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diff --git a/controle-20240410.tex b/controle-20240410.tex new file mode 100644 index 0000000..9e6ad1d --- /dev/null +++ b/controle-20240410.tex @@ -0,0 +1,437 @@ +%% 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} +% +\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.} + +\textcolor{red}{xxx} + +\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 \textcolor{red}{xxx} pages (cover page included). +\else +This exam has \textcolor{red}{xxx} pages (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 decide over which fields $k$ a +Möbius-Kantor configuration exists, and compute the coordinates of its +points. + +We fix a field $k$ and 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 in +$\mathbb{P}^2(k)$, 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, you +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.} + +\textbf{(2)} Compute the coordinates of the lines $\ell_{013}$, +$\ell_{124}$, $\ell_{235}$, $\ell_{560}$ and $\ell_{702}$, and of the +point $p_3$. + +\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 $1$.) + +\textbf{(4)} Now compute the coordinates of the lines $\ell_{346}$ and +$\ell_{457}$, of the point $p_6$, and of the line $\ell_{671}$. + +\textbf{(5)} Write the coordinates of the last remaining point $p_7$ +in two different ways (using two different pairs of lines) and +conclude that $\xi$ must satisfy $1-\xi+\xi^2 = 0$. + +\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 $1-\xi+\xi^2 = 0$. + +\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 you should at least explain +what checks need be done.) + +\textbf{(8)} Give two different examples of fields $k$, one infinite +and one finite, over which a Möbius-Kantor configuration exists, and +similarly two examples over which it does not exist. + + +% +% +% + + +\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? + +\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 $z$. + +\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 +should 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 $b$? + +\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 $\frac{v}{u}$). + +\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. + + +% +% +% + + +\exercise + +This exercise is about the \textbf{Segre embedding}, 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 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 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. + +\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 $\mathbb{A}^1 \times +\mathbb{A}^1$. + +\textbf{(3)} Returning to the case of general $p$ and $q$, show that +the image of $\psi$ is included in $S$. + +\textbf{(4)} Conversely, explain why given a 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 $S$. + +\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 you find this too +difficult, consider the special case $p=q=1$, and at least try to +explain what needs to be checked.) + + + +% +% +% +\end{document} |