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diff --git a/controle-20240410.tex b/controle-20240410.tex new file mode 100644 index 0000000..bce1015 --- /dev/null +++ b/controle-20240410.tex @@ -0,0 +1,767 @@ +%% This is a LaTeX document. Hey, Emacs, -*- latex -*- , get it? +\documentclass[12pt,a4paper]{article} +\usepackage[a4paper,margin=2.5cm]{geometry} +\usepackage[english]{babel} +\usepackage[utf8]{inputenc} +\usepackage[T1]{fontenc} +%\usepackage{ucs} +\usepackage{times} +% A tribute to the worthy AMS: +\usepackage{amsmath} +\usepackage{amsfonts} +\usepackage{amssymb} +\usepackage{amsthm} +% +\usepackage{mathrsfs} +\usepackage{wasysym} +\usepackage{url} +% +\usepackage{graphics} +\usepackage[usenames,dvipsnames]{xcolor} +\usepackage{tikz} +\usetikzlibrary{matrix,calc} +\usepackage{hyperref} +% +%\externaldocument{notes-accq205}[notes-accq205.pdf] +% +\theoremstyle{definition} +\newtheorem{comcnt}{Whatever} +\newcommand\thingy{% +\refstepcounter{comcnt}\smallskip\noindent\textbf{\thecomcnt.} } +\newcommand\exercise{% +\refstepcounter{comcnt}\bigskip\noindent\textbf{Exercise~\thecomcnt.}\par\nobreak} +\renewcommand{\qedsymbol}{\smiley} +\renewcommand{\thefootnote}{\fnsymbol{footnote}} +% +\newcommand{\id}{\operatorname{id}} +\newcommand{\alg}{\operatorname{alg}} +\newcommand{\ord}{\operatorname{ord}} +\newcommand{\divis}{\operatorname{div}} +% +\DeclareUnicodeCharacter{00A0}{~} +\DeclareUnicodeCharacter{A76B}{z} +% +\DeclareMathSymbol{\tiret}{\mathord}{operators}{"7C} +\DeclareMathSymbol{\traitdunion}{\mathord}{operators}{"2D} +% +\DeclareFontFamily{U}{manual}{} +\DeclareFontShape{U}{manual}{m}{n}{ <-> manfnt }{} +\newcommand{\manfntsymbol}[1]{% + {\fontencoding{U}\fontfamily{manual}\selectfont\symbol{#1}}} +\newcommand{\dbend}{\manfntsymbol{127}}% Z-shaped +\newcommand{\danger}{\noindent\hangindent\parindent\hangafter=-2% + \hbox to0pt{\hskip-\hangindent\dbend\hfill}} +% +\newcommand{\spaceout}{\hskip1emplus2emminus.5em} +\newif\ifcorrige +\corrigetrue +\newenvironment{answer}% +{\ifcorrige\relax\else\setbox0=\vbox\bgroup\fi% +\smallbreak\noindent{\underbar{\textit{Answer.}}\quad}} +{{\hbox{}\nobreak\hfill\checkmark}% +\ifcorrige\par\smallbreak\else\egroup\par\fi} +% +% +% +\begin{document} +\ifcorrige +\title{ACCQ205\\Final exam — answer key\\{\normalsize Algebraic curves}} +\else +\title{ACCQ205\\Final exam\\{\normalsize Algebraic curves}} +\fi +\author{} +\date{2024-04-10} +\maketitle + +\pretolerance=8000 +\tolerance=50000 + +\vskip1truein\relax + +\noindent\textbf{Instructions.} + +This exam consists of three completely independent exercises. They +can be tackled in any order, but students must clearly and readably +indicate where each exercise starts and ends. + +\medbreak + +Answers can be written in English or French. + +\medbreak + +Use of written documents of any kind (such as handwritten or printed +notes, exercise sheets or books) is authorized. + +Use of electronic devices of any kind is prohibited. + +\medbreak + +Duration: 2 hours + +\ifcorrige +This answer key has 8 pages (this cover page included). +\else +This exam has 4 pages (this cover page included). +\fi + +\vfill +{\noindent\tiny +\immediate\write18{sh ./vc > vcline.tex} +Git: \input{vcline.tex} +\immediate\write18{echo ' (stale)' >> vcline.tex} +\par} + +\pagebreak + + +% +% +% + + +\exercise + +We say that a set of eight distinct points $p_0,\ldots,p_7$ in the +projective plane $\mathbb{P}^2$ over a field $k$ is a +\textbf{Möbius-Kantor configuration} when the points $p_0,p_1,p_3$ are +aligned, as well as $p_1,p_2,p_4$ and $p_2,p_3,p_5$ and so on +cyclically mod $8$, and no other set of three of the $p_i$ is aligned. +In other words, this means that $p_i,p_j,p_k$ are aligned if and only +if $\{i,j,k\} = \{\ell,\; \ell+1,\; \ell+3\}$ for some $\ell \in +\mathbb{Z}/8\mathbb{Z}$, where the subscripts are understood to be +mod $8$. + +The following figure (which is meant as a \emph{symbolic +representation} of the configuration and not as an actual geometric +figure!) illustrates the setup and can help keep track of which points +are aligned with which: + +\begin{center} +\vskip-7ex\leavevmode +\begin{tikzpicture} +\coordinate (P0) at (2cm,0); +\coordinate (P1) at (1.414cm,1.414cm); +\coordinate (P2) at (0,2cm); +\coordinate (P3) at (-1.414cm,1.414cm); +\coordinate (P4) at (-2cm,0); +\coordinate (P5) at (-1.414cm,-1.414cm); +\coordinate (P6) at (0,-2cm); +\coordinate (P7) at (1.414cm,-1.414cm); +\draw[shorten <=-0.5cm, shorten >=-0.3cm] (P0) -- (P1) .. controls ($2.5*(P1)-1.5*(P0)$) and ($2.5*(P2)-1.5*(P1)$) .. (P3); +\draw[shorten <=-0.5cm, shorten >=-0.3cm] (P1) -- (P2) .. controls ($2.5*(P2)-1.5*(P1)$) and ($2.5*(P3)-1.5*(P2)$) .. (P4); +\draw[shorten <=-0.5cm, shorten >=-0.3cm] (P2) -- (P3) .. controls ($2.5*(P3)-1.5*(P2)$) and ($2.5*(P4)-1.5*(P3)$) .. (P5); +\draw[shorten <=-0.5cm, shorten >=-0.3cm] (P3) -- (P4) .. controls ($2.5*(P4)-1.5*(P3)$) and ($2.5*(P5)-1.5*(P4)$) .. (P6); +\draw[shorten <=-0.5cm, shorten >=-0.3cm] (P4) -- (P5) .. controls ($2.5*(P5)-1.5*(P4)$) and ($2.5*(P6)-1.5*(P5)$) .. (P7); +\draw[shorten <=-0.5cm, shorten >=-0.3cm] (P5) -- (P6) .. controls ($2.5*(P6)-1.5*(P5)$) and ($2.5*(P7)-1.5*(P6)$) .. (P0); +\draw[shorten <=-0.5cm, shorten >=-0.3cm] (P6) -- (P7) .. controls ($2.5*(P7)-1.5*(P6)$) and ($2.5*(P0)-1.5*(P7)$) .. (P1); +\draw[shorten <=-0.5cm, shorten >=-0.3cm] (P7) -- (P0) .. controls ($2.5*(P0)-1.5*(P7)$) and ($2.5*(P1)-1.5*(P0)$) .. (P2); +\fill[black] (P0) circle (2.5pt); +\fill[black] (P1) circle (2.5pt); +\fill[black] (P2) circle (2.5pt); +\fill[black] (P3) circle (2.5pt); +\fill[black] (P4) circle (2.5pt); +\fill[black] (P5) circle (2.5pt); +\fill[black] (P6) circle (2.5pt); +\fill[black] (P7) circle (2.5pt); +\node[anchor=west] at (P0) {$p_0$}; +\node[anchor=south west] at (P1) {$p_1$}; +\node[anchor=south] at (P2) {$p_2$}; +\node[anchor=south east] at (P3) {$p_3$}; +\node[anchor=east] at (P4) {$p_4$}; +\node[anchor=north east] at (P5) {$p_5$}; +\node[anchor=north] at (P6) {$p_6$}; +\node[anchor=north west] at (P7) {$p_7$}; +\end{tikzpicture} +\vskip-7ex\leavevmode +\end{center} + +The goal of this exercise is to determine over which fields $k$ a +Möbius-Kantor configuration exists, and compute the coordinates of its +points. + +We fix a field $k$. The word “point”, in what follows, will refer +to an element of $\mathbb{P}^2(k)$, in other words, a point with +coordinates in $k$ (that is, a $k$-point). + +We shall write as $(x{:}y{:}z)$ the coordinates of a point, and as +$[u{:}v{:}w]$ the line $\{ux+vy+wz = 0\}$. Recall that the line +through $(x_1{:}y_1{:}z_1)$ and $(x_2{:}y_2{:}z_2)$ (assumed distinct) +is given by the formula $[(y_1 z_2 - y_2 z_1) : (z_1 x_2 - z_2 x_1) : + (x_1 y_2 - x_2 y_1)]$, and that the same formula (exchanging +parentheses and square brackets) can also be used to compute the +intersection of two distinct lines. (This may not always be the best +or simplest way\footnote{For example, one shouldn't need this formula + to notice that the line through $(42{:}0{:}0)$ and $(0{:}1729{:}0)$ + is $[0{:}0{:}1]$.} to compute coordinates, however!) + +\emph{We assume for questions (1)–(5) below that $p_0,\ldots,p_7$ is a +Möbius-Kantor configuration of points (over the given field $k$), and +the questions will serve to compute the coordinates of the points.} +We denote $\ell_{ijk}$ the line through $p_i,p_j,p_k$ when it exists. + +\textbf{(1)} Explain why we can assume, without loss of generality, +that $p_0=(1{:}0{:}0)$ and $p_1=(0{:}1{:}0)$ and $p_2=(0{:}0{:}1)$ and +$p_5=(1{:}1{:}1)$. \emph{We shall henceforth do so.} + +\begin{answer} +No three of the four points $p_0,p_1,p_2,p_5$ are aligned, so they are +a projective basis of $\mathbb{P}^2$: thus, there is a unique +projective transformation of $\mathbb{P}^2$ mapping them to the +standard basis $(1{:}0{:}0), \penalty-100 (0{:}1{:}0), \penalty-100 +(0{:}0{:}1), \penalty-100 (1{:}1{:}1)$. Since projective +transformations preserve alignment, we can apply this projective +transformation and assume that $p_0=(1{:}0{:}0)$ and $p_1=(0{:}1{:}0)$ +and $p_2=(0{:}0{:}1)$ and $p_5=(1{:}1{:}1)$. +\end{answer} + +\textbf{(2)} Compute the coordinates of the lines $\ell_{013}$, +$\ell_{124}$, $\ell_{235}$, $\ell_{560}$ and $\ell_{702}$, and of the +point $p_3$. + +\begin{answer} +Denoting $p\vee q$ the line through distinct points $p$ and $q$, we +get $\ell_{013} = p_0 \vee p_1 = [0{:}0{:}1]$ and $\ell_{124} = +p_1\vee p_2 = [1{:}0{:}0]$ and $\ell_{235} = p_5\vee p_2 = +[1{:}{-1}{:}0]$ and $\ell_{560} = p_5\vee p_0 = [0{:}1{:}{-1}]$ and +$\ell_{702} = p_2\vee p_0 = [0{:}1{:}0]$. Denoting by $\ell\wedge m$ +the point of intersection of distinct lines $\ell$ and $m$, we get +$p_3 = \ell_{013} \wedge \ell_{235} = (1{:}1{:}0)$. +\end{answer} + +\textbf{(3)} Explain why we can write, without loss of generality, the +coordinates of $p_4$ in the form $(0{:}\xi{:}1)$ for some $\xi$ +(in $k$). (Note that two things need to be explained here: why the +first coordinate is $0$ and why the last can be taken to be $1$.) + +\begin{answer} +The point $p_4$ is on $\ell_{124} = [1{:}0{:}0]$, so it is of the form +$(0{:}\tiret{:}\tiret)$ (its first coordinate is zero). On the other +hand, it is \emph{not} on $\ell_{013} = [0{:}0{:}1]$, so it is +\emph{not} of the form $(\tiret{:}\tiret{:}0)$ (its last coordinate is +\emph{not} zero). Since homogeneous coordinates are defined up to +multiplication by a common constant, we can divide them by this +nonzero last coordinate, and we get $p_4$ of the form +$(0{:}\tiret{:}1)$, as required. +\end{answer} + +\textbf{(4)} Now compute the coordinates of the line $\ell_{346}$, of +the point $p_6$, and of the lines $\ell_{457}$ and $\ell_{671}$. + +\begin{answer} +We have $\ell_{346} = p_3\vee p_4 = [1{:}{-1}{:}\xi]$. Therefore $p_6 += \ell_{346} \wedge \ell_{560} = (1-\xi : 1 : 1)$. Further, +$\ell_{457} = p_4\vee p_5 = [\xi-1 : 1 : -\xi]$ and $\ell_{671} = +p_1\wedge p_6 = [1{:}0{:}\xi-1]$. +\end{answer} + +\textbf{(5)} Write the coordinates of the last remaining point $p_7$ +and using the fact that we now have three lines on which it lies, +conclude that $\xi$ must satisfy $1-\xi+\xi^2 = 0$. + +\begin{answer} +The point $p_7$ can be written as $\ell_{571} \wedge \ell_{702}$, +giving coordinates $(1-\xi:0:1)$, or as $\ell_{457} \wedge +\ell_{702}$, giving coordinates $(\xi:0:\xi-1)$. That they are equal +gives the relation $\xi + (1-\xi)^2 = 0$ or $1-\xi+\xi^2 = 0$. +Alternatively, we can write $p_7$ as $\ell_{671} \wedge \ell_{457}$ +with coordinates $(1-\xi : 1-\xi+\xi^2 : 1)$, and the fact that it +lies on $\ell_{702}$. we get $1-\xi+\xi^2 = 0$. +\end{answer} + +\textbf{(6)} Deduce from questions (1)–(5) above that, if a +Möbius-Kantor configuration over $k$ exists, then there is $\xi\in k$ +such that $1-\xi+\xi^2 = 0$. + +\begin{answer} +As explained in (1), we can find a projective transformation of +$\mathbb{P}^2$ such giving $p_0,p_1,p_2,p_5$ the coordinates +$(1{:}0{:}0), \penalty-100 (0{:}1{:}0), \penalty-100 (0{:}0{:}1), +\penalty-100 (1{:}1{:}1)$, and as explained in (3) we then get $p_4$ +of the form $(0{:}\xi{:}1)$, and as explained in (5) this $\xi$ must +satisfy $1-\xi+\xi^2 = 0$. So if there is Möbius-Kantor configuration +over $k$ then there is such a $\xi$. +\end{answer} + +\textbf{(7)} Conversely, using the coordinate computations performed +in questions (2)–(5), explain why, if there is $\xi\in k$ such that +$1-\xi+\xi^2 = 0$, then a Möbius-Kantor configuration over $k$ exists. +(A long explanation is not required, but at least explain what checks +need be done.) + +\begin{answer} +Conversely, if a $\xi$ such that $1-\xi+\xi^2 = 0$ exists, then the +coordinates we have computed, namely +\[ +\arraycolsep=1em +\begin{array}{cc} +p_0 = (1 : 0 : 0) & \ell_{013} = [0 : 0 : 1]\\ +p_1 = (0 : 1 : 0) & \ell_{124} = [1 : 0 : 0]\\ +p_2 = (0 : 0 : 1) & \ell_{235} = [1 : {-1} : 0]\\ +p_3 = (1 : 1 : 0) & \ell_{346} = [1 : {-1} : \xi]\\ +p_4 = (0 : \xi : 1) & \ell_{457} = [-1+\xi : 1 : -\xi]\\ +p_5 = (1 : 1 : 1) & \ell_{560} = [0 : 1 : {-1}]\\ +p_6 = (1-\xi : 1 : 1) & \ell_{671} = [1 : 0 : \xi-1]\\ +p_7 = (1-\xi : 0 : 1) & \ell_{702} = [0 : 1 : 0]\\ +\end{array} +\] +define a Möbius-Kantor configuration. To check this, we need to check +that $p_i,p_j,p_k$ lie on $\ell_{ijk}$: most of these checks are +trivial, and the remaining few follow from $1-\xi+\xi^2=0$; but we +also need to check that no other $p_r$ lies on $\ell_{ijk}$: for +example, this requires checking that $\xi \neq 0$ (which follows from +the fact that $0$ certainly does not satisfy $1-\xi+\xi^2=0$) and $\xi +\neq 1$ (similarly). +\end{answer} + +\textbf{(8)} Give examples of fields $k$, at least one infinite and +one finite, over which a Möbius-Kantor configuration exists, and +similarly examples over which it does not exist. + +\begin{answer} +For fields of characteristic $\neq 2$, the usual formula for solving a +quadratic equation shows that a Möbius-Kantor configuration exists +precisely iff $-3$ is a square (since the discriminant of $1-t+t^2$ +is $-3$). This is obviously the case of fields of characteristic $3$ +(with $\xi = -1$). + +Some examples of fields with a Möbius-Kantor configuration are: any +algebraically closed field (e.g., $\mathbb{C}$), the field +$\mathbb{Q}(\sqrt{-3}) = \{u+v\sqrt{-3} : u,v\in\mathbb{Q}\}$, any +field of characteristic $3$ (e.g., $\mathbb{F}_3$), the field +$\mathbb{F}_4$ with $4$ elements (because it is +$\mathbb{F}_2[t]/(1+t+t^2)$), or the field $\mathbb{F}_7$ (because +$\xi = 3$ satisfies $1-\xi+\xi^2 = 0$). + +Some examples of fields without a Möbius-Kantor configuration are: any +subfield of $\mathbb{R}$ (including $\mathbb{Q}$ or $\mathbb{R}$ +itself), since $-3$ is not a square in $\mathbb{R}$, the field +$\mathbb{F}_2$ or the field $\mathbb{F}_5$ (checking for each element +that it does not satisfy $1-\xi+\xi^2 = 0$). + +(In fact, for finite fields, the law of quadratic reciprocity gives us +a complete answer of when a Möbius-Kantor configuration over +$\mathbb{F}_q$ exists: if $q \equiv 1 \pmod{4}$ we have +$\big(\frac{-3}{q}\big) = \big(\frac{-1}{q}\big) \, +\big(\frac{3}{q}\big) = \big(\frac{3}{q}\big) = +\big(\frac{q}{3}\big)$, while if $q \equiv 3 \pmod{4}$ we have +$\big(\frac{-3}{q}\big) = \big(\frac{-1}{q}\big) \, +\big(\frac{3}{q}\big) = -\big(\frac{3}{q}\big) = +\big(\frac{q}{3}\big)$; so if $q$ is neither a power of $2$ nor of $3$ +this is $+1$ iff $q \equiv 1 \pmod{3}$. For $q$ a power of $3$, a +Möbius-Kantor configuration always exists. For $q$ a power of $2$, it +is not hard to check that it exists iff $q$ is an \emph{even} power +of $2$. Putting all cases together, a Möbius-Kantor configuration +exists over $\mathbb{F}_q$ iff either $q$ is a power of $3$ or $q +\equiv 1 \pmod{3}$.) +\end{answer} + + +% +% +% + + +\exercise + +The focus of this exercise is \textbf{Klein's quartic}, namely the +projective algebraic variety $C$ defined by the equation +\[ +x^3 y + y^3 z + z^3 x = 0 +\] +in $\mathbb{P}^2$ with coordinates $(x{:}y{:}z)$. Note the symmetry +of this equation under cyclic permutation of the +coordinates\footnote{To dispel any possible confusion, this means +simultaneously replacing $x$ by $y$, $y$ by $z$ and $z$ by $x$.}, +which will come in handy to simplify some computations. To refer to +it more easily, we shall denote $f := x^3 y + y^3 z + z^3 x$ the +polynomial defining the equation of $C$. + +We shall work over a field $k$ having characteristic $\not\in\{2,7\}$. +For simplicity, we shall also assume $k$ to be algebraically closed +(even though this won't matter at all). + +\textbf{(1)} The following relation holds (this is a straightforward +computation, and it is not required to check it): +\[ +-27xyz\,\frac{\partial f}{\partial x} ++(28x^3-3y^2 z)\,\frac{\partial f}{\partial y} +-9yz^2\,\frac{\partial f}{\partial z} += 28x^6 +\tag{$*$} +\] +What does the relation ($*$), together with the other two obtained by +cyclically permuting coordinates, tell us about the ideal generated by +$\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$ and +$\frac{\partial f}{\partial z}$ in $k[x,y,z]$? What does this imply +on the set of points where $\frac{\partial f}{\partial x}$, +$\frac{\partial f}{\partial y}$ and $\frac{\partial f}{\partial z}$ +all vanish? + +\begin{answer} +The relation $*$ tells us that $28 x^6$, and consequently $x^6$ itself +(since $k$ is of characteristic $\not\in\{2,7\}$), belongs to the +ideal generated by $\frac{\partial f}{\partial x}$, $\frac{\partial + f}{\partial y}$ and $\frac{\partial f}{\partial z}$. By cyclic +permutation of coordinates, this is also the case for $y^6$ and $z^6$: +so this ideal is irrelevant: the set of points in $\mathbb{P}^2$ where +$\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$ and +$\frac{\partial f}{\partial z}$ all vanish is empty (because +$x^6,y^6,z^6$ do not vanish simultaneously). This implies that $C$ is +\emph{smooth}. +\end{answer} + +\smallskip + +\emph{The previous question implies that $C$ is a (plane) curve. The +following picture is a rough sketch of an affine part of $C$ over the +real field.} + +\begin{center} +\begin{tikzpicture} +\begin{scope}[thick] +\clip (-3,-3) -- (3,-3) -- (3,3) -- (-3,3) -- cycle; +\draw (-3.000,5.251) .. controls (-2.667,4.397) and (-2.333,3.618) .. (-2.000,2.946) ; +\draw (-2.000,2.946) .. controls (-1.833,2.610) and (-1.667,2.301) .. (-1.500,2.028); +\draw (-1.500,2.028) .. controls (-1.333,1.755) and (-1.167,1.519) .. (-1.000,1.325) ; +\draw (-1.000,1.325) .. controls (-0.833,1.130) and (-0.667,0.981) .. (-0.500,0.846) ; +\draw (-0.500,0.846) .. controls (-0.417,0.779) and (-0.333,0.716) .. (-0.250,0.638) ; +\draw (-0.250,0.638) .. controls (-0.208,0.600) and (-0.167,0.558) .. (-0.125,0.501) ; +\draw (-0.125,0.501) .. controls (-0.104,0.473) and (-0.083,0.441) .. (-0.062,0.397) ; +\draw (-0.062,0.397) .. controls (0,0.265) and (0,0.133) .. (0,0) ; +\draw (0,0) .. controls (0,-0.133) and (0,-0.265) .. (0.062,-0.397) ; +\draw (0.062,-0.397) .. controls (0.083,-0.441) and (0.104,-0.471) .. (0.125,-0.499) ; +\draw (0.125,-0.499) .. controls (0.167,-0.553) and (0.208,-0.590) .. (0.250,-0.622) ; +\draw (0.250,-0.622) .. controls (0.333,-0.684) and (0.417,-0.720) .. (0.500,-0.741) ; +\draw (0.500,-0.741) .. controls (0.667,-0.783) and (0.833,-0.755) .. (1.000,-0.682) ; +\draw (1.000,-0.682) .. controls (1.167,-0.610) and (1.333,-0.501) .. (1.500,-0.422) ; +\draw (1.500,-0.422) .. controls (1.667,-0.343) and (1.833,-0.288) .. (2.000,-0.248) ; +\draw (2.000,-0.248) .. controls (2.333,-0.168) and (2.667,-0.136) .. (3.000,-0.111) ; +\draw (-3.000,-5.140) .. controls (-2.667,-4.261) and (-2.333,-3.452) .. (-2.000,-2.694) ; +\draw (-2.000,-2.694) .. controls (-1.833,-2.315) and (-1.667,-1.962) .. (-1.500,-1.552) ; +\draw (-1.500,-1.552) .. controls (-1.458,-1.449) and (-1.417,-1.346) .. (-1.375,-1.209) ; +\draw (-1.375,-1.209) .. controls (-1.315,-1.013) and (-1.263,-0.817) .. (-1.375,-0.621) ; +\draw (-1.375,-0.621) .. controls (-1.417,-0.548) and (-1.458,-0.511) .. (-1.500,-0.477) ; +\draw (-1.500,-0.477) .. controls (-1.667,-0.339) and (-1.833,-0.295) .. (-2.000,-0.252) ; +\draw (-2.000,-0.252) .. controls (-2.333,-0.166) and (-2.667,-0.136) .. (-3.000,-0.111) ; +\end{scope} +\draw[->, shorten <=-0.1cm, shorten >=-0.1cm, thin] (-3,0) -- (3,0); +\draw[->, shorten <=-0.1cm, shorten >=-0.1cm, thin] (0,-3) -- (0,3); +\node[anchor=west] at (3,0) {$\scriptstyle x/z \,=:\, u$}; +\node[anchor=south] at (0,3) {$\scriptstyle y/z \,=:\, v$}; +\end{tikzpicture} +\end{center} + +We now define the three points $a := (1{:}0{:}0)$, $b := (0{:}1{:}0)$ +and $c := (0{:}0{:}1)$ (which obviously lie on $C$). + +\textbf{(2)} List all points of $C$ where $x$ vanishes. Do the same +for $y$ and $z$. + +\begin{answer} +If $x$ vanishes on $C$ then $y^3 z = 0$, so $y=0$ or $z=0$. So either +$x=y=0$ and we are at $c$, or $x=z=0$ and we are at $b$; so the set of +points where $x$ vanishes on $C$ is exactly $\{b,c\}$. By cyclic +rotation of coordinates, the set of points of $C$ where $y$ vanishes +is $\{c,a\}$ and the set of points of $C$ where $z$ vanishes is +$\{a,b\}$. +\end{answer} + +\textbf{(3)} Where do the points $a,b,c$ lie on the printed picture? +(If they do not lie on the picture, show the direction in which they +would be.) What is the equation of the affine part of $C$ drawn on +the picture? What is the tangent line at the point $c$? What about +$a$ and $b$? + +\begin{answer} +The point $c$ is at affine coordinates $(u,v) = (0,0)$ where $u = +\frac{x}{z}$ and $v = \frac{y}{z}$, that is, it is at the origin of +the printed picture. The point $a$ is at infinity ($z=0$) on the axis +$y=0$ (or $v=0$ if we prefer), so it is at infinity in the horizontal +direction, whereas $b$ is at infinity on the axis $x=0$ (or $u=0$ if +we prefer), so at infinity in the vertical direction. + +The equation of the affine part of $C$ is obtained by dehomogenizing +$x^3 y + y^3 z + z^3 x = 0$ with respect to $z$, i.e., by dividing by +$z^3$ and replacing $\frac{x}{z}$ by $u$ and $\frac{y}{z}$ by $v$, +giving $u^3 v + v^3 + u = 0$. + +The tangent line at the origin $c$ of the affine part $\{z\neq 0\}$ is +given by $\frac{\partial g}{\partial u}|_{(0,0)}\cdot u + +\frac{\partial g}{\partial v}|_{(0,0)}\cdot v =0$ where $g := u^3 v + +v^3 + u$. This simply gives $u=0$, so it is the vertical axis (as +could be guessed from the figure); as a projective line, this is +$x=0$. By cyclic permutation of coordinates, we get $y=0$ as tangent +line at $a$ and $z=0$ as tangent line at $c$. (Of course, one might +also compute these by taking affine charts around each one of the +points, but this would be more tedious.) +\end{answer} + +\textbf{(4)} Considering $v := \frac{y}{z}$ as a rational function +on $C$, explain why it vanishes at order exactly $1$ at $c$, that +is\footnote{We write $\ord_p(h)$ for the order at a point $p \in C$ of +a rational function $h \in k(C)$. By the way, please note that +$x,y,z$ themselves do not belong to $k(C)$ (they are not functions and +have no value by themselves), so we cannot speak of $\ord_p(x)$.}, +$\ord_c(v) = 1$. Explain why $\ord_c(u) = \ord_c(u^3 v + v^3)$ where +$u := \frac{x}{z}$ and deduce that $\ord_c(u) = 3$. Deduce the order +at $c$ of $\frac{y}{x}$ (which is also $\frac{v}{u}$). + +\begin{answer} +The coordinate $v$ vanishes with order exactly $1$ at the origin $c$ +of the tangent line $u=0$ to $C$ at $c$; therefore it also has order +exactly $1$ at $c$ on $C$. In other words, $\ord_c(v) = 1$. + +Now $u^3 v + v^3 + u = 0$ on $C$, that is $u = -u^3 v - v^3$, so +$\ord_c(u) = \ord_c(u^3 v + v^3)$. This shows that $\ord_c(u) =: k$, +which is $\geq 1$ because $u$ vanishes at $c$, satisfies $k \geq +\min(3k+1,3)$, so $k \geq 3$; but now $3k+1 \geq 10$, so $\ord_c(u^3 +v) = 3k+1 \neq 3 = \ord_c(v^3)$, so in fact $k = \min(3k+1,3) = 3$, as +required. + +Consequently, $\frac{y}{x} = \frac{v}{u}$ has order $\ord_c(v) - +\ord_c(u) = 1 - 3 = -2$ at $c$. +\end{answer} + +\textbf{(5)} By using symmetry, compute the order at each one of the +three points $a,b,c$ of each one of the three functions $\frac{x}{z}$, +$\frac{y}{x}$ and $\frac{z}{y}$. Explain why there are no points +(of $C$) other than $a,b,c$ where any of these functions (on $C$) +vanishes or has a pole. Summarize this by writing the principal +divisors $\divis(\frac{x}{z})$, $\divis(\frac{y}{x})$ and +$\divis(\frac{z}{y})$ associated with these three functions. + +\begin{answer} +We have seen that +\[ +\arraycolsep=1em +\begin{array}{ccc} +\ord_c(\frac{x}{z}) = 3 & +\ord_c(\frac{y}{x}) = -2 & +\ord_c(\frac{z}{y}) = -1 +\end{array} +\] +so by cyclic permutation we get +\[ +\arraycolsep=1em +\begin{array}{ccc} +\ord_a(\frac{x}{z}) = -1 & +\ord_a(\frac{y}{x}) = 3 & +\ord_a(\frac{z}{y}) = -2 +\\ +\ord_b(\frac{x}{z}) = -2 & +\ord_b(\frac{y}{x}) = -1 & +\ord_b(\frac{z}{y}) = 3 +\end{array} +\] +Now we have also pointed out earlier that none of $x,y,z$ vanishes on +$C$ outside possibly of $\{a,b,c\}$: so +$\frac{x}{z},\frac{y}{x},\frac{z}{y}$ have neither zero nor pole on +$C\setminus\{a,b,c\}$, i.e., their order is $0$ everywhere on this +open set. This shows that +\[ +\begin{aligned} +\divis(\frac{x}{z}) &= -[a] -2\,[b] + 3\,[c]\\ +\divis(\frac{y}{x}) &= \hphantom{+}3\,[a] - [b] - 2\,[c]\\ +\divis(\frac{z}{y}) &= -2\,[a] + 3\,[b] - [c] +\end{aligned} +\] +Two sanity checks can be performed: the degree of each of these +divisors (i.e., the sum of the coefficients) is zero, as befits a +principal divisor; and the sum of these three divisors is also zero, +as it should be because it is the divisor of the constant nonzero +function $1$. +\end{answer} + + +% +% +% + + +\exercise + +This exercise is about the \textbf{Segre embedding}\footnote{French: + “plongement de Segre”}, which is a way to map the product +$\mathbb{P}^p \times \mathbb{P}^q$ of two projective spaces to a +larger projective space $\mathbb{P}^n$ (with, as we shall see, $n = +pq+p+q$). + +Assume $k$ is a field. To simplify presentation, assume $k$ is +algebraically closed (even though this won't matter at all). + +Given $p,q\in\mathbb{N}$, the Segre embedding of $\mathbb{P}^p \times +\mathbb{P}^q$ is the map $\psi$ given by: +\[ +\begin{aligned} +\psi\colon & \mathbb{P}^p \times \mathbb{P}^q \to \mathbb{P}^n\\ +&((x_0:\cdots:x_p),\, (y_0:\cdots:y_q)) \mapsto (x_0 y_0 : x_0 y_1 : \cdots +: x_0 y_q : x_1 y_0 : \cdots : x_p y_q)\\ +\end{aligned} +\] +where $n = (p+1)(q+1)-1$ and the coordinates of the endpoint consist +of every product $x_i y_j$ with $0\leq i\leq p$ and $0\leq j\leq q$ +(in some order which doesn't really matter: here we have chosen the +lexicographic ordering). + +Note that with the definitions given in this course, we cannot state +that $\psi$ is a morphism of algebraic varieties (although it +certainly \emph{should} be one), because we did not define a “product +variety”\footnote{In fact, the Segre embedding is one way of doing +this.} $\mathbb{P}^p \times \mathbb{P}^q$. But we can still consider +it as a function. + +Let us label $(z_{0,0} : z_{0,1} : \cdots : z_{p,q})$ the homogeneous +coordinates in $\mathbb{P}^n$ (that is, $z_{i,j}$ with $0\leq i\leq p$ +and $0\leq j\leq q$), so that $\psi$ is given simply by “$z_{i,j} = +x_i y_j$”. + +We finally consider the Zariski closed subset $S$ of $\mathbb{P}^n$, +known as the \textbf{Segre variety}, defined in $\mathbb{P}^n$ by the +equations $z_{i,j} z_{i',j'} = z_{i,j'} z_{i',j}$ for all $0\leq +i,i'\leq p$ and $0\leq j,j'\leq q$. + +\medskip + +\textbf{(1)} Explain why the map $\psi$ is well-defined, i.e., the +definition given above makes sense: carefully list the properties that +need to be checked, and do so. Explain why $S$ is indeed a Zariski +closed subset of $\mathbb{P}^n$: again, carefully state what needs to +be checked before doing so. + +\begin{answer} +For the point $(x_0 y_0 : \cdots : x_p y_q)$ to make sense, we need to +check that not all its coordinates are zero. But we know that at +least one of the $x_i$ is nonzero and at least one of the $y_j$ is +nonzero, so (as we are working over a field) the product $x_i y_j$ is +nonzero. + +For the map $((x_0:\cdots:x_p),\, (y_0:\cdots:y_q)) \mapsto (x_0 y_0 : +\cdots : x_p y_q)$ to make sense, we need to check that $(x_0 y_0 : +\cdots : x_p y_q)$ does not change if we replace the $x_i$ and the +$y_j$ by different coordinates for the same point, in other words, if +we multiply all the $x_i$ by a common nonzero constant, and all the +$y_j$ by a (possibly different) common nonzero constant. This is +indeed the case as $x_i y_j$ will be multiplied by the product of +these two constants. + +Concerning $S$, we need to check that the equations $z_{i,j} z_{i',j'} += z_{i,j'} z_{i',j}$ are homogeneous: this is indeed the case (they +are homogeneous of degree $2$). +\end{answer} + +\textbf{(2)} Consider in this question the special case $p=q=1$ (so +$n=3$). Simplify the definition of $S$ in this case down to a single +equation. Taking $z_{0,0}=0$ as the plane at infinity in +$\mathbb{P}^3$, give the equation of the affine part $S \cap +\mathbb{A}^3$. Similarly taking $x_0=0$ (resp. $y_0=0$) as the point +at infinity in $\mathbb{P}^1$, describe $\psi$ on $\mathbb{A}^1 \times +\mathbb{A}^1$. + +\begin{answer} +When $p=q=1$ the equations of $S$ are all trivial except $z_{0,0} +z_{1,1} = z_{0,1} z_{1,0}$ (or equations trivially equivalent to +this). Taking $z_{0,0} = 0$ as plane at infinity, we get the equation +of the affine part by dehomogenizing $z_{0,0} z_{1,1} = z_{0,1} +z_{1,0}$, which gives $w_{1,1} = w_{0,1} w_{1,0}$ where $w_{i,j}$ +denotes the affine coordinate $z_{i,j}/z_{0,0}$ in $\mathbb{A}^3$. + +Concerning $\psi$, if we call $u = x_1/x_0$ the affine coordinate on +the first $\mathbb{A}^1$ and $v = y_1/y_0$ that on the second, it is +given by taking $(u,v)$, i.e. $((1:u),\, (1:v))$ to $(1:v:u:uv)$, that +is $(v,u,uv)$. +\end{answer} + +\textbf{(3)} Returning to the case of general $p$ and $q$, show that +the image of $\psi$ is contained in $S$, that is, $\psi(\mathbb{P}^p +\times \mathbb{P}^q) \subseteq S$. + +\begin{answer} +If $(z_{0,0} : \cdots : z_{p,q})$ is given by $z_{i,j} = x_i y_j$, we +just need to check that $z_{i,j} z_{i',j'} = z_{i,j'} z_{i',j}$: but +this just says that $x_i y_j x_{i'} y_{j'} = x_i y_{j'} x_{i'} y_j$, +which is obvious by commutativity. +\end{answer} + +\textbf{(4)} Conversely, explain why for each point $(z_{0,0} : \cdots +: z_{p,q})$ in $S$ there is a unique pair of points $((x_0 : \cdots : +x_p), (y_0 : \cdots : y_q))$ in $\mathbb{P}^p \times \mathbb{P}^q$ +which maps to the given point under $\psi$: in other words, show that +$\psi$ is a bijection between $\mathbb{P}^p \times \mathbb{P}^q$ +and $S$. + +(\textit{Hint:} you may wish to observe that if $(z_{0,0} : \cdots : +z_{p,q})$ is in $S$, the point $(z_{0,j_0} : \cdots : z_{p,j_0})$ in +$\mathbb{P}^p$ does not depend on $j_0 \in \{0,\ldots,q\}$ such that +$\exists i.(z_{i,j_0}\neq 0)$; and similarly for $(z_{i_0,0} : \cdots +: z_{i_0,q})$ in $\mathbb{P}^q$.) + +\begin{answer} +Assume $(z_{0,0} : \cdots : z_{p,q})$ is in $S$. By the definition of +$\mathbb{P}^n$, at least one coordinate $z_{i_0,j_0}$ is nonzero. +Define $x^*_i = z_{i,j_0}$ (note that $x^*_{i_0} \neq 0$) and $y^*_j = +z_{i_0,j}$ (note that $y^*_{j_0} \neq 0$): then $x^*_i y^*_j = +z_{i,j_0} z_{i_0,j}$, which, by the equations of $S$, is also +$z_{i_0,j_0} z_{i,j}$: this shows that $((x^*_0 : \cdots : x^*_p), +(y^*_0 : \cdots : y^*_q))$ maps to the given $(z_{0,0} : \cdots : +z_{p,q})$ under $\psi$ (by dividing all coordinates by the nonzero +value $z_{i_0,j_0}$). So $\psi$ surjects to $S$. + +But in fact, if $((x_0 : \cdots : x_p), (y_0 : \cdots : y_q))$ maps to +$(z_{0,0} : \cdots : z_{p,q})$ under $\psi$, then we have $z_{i,j_0} = +y_{j_0} x_i$ so that $(x_0 : \cdots : x_p) = (z_{0,j_0} : \cdots : +z_{p,j_0})$ provided $y_{j_0} \neq 0$, which is tantamount to saying +$z_{i_0,j_0}\neq 0$ for some $i_0$: so we had no other choice than to +take the $(x^*_0 : \cdots : x^*_p)$ of the previous paragraph, and the +same argument holds for $(y^*_0 : \cdots : y^*_q)$. This shows +uniqueness of the points $((x_0 : \cdots : x_p), (y_0 : \cdots : +y_q))$ mapping to $(z_{0,0} : \cdots : z_{p,q})$ under $\psi$. +\end{answer} + +\textbf{(5)} Call $\pi\colon S\to \mathbb{P}^p\times\mathbb{P}^q$ the +inverse bijection of $\psi$, and call $\pi',\pi''$ its two components. +(In other words, if $s = (z_{0,0} : \cdots : z_{p,q})$ is in $S$ then +$\pi'(s) = (x_0:\cdots:x_p) \in \mathbb{P}^p$ and $\pi''(s) = +(y_0:\cdots:y_p) \in \mathbb{P}^q$ are the unique points such that +$(\pi'(s),\pi''(s))$ maps to $s$ under $\psi$.) Show that the maps +$\pi' \colon S \to \mathbb{P}^p$ and $\pi'' \colon S \to \mathbb{P}^q$ +are morphisms of algebraic varieties. (If this seems too difficult, +consider the special case $p=q=1$, and at least try to explain what +needs to be checked.) + +\begin{answer} +Given $j_0 \in \{0,\ldots,q\}$, consider the map $(z_{0,0} : \cdots : +z_{p,q}) \mapsto (z_{0,j_0} : \cdots : z_{p,j_0})$ which selects only +the coordinates $z_{i,j_0}$. This is a partially defined map from +$\mathbb{P}^n$ to $\mathbb{P}^p$, and the components are homogeneous +polynomials of the same degree (here, $1$): the only thing that can go +wrong is that all the $z_{i,j_0}$ are zero, so this is well-defined on +the open set $\{z_{0,j_0}\neq 0\} \cup \cdots \cup \{z_{p,j_0}\neq +0\}$. Now restrict this map to $S$: this gives us a morphism +$\pi^{\prime(j_0)}$ from the open set $U^{(j_0)} := S \cap +(\{z_{0,j_0}\neq 0\} \cup \cdots \cup \{z_{p,j_0}\neq 0\})$ of $S$ +to $\mathbb{P}^p$. + +Note that the union the union of $U^{0)},\ldots,U^{(q)}$ is all of $S$ +because there is always at least one coordinate nonzero. + +Furthermore, we have seen in (4) that if $s = (z_{0,0} : \cdots : +z_{p,q})$ then $\pi'(s)$ is given by $\pi^{\prime(j_0)}(s) = +(z_{0,j_0} : \cdots : z_{p,j_0})$ where $j_0$ is any element of +$\{0,\ldots,q\}$ such that $z_{i_0,j_0} \neq 0$ for some $i_0$, i.e., +$s \in U^{(j_0)}$. This shows that $\pi'$ coincides with +$\pi^{\prime(j_0)}$ on the open set $U^{(j_0)}$ where the latter is +defined, so $\pi'$ is defined by “gluing” the various +$\pi^{\prime(j_0)}$. So $\pi'$ is indeed a morphism (to be clear: it +is simply defined by selecting the coordinates of the form $z_{i,j_0}$ +for any one $j_0$ such that not all of them vanish). + +The same argument, \textit{mutatis mutandis}, works for $\pi''$. +\end{answer} + + + +% +% +% +\end{document} |