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+%% 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}