x^{+}})$,
-            since a function $t \mapsto h(t - x)$ is positively-valued
-            on $(x, x + \epsilon)$ if and only if the function $t \mapsto h(t)$ is positively valued
-            on $(0, \epsilon)$.
-
-            Note that we can also define orderings on $\R(t)$ by setting $f \le_{x^{-}} g$
-            if either $f = g$ or $g - f$ is positively-valued
-            on $(x-\epsilon, x)$, for $\epsilon > 0$ small enough.
-            The substitution homomorphism $h(t) \mapsto h(-t)$ induces an isomorphism
-            of ordered fields
-            $(\R(t), \le_{0^{-}}) \xlongrightarrow{\simeq} (\R(t), \le_{0^{+}})$.
-    \end{enumerate}
-\end{bsp}
-
-\begin{bem}[]
-    The ordered field $(\R(t), \le_{0^{+}})$
-    is non-Archimedean: the element $t$ is
-    \emph{infinitely small with respect to any real $\delta > 0$} in the sense that for all $n \in \N$,
-    $nt < \delta$ (indeed $t \mapsto n t - \delta$ is negatively-valued
-    on $(0, \epsilon)$ for $\epsilon > 0$ small enough). Equivalently, $\frac{1}{t}$
-    is infinitely large with respect to $ 0 < \delta \in \R$ in the sense that
-    $\frac{1}{t} > n \delta$ for all $n \in \N$.
-\end{bem}
-
-\begin{satz}[]
-    Let $(k, \le)$ be an ordered field and $x, y, z \in k$. Then the following properties hold:
-    \begin{enumerate}[(a)]
-        \item $x \ge 0$ or $- x \ge 0$.
-        \item $-1 < 0$ and $1 > 0$.
-        \item $k$ is of characteristic $0$.
-        \item if $x < y$ and $z > 0$, then $x z < y z$.
-        \item if $x < y$ and $z < 0$, then $x z > y z$.
-        \item $x y \ge 0$ if and only if $x$ and $y$ have the same sign.
-        \item $x^2 \ge 0$ and, if $x \neq 0$, then $x$ and $\frac{1}{x}$ have the same sign.
-        \item if $0 < x \le y$, then $0 < \frac{1}{y} \le \frac{1}{x}$.
-    \end{enumerate}
-    \label{satz:ordered-field-basics}
-\end{satz}
-
-\begin{proof}
-    Elementary verifications.
-\end{proof}
-
-It turns out that it is possible to characterise ordered fields without explicitly mentioning
-the order relation, using cones of positive elements.
-
-\begin{definition}
-    Let $k$ be a field. A \emph{cone} in $k$ is a subset $P \subseteq k$ such that
-    for all $x, y \in P$ and $z \in k$:
-    \begin{enumerate}[(i)]
-        \item $x + y \in P$
-        \item $xy \in P$
-        \item $z^2 \in P$
-    \end{enumerate}
-    A cone $P \subseteq k$ is called a \emph{positive cone} if, additionally, one has:
-    \begin{enumerate}[(i)]
-        \setcounter{enumi}{3}
-        \item $-1 \not\in P$
-    \end{enumerate}
-\end{definition}
-
-\begin{satz}[]
-    Let $k$ be a field. Assume that there exists a positive cone $P \subseteq k$. Then:
-    \begin{enumerate}[(i)]
-        \item $0 \in P$ and $1 \in P$.
-        \item $k$ is of characteristic $0$.
-        \item $P \cap (-P) = \{0\}$
-    \end{enumerate}
-\end{satz}
-
-\begin{proof}
-    \begin{enumerate}[(i)]
-        \item $0 = 0^2 \in P$ and $1 = 1^2 \in P$ by axiom (iii).
-        \item Since $1 \in P$, by induction and axiom (i),
-            $n \cdot 1 = \underbrace{1 + \ldots + 1}_{n \text{ times}} \in P$ for all $n \in \N$.
-            Assume that there exists $n \in \N$, such that $n \cdot 1 = 0$ in $k$.
-            Since $1 \neq 0$ in $k$, it follows $n \ge 2$ so,
-            \[
-            -1 = 0 - 1 =  n \cdot 1 - 1 = (n - 1) \cdot 1 \in P
-            ,\] which contradicts axiom (iv).
-        \item Assume that there exists $x \in P \cap (-P) \setminus \{0\}$. In particular
-            $x \neq 0$ and $-x \in P$. So
-            $- x^2 = (-x) x \in P$ by axiom (ii) and $\frac{1}{x^2} = \left( \frac{1}{x} \right)^2 \in P$
-            by axiom (iii). Again by axiom (ii)
-            \[
-            -1 = \frac{1}{x^2} (-x^2) \in P
-            \] which contradicts axiom (iv).
-    \end{enumerate}
-\end{proof}
-
-Given a positive cone $P$ in a field $k$, let us set $P^{+} = P \setminus \{0\} $
-and $P^{-} = (-P) \setminus \{0\} = - P^{+}$. Then we have a disjoint union
-\[
-P^{-} \sqcup \{0\}  \sqcup P^{+} \subseteq k
-.\] 
-Note that $P^{+}$ satisfies axioms (i) and (ii) of the definition of a cone, as well
-as the property that $x \in k \setminus \{0\} \implies x^2 \in P^{+}$.
-
-We now prove that positive curves can be enlarged, that the resulting notion of
-maximal positive cone satisfies $P \cup (-P) = k$, and that
-this defines a structure of ordered field on $k$ by setting $x \le y$ if and only if $y - x \in P$.
-
-\begin{lemma}
-    Assume that $P$ is a positive cone in a field $k$. If $a \in k \setminus P \cup (-P)$, then the set
-    \[
-    P[a] \coloneqq \{ x + a y \in k \colon x, y \in P\}
-    \]
-    is a positive cone in $k$, satisfying $P \subsetneq P[a]$.
-    \label{lemma:positive-cone-extend-by-one-element}
-\end{lemma}
-
-\begin{proof}
-    Let $x, y, x', y' \in P$. Then
-    \[
-        (x + ay) + (x' + a y') = x + x' + a(y + y') \in P[a]
-    \] and
-    \[
-        (x+ay)(x' + ay') = x x' + a^2 y y' + a (x y' + x' y) \in P[a]
-    .\] Moreover $z^2 \in P \subseteq P[a]$ for all $z \in k$.
-
-    Now assume $-1 = x + a y$ for some $x, y \in P$.
-    If $y = 0$, then $-1 = x \in P$ which is a contradiction. Thus $y \neq 0$ and
-    \[
-    - a = \frac{1 + x}{y} = \left( \frac{1}{y} \right)^2 y (1+x) \in P
-    ,\] which contradicts the assumption on $a$. Finally, we have $P \subseteq P[a]$ and,
-    if $P[a] \subseteq P$ then $a \in P$, again contradicting the assumption on $a$. So
-    $P \subsetneq P[a]$.
-\end{proof}
-
-\begin{satz}
-    Let $\mathcal{P}$ be the set of positive cones of a field $k$ ordered
-    by inclusion. If $\mathcal{P} \neq \emptyset$, then
-    $\mathcal{P}$ admits a maximal element and such an element $P$ satisfies
-    $P \cup (-P) = k$.
-    \label{satz:existence-maximal-positive-cones}
-\end{satz}
-
-\begin{proof}
-    To obtain a maximal element of $\mathcal{P}$,
-    by Zorn's lemma, it suffices to show, that every
-    chain $(P_i)_{i \in I}$ in $\mathcal{P}$ has an upper bound. We set
-    \[
-    P = \bigcup_{i \in  I} P_i \subseteq k
-    .\] One verifies immediately that $P$ is a positive cone and an upper bound of the chain $(P_i)_{i \in I}$.
-
-    Let $P$ be such a maximal element. If there exists $a \in k \setminus P \cup (-P)$, then by
-    \ref{lemma:positive-cone-extend-by-one-element} $P \subsetneq P[a]$ contradicts the maximality of $P$. Thus
-    $P \cup (-P) = k$.
-\end{proof}
-
-\begin{satz}
-    Let $k$ be a field and denote by
-    \[
-    \Sigma k^{[2]} \coloneqq
-    \left\{ y \in k  \mid \exists (a_x)_{x \in k} \in \{0, 1\}^{(k)}, y = \sum_{x \in k} a_x x^2 \right\} 
-    \]
-    the set of sums of squares in $k$. Then
-    $\Sigma k^{[2]}$ is a cone and $-1 \not\in \Sigma k^{[2]}$ if and only if
-    for all $x_1, \ldots, x_n \in k$:
-    \[
-    x_1^2 + \ldots + x_n^2 = 0 \implies x_1 = \ldots = x_n = 0
-    .\] 
-    \label{satz:sums-of-squares-cone}
-\end{satz}
-
-\begin{proof}
-    One verifies immediately that $\Sigma k^{[2]}$ is a cone in $k$. If
-    $-1 \in \Sigma k^{[2]}$, then
-    $-1 = x_1^2 + \ldots + x_n^2$ for some $x_i \in k$. Thus
-    \[
-    0 = \sum_{i=1}^{n} x_i^2 + 1
-    \] but $1 = 1^2$ and $1 \neq 0$. Conversely let
-    $0 = \sum_{i=1}^{n} x_i^2$ with $x_1 \neq 0$. Then
-    \[
-    -1 = \frac{1}{x_1^2} \sum_{i=2}^{n} x_i^2 =
-    \sum_{i=2}^{n} \left(\frac{x_i}{x_1}\right)^2
-    \in \Sigma k^{[2]}
-    .\]
-\end{proof}
-
-\begin{definition}
-    A field $k$ is called a \emph{real field} if $-1 \not\in \Sigma k^{[2]}$, or equivalently
-    if $\sum_{k=1}^{n} x_i^2 = 0$ in $k$ implies $x_k = 0$ for all $k$.
-\end{definition}
-
-\begin{korollar}
-    Let $k$ be a field. $k$ is real if and only if $k$ contains
-    a positive cone.
-\end{korollar}
-
-\begin{proof}
-    $(\Rightarrow)$: By \ref{satz:sums-of-squares-cone} $\Sigma k^{[2]}$ is a positive
-    cone.
-    $(\Leftarrow)$: Let $P$ be a positive cone. Since
-    $P$ is closed under addition and for all $z \in k\colon z^2 \in P$,
-    $\Sigma k^{[2]} \subset P$. Since $P$ is positive, $-1 \not\in \Sigma k^{[2]}$.
-\end{proof}
-
-\begin{satz}
-    Let $(k, \le)$ be an ordered field. Then the set
-    \[
-    P \coloneqq \{ x \in k  \mid x \ge 0\}
-    \] is a maximal positive cone in $k$. In particular,
-    $k$ is a real field. Conversely, if $k$ is a real field and $P$ is a maximal
-    positive cone in $k$, then the relation $x \le_P y$ if $y - x \in P$ is an order
-    relation and $(k, \le_P)$ is an ordered field.
-\end{satz}
-
-\begin{proof}
-    $(\Rightarrow)$:
-    Let $(k, \le )$ be an ordered field. Then by
-    definition and \ref{satz:ordered-field-basics}, $P$ is a maximal positive cone.
-
-    $(\Leftarrow)$: Let $P$ be a maximal positive cone in $k$. Since
-    $0 \in P$, we have $x \le_P x$. Suppose that $x \le_P y$ and $y \le_P x$. Then
-    $y - x \in P \cap (-P) = \{0\} $, so $x = y$. Moreover, if $x \le_P y$
-    and $y \le_P z$, then $z - x = (z - y) + (y - x) \in P$. Thus $x \le_P z$, hence
-    $\le_P$ is an order relation. Moreover, it is a total order, because if
-    $x, y \in k$, then $y - x \in k = P \cup (-P)$, so either $x \le_P y$ or $y \le_P x$.
-
-    Finally,
-    this total order on $k$ is compatible with addition and multiplication because
-    $x \le_P y$ and $z \in k$ implies $(y + z) - (x + z) = y - x \in P$, so
-    $x + z \le_P y + z$, and $x \ge_P 0$, $y \ge_P 0$ means that $x \in P$ and $y \in P$, so $xy \in P$,
-    hence $xy \ge_P 0$.
-\end{proof}
-
-\begin{korollar}
-    Let $k$ be a field. Then $k$ admits a structure of ordered field
-    if and only if $k$ is real.
-\end{korollar}
-
-\end{document}

ws2022/rav/lecture/rav17.tex

@@ -1,227 +0,0 @@
-\documentclass{lecture}
-
-\begin{document}
-
-\section{Real-closed fields}
-
-In this section we study real algebraic extensions of real fields.
-
-\begin{lemma}
-    Let $k$ be a real field and $x \in k \setminus \{0\} $. Then
-    $x$ and $-x$ cannot be both sums of squares in $k$.
-    \label{lemma:real-field-only-one-is-square}
-\end{lemma}
-
-\begin{proof}
-    If $x \in \Sigma k^{[2]}$ and $-x \in \Sigma k^{[2]}$, then
-    \[
-    1 = \frac{1}{x^2} (-x) x \in \Sigma k^{[2]}
-    \] contradicting that $k$ is real.
-\end{proof}
-
-\begin{satz}
-    Let $k$ be a real field and $a \in k$ such that $a$ is not a square in $k$.
-    Then the field
-    \[
-    k(\sqrt{a}) = k[t] / (t^2 - a)
-    \] is real if and only if $-a \not\in \Sigma k^{[2]}$.
-    In particular, if $\Sigma k^{[2]} \cup (- \Sigma k^{[2]}) \neq k$,
-    then $k$ admits real quadratic extensions.
-    \label{satz:quadratic-extensions-of-real-field}
-\end{satz}
-
-\begin{proof}
-    Since $a$ is not a square in $k$, $t^2 - a$ is irreducible in $k[t]$, so
-    $k[t] / (t^2 -a)$ is indeed a field. Denote by $\sqrt{a} $ the class of
-    $t$ in the quotient.
-
-    ($\Rightarrow$): $a$ is a square in $k(\sqrt{a})$, thus by
-    \ref{lemma:real-field-only-one-is-square} we have $-a \not\in \Sigma k(\sqrt{a})^2$.
-    But $\Sigma k^{[2]} \subseteq \Sigma k(\sqrt{a})^{[2]}$, thus
-    $-a \not\in \Sigma k(\sqrt{a})^{[2]}$.
-
-    ($\Leftarrow$):
-    $-1 \in \Sigma k(\sqrt{a})^{[2]}$
-    if and only if there exist $x_i, y_i \in k$, such that
-    \[
-    -1 = \sum_{i=1}^{n} (x_i + y_i \sqrt{a})^2
-    = \sum_{i=1}^{n} (x_i^2 + a y_i^2) + 2 \sqrt{a} \sum_{i=1}^{n} x_i y_i
-    .\]
-    Since $(1, \sqrt{a})$ is a basis of the $k$-vector space $k(\sqrt{a})$, the previous equality
-    implies
-    \begin{salign*}
-        -1 &= \sum_{i=1}^{n} x_i^2 + a \sum_{i=1}^{n} y_i^2
-    .\end{salign*}
-    Since $-1 \not\in \Sigma k^{[2]}$, $\sum_{i=1}^{n} y_i^2 \neq 0$, this
-    implies
-    \[
-    -a = \frac{1 + \sum_{i=1}^{n} x_i^2}{\sum_{i=1}^{n} y_i^2}
-    = \frac{\left( \sum_{i=1}^{n} y_i^2 \right)\left( 1 + \sum_{i=1}^{n} x_i^2 \right) }
-    {\left( \sum_{i=1}^{n} y_i^2 \right)^2}
-    \in \Sigma k^{[2]}
-    .\]
-\end{proof}
-
-Simple extensions of odd degree are simpler from the real point of view:
-
-\begin{satz}
-    Let $k$ be a real field and $P \in k[t]$ be an irreducible polynomial of odd degree.
-    Then the field $k[t]/(P)$ is real.
-    \label{satz:odd-real-extension}
-\end{satz}
-
-\begin{proof}
-    Denote by $n$ the degree of $P$. We proceed by induction on $n \ge 1$.
-    If $n = 1$, then $k[t]/(P) \simeq k$ is real. Since $n$ is odd, we
-    may now assume $n \ge 3$.
-    Let $L \coloneqq k[t]/(P)$. Suppose $L$ is not real. Then there exist
-    polynomials $g_i \in k[t]$, of degree at most $n-1$, such that
-    $-1 = \sum_{i=1}^{m} g_i^2$
-    in $L = k[t]/(P)$. Since $k \subseteq L$ and $k$ is real, at least
-    one of the $g_i$ is non-constant.
-    By definition of $L$, there exists $Q \in k[t] \setminus \{0\}$
-    such that
-    \begin{equation}
-    -1 = \sum_{i=1}^{m} g_i^2 + P Q
-    \label{eq:gi-sq+pq}
-    \end{equation}
-    in $k[t]$. Since $k$ is real, in $\sum_{i=1}^{m} g_i^2$ no cancellations
-    of the terms of highest degree can occur. Thus
-    $\sum_{i=1}^{m} g_i^2$ is of positive, even degree at most $2n-2$. By
-    \ref{eq:gi-sq+pq}, it follows that $Q$ is of odd degree at most $n-2$.
-    In particular, $Q$ has at least one irreducible factor $Q_1$ of odd degree at most
-    $n-2$. Since $n \ge 3$, $n-2 \ge 1$. By induction,
-    $M \coloneqq k[t]/(Q_1)$ is real. But \ref{eq:gi-sq+pq} implies
-    \[
-    -1 = \sum_{i=1}^{m} g_i^2
-    \] in $M = k[t] / (Q_1)$ contradicting the fact that $M$ is real.
-\end{proof}
-
-\begin{definition}
-    A \emph{real-closed} field is a real field that
-    has no proper real algebraic extensions.
-\end{definition}
-
-\begin{theorem}
-    Let $k$ be a field. Then the following conditions are equivalent:
-    \begin{enumerate}[(i)]
-        \item $k$ is real-closed.
-        \item $k$ is real and for all $a \in k$, either $a$ or $-a$
-            is a square in $k$ and
-            every polynomial of odd degree in $k[t]$ has a
-            root in $k$.
-        \item the $k$-algebra
-            \[
-            k[i] \coloneqq k[t] / (t^2+1)
-            \] is algebraically closed.
-    \end{enumerate}
-    \label{thm:charac-real-closed}.
-\end{theorem}
-
-\begin{proof}
-    (i)$\Rightarrow$(ii): Let $a \in k$ such that neither $a$ nor $-a$ is a square in $k$. Then
-    by \ref{satz:quadratic-extensions-of-real-field} and (i), $\pm a \in \Sigma k^{[2]}$
-    contradicting
-    \ref{lemma:real-field-only-one-is-square}. Let $P \in k[t]$ be a polynomial
-    of odd degree. $P$ has at least one irreducible factor $P_1$ of odd degree.
-    By \ref{satz:odd-real-extension}, $k[t]/(P_1)$ is a real extension of $k$.
-    Since $k$ is real-closed, $P_1$ must be of degree $1$ and thus $P$ has a root in $k$.
-
-    (ii)$\Rightarrow$(iii): Since $-1$ is not a square in $k$, the polynomial
-    $t^2 + 1$ is irreducible over $k$. Thus $L \coloneqq k[t]/(t^2 + 1)$ is a field. Denote
-    by $i$ the image of $t$ in $L$ and for $x = a + ib \in L = k[i]$, denote
-    by $\overline{x} = a - ib$. This extends to a ring homomorphism $L[t] \to L[t]$. Let
-    $P \in L[t]$ be non-constant. It remains to show, that $P$ has a root in $L$. We
-    first reduce to the case $P \in k[t]$.
-
-    Assume every non-constant polynomial in $k[t]$ has a root in $L$. Let $P \in L[t]$. Then
-    $P \overline{P} \in k[t]$ has a root $x \in L$, thus either $P(x) = 0$
-    or $\overline{P}(x) = 0$. In the first case, we are done.
-    In the second case, we have $P(\overline{x}) = \overline{\overline{P}(x)}
-    = \overline{0} = 0$, so $\overline{x}$ is a root of $P$ in $L$.
-
-    Thus we may assume $P \in k[t]$. Write $d = \text{deg}(P) = 2^{m} n$ with $2 \nmid n$. We
-    proceed by induction on $m$. If $m = 0$, the result is true by (ii). Now assume $m > 0$.
-    Fix an algebraic closure $\overline{k}$ of $k$. Since $k$ is real, it is of characteristic
-    $0$, thus $k$ is perfect and $\overline{k} / k$ is galois.
-    Let $y_1, \ldots, y_d$ be the roots
-    of $P$ in $\overline{k}$. Consider for all $r \in \Z$:
-    \[
-    F_r \coloneqq \prod_{1 \le p < q \le d}^{}
-    \left( t - (y_p + y_q) - r y_p y_q) \right) \in \overline{k}[t]
-    .\] This polynomial with coefficients in $\overline{k}$ is invariant
-    under permutation of $y_1, \ldots, y_d$. Thus its coefficients
-    lie in $\overline{k}^{\text{Gal}(\overline{k} / k)} = k$. Moreover
-    \[
-    \text{deg}(F_r) = \binom{d}{2} = \frac{d(d-1)}{2} = 2^{m-1} n (2^{m} -1)
-    .\] with $n (2^{m} -1)$ odd. So the induction hypothesis applies and,
-    for all $r \in \Z$, there is a pair $p < q$ in $\{1, \ldots, d\} $
-    such that $(y_p + y_q) + r y_p y_q \in L$. Since $\Z$ is infinite,
-    we can find a pair $p < q$ in $\{1, \ldots, d\} $ for which
-    there exists a pair $r \neq r'$ such that
-    \begin{salign*}
-        &(y_p + y_q) + r y_p y_q \in L \\
-        \text{and } & (y_p + y_q) + r' y_p y_q \in L
-    .\end{salign*}
-    By solving the system, we get $y_p + y_q \in L$ and $y_p y_q \in L$. But $y_p, y_q$
-    are roots of the quadratic polynomial
-    \[
-    t^2 - (y_p + y_q)t + y_p y_q \in L[t]
-    \] and since $i^2 = -1$, the roots of this polynomial lie in $L = k[i]$, by (ii) and the
-    usual formulas
-    \[
-        t_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}
-    .\] So $P$ indeed has a root in $k[i]$, which finishes the induction.
-
-    (iii)$\Rightarrow$(i): Denote again by $i$ the image in the algebraically closed field
-    $k[t]/(t^2 + 1)$. We first show that $k^{[2]} = \Sigma k^{[2]}$. Let $a, b \in k$. Then
-    $a + ib = (c + id)^2$ in $k[i]$ for some $c, d \in k$. Thus
-    \[
-    a^2 + b^2 = (a+ib)(a-ib) = (c+id)^2(c-id)^2 = (c^2 + d^2)^2
-    .\] By induction the claim follows. Since $t^2 + 1$ is irreducible,
-    $-1 \not\in k^{[2]} = \Sigma k^{[2]}$ and $k$ is real.
-
-    Let $L$ be a real algebraic extension of $k$. Since $k[i]$ is algebraically closed and contains
-    $k$, there exists a $k$-homomorphism $L \xhookrightarrow{} k[i]$. Since
-    $[ k[i] : k ] = 2$, either $L = k$ or $L = k[i]$, but $k[i]$ is not real, since $i^2 = -1$
-    in $k[i]$. So $L = k$ and $k$ is real-closed.
-\end{proof}
-
-\begin{korollar}
-    A real-closed field $k$ admits a canonical structure of ordered field, in
-    which the cone of positive elements is exactly $k^{[2]}$, the set of squares in $k$.
-\end{korollar}
-
-\begin{proof}
-    This was proven in the implication (i)$\Rightarrow$(ii) of \ref{thm:charac-real-closed}.
-\end{proof}
-
-\begin{bsp}[]
-    \begin{itemize}
-        \item $\R$ is a real-closed field, because $\R[i] = \mathbb{C}$ is algebraically closed.
-        \item The field of real Puiseux series
-            \begin{salign*}
-            \widehat{\R(t)} \coloneqq \bigcup_{q > 0} \R((t ^{\frac{1}{q}}))
-            = \left\{ 
-                \sum_{n=m}^{\infty} a_n t ^{\frac{n}{q}} \colon
-                m \in \Z, q \in \N \setminus \{0\}, a_n \in \R
-            \right\} 
-            \end{salign*}
-            is a real closed field because
-            $\widehat{\R(t)}[i] = \widehat{\R[i][t]} = \widehat{\mathbb{C}[t]}$ is the field
-            of complex Puiseux series, which is algebraically closed by the
-            Newton-Puiseux theorem.
-    \end{itemize}
-\end{bsp}
-
-\begin{bem}[]
-    By \ref{thm:charac-real-closed}, if $k$ is a real-closed field, then the absolute galois
-    group of $k$ is
-    \[
-    \text{Gal}(\overline{k} / k) = \text{Gal}(k[i] / k) \simeq \Z / 2 \Z
-    .\] The Artin-Schreier theorem shows that if $\overline{k} / k$
-    is a non-trivial extension of \emph{finite} degree,
-    then $k$ is real-closed.
-\end{bem}
-
-\end{document}

ws2022/rav/lecture/rav18.tex

@@ -1,153 +0,0 @@
-\documentclass{lecture}
-
-\begin{document}
-
-\section{Extensions of ordered fields}
-
-If a field $k$ admits a structure of ordered field, we will say that $k$ is \emph{orderable}.
-For an \emph{ordered} field $k$, an extension $L / k$ is called \emph{oderable} if the
-field $L$ is orderable such that the induced order on $k$ coincides with the fixed order
-on $k$.
-
-\begin{definition}[]
-    Let $k$ be a field. A quadratic form $q\colon k^{n} \to k$
-    is called \emph{isotropic} if there exists
-    $x \in k \setminus \{0\} $ such that $q(x) = 0$. Otherwise, the quadratic
-    form is called \emph{anisotropic}.
-\end{definition}
-
-\begin{bem}[]
-    Recall that, given a quadratic form $q$ on a finite-dimensional
-    $k$-vector space $E$, there always exists a basis of $E$ in which
-    $q(x_1, \ldots, x_n) = a_1x_1^2 + \ldots + a_r x_r^2$, where
-    $r = \text{rg}(q) \le n = \text{dim } E$ and $a_1, \ldots, a_r \in k$.
-    The form $q$ is non-degenerate on $E$ if and only if $r = \text{dim } E$.
-\end{bem}
-
-\begin{bsp}[]
-    \begin{itemize}
-        \item A field $k$ is real if and only if for all $n \in \N$, the form
-            $x_1^2 + \ldots + x_n^2$ is anisotropic.
-        \item A degenerate quadratic form is isotropic.
-        \item If $k$ is algebraically closed and $n \ge 2$,
-            all quadratic forms on $k^{n}$ are isotropic.
-        \item If $(k, \le )$ is an ordered field and
-            $q(x_1, \ldots, x_n) = a_1 x_1^2 + \ldots + a_n x_n^2$ with
-            $a_i > 0$ for all $i$, then $q$ and $-q$ are anisotropic on $k^{n}$.
-    \end{itemize}
-\end{bsp}
-
-\begin{definition}
-    Let $k$ be a field and $L$ an extension of $k$. A quadratic form
-    $q\colon k^{n} \to k$ induces a quadratic form $q_L \colon L^{n} \to L$. The form
-    $q$ is called \emph{anisotropic over $L$} if $q_L$ is anisotropic.
-\end{definition}
-
-It can be checked that, on an ordered field $(k, \le )$, a quadratic form
-$q$ is anisotropic if and only if it is non-degenerate and of constant sign. The interest
-of this notion for us is given by the following result.
-
-\begin{theorem}
-    \label{thm:charac-orderable-extension}
-    Let $(k, \le )$ be an ordered field and $L$ be an extension of $k$. Then the following
-    conditions are equivalent:
-    \begin{enumerate}[(i)]
-        \item The extension $L / k$ is orderable.
-        \item For all $n \ge 1$ and all $a = (a_1, \ldots, a_n) \in k^{n}$ such that
-            $a_i > 0$ for all $i$, the quadratic form
-            $q(x_1, \ldots, x_n) = a_1x_1^2 + \ldots + a_n x_n^2$ is anisotropic over $L$
-            (i.e. all positive definite quadratic forms on $k$ are anisotropic over $L$).
-    \end{enumerate}
-\end{theorem}
-
-\begin{proof}
-    (i)$\Rightarrow$(ii): Assume that there is an ordering of $L$ that extends
-    the ordering of $k$ and let $n \ge 1$. Let $a = (a_1, \ldots, a_n) \in k^{n}$
-    with $a_i > 0$ for all $i$. Then $a_i > 0$ still holds in $L$. Since
-    squares are non-negative for all orderings, the sum
-    $a_1 x_1^2 + \ldots + a_n x_n^2$ is a sum of positive terms in $L$. Therefore
-    it can only be $0$, if all of its terms are $0$. Since $a_i \neq 0$, it follows
-    $x_i = 0$ for all $i$.
-
-    (ii)$\Rightarrow$(i): Define
-    \[
-    P = \bigcup_{n \ge 1} \left\{ \sum_{i=1}^{n} a_i x_i^2 \colon a_i \in k, a_i > 0, x_i \in L \right\} 
-    .\] The set $P$ is stable by sum and product and contains all squares of $L$,
-    so it is a cone in $L$. Suppose $-1 \in P$. Then there exists
-    $n \ge 1$ and $a = (a_1, \ldots, a_n) \in k^{n}$ with $a_i > 0$
-    and $x = (x_1, \ldots, x_n) \in L^{n}$ such that
-    $-1 = \sum_{i=1}^{n} a_i x_i^2$. So
-    \[
-    a_1 x_1^2 + \ldots + a_n x_n^2 + 1 = 0
-    ,\] meaning that the quadratic form $a_{1} x_1^2 + \ldots + a_n x_n^2 + x_{n+1}^2$
-    is isotropic on $L^{n+1}$, contradicting (ii).
-    Thus $P$ is a positive cone containing all positive elements of $k$. By
-    embedding $P$ in a maximal positive cone, the claim follows.
-\end{proof}
-
-\begin{satz}[]
-    Let $(k, \le )$ be an ordered field and let $c > 0$ be a positive element in $k$.
-    Then $k[\sqrt{c}]$ is an orderable extension of $k$.
-\end{satz}
-
-\begin{proof}
-    If $c$ is a square in $k$, there is nothing to prove. Otherwise, $k[\sqrt{c}]$ is
-    indeed a field. Let $n \ge 1$ and let $a = (a_1, \ldots, a_n) \in k^{n}$
-    with $a_i > 0$ for all $i$. Assume that $x = (x_1, \ldots, x_n) \in k[\sqrt{c}]^{n}$
-    satisfies
-    \[
-    a_1 x_1^2 + \ldots + a_n x_n^2 = 0
-    .\] Since $x_i = u_i + v_i \sqrt{c} $ for some $u_i, v_i \in k$, we can rewrite
-    this equation as
-    \[
-    \sum_{i=1}^{n} a_i (u_i^2 + c v_i^2) + 2 \sum_{i=1}^{n} u_i v_i \sqrt{c} = 0
-    .\] 
-    Since $1$ and $\sqrt{c}$ are linearly independent over $k$, we get
-    $\sum_{i=1}^{n} a_i (u_i^2 + c v_i^2) = 0$, hence
-    $u_i = v_i = 0$ for all $i$, since all terms in the previous sum are non-negative.
-    So $x_i = 0$ for all $i$ and (ii) of \ref{thm:charac-orderable-extension}
-    is satisfied.
-\end{proof}
-
-\begin{satz}
-    Let $(k, \le )$ be an ordered field and let $P \in k[t]$ be an irreducible
-    polynomial of odd degree. Then the field $L \coloneqq k[t]/ (P)$ is an orderable
-    extension of $k$.
-\end{satz}
-
-\begin{proof}
-    Denote by $d$ the degree of $P$ and proceed by induction on $d \ge 1$. If $d = 1$, then
-    $L = k$. Now assume $d \ge 2$. Let $n \ge 1$ and $a_1, \ldots, a_n \in k$ with $a_i > 0$.
-    Denote by $q_L$ the quadratic form
-    \[
-    q_L(x_1, \ldots, x_n) = a_1 x_1^2 + \ldots + a_n x_n^2
-    \] on $L^{n}$. If $q_L$ is isotropic over $L$, then there exist
-    polynomials $g_1, \ldots, g_n \in k[t]$ with $\text{deg}(g_i) < d$
-    and $h \in k[t]$ such that
-    \begin{equation}
-        q_L(g_1, \ldots, g_n) = h P
-        \label{eq:quad-form}
-    \end{equation}
-    Let $g$ be the greatest common divisor of $g_1, \ldots, g_n$. Since $q_L$ is
-    homogeneous of degree $2$,
-    $g^2$ divides $q_L(g_1, \ldots, g_n)$. Since $P$ is irreducible, $g$ divides $h$.
-    We may thus assume that $g = 1$. The leading coefficients of the terms on
-    the left hand side of (\ref{eq:quad-form}) are non-negative, thus
-    the sum has even degree $< 2d$. Since the degree of $P$ is odd,
-    $h$ must be of odd degree $< d$. Therefore, $h$ has an irreducible factor
-    $h_1 \in k[t]$ of odd degree. Let $\alpha$ be a root of $h_1$. By evaluating
-    (\ref{eq:quad-form}) at $\alpha$, we get
-    \[
-    q_{k[\alpha]}(g_1(\alpha), \ldots, g_n(\alpha)) = 0
-    \] in $k[\alpha]$. Since the $gcd(g_1, \ldots, g_n) = 1$ and $k[t]$ is a principal ideal
-    domain, there exist $h_1, \ldots, h_n \in k[t]$ such that
-    \[
-    h_1 g_1 + \ldots + h_n g_n = 1
-    .\] In particular
-    \[
-    h_1(\alpha) g_1(\alpha) + \ldots + h_n(\alpha) g_n(\alpha) = 1
-    ,\] so not all $g_i(\alpha)$ are $0$ in $k[\alpha]$. Thus $q_{k[\alpha]}$
-    is isotropic over $k[\alpha] = k[t] / (h_1)$ contradicting the induction hypothesis.
-\end{proof}
-
-\end{document}

ws2022/rav/lecture/rav19.tex

@@ -1,161 +0,0 @@
-\documentclass{lecture}
-
-\begin{document}
-
-\section{Real closures}
-
-\begin{satz}
-    Let $k$ be a real field. Then there exists a real-closed
-    algebraic orderable extension $k^{r}$ of $k$.
-    \label{satz:existence-alg-closure}
-\end{satz}
-
-\begin{proof}
-    Let $\overline{k}$ be an algebraic closure of $k$ and $E$ be the set of intermediate
-    extensions $k \subseteq L \subseteq \overline{k}$ such that $L$ is real and algebraic over $k$.
-    $E \neq \emptyset$ since $k \in E$. Define $L_1 < L_2$ on $E$ if and only if
-    $L_1 \subseteq L_2$ and $L_2 / L_1$ is ordered, i.e.
-    the order relation on $L_1$ coincides with the on induced by $L_2$.
-    Then
-    every totally ordered familiy $(E_i)_{i \in I}$ has an upper bound, namely
-    $\bigcup_{i \in  I} E_i$. By Zorn, $E$ has a maximal element, which we
-    denote by $k^{r}$ and which is an algebraic extension of $k$. Such
-    a $k^{r}$ is real-closed, because otherwise it would admit a proper real
-    algebraic extension contradicting the maximality of $k^{r}$ as a real algebraic extension of $k$.
-\end{proof}
-
-\begin{definition}[]
-    A real-closed real algebraic extension of a real field $k$ is called
-    a \emph{real closure} of $k$.
-\end{definition}
-
-\begin{bem}
-    By the construction in the proof of \ref{satz:existence-alg-closure},
-    a real closure of a real field $k$ can be chosen as a subfield
-    $k^{r}$ of an algebraic closure of $\overline{k}$.
-    Since $k^{r}[i]$ is algebraically closed and algebraic over $k^{r}$, so also over $k$,
-    it follows $k^{r}[i] = \overline{k}$.
-\end{bem}
-
-\begin{satz}
-    Let $k$ be a real field and $L$ be a real-closed extension of $k$. Let
-    $\overline{k}^{L}$ be the relative algebraic closure of $k$ in $L$, i.e.
-    \[
-    \overline{k}^{L} = \{ x \in L \mid x \text{ algebraic over } k\} 
-    .\] Then $\overline{k}^{L}$ is a real closure of $k$.
-\end{satz}
-
-\begin{proof}
-    It is immediate that $\overline{k}^{L}$ is a real algebraic extension of $k$. Let
-    $x \in \overline{k}^{L}$. Then $x$ or $-x$ is a square in $L$, since
-    $L$ is real-closed. Without loss of generality, assume that
-    $x \in L^{[2]}$. Then $t^2 - x \in \overline{k}^{L}[t]$
-    has a root in $L$. Since this root is algebraic over $\overline{k}^{L}$, hence over $k$,
-    it belongs to $\overline{k}^{L}$. Thus $x$ is in fact a square in $\overline{k}^{L}$. By
-    the same argument every polynomial of odd degree has a root in $\overline{k}^{L}$.
-\end{proof}
-
-\begin{bsp}[]
-    \begin{enumerate}[(i)]
-        \item $\overline{\Q}^{\R} = \overline{\Q}^{\mathbb{C}} \cap \R$
-            is a real closure of $\Q$. In particular, $\overline{\Q}^{\mathbb{C}}
-            = \overline{\Q}^{\R}[i]$ as subfields of $\mathbb{C}$.
-        \item Consider the real field $k = \R(t)$ and the real-closed extension
-            \begin{salign*}
-            \widehat{\R(t)} =
-            \bigcup_{q > 0} \R((t ^{t/q}))
-            .\end{salign*} Then the subfield
-            $\overline{\R(t)}^{\widehat{\R(t)}}$, consisting of all those real
-            Puiseux series that are algebraic over $\R(t)$, is a real closure of $\R(t)$.
-
-            The field of real Puiseux series itself is a real closure of the field $\R((t))$
-            of real formal Laurent series.
-    \end{enumerate}
-\end{bsp}
-
-\begin{lemma}
-    Let $L_1, L_2$ be real-closed fields and let $\varphi\colon L_1 \to L_2$ be a homomorphism
-    of fields. Then $\varphi$ is compatible with the canonical orderings of $L_1$ and $L_2$.
-    \label{lemma:hom-real-closed-fields-respects-orderings}
-\end{lemma}
-
-\begin{proof}
-    It suffices to prove that $x \ge_{L_1} 0$ implies $\varphi(x) \ge_{L_2} 0$
-    for all $x \in L_1$. This follows from the fact that in a real-closed field $L$,
-    for all $x \in L$, $x \ge 0$ if and only if $x$ is a square.
-\end{proof}
-
-If $k$ is a real field and $k^{r}$ is a real closure of $k$, then
-$k$ inherits an ordering from $k^{r}$. However, different real closures may induce
-different orderings on $k$, as the next example shows.
-
-\begin{bsp}[]
-    \label{bsp:different-real-closures-depending-on-ordering}
-    Let $k = \Q(t)$. This is a real field, since $\Q$ is real. Since $\pi$
-    is transcendental over $\Q$, we can embed $\Q(t)$ in $\R$ by sending $t$ to $\pi$.
-    \[
-    i_1\colon \Q(t) \xhookrightarrow{\simeq} \Q(\pi) \subseteq \R
-    .\] Since $\R$ is real-closed, the relative algebraic closure
-    $i_1(\Q(t))^{\R}$ is a real closure of $i_1(\Q(t))$.
-
-    We can also embed $\Q(t)$ in the field $\widehat{\R(t)}$ of real Puiseux series via
-    a homomorphism $i_2$ and then
-    $\overline{i_2(\Q(t))}^{\widehat{\R(t)}}$ is a real closure of $i_2(\Q(t))$.
-    However, the ordering on $\overline{i_1(\Q(t))}^{\R}$
-    is Archimedean, because it is a subfield of $\R$,
-    while the ordering on $\overline{i_2(\Q(t))}^{\widehat{\R(t)}}$
-    is not Archimedean (it contains infinitesimal elements, such as $t$ for instance).
-
-    The fields $\overline{i_1(\Q(t))}^{\R}$
-    and $\overline{i_2(\Q(t))}^{\widehat{\R(t)}}$ cannot be isomorphic as fields.
-    Indeed, when two real-closed fields $L_1, L_2$ are isomorphic as fields,
-    then they are isomorphic as ordered fields, since positivity on a real
-    closed field is defined by the condition of being a square, which is preserved
-    under isomorphisms of fields.
-\end{bsp}
-
-%The next result will be proved later on.
-%
-%\begin{lemma}
-%    Let $(k, \le )$ be an ordered field and $P \in k[t]$ be an irreducible polynomial.
-%    Let $L_1, L_2$ be real-closed extensions of $k$ that are compatible with the ordering of $k$.
-%    Then $P$ has the same number of roots in $L_1$ as in $L_2$.
-%    \label{lemma:number-of-roots-in-real-closed-extension}
-%\end{lemma}
-%
-%\begin{bem}
-%    In particular, if $P \in k[t]$ is an arbitrary polynomial, then if $P$ has a root
-%    in a real-closed extension $L$ of $k$, then it has a root in all real-closed extensions of $k$.
-%
-%    A polynomial with coefficients in an ordered field $(k, \le)$ might not have roots
-%    in any real-closed extensions of $k$.
-%\end{bem}
-
-\begin{lemma}
-    Let $(k, \le)$ be an ordered field, $L / k$ an orderable real-closed extension of $k$
-    and $\varphi\colon k \to L$ a morphism of $k$-algebras.
-    If $E / k$ is a finite, real extension of $k$, then $\varphi$ admits a continuation, i.e.
-    a morphism of $k$-algebras $\varphi'$ such that the following diagram commutes:
-    \[
-    \begin{tikzcd}
-        k \arrow[hook]{d} \arrow{r}{\varphi} & L \\
-        E \arrow[dashed, swap]{ur}{\varphi'}
-    \end{tikzcd}
-    .\] 
-    \label{lemma:continuation-in-real-closed}
-\end{lemma}
-
-\begin{proof}
-    Since $k$ is perfect, $E / k$ is separable. Moreover $E / k$ is finite, thus by the
-    primitive element theorem, $E = k[a]$ for $a \in E$. Let
-    $P \in k[t]$ be the minimal polynomial of $a$ over $k$. Let $E^{r}$ be
-    an orderable real-closure of $E$. Thus $E^{r}$ is
-    a real-closed extension of $k$ that contains a root of $P$. By
-    \ref{lemma:number-of-roots-in-real-closed-extension},
-    $P$ has a root $b \in L$. Now
-    define $\psi\colon k[t] \to L$ by $t \mapsto b$ and
-    $\psi|_k = \varphi$. Since $b$ is a root of $P$,
-    $\psi$ factors through $E = k[a] = k / (P)$ and gives $\varphi'\colon E \to L$.
-\end{proof}
-
-\end{document}

ws2022/rav/lecture/rav2.tex

@@ -1,360 +0,0 @@
-%& -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}

ws2022/rav/lecture/rav20.tex

@@ -1,87 +0,0 @@
-\documentclass{lecture}
-
-\begin{document}
-
-\begin{theorem}
-    Let $(k, \le )$ be an ordered field and
-    $k^{r}$ be a real closure of $k$ that extends the ordering of $k$. Let $L$
-    be a orderable real-closed extension of $k$. Then there exists a unique
-    homomorphism of $k$-algebras $k^{r} \to L$.
-    \label{thm:unique-hom-of-real-closure-in-real-closed}
-\end{theorem}
-
-\begin{proof}
-    Uniqueness: Let $\varphi\colon k^{r} \to L$ be a homomorphism of $k$-algebras and
-    $a \in k^{r}$. Since $a$ is algebraic over $k$, it has
-    a minimal polynomial $P \in k[t]$ over $k$. Denote
-    by $a_1 \le \ldots \le a_n$ the roots of $P$ in $k^{r}$. Since
-    the characteristic of $k$ is $0$, $k$ is perfect, in particular
-    the irreducible polynomial $P$ is separable and thus
-    $a_1 < \ldots < a_n$. Now there exists a unique $1, \ldots, n$ such that
-    $a = a_j$. By \ref{lemma:number-of-roots-in-real-closed-extension}, the polynomial
-    $P$ also has $n$ distinct roots $b_1 < \ldots < b_n$ in the real-closed field $L$.
-    Since $\varphi$ sends roots of $P$ in $k^{r}$ to roots of $P$ in $L$, there is a
-    permutation $\sigma \in S_n$ such that
-    $\varphi(a_i) = b_{\sigma(i)}$. By \ref{lemma:hom-real-closed-fields-respects-orderings},
-    $\varphi$ respects the ordering of the roots and thus $\sigma = \text{id}$
-    and $\varphi(a) = \varphi(a_j) = b_j$.
-
-    Existence: Consider the set $\mathcal{F}$ of all pairs
-    $(E, \psi)$ where $k \subseteq E \subseteq k^{r}$ is a subextension
-    of $k^{r} / k$ and $\psi\colon E \to L$ is a homomorphism of $k$-algebras. Since
-    $(k, k \xhookrightarrow{} L) \in \mathcal{F}$, $\mathcal{F} \neq \emptyset$. Define
-    an inductive ordering on $\mathcal{F}$ by $(E, \psi) \le (E', \psi)$
-    if there is a commutative diagram
-    \[
-    \begin{tikzcd}
-        & E' \arrow{d}{\psi'} \\
-        E \arrow[dashed]{ur} \arrow{r}{\psi} & L
-    \end{tikzcd}
-    \] in the category of $k$-algebras. Then by Zorn, the set
-    $\mathcal{F}$ admits a maximal element $(E, \psi)$. $E$ is real-closed, otherwise
-    it admits a finite real extension $E'$ of $E$. In particular $E' \subseteq k^{r}$.
-    Since $L$ is real-closed, $\psi\colon E \to L$ admits a continuation
-    $\psi'\colon E' \to L$ by \ref{lemma:continuation-in-real-closed}.
-    Thus $(E, \psi) < (E', \psi')$ contradicting the
-    maximality of $(E, \psi)$. Hence $E$ is real-closed
-    and $k^{r} / E$ is real algebraic, thus $E = k^{r}$. So
-    $\psi$ is a homomorphism of $k$-algebras from $k^{r}$ to $L$.
-\end{proof}
-
-\begin{korollar}
-    Let $(k, \le )$ be an ordered field. If $k_1^{r}$ and $k_2^{r}$ are real closures
-    of $k$ whose canonical orderings are compatible with that of $k$, then
-    there exists a unique isomorphism of $k$-algebras
-    $k_1^{r} \xrightarrow{\simeq} k_2^{r}$.
-    \label{kor:unique-iso-of-real-closures}
-\end{korollar}
-
-\begin{proof}
-    By \ref{thm:unique-hom-of-real-closure-in-real-closed} there exist
-    unique homomorphisms of $k$-algebras $\varphi\colon k_1^{r} \to k_2^{r}$
-    and $\psi\colon k_2^{r} \to k_1^{r}$. Then
-    $\psi \circ \varphi$ and $\text{id}_{k_1^{r}}$ are
-    homomorphisms $k_1^{r} \to k_1^{r}$ of $k$-algebras. By uniqueness in
-    \ref{thm:unique-hom-of-real-closure-in-real-closed},
-    $\psi \circ \varphi = \text{id}_{k_1^{r}}$. Similarly, $\varphi \circ \psi = \text{id}_{k_2^{r}}$.
-\end{proof}
-
-\begin{bem}
-    Contrary to the situation of algebraic closures of a field $k$,
-    for ordered fields $(k, \le)$ there is a well-defined notion
-    of the real closure of $k$ whose canonical ordering is compatible with that of $k$.
-    As shown by \ref{bsp:different-real-closures-depending-on-ordering},
-    it is necessary to fix an ordering of the real field $k$ to get the
-    existence of an isomorphism of fields between two orderable real closures of $k$.
-\end{bem}
-
-\begin{korollar}
-    Let $(k, \le )$ be an ordered field and let $k^{r}$ be the real closure of $k$.
-    Then $k^{r}$ has no non-trivial $k$-automorphism.
-\end{korollar}
-
-\begin{proof}
-    Take $k_1^{r} = k_2^{r}$ in \ref{kor:unique-iso-of-real-closures}.
-\end{proof}
-
-\end{document}

ws2022/rav/lecture/rav21.tex

@@ -1,225 +0,0 @@
-\documentclass{lecture}
-
-\begin{document}
-
-\section{Counting real roots}
-
-In this section, we will study \emph{Sturm's method} of counting
-the number of roots of a separable polynomial with coefficients
-in a real-closed field $L$.
-
-\begin{lemma}
-    Let $(k, \le)$ be an ordered field and let $P \in k[t]$ be a separable polynomial.
-    Assume that $P$ has a root $a \in k$. Then there exists $\delta > 0$ such that
-    \begin{enumerate}[(i)]
-        \item for $x \in \;]a- \delta, a + \delta[$ and $x \neq a$, $P(x) \neq 0$.
-        \item for $x \in \;]a, a + \delta[$, $P(x)$ and $P'(x)$ have the same sign.
-        \item for $x \in \;]a- \delta, a[$, $P(x)$ and $P'(x)$ have opposite signs.
-        \item for $x \in \;]0, \delta[$, $P(a+h)$ and $P(a-h)$ have opposite signs.
-    \end{enumerate}
-    \label{lemma:root-signs-separable}
-\end{lemma}
-
-\begin{proof}
-    Since $P$ is separable and $P(a) = 0$, it follows that $P'(a) \neq 0$.
-    By continuity of $P'$, there exists $\delta > 0$ such that
-    $P'$ has constant sign on $] a- \delta, a + \delta[$. Suppose $P'(x) > 0$.
-    Since $k$ is real-closed,
-    this implies that $P$ is strictly increasing on this interval. In particular,
-    $P(x) < P(a) = 0$ for $x \in \;]a - \delta, a[$
-    and $P(x) > P(a) = 0$ for $x \in \;]a, a + \delta[$. The case $P'(x) < 0$ is
-    similar which concludes the proof.
-\end{proof}
-
-\begin{definition}
-    Let $(k, \le )$ be an ordered field. A finite sequence $(P_0, \ldots, P_n)$ of polynomials
-    $P_i \in k[t]$ is called a \emph{Sturm sequence} if it satisfies the following
-    properties:
-    \begin{enumerate}[(i)]
-        \item $P_1 = P_0'$
-        \item for all $x \in k$ and $i \in \{0, \ldots, n\}$, if $P_i(x) = 0$, then
-            $P_{i+1}(x) \neq 0$.
-        \item for all $x \in k$ and all $i \in \{1, \ldots, n-1\} $,
-            if $P_i(x) = 0$ then $P_{i-1}(x) P_{i+1}(x) < 0$.
-        \item $P_n \in k^{\times}$.
-    \end{enumerate}
-\end{definition}
-
-\noindent If $P \in k[t]$ is separable and $k$ has characteristic $0$, then the greatest
-common divisor of $P$ and $P'$ is $1$. To determine a Bézout relation between $P$ and $P'$,
-one proceeds by successive Euclidean divisions:
-
-First set $P_0 = P$ and $P_1 = P'$, next define $P_2$ such that $P_0 = P_1 Q_1 - P_2$
-and $\text{deg}(P_2) < \text{deg}(P_1)$. Inductively, this
-defines $P_i = P_{i+1} Q_{i+1} - P_{i+2}$ with $\text{deg}(P_{i+2}) < \text{deg}(P_{i+1})$.
-This algorithm stops after at most $\text{deg}(P_0) $ steps
-with $P_{n-1} = P_n Q_n$ and $P_n \neq 0$.
-Then $P_n$ is a greatest common divisor of $P = P_0$ and $P' = P_1$. Since $P$ and
-$P'$ are coprime, $P_n$ is a non-zero constant.
-
-\begin{korollar}
-    The sequence of polynomials $(P_0, \ldots, P_n)$ is a Sturm sequence.
-    This is called the to $P$ associated Sturm sequence.
-\end{korollar}
-
-\begin{proof}
-    (i) and (iv) are clear. For (ii) observe that if there exists $x \in k$ and
-    $i \in \{0, \ldots, n\} $ such that $P_i(x) = P_{i+1}(x) = 0$, then
-    $P_j(x) = 0$ for all $j \ge i$ which contradicts $P_n(x) = P_n \neq 0$.
-    Finally for (iii), if $P_i(x) = 0$, then $P_{i-1}(x) = - P_{i+1}(x)$, so $P_{i-1}(x)$
-    and $P_{i+1}(x)$ have opposite signs.
-\end{proof}
-
-\begin{bem}
-    Let $(k, \le)$ be an ordered field.
-    For a finite sequence of elements $(a_0, \ldots, a_n)$ in $k$ with $a_0 \neq 0$,
-    the number of \emph{sign changes} in this sequence is the number of pairs
-    $(i, j)$ such that $i < j$, $a_i \neq 0$ and $a_i a_j < 0$ with either $j = i+1$ or
-    $j > i+1$ and $a_{i+1} = \ldots = a_{j-1} = 0$.
-\end{bem}
-
-\begin{theorem}[Sturm's algorithm]
-    Let $k$ be a real-closed field equipped with its canonical ordering and let $P \in k[t]$
-    be a separable polynomial. Let $(P_0, \ldots, P_n)$ be the associated Sturm sequence.
-    For all $a \in k$, we denote by $\nu(a)$ the number of sign changes
-    in the sequence $(P_0(a), \ldots, P_n(a))$. Then, for all pair $a, b \in k$ such that
-    $a < b$ and $P_i(a)P_i(b) \neq 0$ for all $i$, the number of roots of $P$ in the interval
-    $[a, b]$ is equal to $\nu(a) - \nu(b)$.
-    \label{thm:sturm}
-\end{theorem}
-
-\begin{proof}
-    Let $x_1 < \ldots < x_m$ be the elements of the finite set
-    \[
-    E = \{ x \in \; ]a, b[  \mid \exists i \in \{0, \ldots, n\} , P_i(x) = 0\} 
-    .\]
-    %For all $x \in E$, we can choose $\delta > 0$ such that
-    %$] x - \delta, x + \delta[ \cap E = \{x\} $, i.e.
-    %$]x - \delta , x + \delta [$ contains no other root of one of the $P_i$'s.
-    There exists a partition of $[a,b]$ in subintervals
-    $[\alpha_j, \alpha_{j+1}]$ where $\alpha_0 = a$, $\alpha_m = b$,
-    and for all $j \in \{0, \ldots, m-1\} $, $\alpha_j \not\in E$,
-    $[\alpha_j, \alpha_{j+1}] \cap E = \{x_j\} $.
-    Also
-    \[
-    \sum_{j=0}^{m-1} (\nu(\alpha_j) - \nu(\alpha_{j+1}))
-    = \nu(\alpha_0) - \nu(\alpha_1) + \nu(\alpha_1) - \ldots - \nu(\alpha_m)
-    = \nu(a) - \nu(b)
-    .\] Thus it suffices to show that for fixed $j \in \{0, \ldots, m-1\}$, the number of roots
-    of $P$ in $[\alpha_j, \alpha_{j+1}]$ is equal to $\nu(\alpha_j) - \nu(\alpha_{j+1})$.
-    By construction,
-    $P$ has at most one root in $[\alpha_j, \alpha_{j+1}]$, at $x_j$, thus
-    we want to show
-    \[
-    \nu(\alpha_j) - \nu(\alpha_{j+1}) = \begin{cases}
-        0 & P(x_j) \neq 0 \\
-        1 & P(x_j) = 0
-    \end{cases}
-    .\] 
-    If $P(x_j) = 0$, then $P(\alpha_j)$ and $P(\alpha_{j+1})$ must have opposite sign. Indeed,
-    by \ref{lemma:root-signs-separable} $P(x_j + h) P(x_j - h) < 0$ for all $h > 0$ small
-    enough, but $P$ cannot change sign on $[\alpha_j, x_j -h]$ nor on
-    $[x_j + h, \alpha_{j+1}]$, for otherwise the intermediate value theorem would
-    imply the existence of a root $x \neq x_j$ in $[\alpha_j, \alpha_{j+1}]$.
-    So $P(\alpha_j) P(\alpha_{j+1}) < 0$.
-    If $P(\alpha_j) > 0$, then $P(\alpha_{j+1}) < 0$. With $P_1 = P'$ and
-    \ref{lemma:root-signs-separable}, it follows that $P_1(x) < 0$ for
-    $x$ close to $x_j$. But $P_1$ cannot change sign in $[\alpha_j, \alpha_{j+1}]$,
-    otherwise its root in that interval would be $x_j$. Since $P$ is separable
-    and $P_1 = P'$, this is impossible. Thus $P' < 0$
-    and $P$ is strictly decreasing on $[\alpha_j, \alpha_{j+1}]$. So
-    the sequence of signs in the sequence
-    $(P_0(\alpha_j), P_1(\alpha_j), \ldots, P_n(\alpha_j))$ starts
-    with $(+, -, \ldots)$ while the one at $\alpha_{j+1}$ starts
-    with $(-, -, \ldots)$. Similarly, if $P(\alpha_j) < 0$, then
-    the sequences are $(-, +, \ldots)$ and $(+, +, \ldots)$. In either case,
-    there is one more sign change in the sequence corresponding to $\alpha_j$,
-    so $\nu(\alpha_j) - \nu(\alpha_{j+1}) = 1$.
-
-    Now suppose $P(x_j) \neq 0$. Observe that $P_0(\alpha_j)$ and
-    $P_0(\alpha_{j+1})$ have the same sign, otherwise by the intermediate value theorem
-    and the construction, $P_0(x_j) = 0$. Also a difference between
-    $\nu(\alpha_j)$ and $\nu(\alpha_{j+1})$ only occurs if there exists
-    $i \in \{0, \ldots, n-1\} $ such that $P_i(\alpha_j) P_i(\alpha_{j+1}) < 0$. In this
-    case, again by the intermediate value theorem, we have $P_i(x_j) = 0$. By the definition
-    of a Sturm sequence, we have $P_{i-1}(x_j)P_{i+1}(x_j) < 0$. If $P_{i-1}(x_j) < 0$
-    then $P_{i-1} < 0$ on $[\alpha_j, \alpha_{j+1}]$, because $x_j$ is
-    the only possible root for $P_{i-1}$ in $[\alpha_j, \alpha_{j+1}]$,
-    so $P_{i-1}$ cannot change sign on that interval. Likewise,
-    $P_{i+1}$ has the same sign on $[\alpha_j, \alpha_{j+1}]$ as it does at $x_j$. Proceeding
-    similarly when $P_{i-1}(x_j) > 0$, we arrive at the following possibilities
-    for the sign sequences of $P_{i-1}(\alpha_j) P_i(\alpha_j)P_{i+1}(\alpha_j)$
-    and $P_{i-1}(\alpha_{j+1})P_i(\alpha_{j+1})P_{i+1}(\alpha_{j+1})$:
-
-    \begin{figure}[h!]
-        \centering
-        \begin{subfigure}[c]{0.4\textwidth}
-        \begin{tabular}{c|c|c}
-            & $P_i(\alpha_j) < 0$ & $P_i(\alpha_j) > 0$ \\ \hline
-            $P_{i-1}(x_j) < 0$ & $- - +$ & $- + + $ \\ \hline
-            $P_{i-1}(x_j) > 0$ & $+ - -$ & $+ + -$
-        \end{tabular}
-        \subcaption{Sign sequence at $\alpha_{j}$}
-        \end{subfigure}
-        \hspace{1cm}
-        \begin{subfigure}[c]{0.4\textwidth}
-        \begin{tabular}{c|c|c}
-            & $P_i(\alpha_j) < 0$ & $P_i(\alpha_j) > 0$ \\ \hline
-            $P_{i-1}(x_j) < 0$ & $- + +$ & $- - + $ \\ \hline
-            $P_{i-1}(x_j) > 0$ & $+ + -$ & $+ - -$
-        \end{tabular}
-        \subcaption{Sign sequence at $\alpha_{j+1}$}
-        \end{subfigure}
-    \end{figure}
-    Since sign sequences located in cells of the two tables corresponding to the same case have
-    the same number of sign changes, equal to $1$, we see that
-    $\nu(\alpha_{j}) - \nu(\alpha_{j+1}) = 0$.
-\end{proof}
-
-We deduce from the previous result, this important result:
-
-\begin{korollar}
-    Let $(k, \le )$ be an ordered field and let $L_1, L_2$ be real-closed, orderable extensions
-    of $k$. Let $P \in k[t]$ be an irreducible polynomial over $k$. Then $P$ has the same
-    number of roots in $L_1$ as it does in $L_2$.
-    \label{lemma:number-of-roots-in-real-closed-extension}
-\end{korollar}
-
-\begin{proof}
-    For a polynomial $Q = c_n t^{n} + c_{n-1} t ^{n-1} + \ldots + c_0 \in k[t]$ with
-    $c_n \neq 0$, the roots
-    of $Q$ in an ordered real-closed extension $L$ of $k$ are bounded by
-    \begin{salign*}
-    M
-    = 1 + \left| \frac{c_{n-1}}{c_n} \right|_L + \ldots + \left| \frac{c_0}{c_n} \right|_L
-    = 1 + \left| \frac{c_{n-1}}{c_n} \right|_k + \ldots + \left| \frac{c_0}{c_n} \right|_k
-    .\end{salign*}
-    Note that $M$ is independent from $L$.
-    So given $P \in k[t]$ irreducible and the associated Sturm sequence
-    $(P_0, P_1, \ldots, P_n)$, there exists $M \in k$ such that all roots
-    of all $P_i$'s in $L$ are contained in the interval $[-M, M] \subseteq L$. Since
-    $\text{char } k = 0$, $P$ is separable, by \ref{thm:sturm}
-    the number of roots of $P$ in $[-M, M] \subseteq L$ is equal
-    to $\nu(-M) - \nu(M)$. Since $\pm M \in k$,
-    all $P_i \in k[t]$ and the ordering of $L$ extends the one of $k$, the number
-    of sign changes $\nu(\pm M)$ in the sequences
-    $(P_0(-M), P_1(-M), \ldots, P_n(-M))$
-    and $(P_0(M), P_1(M), \ldots, P_n(M))$ does not depend on $L$.
-\end{proof}
-
-\begin{bem}
-    \begin{enumerate}[(i)]
-    \item 
-    In particular, if $P \in k[t]$ is an arbitrary polynomial, then if $P$ has a root
-    in a real-closed extension $L$ of $k$, then it has a root in all real-closed extensions of $k$.
-
-    A polynomial with coefficients in an ordered field $(k, \le)$ might not have roots
-    in any real-closed extensions of $k$.
-        \item 
-    There is a proof of Sturm's algorithm that does not require $P$ to
-    be separable. As a consequence \ref{lemma:number-of-roots-in-real-closed-extension}
-    holds for all $P \in k[t]$, not only the irreducible ones.
-    \end{enumerate}
-\end{bem}
-
-\end{document}

ws2022/rav/lecture/rav22.tex

@@ -1,75 +0,0 @@
-\documentclass{lecture}
-
-\begin{document}
-
-\section{The real Nullstellensatz}
-
-When $k$ is algebraically closed, Hilbert's Nullstellensatz implies
-$\mathcal{I}(\mathcal{V}_{k^{n}}(I)) = \sqrt{I}$ for all ideal
-$I \subseteq k[T_1, \ldots, T_n]$. In this section we try to compute
-$\mathcal{I}(\mathcal{V}_{k^{n}}(I))$ when $k$ is a real-closed field.
-
-\begin{definition}[]
-    Let $(k, \le)$ be an ordered field and let $A$ be a commutative $k$-algebra with unit.
-    An ideal $I \subseteq A$ is called a \emph{real ideal} if it satisfies the following condition: If
-    $\lambda_1, \ldots, \lambda_r > 0$ in $k$ and $a_1, \ldots, a_r \in A$ satisfy
-    \[
-    \sum_{j=1}^{r} \lambda_j a_j^2 \in I
-    ,\] then $a_j \in I$ for all $j$.
-    $A$ is a \emph{real algebra} if the zero ideal in $A$ is
-    a real ideal.
-\end{definition}
-
-\begin{satz}
-    Let $(k, \le)$ be an ordered field and let $Z \subseteq k^{n}$ be a subset. Then the ideal
-    $\mathcal{I}(Z)$ is a real ideal.
-\end{satz}
-
-\begin{proof}
-    If $Z = \emptyset$, then $\mathcal{I}(Z) = \mathcal{I}(\emptyset) = k[T_1, \ldots, T_n]$ is
-    a real ideal. Now assume $Z \neq \emptyset$. In this case, if $P_1, \ldots, P_r \in k[T_1, \ldots, T_n]$
-    and $\lambda_1, \ldots, \lambda_r > 0$ in $k$ are such that
-    $\sum_{j=1}^{r} \lambda_j P_j^2 \in \mathcal{I}(Z)$, then
-    for all $x \in Z$, $\sum_{j=1}^{r} \lambda_j P_j^2(x) = 0$ in $k$. Since
-    $k$ is an ordered field and $\lambda_j > 0$ for all $j$, this implies
-    that for all $j$, $P_j(x) = 0$, i.e. $P_j \in \mathcal{I}(Z)$.
-\end{proof}
-
-Recall that if $k$ is an arbitrary field and $I \subsetneq k[T_1, \ldots, T_n]$ is a proper ideal,
-then finding a common zero $x \in L^{n}$ to all polynomials $P \in I$ for some extension $L$ of $k$
-is equivalent to finding a homomorphism of $k$-algebras
-\[
-\varphi\colon k[T_1, \ldots, T_n]/I \longrightarrow L
-.\] Indeed, the correspondence is obtained by sending such a $\varphi$
-to $x = (x_1, \ldots, x_n)$ where $x_i = \varphi(T_i \text{ mod } I)$. The basic
-result should be about giving sufficient conditions for such homomorphisms to exist.
-
-\begin{theorem}[Real Nullstellensatz I]
-    Let $(k, \le )$ be an ordered field and let $k^{(r)}$ be the real closure of $k$. Let
-    $I \subseteq k[T_1, \ldots, T_n]$ be a real ideal. Then there exists a homomorphism
-    of $k$-algebras
-    \[
-    k[T_1, \ldots, T_n] / I \longrightarrow k^{(r)}
-    .\] In particular, if $I \subsetneq k[T_1, \ldots, T_n]$ is a proper real ideal, then
-    $\mathcal{V}_{k^{r}}(I) \neq \emptyset$.
-    \label{thm:real-nullstellensatz}
-\end{theorem}
-
-Let $(k, \le)$ be an ordered field. For the proof of \ref{thm:real-nullstellensatz}, we need
-two lemmata:
-
-\begin{lemma}
-    Let $I \subseteq k[T_1, \ldots, T_n]$ be a real ideal. Then $\sqrt{I} = I$. Moreover,
-    if $\mathfrak{p} \supset I$ is a minimal prime ideal containing $I$, then
-    $\mathfrak{p}$ is real.
-\end{lemma}
-
-\begin{lemma}
-    Let $\mathfrak{p} \subseteq k[T_1, \ldots, T_n]$ be a prime ideal. Then the fraction field
-    \[
-    K \coloneqq \operatorname{Frac}\left( k[T_1, \ldots, T_n]/\mathfrak{p} \right) 
-    \] is a real field if and only if the prime ideal $\mathfrak{p}$ is real. In that case
-    $K$ can be ordered in a way that extends the order of $k$.
-\end{lemma}
-
-\end{document}

ws2022/rav/lecture/rav3.tex

@@ -1,263 +0,0 @@
-\documentclass{lecture}
-
-\begin{document}
-
-\section{Plane algebraic curves}
-
-\begin{theorem}
-    If $f \in k[x,y]$ is an irreducible polynomial such that $\mathcal{V}(f)$
-    is infinite, then $\mathcal{I}(\mathcal{V}(f)) = (f)$. In particular,
-    $\mathcal{V}(f)$ is irreducible in this case.
-    \label{thm:plane-curve-ivf=f}
-\end{theorem}
-
-\begin{bem}[]
-    \begin{enumerate}[(i)]
-        \item If $k$ is algebraically closed and $n \ge 2$, then for all $f \in k[x_1, \ldots, x_n]$
-            non-constant, the zero set $\mathcal{V}(f)$ is necessarily infinite.
-        \item The assumption $\mathcal{V}(f)$ infinite is necessary for the conclusion of
-            \ref{thm:plane-curve-ivf=f} to hold:
-            The polynomial
-            \[
-            f(x,y) = (x^2 - 1)^2 + y^2
-            \]
-            is irreducible because, as a polynomial in $y$, it is monic and does not have a root
-            in $\R[x]$ (for otherwise there would be a polynomial $P(x) \in \R[x]$
-            such that $P(x)^2 = -(x^2-1)^2$)
-            and the zero set of $f$ is
-            \[
-            \mathcal{V}(f) = \{ (1, 0)\} \cup \{(-1, 0)\} 
-            ,\] which is reducible.
-        \item \ref{thm:plane-curve-ivf=f} does not hold in this form for hypersurfaces of $k^{n}$ for $n \ge 3$.
-            For instance, the polynomial
-            \[
-            f(x,y,z) = x^2 y^2 + z^{4} \in \R[x,y,z]
-            \] is irreducible and the hypersurface
-            \[
-            \mathcal{V}(f) = \{ (0, y, 0)\colon y \in \R\} \cup \{(x, 0, 0)\colon x \in \R\} 
-            \] is infinite. However, the function
-            \[
-            P\colon (x,y,z) \mapsto xy
-            \] belongs to $\mathcal{I}(\mathcal{V}(f))$ but not to $(f)$. Moreover,
-            $P \in \mathcal{I}(\mathcal{V}(f))$ but neither $x$ nor $y$ are in $\mathcal{I}(\mathcal{V}(f))$,
-            so this ideal is not prime.
-        \item Take $f(x,y) = (x-a)^2 + y^2 \in \R[x,y]$ which is irreducible. Then
-            $\mathcal{V}(f) = \{ (a, 0) \} $ is irreducible, and
-            $\mathcal{I}(\mathcal{V}(f)) = (x-a, y) \supsetneq (f)$. In particular, $(f)$ is a non-maximal
-            prime ideal.
-    \end{enumerate}
-\end{bem}
-
-We need a special case of the famous Bézout theorem, for which we need a result from algebra.
-For an integral domain $R$ denote by $Q(R)$ its fraction field. If $R$ is a factorial ring then
-$q \in R[T]$ is called \emph{primitve} if it is non-constant and its
-coefficients are coprime in $R$.
-
-\begin{satz}[Gauß]
-    Let $R$ be a factorial ring. Then $R[T]$ is also factorial. A polynomial
-    $q \in R[T]$ is prime in $R[T]$ if and only if
-    \begin{enumerate}[(i)]
-        \item $q \in R$ and $q$ is prime in $R$, or
-        \item $q$ is primitve in $R[T]$ and prime in $Q(R)[T]$
-    \end{enumerate}
-    \label{satz:gauss}
-\end{satz}
-
-\begin{proof}
-    Any algebra textbook.
-\end{proof}
-
-\begin{satz}
-    Let $R$ be a factorial ring and $f,g \in R[X]$ coprime. Then $f$ and $g$ are
-    coprime in $Q(R)[X]$.
-    \label{satz:coprime-in-r-is-coprime-in-qr}
-\end{satz}
-
-\begin{proof}
-    Let $h = \frac{a}{b} \in Q(R)[X]$ be a common irreducible factor of $f$ and $g$ with
-    $a \in R[X]$ and $b \in R \setminus 0$. By Gauß $R[X]$ is factorial, thus we
-    may assume $a$ irreducible. Then
-    \[
-    \frac{f}{1} = \frac{p_1}{q_1} \frac{a}{b} \text{ and } \frac{g}{1} = \frac{p_2}{q_2} \frac{a}{b}
-    \] for some $p_1, p_2 \in R[X]$ and $q_1, q_2 \in R \setminus 0$.
-    So $p_1 a = f q_1 b$ and $p_2 a = g q_2 b$. $a$ neither divides $q_1$, $q_2$ nor $b$, for otherwise
-    $a \in R \setminus 0$ by the degree formula for polynomials and $h$ is a unit.
-    Since $a$ divides $fq_1 b$ and
-    $g q_2 b$ and, since $R[X]$ is factorial, $a$ is prime in $R[X]$ and thus
-    $a  \mid f$ and $a  \mid g$.
-\end{proof}
-
-\begin{lemma}[Special case of Bézout]
-    Let $f, g \in k[x,y]$ be two polynomials without common factors in $k[x,y]$. Then the set
-    $\mathcal{V}(f) \cap \mathcal{V}(g)$ is finite.
-    \label{lemma:coprime-finite-zero-locus}
-\end{lemma}
-
-\begin{proof}
-    %\begin{enumerate}[(i)]
-        %\item Claim: $f$ and $g$ have no common factors in $k(x)[y]$. Indeed, if
-        %    $h(x,y) = \frac{H(x,y)}{L(x)} \in k(x)[y]$ is a common irreducible factor of $f$ and $g$,
-        %    then we may assume $H$ and $L$ coprime in $k[x,y]$, with $L$ irreducible in $k[x]$
-        %    and $H$ irreducible in $k[x,y]$. Thus we can write
-        %    \begin{salign*}
-        %        f(x,y) &= \frac{A(x,y)}{M(x)} \frac{H(x,y)}{L(x)}
-        %        \intertext{and}
-        %        g(x,y) &= \frac{B(x,y)}{N(x)} \frac{H(x,y)}{L(x)}
-        %    \end{salign*}
-        %    with $A$ and $M$ coprime, as well as $B$ and $N$ coprime in $k[x,y]$.
-        %    So $A(x,y) H(x,y) = M(x) L(x) f(x,y)$
-        %    and $B(x,y) H(x,y) = N(x) L(x) g(x,y)$.
-        %    But $H(x,y)$ cannot divide $L(x), M(x)$ nor $N(x)$ in $k[x,y]$, for otherwise
-        %    $H(x,y) \in k[x]$, making $h(x,y) = \frac{H(x,y)}{L(x)}$ a unit in $k(x)[y]$. But
-        %    $H(x,y)$ is irreducible in $k[x,y]$ and divides both $M(x)L(x)f(x,y)$ and
-        %    $N(x)L(x)g(x,y)$, so $H(x,y)$ divides $f(x,y)$ and $g(x,y)$ in $k[x,y]$. Contradiction.
-        %\item
-    Since $k(x)[y]$ is a principal ideal domain, \ref{satz:coprime-in-r-is-coprime-in-qr} implies
-            $(f,g) = k(x)[y]$, hence the existence of $A(x,y), B(x,y), M(x), N(x)$ such that
-            \[
-            f(x,y) A(x,y) + g(x,y) B(x,y) = \underbrace{M(x) N(x)}_{=: D(x)}
-            \] with $D(x) \in k[x]$. Since a common zero $(x,y)$ of $f$ and $g$ gives a zero of
-            $D$, and $D$ has finitely many zeros, there are only finitely many $x$ such that
-            $(x,y)$ is a zero of both $f$ and $g$. But, for fixed $x \in k$, the polynomial
-            \[
-            y \mapsto f(x,y) - g(x,y)
-            \] has only finitely many zeros in $k$. So $\mathcal{V}(f) \cap \mathcal{V}(g)$ is finite.
-    %\end{enumerate}
-\end{proof}
-
-\begin{proof}[Proof of \ref{thm:plane-curve-ivf=f}]
-    Let $f \in k[x,y]$ be irreducible such that $\mathcal{V}(f) \subseteq k^2$ is infinite.
-    Since $f \in \mathcal{I}(\mathcal{V}(f))$, it suffices to show that
-    $\mathcal{I}(\mathcal{V}(f)) \subseteq (f)$.
-    Let
-    $g \in \mathcal{I}(\mathcal{V}(f))$. Then $\mathcal{V}(f) \subseteq \mathcal{V}(g)$. Thus
-    \[
-    \mathcal{V}(f) \cap \mathcal{V}(g) = \mathcal{V}(f)
-    \] which is infinite by assumption. Thus by \ref{lemma:coprime-finite-zero-locus},
-    $f$ and $g$ have a common factor. Since $f$ is irreducible, this implies that $f  \mid g$, i.e.
-    $g \in (f)$.
-\end{proof}
-
-We can use \ref{thm:plane-curve-ivf=f} to find the irreducible components of a
-hypersurface $\mathcal{V}(P) \subseteq k^2$.
-
-\begin{korollar}
-    Let $P \in k[x,y]$ be non-constant and $P = u P_1^{n_1} \cdots P_r^{n_r}$ be the decomposition
-    into irreducible factors. If each $\mathcal{V}(P_i)$ is infinite, then the algebraic sets
-    $\mathcal{V}(P_i)$ are the irreducible components of $\mathcal{V}(P)$.
-\end{korollar}
-
-\begin{proof}
-    Note that
-    \[
-    \mathcal{V}(P) = \mathcal{V}(P_1^{n_1} \cdots P_r^{n_r}) = \mathcal{V}(P_1) \cup \ldots \cup \mathcal{V}(P_r)
-    .\] Since $P_i$ is irreducible and $\mathcal{V}(P_i)$ is infinite for all $i$,
-    by \ref{thm:plane-curve-ivf=f} $\mathcal{V}(P_i)$ is irreducible and for $i \neq j$
-    $\mathcal{V}(P_i) \not\subset \mathcal{V}(P_j)$, for otherwise
-    \[
-        (P_i) = \mathcal{I}(\mathcal{V}(P_i)) \supset \mathcal{I} (\mathcal{V}(P_j)) = (P_j)
-    \] which is impossible for distinct irreducible elements $P_i, P_j$.
-\end{proof}
-
-\begin{bsp}[Real plane cubics]
-    Let $P(x,y) = y^2 - f(x)$ with $\text{deg}_xf = 3$ in $k[x]$. Since $\text{deg}_y P \ge 1 $
-    and the leading coefficient of $P$ is $1$, the polynomial $P$ is primitive in $k[x][y]$.
-    It is reducible in $k(x)[y]$ if and only if there exists $a(x), b(x) \in k(x)$ such that
-    $(y-a)(y-b) = y^2 - f$, i.e. $b = -a$ and $f = a^2$ in $k(x)$, therefore also in $k[x]$.
-    Since $\text{deg}_xf = 3 $, this cannot happen. So, $P$ is irreducible
-    by \ref{satz:gauss}.
-
-    Moreover, when $k = \R$, the
-    cubic polynomial $f(x)$ takes on an infinite number of positive values,
-    so $\mathcal{V}(y^2 - f(x)) = \mathcal{V}(P)$ is infinite. In conclusion,
-    real cubics of the form $y^2 - f(x) = 0$ are irreducible algebraic sets in $\R^2$
-    by \ref{thm:plane-curve-ivf=f}.
-\end{bsp}
-
-\begin{figure}
-    \centering
-    \begin{tikzpicture}
-        \begin{axis}[
-            xmin = -1
-            ]
-        \algebraiccurve[red][$y^2 = x^3$]{y^2 - x^3}
-        \end{axis}
-    \end{tikzpicture}
-    \caption{the cuspidal cubic}
-\end{figure}
-
-\begin{figure}
-    \centering
-    \begin{tikzpicture}
-        \begin{axis}[
-            ]
-        \algebraiccurve[red][$y^2 = x^2(x+1)$][-2:2][-2:2]{y^2 - x^2*(x+1)}
-        \end{axis}
-    \end{tikzpicture}
-    \caption{the nodal cubic}
-\end{figure}
-
-\begin{figure}
-    \centering
-    \begin{tikzpicture}[scale=0.9]
-        \begin{axis}[
-            xmin = -1
-            ]
-        \algebraiccurve[red][$y^2 = x(x^2+1)$][-2:2][-2:2]{y^2 - x*(x^2+1)}
-        \end{axis}
-    \end{tikzpicture}
-    \hspace{.05\textwidth}
-    \begin{tikzpicture}[scale=0.9]
-        \begin{axis}[
-            ]
-        \algebraiccurve[red][$y^2 = x(x^2-1)$][-2:2][-2:2]{y^2 - x*(x^2-1)}
-        \end{axis}
-    \end{tikzpicture}
-    \caption{the smooth cubics: the second curve demonstrates that the notion of connectedness in
-        the Zariski topologoy of $\R^2$ is very different from the one in the usual topology of $\R^2$.}
-\end{figure}
-
-\begin{satz}
-    Let $k$ be an algebraically closed field and let $P \in k[T_1, \ldots, T_n]$ be a non-constant polynomial
-    with $n \ge 2$. Then $\mathcal{V}(P)$ is infinite.
-\end{satz}
-
-\begin{proof}
-    Since $P$ is non-constant, we may assume that $\text{deg}_{x_1} P \ge 1$. Write
-    \[
-    P(T_1, \ldots, T_n) = \sum_{i=1}^{d} g_i(T_2, \ldots, T_n) T_1^{i}
-    ,\]
-    with $d \ge 1$ and $g_d \neq 0$. Then $D_{k^{n-1}}(g_d)$ is infinite: Since $g_d \neq 0$ and
-    $k$ infinite, it is non-empty. Thus let $(a_2, \ldots, a_n) \in k^{n-1}$ such that
-    $g_d(a) \neq 0$. Then $g_d(ta) = g_d(ta_2, \ldots, ta_n) \in k[t]$ is a non-zero polynomial and thus
-    has only finitely many zeros in $k$. In particular $D_{k^{n-1}}(g_d)$ is infinite.
-
-    For $(a_2, \ldots, a_{n-1}) \in D_{k^{n-1}}(g_d)$, $P(T_1, a_2, \ldots, a_n) \in k[T_1]$ is non-constant
-    and thus has a root $a_1$ in the algebraically closed field $k$. Hence
-    $(a_1, \ldots, a_n) \in \mathcal{V}(P)$.
-\end{proof}
-
-We finally give a complete classification of irreducible algebraic sets in the affine plane $k^2$ for
-an infinite field $k$.
-
-\begin{satz}
-    Let $k$ be an infinite field. Then the irreducible algebraic subsets of $k^2$ are:
-    \begin{enumerate}[(i)]
-        \item the whole affine plane $k^2$
-        \item single points $\{ (a, b) \} \subseteq k^2$
-        \item infinite algebraic sets defined by an irreducible polynomial $f \in k[x,y]$.
-    \end{enumerate}
-    \label{satz:classification-irred-alg-subsets-plane}
-\end{satz}
-
-\begin{proof}
-    Let $V \subseteq k^2$ be an irreducible algebraic subset of the affine plane. If $V$ is finite,
-    it reduces to a point. So we may assume $V$ infinite. If $\mathcal{I}(V) = (0)$, then $V = k^2$.
-    Otherwise, there is a non-constant polynomial $P \in k[x,y]$ such that $P$ vanishes on $V$. Since
-    $V$ is irreducible, $\mathcal{I}(V)$ is prime, so it contains an irreducible factor $f$ of $P$.
-    Let $g \in \mathcal{I}(V)$. Then $V \subseteq \mathcal{V}(f) \cap \mathcal{V}(g)$, but since
-    $V$ is infinite, $f$ and $g$ must have a common factor. By irreducibility of $f$, it follows
-    $f  \mid g$, i.e. $g \in (f)$. Hence $\mathcal{I}(V) = (f)$ and $V = \mathcal{V}(f)$.
-\end{proof}
-
-\end{document}

ws2022/rav/lecture/rav4.tex

@@ -1,181 +0,0 @@
-\documentclass{lecture}
-
-\begin{document}
-
-\section{Prime ideals in $k[x,y]$}
-
-\begin{satz}
-    Let $A$ be a principal ideal domain. Let $\mathfrak{p} \subseteq A[X]$ be a prime ideal. Then
-    $\mathfrak{p}$ satisfies exactly one of the following three mutually exclusive possibilities:
-    \begin{enumerate}[(i)]
-        \item $\mathfrak{p} = (0)$
-        \item $\mathfrak{p} = (f)$, where $f \in A[X]$ is irreducible
-        \item $\mathfrak{p} = (a, q)$, where $a \in A$ is irreducible and
-            $q \in A[X]$ such that its reduction modulo $a A$ is an irreducible element
-            in $A / aA [X]$. In this case, $\mathfrak{p}$ is a maximal ideal.
-    \end{enumerate}
-    \label{thm:class-prim-pol-pid}
-\end{satz}
-
-\begin{proof}
-    Let $\mathfrak{p} \subseteq A[X]$ be a prime ideal. If $\mathfrak{p}$ is principal, then
-    $\mathfrak{p} = (f)$ for some $f \in A[X]$. If $f = 0$, we are done. Otherwise,
-    since $A[X]$ is factorial by Gauß and $\mathfrak{p}$ is prime, $f$ is irreducible.
-
-    Let now $\mathfrak{p}$ not be principal. Then there exist $f, g \in \mathfrak{p}$ without
-    common factors in $A[X]$. By \ref{satz:coprime-in-r-is-coprime-in-qr}, they
-    also have no common factors in the principal ideal domain $Q(A)[X]$, so
-    $Mf + Ng = 1$ for some $M, N \in Q(A)[X]$. By multiplying with the denominators, we obtain
-    $Pf + Qg = b$ for some $b \in A$ and $P, Q \in A[X]$. So $b \in (f, g) \subseteq \mathfrak{p}$,
-    thus there is an irreducible factor $a$ of $b$ in $A$ such that $a \in \mathfrak{p}$.
-    Moreover, $a A[X] \subsetneq \mathfrak{p}$ since $\mathfrak{p}$ is not principal. Now consider
-    the prime ideal
-    \[
-    \mathfrak{p} / a A[X] \subset  A[X]/aA[X] \simeq \left( A / aA \right)[X]
-    .\] Since $A$ is a PID and $a$ is irreducible, $A/aA$ is a field and $(A / aA)[X]$ a PID.
-    So $\mathfrak{p}/aA[X]$ is generated by an irreducible element $\overline{q} \in (A/aA)[X]$
-    for some $q \in A[X]$. Thus $\mathfrak{p} = (a, q)$. Moreover
-    \[
-    \faktor{A[X]}{\mathfrak{p}} \simeq
-    \faktor{\left(\faktor{A}{aA}\right)[X]}{\left(\faktor{\mathfrak{p}}{aA}\right)[X]}
-    =
-    \faktor{\left( \faktor{A}{aA} \right)[X] }
-    {\overline{q} \left( \faktor{A}{aA} \right)[X] }
-    \] which is a field since $\left( \faktor{A}{aA} \right)[X]$ is a PID. So $\mathfrak{p}$ is maximal in
-    $A[X]$.
-\end{proof}
-
-Using \ref{thm:class-prim-pol-pid} we can give a simple proof for the classification of maximal ideals
-of $k[T_1, \ldots, T_n]$ when $k$ is algebraically closed and $n=2$.
-
-\begin{korollar}
-    If $k$ is algebraically closed, a maximal ideal $\mathfrak{m}$ of $k[x,y]$ is of the form
-    $\mathfrak{m} = (x-a, y-b)$ with $(a, b) \in k^2$. In particular, principal ideals are never maximal.
-    \label{kor:max-ideals-alg-closed-k2}
-\end{korollar}
-
-\begin{proof}
-    Since $\mathfrak{m}$ is maximal, it is prime and $\mathfrak{m} \neq (0)$. By
-    \ref{thm:class-prim-pol-pid}, $\mathfrak{m} = (P, f)$
-    with $P \in k[x]$ irreducible and $f \in k[x,y]$ such that
-    its image $\overline{f}$ in $(k[x]/(P))[y]$ is irreducible or
-    $\mathfrak{m} = (f)$ for $f \in k[x,y]$ irreducible.
-
-    \begin{enumerate}[(1)]
-        \item $\mathfrak{m} = (P, f)$. Since $k$ is algebraically closed and $P \in k[x]$ is irreducible,
-            $P = x - a$ for some $a \in k$.
-            \[
-            k[x]/(P) = k[x]/(x-a) \simeq k
-            .\] Since $\overline{f} \in k[y]$ is also irreducible, $\overline{f} = y - b$ for some $b \in k$.
-        \item $\mathfrak{m} = (f)$. Since $k = \overline{k}$, $\mathcal{V}(f)$ is infinite, in particular
-            $\mathcal{V}(f) \neq \emptyset$. Then if $(a,b) \in \mathcal{V}(f)$,
-            \[
-                (x-a, y-b) = \mathcal{I}(\{(a, b)\})
-                \supset \mathcal{I}(\mathcal{V}(f)) \supset (f)
-            .\] Since $(f)$ is maximal, it follows that $(f) = (x-a, y-b)$, which is impossible since
-            $x -a $ and $y-b$ habe no common factors in $k[x,y]$.
-    \end{enumerate}
-\end{proof}
-
-\begin{bem}[]
-    The ideal $(x^2 + 1, y)$ is maximal in $\R[x,y]$ and is not of the form $(x-a, y-b)$ for $(a,b) \in \R^2$.
-    Indeed,
-    \[
-    \faktor{\R[x,y]}{(x^2 + 1, y)}
-    \simeq
-    \faktor{\left( \R[y]/y\R[y] \right)[x]}{(x^2 + 1)}
-    \simeq \R[x]/(x^2 + 1)
-    \simeq \mathbb{C}
-    .\] 
-\end{bem}
-
-\begin{satz}[]
-    Let $k$ be an algebraically closed field. Then the maps $V \mapsto \mathcal{I}(V)$
-    and $I \mapsto \mathcal{V}(I)$ induce a bijection
-    \begin{salign*}
-    \{ \text{irreducible algebraic subsets of } k^2\}
-    &\longleftrightarrow \{ \text{prime ideals in } k[x,y]\} 
-    \intertext{through wich we have correspondences}
-        \text{points } (a, b) \in k^2 &\longleftrightarrow \text{maximal ideals } (x-a, y-b) \text{ in }k[x,y] \\
-        \text{proper, infinite, irreducible algebraic sets}
-                                      &\longleftrightarrow \text{prime ideals } (f) \subseteq k[x,y]
-                                      \text{ with } f \text{ irreducible} \\
-        k^2 &\longleftrightarrow (0)
-    .\end{salign*} 
-    \label{satz:correspondence-irred-subsets-prime-ideals}
-\end{satz}
-
-\begin{proof}
-    Let $V \subseteq k^2$ be an irreducible algebraic set. By
-    \ref{satz:classification-irred-alg-subsets-plane} we
-    can distinguish the following cases:
-    \begin{enumerate}[(i)]
-        \item If $V = k^2$, then $\mathcal{I}(V) = (0)$ since $k$ is infinite and
-            $\mathcal{I}(\mathcal{V}(0)) = (0)$.
-        \item If $V = \{(a,b)\} $, then $\mathcal{I}(V) \supset (x-a, y-b) \eqqcolon \mathfrak{m} $. Since
-            $\mathfrak{m}$ is maximal, $\mathcal{I}(V) = \mathfrak{m}$. Since $V = \mathcal{V}(\mathfrak{m})$,
-            this also shows $\mathcal{I}(\mathcal{V}(\mathfrak{m})) = \mathfrak{m}$.
-        \item If $V = \mathcal{V}(f)$ where $f \in k[x,y]$ is irreducible, 
-            then by \ref{thm:plane-curve-ivf=f} $\mathcal{I}(\mathcal{V}(f)) = (f)$.
-    \end{enumerate}
-    So, every irreducible algebraic set $V \subseteq k^2$ is of the form
-    $\mathcal{V}(\mathfrak{p})$ for some prime ideal $\mathfrak{p} \subseteq k[x,y]$. Moreover,
-    \[
-    \mathcal{I}(\mathcal{V}(\mathfrak{p})) = \mathfrak{p}
-    .\]
-    Let now $\mathfrak{p}$ be a prime ideal in $k[x,y]$. By \ref{thm:class-prim-pol-pid} we can dinstiguish
-    the following cases:
-    \begin{enumerate}[(i)]
-        \item $\mathfrak{p} = (0)$: Then $\mathcal{V}(\mathfrak{p}) = k^2$ and
-            since $k$ is infinite, $k^2$ is irreducible.
-        \item $\mathfrak{p}$ maximal: Then by \ref{kor:max-ideals-alg-closed-k2},
-            $\mathfrak{p} = (x-a, y-b)$ for some $(a, b) \in k^2$. So $\mathcal{V}(m) = \{(a, b)\}$
-            is irreducible.
-        \item $\mathfrak{p} = (f)$ with $f \in k[x,y]$ irreducible. Since $k = \overline{k}$,
-            $\mathcal{V}(f)$ is infinite and hence by \ref{thm:plane-curve-ivf=f} irreducible.
-    \end{enumerate}
-    Thus the maps in the proposition are well-defined, mutually inverse and induce the stated
-    correspondences.
-\end{proof}
-
-\begin{korollar}
-    Assume that $k$ is algebraically closed and let $\mathfrak{p} \subseteq k[x,y]$ be a prime ideal.
-    Then
-    \[
-    \mathfrak{p} = \bigcap_{\mathfrak{m} \; \mathrm{maximal}, \mathfrak{m} \supset \mathfrak{p}} 
-    \mathfrak{m}
-    .\] 
-\end{korollar}
-
-\begin{proof}
-    If $\mathfrak{p}$ is maximal, there is nothing to prove. If $\mathfrak{p} = (0)$, $\mathfrak{p}$
-    is contained in $(x-a, y-b)$ for $(a, b) \in k^2$. Since $k$ is infinite, the intersection
-    of these ideals is $(0)$. Otherwise, by \ref{satz:correspondence-irred-subsets-prime-ideals},
-    $\mathfrak{p} = (f)$ for some $f \in k[x,y]$ irreducible. Then, since $k = \overline{k}$,
-    $\mathcal{V}(f)$ is infinite and with \ref{thm:plane-curve-ivf=f}:
-    \[
-    \mathfrak{p} = (f) = \mathcal{I}(\mathcal{V}(f))
-    = \mathcal{I}\left( \bigcup_{(a, b) \in \mathcal{V}(f)}  \{(a, b)\} \right)
-    \supset \bigcap_{(a,b) \in \mathcal{V}(f)}
-    \mathcal{I}(\{(a,b)\})
-    \supset (f) = \mathfrak{p}
-    .\] By \ref{satz:correspondence-irred-subsets-prime-ideals}, the ideals
-    $\mathcal{I}(\{(a,b)\})$ for $(a, b) \in \mathcal{V}(f)$ are exactly the
-    maximal ideals containing $(f) = \mathfrak{p}$.
-\end{proof}
-
-\begin{korollar}
-    Let $\mathfrak{p} \subseteq k[x,y]$ be a non-principal prime ideal.
-    Then $\mathcal{V}(\mathfrak{p}) \subseteq k^2$ is finite.
-\end{korollar}
-
-\begin{proof}
-    Since $\mathfrak{p}$ is not principal, there exist $f, g \in \mathfrak{p}$ without common factors. Since
-    $(f, g) \subset \mathfrak{p}$, we have
-    \[
-    \mathcal{V}(f) \cap \mathcal{V}(g) =
-    \mathcal{V}(f, g) \supset \mathcal{V}(\mathfrak{p})
-    \] and the left hand side is finite by \ref{lemma:coprime-finite-zero-locus}.
-\end{proof}
-
-\end{document}

ws2022/rav/lecture/rav5.tex

@@ -1,236 +0,0 @@
-\documentclass{lecture}
-
-\begin{document}
-
-\chapter{Algebraic varieties}
-
-\section{Spaces with functions}
-
-\begin{definition}[]
-    Let $k$ be a field. A \emph{space with functions over $k$} is a pair
-    $(X, \mathcal{O}_x)$ where $X$ is a topological space and
-    $\mathcal{O}_X$ is a subsheaf of the sheaf of $k$-valued functions, seen as 
-    a sheef of $k$-algebras, and satisfying the following condition:
-
-    If $U \subseteq X$ is an open set and $f \in \mathcal{O}_X(U)$, then
-    the set
-    \[
-    D_U(f) \coloneqq \{ x \in U \mid f(x) \neq 0\}
-    \] is open in $U$ and the function $\frac{1}{f}\colon D_U(f) \to k$, 
-    $x \mapsto \frac{1}{f(x)}$ belongs to $\mathcal{O}_X(D_U(f))$.
-\end{definition}
-
-\begin{bem}[]
-    Concretely, it means that there is for each open set $U \subseteq X$ a
-    $k$-Algebra $\mathcal{O}_X(U)$ of ,,regular`` functions such that
-    \begin{enumerate}[(i)]
-        \item the restriction of a regular function $f\colon  U \to k$ to
-            a sub-open $U' \subseteq U$ is regular on $U'$.
-        \item if $f\colon U \to k$ is a function and $(U_{\alpha})_{\alpha \in A}$ is
-            an open cover of $U$ such that $f|_{U_{\alpha}}$ is regular on
-            $U_{\alpha}$, then $f$ is regular on $U$.
-        \item if $f$ is regular on $U$, the set $\{f \neq 0\} $ is open in $U$ and
-            $\frac{1}{f}$ is regular wherever it is defined.
-    \end{enumerate}
-\end{bem}
-
-\begin{bem}[]
-    If $\{0\} $ is closed in $k$ and $f\colon U \to k$ is continuous, then
-    $D_U(f)$ is open in $U$. So, this conditions is often automatically met in practice.
-\end{bem}
-
-\begin{bsp}
-
-\begin{enumerate}[(i)]
-    \item $(X, \mathcal{C}_X)$ a topological space endowed with its sheaf of $\R$-valued
-        (or $\mathbb{C}$-valued) continuous functions, the fields $\R$ and $\mathbb{C}$
-        being endowed here with their classical topology.
-    \item $(V, \mathcal{O}_V)$ where
-        $V = \mathcal{V}(P_1, \ldots, P_m)$ is an algebraic subset of $k^{n}$
-        (endowed with the Zariski topology) and, for all $U \subseteq V$ open,
-        \[
-        \mathcal{O}_V(U) \coloneqq
-        \left\{ f \colon U \to k\ \middle \vert
-            \begin{array}{l}
-            \forall x \in U \exists x \in U_x \text{ open},
-        P, Q \in k[x_1, \ldots, x_n] \text{ such that }\\ \text{for } z \in U \cap U_x,
-        Q(z) \neq 0 \text{ and } f(z) = \frac{P(z)}{Q(z)}
-            \end{array}
-        \right\}
-        .\] 
-    \item $(M, \mathcal{C}^{\infty}_M)$ where
-        $M = \varphi^{-1}(0)$ is a non-singular level set of a $\mathcal{C}^{\infty}$
-        map $\varphi\colon \Omega \to \R^{m}$ where
-        $\Omega \subseteq \R^{p+m}$ is an open set
-        (in the usual topology of $\R^{p+m}$)
-        and, for all $U \subseteq M$ open,
-        $\mathcal{C}^{\infty}_M(U)$ locally smooth maps.
-        %\[
-        %\mathcal{C}^{\infty}_M(U)
-        %\coloneqq \{ f \colon U \to \R\} 
-        %.\] 
-\end{enumerate}
-
-\end{bsp}
-
-\begin{aufgabe}[]
-    Let $(X, \mathcal{O}_X)$ be a space with functions and let $U \subseteq X$ be
-    an open subset. Define, for all $U' \subseteq U$ open,
-    \[
-    \mathcal{O}_X|_{U}(U') \coloneqq \mathcal{O}_X(U')
-    .\] Then $(U, \mathcal{O}_X|_U)$ is a space with functions.
-\end{aufgabe}
-
-\begin{bsp}[]
-
-\begin{enumerate}[(i)]
-    \item $(V, \mathcal{O}_V)$ an algebraic subset of $k^{n}$,
-        $f\colon V \to k$ a polynomial function,
-        $U \coloneqq D_V(f)$ is open in $V$ and the sheaf
-        of regular functions that we defined on the locally closed subset
-        $D_V(f) = D_{k^{n}}(f) \cap V$ coincides with
-        the restriction to $D_V(f)$ of the sheaf of regular functions on $V$.
-    \item $B \subseteq \R^{n}$ or $\mathbb{C}^{n}$ an open ball
-        (with respect to the usual topology), equipped with the sheaf of
-        $\mathcal{C}^{\infty}$ or holomorphic functions.
-\end{enumerate}
-    
-\end{bsp}
-
-\section{Morphisms}
-
-\begin{bem}[]
-    Note that if $f\colon X \to Y$ is a map and
-    $h\colon U \to k$ is a function defined on a subset $U \subseteq Y$, there
-    is a pullback map $f_U^{*}$ taking
-    $h\colon U \to k$ to the function
-    $f_U^{*} \coloneqq h \circ f \colon f^{-1}(U) \to k$. This map is a homomorphism of $k$-algebras.
-    Moreover given a map $g\colon Y \to Z$ and a subset $V \subseteq Z$ such that
-    $g^{-1}(V) \subseteq U$, we have, for all $h\colon V \to k$,
-    \[
-    f_U^{*}(g_V^{*}(h)) = f_U^{*}(h \circ g) = (h \circ g) \circ f = h \circ (g \circ f)
-    = (g \circ f)_V^{*}(h)
-    .\]
-\end{bem}
-
-\begin{definition}[]
-    Let $(X, \mathcal{O}_X)$ and $(Y, \mathcal{O}_Y)$ be two spaces with functions over a field
-    $k$. A \emph{morphism of spaces with functions} between $(X, \mathcal{O}_X)$
-    and $(Y, \mathcal{O}_Y)$ is a
-    continuous map $f\colon X \to Y$ such that, for all open set $U \subseteq Y$, the
-    pullback map $f_U^{*}$ takes a regular function on the open set $U \subseteq Y$ to
-    a regular function on the open set $f^{-1}(U) \subseteq X$.
-\end{definition}
-
-\begin{bem}[]
-    Then, given open sets $U' \subseteq U$ in $Y$, we have compatible homomorphisms of $k$-algebras:
-
-    In other words, we have a morphism of sheaves on $Y$
-    $f^{*}\colon \mathcal{O}_Y \to f_{*} \mathcal{O}_X$, where
-    by definition $(f_{*}\mathcal{O}_X)(U) = \mathcal{O}_X(f^{-1}(U))$.
-\end{bem}
-
-\begin{aufgabe}[]
-    Given $g\colon Y \to Z$, show that $(g \circ f)_{*}\mathcal{O}_X
-    = g_{*}(f_{*} \mathcal{O}_X)$ and that
-    $g_{*}$ is a functor from sheaves on $Y$ to sheaves on $Z$.
-\end{aufgabe}
-
-\begin{bem}
-    If $f\colon (X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$
-    and $g\colon (Y, \mathcal{O}_Y) \to (Z, \mathcal{O}_Z)$ are morphisms,
-    so is the composed map $g \circ f\colon X \to Z$.
-\end{bem}
-
-\begin{satz}[]
-    Let $(X, \mathcal{O}_X)$ and $(Y, \mathcal{O}_Y)$ be locally closed subsets
-    of an affine space $(X \subseteq k^{n}, Y \subseteq K^{m})$ equipped with
-    their respective sheaves of regular functions. Then a map $f\colon X \to Y$
-    is a morphism of spaces with functions if and only if $f = (f_1, \ldots, f_m)$ with
-    each $f_i\colon X \to k$ a regular function on $X$.
-\end{satz}
-
-\begin{proof}
-    The proof that if each of the $f_i$'s is a regular function, then $f$ is a morphism
-    is similar to point (i) of the previous example: it holds because the pullback
-    of a regular function (in particular, the pullback of a polynomial) by a regular function
-    is a regular function, and because an equation of the form $h(x) = 0$ for $h$ a regular
-    function is locally equivalent to a polynomial equation $P(x) = 0$.
-
-    Conversely, if $f\colon X \to Y \subseteq k^{m}$ is a morphism, then the pullback of
-    the $i$-th projection $p_i\colon k^{m} \to k$ is a regular function
-    on $X$. Since $f^{*}p_i = f_i$, the proposition is proved.
-\end{proof}
-
-\begin{bem}[]
-    In the proof of the previous proposition, we used that if the
-    $(f_i\colon X \to k)_{1 \le i \le m}$ are regular functions on the locally closed
-    subset $X \subseteq k^{n}$, then the map
-    \begin{salign*}
-        f\colon X &\to k^{m} \\
-        x &\mapsto (f_1(x), \ldots, f_m(x))
-    \end{salign*} is continuous on $X$. This is because
-    the pre-image of $f^{-1}(V)$ of an algebraic subset
-    $V = V(P_1, \ldots, P_r) \subseteq k^{m}$ is the intersection
-    of $X$ with the zero set
-    \[
-    W = V(P_1 \circ f, \ldots, P_r \circ f) \subseteq k^{n}
-    \] which is indeed an algebraic set, because $P_j \circ f$ is a regular function
-    so the equation $P_j \circ f = 0$ is equivalent to a polynomial equation.
-
-    Beware, however, that if the $(f_i)_{1 \le i \le m}$ are only continuous maps, then
-    $W$ is no longer an algebraic set, so we would need another argument in order to prove
-    the continuity of $f$. Typically, in general topology, we
-    say that $f\colon X \to k^{m}$ is continuous because its components $(f_1, \ldots, f_m)$ are
-    continuous. This argument is valid when the topology used on $k^{m}$ is the
-    product topology of the topologies on $k$. However, this does not hold in general
-    for the Zariski topology, which is strictly larger than the product topology when $k$ is
-    infinite.
-\end{bem}
-
-\begin{bsp}
-
-\begin{enumerate}[(i)]
-    \item The projection map
-        \begin{salign*}
-            \mathcal{V}_{k^{2}}(y - x^2) &\to k \\
-            (x,y) &\mapsto x
-        \end{salign*}
-        is a morphism of spaces with functions, because it is a regular function
-        on $\mathcal{V}_{k^2}(y - x^2)$. It is actually an isomorphism, whose inverse
-        is the morphism
-        \begin{salign*}
-            k &\to \mathcal{V}(y - x^2) \\
-            x &\mapsto (x, x^2)
-        .\end{salign*}
-        Note that $\mathcal{V}_{k^2}(y-x^2)$ is the graph of the polynomial function
-        $x \mapsto x^2$.
-    \item Let $k$ be an infinite field. The map
-        \begin{salign*}
-            k &\to \mathcal{V}_{k^2}(y^2 - x^{3}) \\
-            t &\mapsto (t^2, t ^{3})
-        \end{salign*}
-        is a morphism and a bijection, but it is not an isomorphism, because its inverse
-        \begin{salign*}
-            \mathcal{V}_{k^2}(y^2 - x^{3}) &\to k \\
-            (x, y) &\mapsto \begin{cases}
-                \frac{y}{x} & (x,y) \neq (0,0) \\
-                0 & (x,y) = (0,0)
-            \end{cases}
-        \end{salign*}
-        is not a regular map (this is where we use that $k$ is infinite).
-    \item Consider the groups $G = \mathrm{GL}(n; k)$, $\mathrm{SL}(n; k)$,
-        $\mathrm{O}(n ; k)$, $\mathrm{SO}(n;k)$ etc. as locally closed subsets in
-        $k^{n^2}$ and equip them with their sheaves of regular functions. Then the multiplication
-        $\mu\colon G x G \to G, (g_1, g_2) \mapsto g_1g_2$ and
-        and inversion $\iota\colon G \to G, g \mapsto g^{-1}$
-        are morphisms (here $G\times G$ is viewed as a locally closed subset of
-        $k^{n^2} \times k^{n^2} \simeq k^{2n^2}$, equipped with its Zariski topology), since
-        they are given by regular functions in the coefficients of the matrices.
-
-        Such groups will  later be called \emph{affine algebraic groups}.
-\end{enumerate}
-
-\end{bsp}
-
-\end{document}

ws2022/rav/lecture/rav6.tex

@@ -1,247 +0,0 @@
-\documentclass{lecture}
-
-\begin{document}
-
-\section{Abstract affine varieties}
-
-Recall that an isomorphism of spaces with functions is a morphism
-$f\colon (X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ that admits an inverse morphism.
-
-\begin{bem}[]
-    As we have seen, a bijective morphism is not necessarily an isomorphism.
-\end{bem}
-
-\begin{bem}
-    Somewhat more formally, one could also define a morphism of spaces
-    with functions (over $k$) to be a pair $(f, \varphi)$ such that
-    $f\colon X \to Y$ is a continuous map and $\varphi\colon \mathcal{O}_Y \to f_{*}\mathcal{O}_X$
-    is the morphism of sheaves $f^{*}$. The question then arises how to define
-    properly the composition $(g, \psi) \circ (f, \varphi)$. The formal answer is
-    $(g \circ f, f_{*}(\varphi) \circ \psi)$.
-\end{bem}
-
-\begin{definition}[]
-    Let $k$ be a field. An (abstract) \emph{affine variety over $k$}
-    (also called an affine $k$-variety)
-    is a space with functions $(X, \mathcal{O}_X)$
-    over $k$ that is isomorphic to the space with functions $(V, \mathcal{O}_V)$, where
-    $V$ is an algebraic subset of some affine space $k^{n}$ and $\mathcal{O}_V$ is the
-    sheaf of regular functions on $V$.
-
-    A morphism of affine $k$-varieties is a morphism of the underlying spaces with functions.
-\end{definition}
-
-\begin{bsp}[]
-
-\begin{enumerate}[(i)]
-    \item An algebraic subset $V \subseteq k^{n}$, endowed with its sheaf of regular functions
-        $\mathcal{O}_V$, is an affine variety.
-    \item It is perhaps not obvious at first, but a standard open set
-        $D_V(f)$, where $f\colon V \to k$ is a regular function on an algebraic set
-        $V \subseteq k^{n}$, defines an affine variety. Indeed, when
-        equipped with its sheaf of regular functions,
-        $D_V(f) \simeq \mathcal{V}_{k^{n+1}}(tf(x) - 1)$.
-\end{enumerate}
-
-\end{bsp}
-
-\begin{bem}[]
-    Let $(X, \mathcal{O}_X)$ be a space with functions. An open subset $U \subseteq X$ defines
-    a space with functions $(U, \mathcal{O}_U)$. If
-    $(U, \mathcal{O}_U)$ is isomorphic to some standard open set
-    $D_V(f)$ of an algebraic set $V \subseteq k^{n}$, we will call
-    $U$ an \emph{affine open set}.
-
-    Then the observation is the following: since an algebraic set $V \subseteq k^{n}$
-    is a finite union of standard open sets, every point $x$ in an affine variety $X$
-    has an affine open neighbourhood.
-
-    Less formally, an affine variety $X$, locally ,,looks like`` a standard open set
-    $D_V(f) \subseteq k^{n}$, where $V \subseteq k^{n}$ is an algebraic set. In particular,
-    open subsets of an affine variety also locally look like standard open sets. In fact,
-    they are finite unions of such sets.
-\end{bem}
-
-\begin{bsp}[]
-    The algebraic group $\mathrm{GL}(n ; k)$ is an affine variety over $k$.
-\end{bsp}
-
-\begin{bem}[]
-    An algebraic set $(V, \mathcal{O}_V)$ is a subset $V \subseteq k^{n}$ defined
-    by polynomial equations and equipped with its sheaf of regular functions.
-    An affine variety $(X, \mathcal{O}_X)$ is
-    ,,like an algebraic set`` but without a reference to a particular
-    ,,embedding`` in affine space. This is similar to having a finitely generated $k$-Algebra $A$
-    without specifying a particular isomorphism
-    \[
-    A \simeq k[X_1, \ldots, X_n] / I
-    .\] The next example will illustrate precisely this fact.
-\end{bem}
-
-\begin{bsp}[]
-    Let us now give an abstract example of an affine variety.
-    We consider a finitely generated $k$-algebra $A$ and define
-    $X \coloneqq \operatorname{Hom}_{k-\mathrm{Alg}}(A, k)$. The idea is to think
-    of $X$ as points on which we can evaluate elements of $A$, which are thought of
-    as functions on $X$. For $x \in \operatorname{Hom}_{k}(A, k)$ and
-    $f \in A$ we set $f(x) \coloneqq x(f) \in k$.
-    \begin{itemize}
-        \item Topology on $X$: for all ideal $I \subseteq A$, set
-            \[
-            \mathcal{V}_X(I) \coloneqq \{ x \in X  \mid \forall x \in I\colon f(x) = 0\}
-            .\] These subsets of $X$ are the closed sets of a topology on $X$, which
-            we may call the Zariski topology.
-        \item Regular functions on $X$: if $U \subseteq X$ is open,
-            a function $h\colon U \to k$ is called regular at $x \in U$ if
-            there it exists an open set $x \in U_x$ and elements
-            $P, Q \in A$ such that for $y \in U_x$, $Q(y) \neq 0$ and
-            $h(y) = \frac{P(y)}{Q(y)}$ in $k$.
-
-            The function $h$ is called regular on $U$
-            iff it is regular at $x \in U$. Regular functions then form a sheaf of
-            $k$-algebras on $X$.
-
-            Moreover, if $h\colon U \to k$ is regular on $X$, the
-            set $D_X(h) \coloneqq \{ x \in X  \mid h(x) \neq 0\} $ is open in $X$
-            and the function $\frac{1}{h}$ is regular on $D_X(h)$.
-    \end{itemize}
-    So, we have defined a space with functions $(X, \mathcal{O}_X)$, at least
-    whenever $X \neq \emptyset$. We show that $X$ is an affine variety.
-
-    \begin{proof}
-        Fix a system of generators of $A$, i.e.
-        \[
-        A \simeq k[t_1, \ldots, t_n] / I
-        \] where $k[t_1, \ldots, t_n]$ is a polynomial algebra. We denote
-        by $\overline{t_1}, \ldots, \overline{t_n}$ the images of $t_1, \ldots, t_n$ in $A$
-        and we define
-        \begin{salign*}
-            \varphi\colon X = \operatorname{Hom}_{k}(A, k)& \to k^{n} \\
-            x &\mapsto (x(\overline{t_1}), \ldots, x(\overline{t_n}))
-        .\end{salign*}
-        Let $P \in I$ and $x \in X$. Then
-        \[
-        P(\varphi(x)) = P(x(\overline{t_1}), \ldots, x(\overline{t_n}))
-        = x(\overline{P}) = 0
-        .\] Thus $\varphi(x) \in \mathcal{V}_{k^{n}}(I)$.
-        Conversely let $a = (a_1, \ldots, a_n) \in \mathcal{V}_{k^{n}}(I)$, then
-        we can define a morphism of $k$-algebras
-        \[
-        x_a\colon A \to A / (\overline{t_1} -a_1, \ldots, \overline{t_n} - a_n)
-        \simeq k
-        \] which satisfies $x_a(\overline{t_i}) = a_i$ for all $i$. So
-        $(a_1, \ldots, a_n) = \varphi(x_a) \in \text{im } \varphi$.
-
-        In particular, we have defined a map
-        \begin{salign*}
-            \psi\colon \mathcal{V}_{k^{n}}(I) &\to X = \operatorname{Hom}_k(A, k) \\
-            a &\mapsto x_a
-        \end{salign*} such that $\varphi \circ \psi = \text{Id}_{\mathcal{V}_{k^{n}}(I)}$. In fact,
-        we also have $\psi \circ \varphi = \text{Id}_X$.
-
-        It remains to check that $\varphi$ and $\psi$ are morphisms of spaces with functions, which
-        follows from the definition of the topology and the notion of regular function on $X$.
-    \end{proof}
-
-    The elements of $X \coloneqq \operatorname{Hom}_k(A, k)$ are also called the
-    \emph{characters} of the $k$-algebra $A$, and this is sometimes denoted
-    by $\hat{A} \coloneqq \operatorname{Hom}_{k-\text{alg}}(A, k)$. Note that
-    $\hat{A}$ is a $k$-subalgebra of the algebra of all functions $f\colon A \to k$.
-
-    The character $x_a$ introduced above and associated to an alemenet $a \in A$ is then
-    denoted by $\hat{a}$ and called the \emph{Gelfand transform} of $a$. The
-    \emph{Gelfand transformation} is the morphism of $k$-algebras
-    \begin{salign*}
-        A &\to \hat{A} \\
-        a &\mapsto \hat{a}
-    .\end{salign*}
-\end{bsp}
-
-\begin{aufgabe}
-    Let $A$ be a finitely generated $k$-algebra and let
-    $X = \operatorname{Hom}_{k\text{-alg}}(A, k)$. Show that the map
-    $x \mapsto \text{ker } x$ induces a bijection
-    \[
-    X \simeq \{ \mathfrak{m} \in \operatorname{Spm} A  \mid A / \mathfrak{m} \simeq k\}
-    .\] 
-\end{aufgabe}
-
-\begin{bem}[]
-    Note that we have not assumed $A$ to be reduced and that, if we
-    set $A_{\text{red}} \coloneqq A / \sqrt{(0)}$, then
-    $A_{\text{red}}$ is reduced and
-    $\hat{A_{\text{red}}} = \hat{A}$, because a maximal ideal of $A$ necessarily
-    contains $\sqrt{(0)}$ and the quotient field is ,,the same``.
-\end{bem}
-
-\begin{bem}
-    Let $(X, \mathcal{O}_X)$ be an affine variety. One can associate the $k$-algebra
-    $\mathcal{O}_X(X)$ of globally defined regular functions on $X$:
-    \[
-    \mathcal{O}_X(X) = \{ f \colon X \to k  \mid f \text{ regular on } X\} 
-    .\]
-    Moreover, if $\varphi\colon (X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$ is
-    a morphism between two affine varieties, we have a $k$-algebra homomorphism
-    \begin{salign*}
-        \varphi^{*}\colon \mathcal{O}_Y(Y) &\to \mathcal{O}_X(X) \\
-        f &\mapsto f \circ \varphi
-    .\end{salign*}
-    Also, $(\text{id}_X)^{*} = \text{id}_{\mathcal{O}_X(X)}$ and
-    $(\psi \circ \varphi)^{*} = \varphi^{*} \circ \psi^{*}$ whenever
-    $\psi\colon (Y, \mathcal{O}_Y) \to (Z, \mathcal{O}_Z)$ is a morphism of
-    affine varieties. In other words, we have defined a (contravariant) functor
-    $k$-Aff $\to k$-Alg.
-\end{bem}
-
-\begin{satz}
-    Let $k$ be a field. The functor
-    \begin{salign*}
-        k\text{-Aff} &\to k\text{-Alg} \\
-        (X, \mathcal{O}_X) &\mapsto \mathcal{O}_X(X)
-    \end{salign*}
-    is fully faithful.
-\end{satz}
-
-\begin{proof}
-    Since $X$ and $Y$ are affine, we may assume $X = V \subseteq k^{n}$
-    and $Y = W \subseteq k^{m}$. Then $\varphi\colon V \to W$
-    is given by $m$ regular functions $(\varphi_1, \ldots, \varphi_m)$
-    on $V$. On $k^{m}$, let us denote by $y_i$ the projection to the $i$-th factor.
-    Its restriction to $W$ is a regular function
-    \[
-    y_i|_W \colon W \to k
-    \] that satisfies $\varphi^{*}(y_i|_W) = \varphi_i$.
-
-    Since for all regular functions $f\colon W \to k$ one has
-    \[
-    \varphi^{*}f = f \circ \varphi = f(\varphi_1, \ldots, \varphi_m)
-    ,\] we see that the morphism
-    \[
-    \varphi^{*}\colon \mathcal{O}_W(W) \to \mathcal{O}_V(V)
-    \] is entirely determined by the $m$ regular functions $\varphi^{*}(y_i|_W) = \varphi_i$
-    on $V$. In particular, if $\varphi^{*} = \psi^{*}$, then
-    $\varphi_i = \varphi^{*}(y_i|_W) = \psi^{*}(y_i|_W) = \psi_i$, so $\varphi = \psi$,
-    which proves that $\varphi \mapsto \varphi^{*}$ is injective.
-
-    Surjectivity: Let $h\colon \mathcal{O}_W(W) \to \mathcal{O}_V(V)$ be a morphism
-    of $k$-algebras. Let
-    \[
-    \varphi \coloneqq (h(y_1|_W), \ldots, h(y_m|_W))
-    \] which is a morphism from $V$ to $k^{m}$, because its components are regular functions
-    on $V$. It satisfies $\varphi^{*}(y_i|_W) = \varphi_i = h(y_i|_W)$, so $\varphi^{*} = h$.
-
-    It remains to show, that $\varphi(V) \subseteq W$. Let $W = \mathcal{V}(P_1, \ldots, P_r)$
-    with $P_j \in k[Y_1, \ldots, Y_m]$. Then for all $j \in \{1, \ldots, r\} $
-    and $x \in V$
-    \[
-    P_j(\varphi(x)) = P_j(h(y_1|_W), \ldots, h(y_m|_W))(x)
-    .\] Since $h$ is a morphism of $k$-algebras and $P_j$ is a polynomial, we have
-    \[
-    P_j(h(y_1|_W), \ldots, h(y_m|_W)) = h(P_j(y_1|_W), \ldots, P_j(y_m|_W))
-    .\] But $P_j \in \mathcal{I}(W)$, so
-    \[
-    P_j(y_1|_W, \ldots, y_m|_W) = P_j(y_1, \ldots, y_m)|_W = 0
-    ,\] which proves that for $x \in V$, $\varphi(x) \in W$.
-\end{proof}
-
-\end{document}

ws2022/rav/lecture/rav7.tex

@@ -1,227 +0,0 @@
-\documentclass{lecture}
-
-\begin{document}
-
-\section{Geometric Noether normalisation}
-
-Consider a plane algebraic curve $\mathcal{C}$, defined by the equation $f(x,y) = 0$.
-If we fix $x = a$, then the polynomial equation $f(a, y) = 0$ has only finitely many solutions
-(at most $\text{deg}_y f$). This means that the map
-\begin{salign*}
-    \mathcal{C} \coloneqq \mathcal{V}(f) &\to k
-    (x,y) \mapsto x
-\end{salign*}
-has finite fibres. A priori, such a map is not surjective, e.g. for $f(x,y) = xy - 1$. If
-$k$ is algebraically closed, one can always find such a surjective projection.
-
-\begin{theorem}
-    Let $k$ be an algebraically closed field and $f \in k[x_1, \ldots, x_n]$ be a polynomial
-    of degree $d \ge 1$. Then there is a morphism of affine varieties
-    \[
-    \pi\colon \mathcal{V}_{k^{n}}(f) \to k^{n-1}
-    \] 
-    such that:
-    \begin{enumerate}[(i)]
-        \item $\pi$ is surjective
-        \item for $t \in k^{n-1}$, the fibre $\pi^{-1}(\{t\}) \subseteq \mathcal{V}(f)$ consists
-            of at most $d$ points.
-    \end{enumerate}
-    \label{thm:geom-noether-norm}
-\end{theorem}
-
-\begin{proof}
-    Let $f \in k[x_1, \ldots, x_n]$ be of degree $d$. We construct a change of variables
-    of the form $(x_i \mapsto x_i + a_i x_n)_{1 \le i \le n-1}$ and
-    $x_n \mapsto x_n$, such that the term of degree $d$ of
-    $f(x_1 + a_1x_n, \ldots, x_{n-1} + a_{n-1}x_n, x_n)$ becomes
-    $c x_n^{d}$ with $c \in k^{\times }$. Since
-    \begin{salign*}
-    f(x_1 + a_1 x_n, \ldots, x_{n-1} + a_{n-1} x_n, x_n)
-    =
-    \sum_{(i_1, \ldots, i_n) \in \N^{n}} \alpha_{i_1, \ldots, i_n}
-    (x_1 + a_1 x_n)^{i_1} \cdots (x_{n-1} + a_{n-1} x_n)^{i_{n-1}} x_n^{i_n}
-    ,\end{salign*}
-    the coefficient of $x_n^{d}$ in the above equation is obtained by considering all
-    $(i_1, \ldots, i_n)$ such that $i_1 + \ldots + i_n = d$, and keeping only the term
-    in $x_n^{i_j}$ when expanding $(x_j + a_j x_n)^{i_j}$, so we get
-    \[
-    \sum_{(i_1, \ldots, i_n) \in \N \\ i_1 + \ldots + i_n = d}
-    \alpha_{i_1, \ldots, i_n} a_1^{i_1} \cdots a_{n-1}^{i_{n-1}}
-    ,\] which is equal to $f_d(a_1, \ldots, a_{n-1}, 1)$, where
-    $f_d$ is the (homogeneous) degree $d$ part of $f$.
-
-    Claim: There exist $a_1, \ldots, a_{n-1} \in k$ such that $f_d(a_1, \ldots, a_{n-1}, 1) \neq 0$.
-    Proof of claim by induction: if $n = 1$, $f_d = c x_1^{d}$ for some $c \neq 0$, so
-    $f_d(1) = c \neq 0$. If $n \ge 2$, we can write
-    \[
-    f_d(x_1, \ldots, x_n) = \sum_{i=0}^{d} h_i(x_2, \ldots, x_n) x_1^{i}
-    \] where $h_i \in k[x_2, \ldots, x_n]$ is homogeneous of degree $d-i$.
-    Since $f_d \neq 0$, there is at least one $i_0$ such that $h_{i_0} \neq 0$. By induction,
-    we can find $(a_2, \ldots, a_{n-1}) \in k^{n-2}$ such that
-    $h_{i_0}(a_2, \ldots, a_{n-1}, 1) \neq 0$. But then
-    $f(\cdot, a_2, \ldots, a_{n-1}, 1) \in k[x_1]$ is a non zero polynomial, so it has
-    only finitely many roots. As $k$ is infinite, there exists $a_1 \in k$, such that
-    $f(a_1, \ldots, a_{n-1}, 1) \neq 0$.
-
-    Then
-    \[
-    \varphi\colon \begin{cases}
-        x_i \mapsto x_i + a_i x_n & 1 \le i \le n-1\\
-        x_n \mapsto x_n
-    \end{cases}
-    \] is a invertible linear transformation $k^{n} \to k^{n}$, such that
-    \[
-        (f \circ \varphi^{-1})(y_1, \ldots, y_n)
-        = c (y_n^{d} + g_1(y_1, \ldots, y_n) y_n^{d-1} + \ldots + g_d(y_1, \ldots, y_{n-1})
-    \] for $c \neq 0$. This induces an isomorphism of affine varieties
-    \begin{salign*}
-        \mathcal{V}(f) &\to \mathcal{V}(f \circ \varphi^{-1}) \\
-    x &\mapsto \varphi(x)
-    \end{salign*}
-    such that
-    \[
-    \begin{tikzcd}
-        \mathcal{V}(f) \arrow[hookrightarrow]{r}{\varphi} \arrow[dashed]{dr}{\pi} & \arrow{d} k^{n} = k^{n-1} \times k \\
-                       & k^{n-1}
-    \end{tikzcd}
-    \] defines the morphism $\pi$ with the desired properties. Indeed:
-    Let $(x_1, \ldots, x_n) \in k^{n}$ and set $y_i \coloneqq \varphi(x_i)$. Then
-
-    $(x_1, \ldots, x_n) \in \mathcal{V}(f)$ iff $x_n = y_n$
-    is a root of the polynomial
-    \[
-    t ^{d} + \sum_{j=1}^{d} g_j(y_1, \ldots, y_{n-1}) t ^{d-j}
-    .\] Therefore for all $t = (y_1, \ldots, y_{n-1}) \in k^{n-1}$,
-    $\pi^{-1}(\{t\}) \neq \emptyset$ (because $\overline{k} = k$) and
-    $\pi^{-1}(\{t\})$ has at most $d$ points.
-\end{proof}
-
-\begin{definition}
-    Let $f \in k[x_1, \ldots, x_n]$ be a polynomial of degree $d$.
-    As in the proof of \ref{thm:geom-noether-norm}, ther exists a linear coordinate transformation
-    $\varphi\colon k^{n} \to k^{n}$, such that
-    $f \circ \varphi^{-1}(y_1, \ldots, y_n) = c y_n^{d} + \sum_{j=1}^{d} g_j(y_1, \ldots, y_{n-1})y_n^{d-j}$. For a point $x \in \pi^{-1}(y_1, \ldots, y_{n-1}) \subseteq \mathcal{V}(f)$,
-    the \emph{multiplicity} of $x$ is the multiplicity of $y_n$ as a root of that polynomial.
-
-    A point with multiplicity $\ge 2$ are called \emph{ramification point} and
-    its image lies in the \emph{discriminant locus} of $\pi$.
-\end{definition}
-
-With this vocabulary, we can refine the statement of \ref{thm:geom-noether-norm}.
-
-\begin{definition}[Geometric Noether normalisation]
-    Assume $k = \overline{k}$. If $f \in k[x_1, \ldots, x_n]$ is polynomial
-    of degree $d$, a morphism of affine varieties
-    \[
-    \pi\colon \mathcal{V}_{k^{n}}(f) \to k^{n-1}
-    \] such that
-    \begin{enumerate}[(i)]
-        \item $\pi$ is surjective
-        \item for $t \in k^{n-1}$, the number of elements in $\pi^{-1}(\{t\})$, counted
-            with their respective multiplicities, is exactly $d$,
-    \end{enumerate}
-    is called a \emph{geometric Noether normalisation}.
-\end{definition}
-
-\begin{korollar}[Geometric Noether normalisation for hypersurfaces]
-    Let $k$ be an algebraically closed field and $f \in k[x_1, \ldots, x_n]$ be a polynomial
-    of degree $d \ge 1$. Then there exists a geometric Noether normalisation.
-\end{korollar}
-
-\begin{bsp}
-    Let $f(x,y) = y^2 - x^{3} \in \mathbb{C}[x,y]$. Then the map
-    \begin{salign*}
-        \mathcal{V}_{\mathbb{C}^2}(y^2 - x^{3}) &\to \mathbb{C}
-        (x,y) &\mapsto y
-    \end{salign*}
-    is a geometric Noether normalisation, but
-    $(x,y) \mapsto x$ is not (the fibres of the latter have degree $2$, while $\text{deg } f = 3$).
-\end{bsp}
-
-\begin{bem}
-    In the proof of \ref{thm:geom-noether-norm}, to construct $\varphi$ and
-    the $g_j$, we only used that $k$ is infinte. Thus the statement, that
-    for all $f \in k[x_1, \ldots, x_n]$ there exists a linear automorphism
-    $\varphi\colon k^{n} \to k^{n}$ such that
-    \[
-    f \circ \varphi^{-1}(y_1, \ldots, y_n)
-    = c \left(y_n^{d} + \sum_{j=1}^{d} g_j(y_1, \ldots, y_{n-1}) y_n^{d-j}\right)
-    \] is valid over $k$ if $k$ is infinite. The resulting map
-    \[
-    \pi\colon \mathcal{V}_{k^{n}}(f) \to k^{n-1}
-    \] still has finite fibres, but it is no longer surjective in general, as
-    the example $f(x,y) = x^2 + y^2 - 1$ shows.
-
-    However, it induces a surjective map with finite fibres
-    \[
-    \hat{\pi}\colon \mathcal{V}_{\overline{k}^{n}}(f) \to \overline{k}^{n-1}
-    \] which moreover commutes with the action of $\text{Gal}(\overline{k} / k)$.
-\end{bem}
-
-\begin{theorem}
-    Let $k$ be an infinite field and $\overline{k}$ an algebraic closure of $k$. Let
-    $f \in k[x_1, \ldots, x_n]$ be a polynomial of degree $d \ge 1$. Then there exists
-    a $\text{Gal}(\overline{k} / k)$-equivariant geometric Noether normalisation map
-    $\pi\colon \mathcal{V}_{\overline{k}^{n}}(f) \to \overline{k}^{n-1}$.
-\end{theorem}
-
-\begin{bsp}[]
-    Let $f(x,y) = y^2 - x^{3} \in \R[x,y]$. Then the map
-    \begin{salign*}
-        \pi\colon \mathcal{V}_{\mathbb{C}^2}(y^2 - x^{3}) &\to \mathbb{C} \\
-        (x,y) &\mapsto y
-    .\end{salign*}
-    is a geometric Noether normalisation map and it is Galois-invariant:
-    \[
-    \pi(\overline{(x,y)}) = \pi(\overline{x}, \overline{y}) = \overline{y} = \overline{\pi(x,y)}
-    .\] 
-\end{bsp}
-
-\begin{aufgabe}[]
-    Show that if $y \in \R$, the group $\text{Gal}(\mathbb{C} / \R)$ acts on $\pi^{-1}(\{y\})$,
-    and that the fixed point set of that action is in bijection with
-    $\{x \in \R \mid y^2 - x^{3} = 0\} $.
-\end{aufgabe}
-
-Next, we want to generalise the results above beyond the case of hypersurfaces.
-
-\begin{theorem}
-    Assume $k$ is algebraically closed. Let $V \subseteq k^{n}$ be an algebraic set.
-    Then there exists a natural number $r \le n$ and a morphism of algebraic sets
-    \[
-    p\colon V \to k^{r}
-    \] such that $p$ is surjective and has finite fibres.
-    \label{thm:geom-noether-norm-general}
-\end{theorem}
-
-\begin{proof}[Sketch of proof]
-    If $V = k^{n}$, we take $r = n$ and $p = \text{id}_{k^{n}}$. Otherwise
-    $V = \mathcal{V}(I)$ with $I \subseteq k[x_1, \ldots, x_n]$ a non-zero ideal.
-    Take $f \in I \setminus \{0\} $. Then there exists a geometric Noether normalisation
-    \[
-    p_1\colon \mathcal{V}(f) \to k^{n-1}
-    .\]
-    One can now show that $V_1 \coloneqq p_1(V)$ is an algebraic set in $k^{n-1}$. Thus there are
-    two cases:
-    \begin{enumerate}[(1)]
-        \item $p_1(V) = k^{n-1}$. Thus $p_1|_V\colon V \to k^{n-1}$ is surjective with finite fibres
-            and we are done.
-        \item $p_1(V) \subsetneq k^{n-1}$. In this case
-            $p_1(V) = \mathcal{V}(I_1)$ with $I_1 \subseteq k[x_1, \ldots, x_{n-1}]$ a
-            non-zero ideal. So we can repeat the argument.
-    \end{enumerate}
-    After $r \le n$ steps, the above algorithm terminates, and this happens precisely when
-    $V_r = k^{n-r}$. If we set
-    \[
-    p\coloneqq p_r \circ \ldots \circ p_1 \colon V \to k^{n-r}
-    \] then $p$ is surjective with finite fibres because $p(V) = V_r = k^{n-r}$ and
-    each $p_i$ has finite fibres.
-\end{proof}
-
-\begin{bem}[]
-    By the fact used in the proof of \ref{thm:geom-noether-norm-general}, $p$ is in fact
-    a closed map. Note that when $r = n$, $V = p^{-1}(\{0\})$ is actually finite, in which case
-    $\text{dim }V$ should indeed be $0$.
-\end{bem}
-
-\end{document}

ws2022/rav/lecture/rav8.tex

@@ -1,289 +0,0 @@
-\documentclass{lecture}
-
-\begin{document}
-
-\section{Gluing spaces with functions}
-
-We present a general technique to construct spaces with functions by
-,,patching together`` other spaces with functions ,,along open subsets``. This
-will later be used to argue that, in order to define a structure of variety on a
-topological sapce (or even a set), it suffices to give one atlas.
-
-\begin{theorem}[Gluing theorem]
-    Let $(X_i, \mathcal{O}_{X_i})_{i \in I}$ be a family of spaces with functions. For
-    all pair $(i, j)$, assume that the following has been given
-    \begin{enumerate}[(a)]
-        \item an open subset $X_{ij} \subseteq X_i$
-        \item an isomorphism of spaces with functions
-            \[
-            \varphi_{ji}\colon (X_{ij}, \mathcal{O}_{X_{ij}})
-            \to (X_{ji}, \mathcal{O}_{X_{ji}})
-            \]
-    \end{enumerate}
-    subject to the following compatibility conditions
-    \begin{enumerate}[(1)]
-        \item for all $i$, $X_{ii} = X_i$ and $\varphi_{ii} = \text{id}_{X_i}$
-        \item for all pair $(i, j)$, $\varphi_{ij} = \varphi_{ji}^{-1}$
-        \item for all triple $(i, j, k)$, $\varphi_{ji}(X_{ik} \cap X_{ij}) = X_{jk} \cap X_{ji}$
-            and $\varphi_{kj} \circ \varphi_{ji} = \varphi_{ki}$
-            on $X_{ik} \cap X_{ij}$.
-    \end{enumerate}
-
-    Then there exists a space with functions $(X, \mathcal{O}_X)$ equipped with a family of
-    open sets $(U_i)_{i \in I}$
-    and isomorphisms of spaces with functions
-    \begin{enumerate}[(A1)]
-        \item $\varphi_i \colon (U_i, \mathcal{O}_X|_{U_i}) \to (X_i, \mathcal{O}_{X_i})$,
-    \end{enumerate}
-    such that $\bigcup_{i \in  I} U_i = X$ and, for all pair $(i, j)$,
-    \begin{enumerate}[(A1)]
-        \setcounter{enumi}{1}
-        \item $\varphi_i(U_i \cap U_j) = X_{ij}$, and
-        \item $\varphi_j \circ \varphi_i^{-1} = \varphi_{ji}$ on $X_{ij}$.
-    \end{enumerate}
-    Such a familiy $(U_i, \varphi_i)_{i \in I}$ is called
-    an atlas for $(X, \mathcal{O}_X)$.
-
-    Moreover, if $(Y, \mathcal{O}_Y)$ is a space with functions equipped with an atlas
-    $(V_i, \psi_i)_{i \in I}$ satisfying conditions (A1), (A2) and (A3), then
-    the isomorphisms $\psi_i^{-1} \circ \varphi_i \colon U_i \to V_i$ induce
-    an isomorphism $(X, \mathcal{O}_X) \to (Y, \mathcal{O}_Y)$.
-\end{theorem}
-
-\begin{proof}
-    Uniqueness up to canonical isomorphism: Let $(U_i, \varphi_i)_{i \in I}$
-    and $(V_i, \psi_i)_{i \in I}$ be two atlases modelled on the same gluing data,
-    then for all pair $(i, j)$,
-    \begin{salign*}
-    \psi_j^{-1} \circ \varphi_j \Big|_{U_i \cap U_j}
-    &= \psi_j^{-1} \circ \underbrace{(\varphi_j \circ \varphi_i^{-1})}_{= \varphi_{ji}}
-    \circ \varphi_i \Big|_{U_i \cap U_j} \\
-    &= \psi_j^{-1} \circ \underbrace{(\psi_j \circ \psi_i^{-1})}_{= \varphi_{ji}}
-    \circ \varphi_i \Big|_{U_i \cap U_j} \\
-    &= \psi_i^{-1} \circ \varphi_i \Big|_{U_i \cap U_j}
-    \end{salign*}
-    so there is a well-defined map
-    \begin{salign*}
-        f\colon X = \bigcup_{i \in  I} U_i &\to \bigcup_{i \in  I} V_i = Y \\
-        (x \in U_i) &\mapsto (\psi_i^{-1} \circ \varphi_i(x) \in V_i)
-    \end{salign*}
-    which induces an isomorphism
-    of spaces with functions.
-
-    Existence: Define $\tilde{X} \coloneqq \bigsqcup_{i \in I} X_i$ and let the
-    topology be the final topology with respect to the canonical maps
-    $(X_i \to \tilde{X})_{i \in I}$. Then define
-    $X \coloneqq \tilde{X} / \sim $ where
-    $(i, x) \sim (j, y)$ in $\tilde{X}$ if $x = \varphi_{ij}(y)$. Conditions
-    (1), (2) and (3) show that $\sim $ is reflexive, symmetric and transitive.
-    We equip $X$ with the quotient topology and denote by
-    \[
-    p\colon \tilde{X} \to X
-    \] the canonical continuous projection. Let $U_i \coloneqq p(X_i)$. Since
-    $p^{-1}(U_i) = \bigsqcup_{j \in I} X_{ji}$
-    is open in $\tilde{X}$, $U_i$ is open in $X$. Moreover,
-    $\bigcup_{i \in  I} U_i = X$, so we have an open covering of $X$. We
-    put $p_i \coloneqq p|_{X_i}$ and we define a sheaf on $X$ by setting
-    \[
-    \mathcal{O}_X(U) \coloneqq \{ f \colon U \to k  \mid \forall i \in I, f \circ p_i
-    \in \mathcal{O}_{X_i}(p_i^{-1}(U)) \} 
-    \] for all open sets $U \subseteq X$. This defines a sheaf on $X$, with
-    respect to which $(X, \mathcal{O}_X)$ is a space with functions.
-    Finally, $p_i\colon X_i \to U_i$ is a homeomorphism and, by construction
-    $\mathcal{O}_{U_i} \simeq (p_i)_{*} \mathcal{O}_{X_i}$ via pullback by $p_i$.
-    We have thus constructed a space with functions $(X, \mathcal{O}_X)$,
-    equipped with an open covering $(U_i)_{i \in I}$ and local charts
-    \[
-    \varphi_i \coloneqq p_i^{-1}\colon (U_i, \mathcal{O}_X|_{U_i})
-    \stackrel{\sim }{\longrightarrow }
-    (X_i, \mathcal{O}_{X_i})
-    .\] It remains to check that
-    $\varphi_i(U_i \cap U_j) = X_{ij}$ and
-    $\varphi_j \circ \varphi_i^{-1} = \varphi_{ji}$ on $X_{ij}$, but
-    this follows from the construction of
-    $\displaystyle{X = \bigsqcup_{i \in I} X_i / \sim }$ and
-    the definition of the $\varphi_i$'s as $p|_{X_i}^{-1}$.
-\end{proof}
-
-\begin{bsp}[]
-    Take $k = \R$ or $\mathbb{C}$ equipped with either the Zariski or the usual topology. Consider
-    the spaces with functions $X_1 = k$, $X_2 = k$ and the open sets
-    $X_{12} = k \setminus \{0\} \subseteq X_1$ and
-    $X_{21} = k \setminus \{0\} \subseteq X_2$. Finally, set
-    \begin{salign*}
-        \varphi_{21}\colon X_{12} &\to X_{21} \\
-        t &\mapsto \frac{1}{t}
-    .\end{salign*}
-    Since this is an isomorphism of spaces with functions, we can glue
-    $X_1$ and $X_2$ along $X_{12} \xlongrightarrow[\varphi_{21}]{\sim } X_{21} $
-    and define a space with functions $(X, \mathcal{O}_X)$ with
-    an atlas modelled on $(X_1, X_2, \varphi_{21})$. We will now identify this
-    space $X$ with the projective line $k \mathbb{P}^{1}$. By definition, the latter
-    is the set of $1$-dimensional vector subspaces (lines) of $k^2$:
-    \begin{salign*}
-    k \mathbb{P}^{1} \coloneqq (k^2 \setminus \{0\}) / k^{\times }
-    .\end{salign*}
-    Then, we have a covering
-    $U_1 \cup U_2 = k \mathbb{P}^{1}$, where
-    $U_1 = \{ [x_1 : x_2]  \mid x_1 \neq 0\} $
-    and $U_2 = \{ [x_1 : x_2 ]  \mid x_2 \neq 0\} $, and we can define charts
-    \begin{salign*}
-        \varphi_1\colon U_1 &\xlongrightarrow{\sim } k \\
-        [x_1 : x_2 ] &\longmapsto x_2 / x_1 \\
-        [1:w] & \longmapsfrom w
-    \end{salign*}
-    and $\varphi_2\colon U_2 \to k$ likewise. Then, on the intersection
-    \[
-    U_1 \cap U_2 = \{ [x_1 : x_2 ]  \mid x_1 \neq 0, x_2 \neq 0\}
-    \] we have a commutative diagram
-    \[
-    \begin{tikzcd}
-        U_1 \cap U_2 \arrow{d}{\varphi_1} \arrow{dr}{\varphi_2} & \\
-        X_1 \arrow{r}{\varphi_{21}} & X_2
-    \end{tikzcd}
-    \] with $\varphi_i(U_1 \cap U_2)$ open in $X_i$. In view of
-    the gluing theorem, we can use this to set up a bijection
-    $k \mathbb{P}^{1} \to X$ where $\displaystyle{X \coloneqq (X_1 \sqcup X_2) / \sim_{\varphi_{12}}}$
-    and define a topology and a sheaf of regular functions on
-    $k \mathbb{P}^{1}$ via this identification. Note that this was done without putting 
-    a topology on $k \mathbb{P}^{1}$: the latter is obtained using the bijection
-    $k \mathbb{P}^{1} \to X$ constructed above. We now spell out the notion of regular functions
-    thus obtained on $k \mathbb{P}^{1}$.
-\end{bsp}
-
-\begin{satz}
-    With the identification
-    \[
-    k \mathbb{P}^{1} = X_1 \sqcup X_2 / \sim
-    \] constructed above, a function $f\colon U \to k$ defined on 
-    an open subset $U \subseteq k \mathbb{P}^{1}$ is an element of $\mathcal{O}_X(U)$ if
-    and only if, for each local chart $\varphi_i \colon U_i \to k$, the function
-    \[
-    f \circ \varphi_i^{-1} \colon \varphi_i(U_i \cap U) \to k
-    \] is regular on the open set $\varphi_i(U_i \cap U) \subseteq k$.
-\end{satz}
-
-\begin{definition}[]
-    Let $k$ be a field. An \emph{algebraic $k$-prevariety} is a space
-    with functions $(X, \mathcal{O}_X)$ such that
-    \begin{enumerate}[(i)]
-        \item $X$ is quasi-compact.
-        \item $(X, \mathcal{O}_X)$ is locally isomorphic to an affine variety.
-    \end{enumerate}
-\end{definition}
-
-\begin{bem}[]
-    Saying that $(X, \mathcal{O}_X)$ is locally isomorphic to an affine variety means
-    that for $x \in X$, it exists an open neighbourhood $x \in U$ such that
-    $(U, \mathcal{O}_X|_U)$ is isomorphic to an open subset of an affine variety. Since
-    such an open set is a union of principal open sets, which are themselves affine, one can
-    equivalently ask that $(U, \mathcal{O}_U)$ be affine. Thus:
-\end{bem}
-
-\begin{satz}
-    A space with functions $(X, \mathcal{O}_X)$ is an algebraic prevariety, if and only if
-    there exists a finite open covering
-    \[
-    X = U_1 \cup \ldots \cup U_n
-    \] such that $(U_i, \mathcal{O}_X|_{U_i})$ is an affine variety.
-\end{satz}
-
-\begin{bem}[]
-    As a consequence of the gluing theorem, in order to either construct an algebraic
-    prevariety or put a structure of an algebraic prevariety on a set, it suffices to either
-    define $X$ from certain gluing data $(X_i, X_{ij}, \varphi_{ij})_{(i,j)}$ satisfying
-    appropriate compatibility conditions, or find a covering
-    $(U_i)_{i \in I}$ of a set $X$ and local charts $\varphi_i \colon U_i \to X_i$ such that
-    $X_{ij} = \varphi_i (U_i \cap U_j)$ is open in $X_i$ and
-    $\varphi_j \circ \varphi_i^{-1}$ is an isomorphism of spaces with functions.
-
-    In practice, $X$ is sometimes given as a topological space, and
-    $(U_i)_{i \in I}$ is an open covering, with local charts $\varphi_i\colon U_i \to X_i$ that
-    are homeomorphisms. So the condition that $X_{ij}$ be open in $X_i$ is automatic
-    in this case and one just has to check that
-    \[
-    \varphi_{j} \circ \varphi_i^{-1} \colon X_{ij} \to X_{ji}
-    \] induces an isomorphism of spaces with functions. In the present context where
-    $X_i$ and $X_j$ are affine varieties, this means a map
-    \[
-    X_{ij} \subseteq k^{n} \to X_{ji} \subseteq k^{m}
-    \] between locally closed subsets of $k^{n}$ and $k^{m}$ whose components are regular functions.
-\end{bem}
-
-\begin{bsp}[Projective sets]
-    We have already seen that projective spaces $k \mathbb{P}^{n}$ are algebraic pre-varieties.
-    Let $P \in k[x_0, \ldots, x_n]_d$ be a homogeneous polynomial of degree $d \ge 0$. Although
-    $P$ cannot be evaluated at a point
-    $[x_0 : \ldots : x_n] \in k \mathbb{P}^{n}$, the condition
-    $P(x_0, \ldots, x_n) = 0$ can be tested, because for $\lambda \in k^{x}$,
-    \begin{salign*}
-    P(x_0, \ldots, x_n) = 0 \iff 0 = \lambda ^{d} P(x_0, \ldots, x_n)
-    = P(\lambda x_0, \ldots, \lambda x_n)
-    .\end{salign*}
-    We use this to define the following \emph{projective sets}
-    \[
-    \mathcal{V}_{k \mathbb{P}^{n}}(P_1, \ldots, P_m)
-    = \{ [x_0 : \ldots : x_n] \in k \mathbb{P}^{n}  \mid  P_i(x_0, \ldots, x_n) = 0 \quad \forall i\} 
-    \] for homogeneous polynomials in $(x_0, \ldots, x_n)$.
-
-    We claim that these projective sets are the clsoed sets of a topology on
-    $k \mathbb{P}^{n}$, called the Zariski topology. A basis for that topology
-    is provided by the principal open sets
-    $D_{k \mathbb{P}^{n}} (P)$ where $P$ is a homogeneous polynomial. By definition, a regular
-    function on a locally closed subset of $k \mathbb{P}^{n}$ is locally given by the restriction
-    of a ration fraction of the form
-    \[
-    \frac{P(x_0, \ldots, x_n)}{Q(x_0, \ldots, x_n)}
-    \] where $P$ and $Q$ are homogeneous polynomials of the same degree.
-    This defines a sheaf of regular functions on any given locally closed subset
-    $X$ of $k \mathbb{P}^{n}$.
-\end{bsp}
-
-\begin{satz}
-    A Zariski-closed subset $X$ of $k \mathbb{P}^{n}$ equipped with its
-    sheaf of regular functions, is an algebraic pre-variety. The same holds
-    for all open subsets $U \subseteq X$.
-\end{satz}
-
-\begin{proof}
-    Consider the open covering
-    \begin{salign*}
-        X &= \bigcup_{i = 0} ^{n} X \cap U_i \\
-          &= \bigcup_{i = 0}^{n} \{ [x_0 : \ldots : x_n ] \in X  \mid x_i \neq 0\} 
-    .\end{salign*}
-    Then the restriction to $X \cap U_i$ of the local chart
-    \begin{salign*}
-        \varphi_i \colon U_i &\longrightarrow k^{n} \\
-        x = [x_0 : \ldots : x_n] &\longmapsto
-        \underbrace{\left( \frac{x_0}{x_i}, \ldots, \hat{\frac{x_i}{x_i}}, \ldots, \frac{x_n}{x_i} \right)}_{w = (w_0, \ldots, \hat{w}_i, \ldots, w_n)}
-    \end{salign*}
-    sends an $x$ such that $P_1(x) = \ldots = P_m(x) = 0$ to a $w$ such that
-    $Q_1(w) = \ldots = Q_m(w) = 0$ where, for all $j$,
-    \begin{salign*}
-        Q_j(w) &= P_j(w_0, \ldots, w_{i-1}, 1, w_{i+1}, \ldots, w_n) \\
-               &= P_j(x_0, \ldots, x_{i-1}, x_i, x_{i+1}, \ldots, x_n)
-    \end{salign*}
-    is the dehomogeneisation of $P_j$. So
-    $\varphi_i(X \cap U_i) = \mathcal{V}_{k^{n}}(Q_1, \ldots, Q_m) \eqqcolon X_i$
-    is an algebraic subset of $k^{n}$, in particular an affine variety. It remains
-    to check that $\varphi_i|_{X \cap U_i}$ pulls back regular functions on $X_i$ to
-    regular functions on $X \cap U_i$, and similarly for $(\varphi_i|_{X \cap U_i})^{-1}$.
-    But if $f$ and $g$ are polynomials in $(w_0, \ldots, \hat{w}_i, \ldots, w_n)$,
-    \begin{salign*}
-        \left(\varphi_i^{*} \frac{f}{g}\right)(x)
-        &= \frac{f(\varphi_i(x))}{g(\varphi_i(x))} \\
-        &= \frac{f\left( \frac{x_0}{x_i}, \ldots, \hat{\frac{x_i}{x_i}}, \ldots, \frac{x_n}{x_i} \right) }{g\left( \frac{x_0}{x_i}, \ldots, \hat{\frac{x_i}{x_i}}, \ldots, \frac{x_n}{x_i} \right) }
-    \end{salign*}
-    which can be rewritten as a quotient of two homogeneous polynomials of the same
-    degree by multiplying the numerator and denominator
-    by $x_i^{r}$ with $r \ge \text{max}(\text{deg}(f) , \text{deg}(g))$. The computation
-    is similar but easier for $\left( \varphi_i |_{X \cap U_i} \right)^{-1}$.
-\end{proof}
-
-\begin{definition}
-    A space with functions $(X, \mathcal{O}_X)$ which is isomorphic to a
-    Zariski-closed subset of $k \mathbb{P}^{n}$ is called a
-    \emph{projective $k$-variety}.
-\end{definition}
-
-\end{document}

ws2022/rav/lecture/rav9.tex

@@ -1,278 +0,0 @@
-\documentclass{lecture}
-
-\begin{document}
-
-\begin{lemma}
-    The category of affine varieties admits products.
-    \label{lemma:aff-var-prod}
-\end{lemma}
-
-\begin{proof}
-    Let $(X, \mathcal{O}_X)$, $(Y, \mathcal{O}_Y)$ be affine varieties. Choose embeddings
-    $X \subseteq k^{n}$ and $Y \subseteq k^{p}$ for some $n$ and $p$. Then
-    $X \times Y \subseteq k^{n+p}$ is an affine variety, endowed with two morphisms
-    of affine varieties $\text{pr}_1\colon X \times Y \to X$ and
-    $\text{pr}_2\colon X \times Y \to Y$. We will prove that
-    the triple $(X \times Y, \text{pr}_1, \text{pr}_2)$ satisfies the universal property of
-    the product of $X$ and $Y$.
-
-    Let $f_X\colon Z \to X$ and $f_Y\colon Z \to Y$ be morphisms of affine varieties.
-    Then define $f = (f_x, f_y)\colon Z \to X \times Y$. This satisfies
-    $\text{pr}_1 \circ f = f_X$ and $\text{pr}_2 \circ f = f_Y$.
-    If we embed $Z$ into some $k^{m}$,
-    the components of $f_X$ and $f_Y$ are regular functions from
-    $k^{m}$ to $k^{n}$ and $k^{p}$. Thus the components of
-    $f = (f_X, f_Y)$ are regular functions $k^{m} \to k^{n+p}$, i.e. $f$ is a morphism.
-\end{proof}
-
-\begin{theorem}
-    The category of algebraic pre-varieties admits products.
-\end{theorem}
-
-\begin{proof}
-    Let $(X, \mathcal{O}_X), (Y, \mathcal{O}_Y)$ algebraic pre-varieties. Let
-    \[
-    X = \bigcup_{i=1} ^{r} X_i \text{ and } Y = \bigcup_{j=1}^{s} Y_j
-    \] be affine open covers. Then, as a set,
-    \[
-    X \times Y = \bigcup_{i,j} X_i \times Y_j
-    .\]
-    By \ref{lemma:aff-var-prod}, each
-    $X_i \times Y_j$ has a well-defined structure of affine variety. Moreover,
-    if $X_i' \subseteq X_i$ and $Y_j' \subseteq Y_j$ are open sets, then
-    $X_i' \times Y_j'$ is open in $X_i \times Y_j$.
-
-    So we can use the identity morphism to glue $X_{i_1} \times Y_{j_1}$
-    to $X_{i_2} \times Y_{j_2}$ along the common open subset
-    $(X_{i_1} \cap X_{i_2}) \times (Y_{j_1} \cap Y_{j_2})$. This defines
-    an algebraic prevariety $P$ whose underlying set is $X \times Y$. Also,
-    the canonical projections
-    $X_i \times Y_j \to X_i$ and $X_i \times X_j \to X_j$
-    glue together to give morphisms
-    $p_X \colon X \times Y \to X$ and $p_Y \colon X \times Y \to Y$, which
-    coincide with $\text{pr}_1$ and $\text{pr}_2$.
-
-    There only remains to prove the universal property. Let $f_x\colon Z \to X$ and
-    $f_Y\colon Z \to Y$ be morphisms of algebraic prevarieties and set
-    $f = (f_x, f_y)\colon Z \to X \times Y$. In particular,
-    $\text{pr}_1 \circ f = f_X$ and $\text{pr}_2 \circ f = f_Y$ as maps between sets.
-    To prove that $f$ is a morphisms of algebraic prevarieties, it suffices to show
-    that this is locally the case. $Z$ is covered by the open subsets
-    $f_X^{-1}(X_i) \cap f_Y^{-1}(Y_j)$, each of which can be covered by affine open subsets
-    $(W_{l}^{ij})_{1 \le l \le q(i, j)}$. By construction,
-    $f(W_{l}^{ij}) \subseteq X_i \times Y_j$. So, by the universal property of the affine
-    variety $X_i \times Y_j$, the map $f|_{W_l^{ij}}$ is a morphism of affine varieties.
-\end{proof}
-
-\begin{definition}[algebraic variety]
-    Let $(X, \mathcal{O}_X)$ be an algebraic pre-variety and
-    $X \times X$ the product in the category of algebraic pre-varieties. If the subset
-    \[
-    \Delta_X \coloneqq \{ (x, y) \in X \times X  \mid x = y\}
-    \]
-    is closed in $X \times X$, then $(X, \mathcal{O}_X)$ is said to be an
-    \emph{algebraic variety}. A morphism of algebraic varieties $f\colon X \to Y$
-    is a morphism of the underlying pre-varieties.
-\end{definition}
-
-\begin{bsp}[of a non-seperated algebraic prevariety]
-    We glue two copies $X_1, X_2$ of $k$ along the open subsets $k \setminus \{0\} $ using
-    the isomorphism of spaces with functions $t \mapsto t$. The resulting
-    algebraic prevariety is a ,,line with two origins'', denoted by $0_1$ and $0_2$. For
-    this prevariety $X$, the diagonal $\Delta_X$ is not closed in $X \times X$.
-
-    Indeed, if $\Delta_X$ were closed in $X \times X$, then its pre-image in $X_1 \times X_2$
-    under the morphism $f\colon X_1 \times X_2 \to X\times X$ defined by
-    \[
-    \begin{tikzcd}
-        X_1 \times X_2 \arrow[dashed]{dr} \arrow[bend right=20, swap]{ddr}{i_2 \circ \text{pr}_2}
-        \arrow[bend left=20]{drr}{i_1 \circ \text{pr}_1} & & \\
-                                  & X \times X \arrow{r} \arrow{d} & X \\
-                                  & X & \\
-    \end{tikzcd}
-    \] where $i_j\colon X_j \xhookrightarrow{} X$ is the canonical inclusion of $X_j$
-    into $X = \left( X_1 \sqcup X_2 \right) / \sim $,
-    would be closed in $X_1 \times X_2$. But
-    \begin{salign*}
-        f^{-1}(\Delta_X) &= \{ (x_1, x_2) \in X_1 \times X_2  \mid i_1(x_1) = i_2(x_2) \} \\
-                         &= \{ (x_1, x_2) \in X_1 \times X_2  \mid x_j \neq 0 \text{ and } x_1 = x_2 \text{ in } k\} \\
-                         &= \{ (x, x) \in k \times k  \mid x \neq 0\}
-                         \subseteq k \times k = X_1 \times X_2
-    \end{salign*}
-    which is not closed in $X_1 \times X_2$. In fact,
-    $f^{-1}(\Delta_X) = \Delta_k \setminus \{ (0, 0) \} \subseteq k \times k$.
-\end{bsp}
-
-\begin{korollar}
-    Let $(X, \mathcal{O}_X)$, $(Y, \mathcal{O}_Y)$ be algebraic varieties, then
-    the product in the category of algebraic pre-varieties is an algebraic variety. In particular
-    the category of algebraic varieties admits products.
-\end{korollar}
-
-\begin{proof}
-    $\Delta_{X \times Y} \simeq \Delta_X \times \Delta_Y \subseteq (X \times X) \times (Y \times Y)$.
-\end{proof}
-
-\begin{satz}
-    Affine varieties are algebraic varieties.
-\end{satz}
-
-\begin{proof}
-    Let $X$ be an affine variety. We choose an embedding $X \subseteq k^{n}$. Then
-    $\Delta_X = \Delta_{k^{n}} \cap (X \times X)$. But
-    \[
-    \Delta_{k^{n}} = \{ (x_i, y_i)_{1 \le i \le n} \in k^{2n}  \mid x_i - y_i = 0\}
-    \] is closed in $k^{2n}$. Therefore,
-    $\Delta_X$ is closed in $X \times X$ (note that the prevariety topology of $X \times X$
-    coincides with its induced topology as a subset of $k^{2n}$ by construction
-    of the product prevariety $X \times X$).
-\end{proof}
-
-\begin{aufgabe}
-    \label{exc:closed-subs-of-vars}
-    Let $(X, \mathcal{O}_X)$ be an algebraic pre-variety and let $Y \subseteq X$ be
-    a closed subset. For all open subsets $U \subseteq Y$, we set
-    \[
-    \mathcal{O}_Y(U) \coloneqq \left\{ h \colon U \to k  \mid \forall x \in U \exists x \in \hat{U} \subseteq X \text{ open, } g \in \mathcal{O}_X(\hat{U}) \text{ such that } g|_{\hat{U} \cap U} = h|_{\hat{U} \cap U} \right\} 
-    .\]
-    \begin{enumerate}[(a)]
-        \item Show that this defines a sheaf of regular functions on $Y$ and that
-            $(Y, \mathcal{O}_Y)$ is an algebraic prevariety.
-        \item Show that the canonical inclusion
-            $i_Y\colon Y \xhookrightarrow{} X$
-            is a morphism of algebraic prevarieties and that if $f\colon Z \to X$ is
-            a morphism of algebraic prevarieties such that
-            $f(Z) \subseteq Y$, then $f$ induces a morphism $\tilde{f}\colon Y \to Z$ such that
-            $i_{Y} \circ \tilde{f} = f$.
-        \item Show that, if $X$ is an algebraic variety, then $Y$ is also an algebraic variety.
-    \end{enumerate}
-\end{aufgabe}
-
-Recall that $k \mathbb{P}^{n}$ is the projectivisation
-of the $k$-vector space $k^{n+1}$:
-\begin{salign*}
-k \mathbb{P}^{n} = P(k^{n+1}) (k^{n+1} \setminus \{0\} ) / k^{\times }
-.\end{salign*}
-
-\begin{satz}[Segre embedding]
-    The $k$-bilinear map
-    \begin{salign*}
-        k^{n+1} \times k^{m+1} &\longrightarrow k^{n+1} \otimes_k k^{m+1} \simeq k^{(n+1)(m+1)} \\
-        (x,y) &\longmapsto x \otimes y
-    \end{salign*}
-    induces an isomorphism of algebraic pre-varieties
-    \begin{salign*}
-        P(k^{n+1}) \times P(k^{m+1}) &\xlongrightarrow{f}
-        \zeta \subseteq P\left(k^{(n+1)(m+1)}\right) = k \mathbb{P}^{nm + n + m}\\
-        ([x_0 : \ldots : x_n], [y_0 : \ldots : y_m]) &\longmapsto
-        [x_0 y_0 : \ldots x_0 y_m : \ldots : x_n y_0 : \ldots : x_n y_m ]
-    \end{salign*}
-    where $\zeta$ is a Zariski-closed subset of $k \mathbb{P}^{nm + n + m}$.
-    \label{prop:segre-embed}
-\end{satz}
-\begin{proof}
-     It is clear that
-$f$ is well-defined. Let us denote by $(z_{ij})_{0 \le i \le n, 0 \le j \le m}$ the
-homogeneous coordinates on $k \mathbb{P}^{nm + n + m}$, and call them
-\emph{Segre coordinates}. Then $f(k \mathbb{P}^{n} \times k \mathbb{P}^{m})$
-is contained in the projective variety
-\begin{salign*}
-    \zeta &= \mathcal{V}\left( \left\{ z_{ij}z_{kl} - z_{kj}z_{il} \mid 0 \le i, k \le n, 0 \le j, l \le m \right\}  \right)  \\
-          &\subseteq P\left( k^{(n+1)(m+1)} \right)
-\end{salign*}
-as can be seen by writing
-\begin{salign*}
-    f([x], [y]) = \begin{bmatrix} x_0 y_0 : & \ldots & : x_0y_m \\
-        \vdots &  & \vdots \\
-        x_n y_0 : & \ldots & : x_n y_m
-
-    \end{bmatrix}
-\end{salign*}
-so that
-\[
-z_{ij} z_{kl} - z_{kj} z_{il} =
-\begin{vmatrix}
-    x_i y_j & x_i y_l \\
-    x_k y_j & x_k y_l
-\end{vmatrix}
-= 0
-.\] 
-The map $f$ is injective because, if $z \coloneqq f([x], [y]) = f([x'], [y'])$ then
-there exists $(i, j)$ such that $z \in W_{ij} \coloneqq \{ z \in k \mathbb{P}^{nm + n + m}  \mid z_{ij} \neq 0\} $
-so $x_i y_j = x_i'y_j' \neq 0$. In particular
-$\frac{x_i}{x_i'} = \frac{y_j'}{y_j} = \lambda \neq 0$. Since
-\[
-[x_0 y_0 : \ldots : x_n y_m ] = [x_0' y_0' : \ldots : x_n' y_m' ]
-\] means that there exists $\mu \neq 0$ such that, for all $(k, l)$,
-$x_k y_l = \mu x_k'y_l'$. Taking $k = i$ and $l = j$, we get that $\mu = 1$
-and hence, for all $k$, $x_k y_j = x_k' y_j'$, so
-$x_k = \frac{y_j'}{y_j} x_k' = \lambda x_k'$. Likewise, for all $l$,
-$x_i y_l = x_i' y_l'$, so $y_l = \frac{1}{\lambda} y_l'$. As a consequence
-$[x_0 : \ldots : x_n ] = [ x_0' : \ldots : x_n' ]$ and
-$[y_0 : \ldots : y_m ] = [y_0' : \ldots : y_m' ]$, thus
-proving that $f$ is injective. Note that we have proven that
-\[
-    f^{-1}(W_{ij}) = U_i \times V_j
-\]
-where $U_i = \{ [x] \in k \mathbb{P}^{n}  \mid x_i \neq 0\} $
-and $V_j = \{ [y] \in k\mathbb{P}^{m}  \mid y_j \neq 0\} $.
-
-For simplicity, let us assume that $i = j = 0$. The open sets $U_0, V_0, W_0$ are affine charts,
-in which $f$ is equivalent to
-\begin{salign*}
-    k^{n} \times k^{m} &\longrightarrow k^{nm + n + m} \\
-    (u, v) &\longmapsto (v_1, \ldots, v_m, u_1, u_1v_1, \ldots, u_1v_m, \ldots, u_n, u_n v_1, \ldots, v_n v_m)
-\end{salign*}
-which is clearly regular. In particular $f \mid U_0 \times V_0$ is a morphism of algebraic
-pre-varieties.
-
-$\text{im }f = \zeta$: Let $[z] \in \zeta$. Since the $W_{ij}$ cover
-$k \mathbb{P}^{nm + n + m}$, we can assume without loss of generality, $z_{00} \neq 0$. Then
-by definition of $\zeta$, $z_{kl} = \frac{z_{k_0} z _{0l}}{z_{00}}$ for all $(k, l)$. If we
-set
-\begin{salign*}
-    ([x_0 : \ldots : x_n ] , [y_0 : \ldots : y_m])
-    &= \left( \left[ 1 : \frac{z_{10}}{z_{00}} : \ldots : \frac{z_{n_0}}{z_{00}}\right],
-    \left[1 : \frac{z_{01}}{z_{00}} : \ldots : \frac{z_{0m}}{z_{00}}\right]\right)
-\end{salign*}
-we have a well defined point $([x], [y]) \in U_0 \times V_0 \subseteq k\mathbb{P}^{n} \times k \mathbb{P}^{m}$, which satisfies $f([x], [y]) = [z]$.
-
-Thus $f^{-1}\colon \zeta \to k \mathbb{P}^{n} \times k \mathbb{P}^{m}$ is defined and
-a morphism of algebraic pre-varieties because, in affine charts
-$W_0 \xlongrightarrow{f^{-1}|_{W_0}} U_0 \times V_0$ as above, it is the regular map
-$(u_{ij})_{(i,j)} \mapsto \left( (u_{i_0})_i, (u_{0j})_j \right) $.
-\end{proof}
-
-\begin{korollar}
-    Projective varieties are algebraic varieties.
-\end{korollar}
-
-\begin{proof}
-    By \ref{exc:closed-subs-of-vars} it suffices to show that
-    $k \mathbb{P}^{n}$ is an algebraic variety. Let
-    $f\colon k \mathbb{P}^{n} \times k \mathbb{P}^{n} \to k \mathbb{P}^{n^2 + 2n}$
-    be the Segre embedding. For $[x] \in k \mathbb{P}^{n}$:
-    \begin{salign*}
-        f([x], [x]) &=
-        \begin{bmatrix} 
-            x_0x_0 : & \ldots &  : x_0 x_m \\
-            \vdots & & \vdots \\
-            x_n x_0 : & \ldots & : x_n x_m
-        \end{bmatrix}
-    .\end{salign*}
-    Thus $f([x], [x])_{ij} = f([x], [x])_{ji}$. Let now
-    $[z] \in \zeta \subseteq k \mathbb{P}^{n^2 + 2n}$, where $\zeta$ is defined
-    in the proof of \ref{prop:segre-embed}, and
-    such that, in Segre coordinates, $z_{ij} = z_{ji}$. Without loss of generality,
-    we can assume $z_{00} = 1$. Set $x_j \coloneqq z_{0j}$ for $1 \le j \le n$. Thus
-    for all $(i, j)$
-    \begin{salign*}
-    f([x], [y])_{ij} = x_i x_j = z_{0i} z_{0j} = z_{i0} z_{0j} = z_{ij} z_{00} = z_{ij}
-    ,\end{salign*} i.e.
-    \[
-    \Delta_{k \mathbb{P}^{n}} \simeq
-    \{ [z] \in \zeta  \mid z_{ij} = z_{ji}\}
-    \] which is a projective and thus closed set of $k \mathbb{P}^{n} \times k \mathbb{P}^{n}$.
-\end{proof}
-
-\end{document}