diff --git a/ws2022/rav/lecture/lecture.cls b/ws2022/rav/lecture/lecture.cls new file mode 100644 index 0000000..d9f8071 --- /dev/null +++ b/ws2022/rav/lecture/lecture.cls @@ -0,0 +1,246 @@ +\ProvidesClass{lecture} +\LoadClass[a4paper]{book} + +\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, + default 2d plot/.style={% + grid=both, + minor tick num=4, + grid style={line width=.1pt, draw=gray!10}, + major grid style={line width=.2pt,draw=gray!50}, + axis lines=middle, + enlargelimits={abs=0.2} + }, +} + +\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} + +% 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{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}} diff --git a/ws2022/rav/lecture/rav.pdf b/ws2022/rav/lecture/rav.pdf new file mode 100644 index 0000000..44f45c0 Binary files /dev/null and b/ws2022/rav/lecture/rav.pdf differ diff --git a/ws2022/rav/lecture/rav.tex b/ws2022/rav/lecture/rav.tex new file mode 100644 index 0000000..3345c7f --- /dev/null +++ b/ws2022/rav/lecture/rav.tex @@ -0,0 +1,28 @@ +\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{rav5.tex} +\input{rav6.tex} +\input{rav7.tex} +\input{rav8.tex} +\input{rav9.tex} + +\end{document} diff --git a/ws2022/rav/lecture/rav5.pdf b/ws2022/rav/lecture/rav5.pdf new file mode 100644 index 0000000..1c10ca7 Binary files /dev/null and b/ws2022/rav/lecture/rav5.pdf differ diff --git a/ws2022/rav/lecture/rav5.tex b/ws2022/rav/lecture/rav5.tex new file mode 100644 index 0000000..731768f --- /dev/null +++ b/ws2022/rav/lecture/rav5.tex @@ -0,0 +1,234 @@ +\documentclass{lecture} + +\begin{document} + +\chapter{Affine 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, + \[ + \mathcal{O}_V(U) \coloneqq + \{ f \colon U \to k \mid \forall x \in U \exists U_x \subseteq + \text{ open neighbourhood of $x$ and polynomials} + P, Q \text{ sucht that } \forall z \in U \cap U_x, + Q(z) \neq 0 \text{ and } f(z) = \frac{P(z)}{Q(z)} + \} + .\] + \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} diff --git a/ws2022/rav/lecture/rav6.pdf b/ws2022/rav/lecture/rav6.pdf new file mode 100644 index 0000000..972c1f4 Binary files /dev/null and b/ws2022/rav/lecture/rav6.pdf differ diff --git a/ws2022/rav/lecture/rav6.tex b/ws2022/rav/lecture/rav6.tex new file mode 100644 index 0000000..763c7a5 --- /dev/null +++ b/ws2022/rav/lecture/rav6.tex @@ -0,0 +1,247 @@ +\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} diff --git a/ws2022/rav/lecture/rav7.pdf b/ws2022/rav/lecture/rav7.pdf new file mode 100644 index 0000000..20755aa Binary files /dev/null and b/ws2022/rav/lecture/rav7.pdf differ diff --git a/ws2022/rav/lecture/rav7.tex b/ws2022/rav/lecture/rav7.tex new file mode 100644 index 0000000..a8c58e1 --- /dev/null +++ b/ws2022/rav/lecture/rav7.tex @@ -0,0 +1,227 @@ +\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} diff --git a/ws2022/rav/lecture/rav8.pdf b/ws2022/rav/lecture/rav8.pdf new file mode 100644 index 0000000..b1d1f46 Binary files /dev/null and b/ws2022/rav/lecture/rav8.pdf differ diff --git a/ws2022/rav/lecture/rav8.tex b/ws2022/rav/lecture/rav8.tex new file mode 100644 index 0000000..4bd39b8 --- /dev/null +++ b/ws2022/rav/lecture/rav8.tex @@ -0,0 +1,289 @@ +\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} diff --git a/ws2022/rav/lecture/rav9.pdf b/ws2022/rav/lecture/rav9.pdf new file mode 100644 index 0000000..e4300ba Binary files /dev/null and b/ws2022/rav/lecture/rav9.pdf differ diff --git a/ws2022/rav/lecture/rav9.tex b/ws2022/rav/lecture/rav9.tex new file mode 100644 index 0000000..3f9fe9c --- /dev/null +++ b/ws2022/rav/lecture/rav9.tex @@ -0,0 +1,278 @@ +\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}