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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}