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ws2022/rav/lecture/lecture.cls
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\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
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% \setheadrule{.55pt}%
% \sethead[\thepage]% even-left
% []% even-center
% [\thechapter~\chaptertitle]% even-right
% {\thesection~\sectiontitle}% odd-left
% {}% odd-center
% {\thepage}% odd-right
%}
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\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
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commentstyle=\color{gray},
keywordstyle=\color{blue},
numberstyle=\tiny\color{gray},
stringstyle=\color{black},
basicstyle=\ttfamily\footnotesize,
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captionpos=b,
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numbers=left,
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showspaces=false,
showstringspaces=false,
showtabs=false,
tabsize=2
}
% activate my colour style
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\renewcommand{\stackrel}[3][]{%
\oldstackrel[\mathclap{#1}]{\mathclap{#2}}{#3}
}%
% integral d sign
\makeatletter \renewcommand\d[2][]{\ensuremath{%
\,\mathrm{d}^{#1}#2\@ifnextchar^{}{\@ifnextchar\d{}{\,}}}}
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% remove page before chapters
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%josua
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% contradiction
\newcommand{\contr}{\text{\Large\lightning}}
% people seem to prefer varepsilon over epsilon
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\ExplSyntaxOn
% S-tackrelcompatible ALIGN environment
% some might also call it the S-uper ALIGN environment
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% to put additional padding on both sides of relation symbols
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{
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\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
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\dim_new:N \l_tmp_dim_b
\dim_new:N \l_tmp_dim_c
\dim_new:N \l_tmp_dim_needed
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% get all stackrels in this align environment
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\int_set:Nn \l_idx_int { 1 }
\dim_set:Nn \l_tmp_dim_needed { 0pt }
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{
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\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
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% 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
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\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}}
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\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}
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\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}
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\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}
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\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}
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\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}
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\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}