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diff --git a/controle-20220413.tex b/controle-20220413.tex new file mode 100644 index 0000000..699743d --- /dev/null +++ b/controle-20220413.tex @@ -0,0 +1,377 @@ +%% This is a LaTeX document. Hey, Emacs, -*- latex -*- , get it? +\documentclass[12pt,a4paper]{article} +\usepackage[francais]{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}{Tout} +\newcommand\thingy{% +\refstepcounter{comcnt}\smallskip\noindent\textbf{\thecomcnt.} } +\newcommand\exercice{% +\refstepcounter{comcnt}\bigskip\noindent\textbf{Exercice~\thecomcnt.}\par\nobreak} +\renewcommand{\qedsymbol}{\smiley} +% +\newcommand{\id}{\operatorname{id}} +\newcommand{\alg}{\operatorname{alg}} +\newcommand{\ord}{\operatorname{ord}} +\newcommand{\val}{\operatorname{val}} +% +\DeclareUnicodeCharacter{00A0}{~} +% +\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{corrige}% +{\ifcorrige\relax\else\setbox0=\vbox\bgroup\fi% +\smallbreak\noindent{\underbar{\textit{Corrigé.}}\quad}} +{{\hbox{}\nobreak\hfill\checkmark}% +\ifcorrige\par\smallbreak\else\egroup\par\fi} +% +% +% +\begin{document} +\ifcorrige +\title{ACCQ205\\Contrôle de connaissances — Corrigé\\{\normalsize Courbes algébriques}} +\else +\title{ACCQ205\\Contrôle de connaissances\\{\normalsize Courbes algébriques}} +\fi +\author{} +\date{13 avril 2022} +\maketitle + +\pretolerance=8000 +\tolerance=50000 + +\vskip1truein\relax + +\noindent\textbf{Consignes.} + +Les exercices sont totalement indépendants. Ils pourront être traités +dans un ordre quelconque, mais on demande de faire apparaître de façon +très visible dans les copies où commence chaque exercice. + +La longueur du sujet ne doit pas effrayer : l'énoncé du dernier +exercice est long parce que beaucoup de rappels ont été faits et que +la rédaction des questions cherche à donner tous les éléments +nécessaires pour passer d'une question aux suivantes. + +La difficulté des questions étant variée, il vaut mieux ne pas rester +bloqué trop longtemps. + +Si on ne sait pas répondre rigoureusement, une réponse informelle peut +valoir une partie des points. + +\medbreak + +L'usage de tous les documents (notes de cours manuscrites ou +imprimées, feuilles d'exercices, livres) est autorisé. + +L'usage des appareils électroniques est interdit. + +\medbreak + +Durée : 2h + +\ifcorrige +Ce corrigé comporte 9 pages (page de garde incluse). +\else +Cet énoncé comporte 5 pages (page de garde incluse). +\fi + +\vfill +{\noindent\tiny +\immediate\write18{sh ./vc > vcline.tex} +Git: \input{vcline.tex} +\immediate\write18{echo ' (stale)' >> vcline.tex} +\par} + +\pagebreak + + +% +% +% + +\exercice + +Soit $k$ un corps de caractéristique $\neq 2$. Soit $C$ le fermé de +Zariski de $\mathbb{A}^2$ sur $k$ d'équation $x^2 + y^2 = 2$ (ainsi, +pour $k = \mathbb{R}$, les points réels de $C$ forment un cercle +euclidien de rayon $\sqrt{2}$). + +(1) Décrire la complétée projective $C^+$ de $C$ (c'est-à-dire +l'adhérence de $C$ dans $\mathbb{P}^2$ où on identifie comme +d'habitude $\mathbb{A}^2$ à l'ouvert $T\neq 0$ du $\mathbb{P}^2$ de +coordonnées $(T:X:Y)$ en envoyant $(x,y)$ sur $(1:x:y)$). + +(2) En remarquant que $P := (1,1)$ est un $k$-point de $C$ et en +considérant une droite $D_t$ de pente $t$ variable passant par $P$, +construire un morphisme d'un ouvert\footnote{C'est-à-dire qu'il peut + admettre un nombre fini de points (géométriques) où il n'est pas + défini.} de $\mathbb{A}^1$ vers $C$ (défini sur $k$), en envoyant +$t$ sur le point d'intersection autre que $P$ de $C$ avec la +droite $D_t$. + +(3) En déduire un morphisme $\mathbb{P}^1 \to C^+$ (défini sur $k$) en +prolongeant le morphisme de la question précédente. + +(4) Donner un exemple de solution entière $(u,v,w) \in \mathbb{Z}^3$ +de l'équation $u^2 + v^2 = 2w^2$ autre que $(0,0,0)$ et $(\pm 1, \pm +1, \pm 1)$. + + +% +% +% + +\exercice + +Sur un corps $k$ quelconque, considérons l'application $\varphi$ +définie sur une partie de $\mathbb{P}^2$ et à valeurs dans +$\mathbb{P}^2$ qui envoie le point de coordonnées homogènes $(X:Y:Z)$ +sur $(YZ:XZ:XY)$ si défini. + +(1) Quel est l'ouvert de Zariski $U$ de définition de $\varphi$ ? +Exprimer celui-ci comme le complémentaire de trois points de +$\mathbb{P}^2$ dont on précisera les coordonnées. + +(2) Quel est l'ouvert de Zariski $V$ des points (de $U$) dont l'image +par $\varphi$ appartient à $U$ ? Exprimer celui-ci comme le +complémentaire de trois droites de $\mathbb{P}^2$ dont on précisera +les équations. + +(3) Que vaut $\varphi\circ\varphi$ sur $V$ ? + + +% +% +% + +%% \exercice + +%% On définit deux suites de polynômes $(T_n)$ et $(U_n)$ +%% dans $\mathbb{Z}[x]$ (polynômes de Čebyšëv de première et seconde +%% espèce) par les formules de récurrence suivantes : +%% \[ +%% \left\{\begin{aligned} +%% T_0(x) &= 1\\ +%% T_1(x) &= x\\ +%% T_{n+1}(x) &= 2x\, T_n(x) - T_{n-1}(x)\\ +%% \end{aligned}\right. +%% \;\;\;\hbox{~et~}\;\;\; +%% \left\{\begin{aligned} +%% U_{-1}(x) &= 0\\ +%% U_0(x) &= 1\\ +%% U_{n+1}(x) &= 2x\, U_n(x) - U_{n-1}(x)\\ +%% \end{aligned}\right. +%% \] + + +% +% +% + +\exercice + +Soit $k$ un corps de caractéristique $0$ et qu'on supposera +algébriquement clos pour simplifier. Soient $\xi_1,\ldots,\xi_5 \in +k$ deux à deux distincts : on appelle $p(x) = (x-\xi_1)\cdots(x-\xi_5) +\in k[x]$ le polynôme unitaire ayant les $\xi_i$ pour racines. On +appelle $C^+$ la courbe (dite « hyperelliptique ») obtenue en ajoutant +un point à l'infini\footnote{Pour être tout à fait exact, il ne s'agit + pas de la complétée projective de $C$ dans $\mathbb{P}^2$, mais + d'une « désingularisation » de celle-ci (qui a cependant un unique + point en plus de ceux de $C$ comme la complétée projective). Les + questions qui suivent ont été rédigées de manière à ce que cette + subtilité ne pose pas de problème.} noté $\infty$ à la variété +algébrique affine $C$ d'équation $y^2 = p(x)$ dans $\mathbb{A}^2$. + +On admettra sans justification les faits suivants : +\begin{itemize} +\item Que son corps des fonctions $K := k(C^+)$ peut se voir comme le + quotient $k(x)[y]/(y^2 - p(x))$ de l'anneau $k(x)[y]$ des polynômes + en l'indéterminée $y$ sur le corps $k(x)$ des fractions rationnelles + en une indéterminée $x$ sur $k$ par le polynôme $y^2 - p(x)$ + définissant $C$, c'est-à-dire, concrètement : +\item que tout élément de $K$ peut s'écrire de façon unique $g_0 + + g_1\,y$ où $g_0,g_1 \in k(x)$ sont deux fractions rationnelles + en $x$, l'addition se calculant en ajoutant terme à terme, et la + multiplication en développant le produit et en remplaçant $y^2$ par + $p(x)$. +\end{itemize} + +On rappelle par ailleurs qu'une \emph{valuation discrète} sur $K$ +au-dessus de $k$ et une fonction $v\colon K\to +\mathbb{Z}\cup\{+\infty\}$ qui vérifie les propriétés suivantes : +\textbf{(o)} $v(f) = +\infty$ si et seulement si $f=0$,\quad +\textbf{(i)} $v(f_1 + f_2) \geq \min(v(f_1), v(f_2))$ (avec +automatiquement l'égalité lorsque $v(f_1) \neq v(f_2)$),\quad +\textbf{(ii)} $v(f_1 f_2) = v(f_1) + v(f_2)$,\quad \textbf{(k)} $v(c) += 0$ si $c\in k$,\quad et enfin \textbf{(n)} il existe $f\in K$ telle +que $v(f) = 1$. De plus, on rappelle que pour chaque point $P$ +de $C^+$ il existe une unique telle valuation discrète $v =: \ord_P$ +vérifiant en outre \textbf{(r)} $v(f) \geq 0$ si $f$ est régulière +en $P$ (et automatiquement, $v(f) > 0$ si $f$ s'annule en $P$). Et +réciproquement, toute valuation discrète de $K$ au-dessus de $k$ est +de cette forme (est un $\ord_P$ pour un certain $P$). + +\smallbreak + +(1) Si $v$ est une valuation discrète de $K$ au-dessus de $k$, +expliquer pourquoi sa restriction à $k(x)$ vérifie encore les +propriétés (o), (i), (ii) et (k) de la définition d'une valuation +discrète. En déduire qu'elle est de la forme $e\cdot v'$ où $v'$ est +une valuation discrète de $k(x)$ au-dessus de $k$, et où $e\geq 1$ est +entier. + +\smallbreak + +On rappelle que toute valuation discrète de $k(x)$ au-dessus de $k$ +est de la forme $\val_{x_0}$ pour $x_0 \in k$ ou bien $\val_\infty$, +où $\val_{x_0}(g)$ est l'ordre d'annulation\footnote{C'est-à-dire que + $\val_{x_0}(g)$ est l'exposant de la plus grande puissance de + $x-x_0$ qui divise $g$ si $g \in k[x]$, et $\val_{x_0}(g/h) = + \val_{x_0}(g) - \val_{x_0}(h)$ en général.} de la fraction +rationnelle $g$ en $x_0$, et $\val_\infty(g)$ est le degré du +dénominateur moins le degré du numérateur. (NB : c'est seulement pour +éviter la confusion entre valuations sur $K$ et sur $k(x)$ qu'on a +écrit $\ord_P$ pour l'ordre d'annulation d'une fonction sur $C^+$ en +un point $P$ de $C^+$ et $\val_Q$ pour l'ordre d'annulation d'une +fonction sur $\mathbb{P}^1$ en un point $Q$ de $\mathbb{P}^1$. Il +s'agit de la même construction sur deux courbes différentes.) + +\smallbreak + +(2) Soit $P_i$ le point $(\xi_i,0)$ de $C$ (pour $1\leq i\leq 5$ +fixé). On cherche à calculer $\ord_{P_i}(g_0 + g_1\,y)$. Montrer que +$\ord_{P_i}(g) = e\, \val_{\xi_i}(g)$ si $g\in k(x)$, où $e\geq 1$ est +un entier restant à déterminer. En déduire que $\ord_{P_i}(y) = +\frac{e}{2}$. En déduire que $\ord_{P_i}(g_0 + g_1\,y) = +e\,\min(\val_{\xi_i}(g_0),\; \val_{\xi_i}(g_1)+\frac{1}{2})$. En +déduire que $e=2$ exactement, et donc que $\ord_{P_i}(g_0 + g_1\,y) = +\min(2\val_{\xi_i}(g_0),\; 2\val_{\xi_i}(g_1)+1)$. + +\smallbreak + +(3) Soit $\infty$ le point à l'infini de $C^+$ (non situé sur $C$). +On cherche à calculer $\ord_\infty(g_0 + g_1\,y)$ de façon analogue à +la question précédente. Montrer que $\ord_\infty(g) = e\, +\val_\infty(g)$ si $g\in k(x)$, où $e\geq 1$ est un entier restant à +déterminer (\textit{a priori} sans lien avec celui de la question +précédente). En déduire que $\ord_\infty(y) = -\frac{5e}{2}$. En +déduire que $\ord_\infty(g_0 + g_1\,y) = e\,\min(\val_\infty(g_0),\; +\val_\infty(g_1)-\frac{5}{2})$. En déduire que $e=2$ exactement, et +donc que $\ord_\infty(g_0 + g_1\,y) = \min(2\val_\infty(g_0),\; +2\val_\infty(g_1)-5)$. + +\smallbreak + +(4) Soit $Q := (x_Q,y_Q)$ un point de $C$ avec $y_Q \neq 0$ (ou, ce +qui revient au même, $x_Q \not\in \{\xi_1,\ldots,\xi_5\}$) ; on notera +$Q' := (x_Q,-y_Q)$ son symétrique. En quels points de $C^+$ la +fonction $h := x - x_Q$ a-t-elle un zéro ? En utilisant le fait que +$\sum_{Q\in C^+} \ord_Q(h) = 0$, montrer que $\ord_Q(x-x_Q) = +\ord_{Q'}(x-x_Q) = 1$. En déduire que $\ord_Q(g) = \val_{x_Q}(g)$ +pour tout $g \in k(x)$. En déduire que $\ord_Q(y - y_Q) = 1$ (on +pourra remarquer que $y^2 - y_Q^2 = p(x) - p(x_Q)$ et que $p'(x_Q) +\neq 0$). Montrer que si $f := g_0 + g_1\,y \in K$ n'a pas de pôle en +$Q$ ni en $Q'$, alors $g_0,g_1$ n'ont pas de pôle en $x_P$ (on pourra +écrire $g_0 = \frac{1}{2}(f+\tilde f)$ et $g_1 = \frac{1}{2y}(f-\tilde +f)$ où $\tilde f = g_0 - g_1\,y$ est la composée de $f$ par la +symétrie $(x,y) \mapsto (x,-y)$). + +\smallbreak + +(5) Pour $n \in \mathbb{N}$, on s'intéresse à l'espace vectoriel +$\mathscr{L}(n[\infty])$ des fonctions rationnelles $f = g_0 + g_1\,y$ +sur $C^+$ ayant au plus un pôle d'ordre $\leq n$ en $\infty$ +(c'est-à-dire $\ord_\infty(f) \geq -n$) et aucun pôle ailleurs +(c'est-à-dire $\ord_Q(f) \geq 0$ pour tout $Q \in C$). Montrer que +cela équivaut à : $g_0 \in k[x]$ polynôme de degré $\leq\frac{n}{2}$ +et $g_1 \in k[x]$ polynôme de degré $\leq\frac{n-5}{2}$. En déduire +que la dimension $\ell(n[\infty])$ de $\mathscr{L}(n[\infty])$ vaut +$\max(0,\lfloor\frac{n}{2}\rfloor+1) + +\max(0,\lfloor\frac{n-5}{2}\rfloor+1)$ où $\lfloor v\rfloor$ désigne +la partie entière de $v$. En déduire que +\[ +\ell(n[\infty]) = +\left\{ +\begin{array}{ll} +1,1,2&\hbox{~si $n=0,1,2$ respectivement}\\ +n-1&\hbox{~si $n\geq 3$}\\ +\end{array} +\right. +\] +(on pourra par exemple calculer les valeurs pour $n=0,1,2,3,4,5$ +séparément et, pour $n\geq 5$, distinguer $n$ pair et $n$ impair). On +rappelle que le théorème de Riemann-Roch prédit $\ell(n[\infty]) = n + +1 - g$ si $n$ est assez grand, où $g$ est le genre de la courbe : que +vaut $g$ ici ? + +\smallbreak + +Pour la question suivante, on rappelle que la différentielle $df$ +d'une fonction $f$ a pour ordre $\ord_Q(df) = \ord_Q(f) - 1$ si +$\ord_Q(f) \neq 0$, et $\ord_Q(df) \geq 0$ dès que $\ord_Q(f) \geq 0$. +On rappelle par ailleurs que $f \mapsto df$ est $k$-linéaire et que +$d(ff') = f\,df' + f'\,df$. + +\smallbreak + +(6) Calculer $\ord_Q(dx)$ en tout $Q \in C^+$ (y compris $\infty$ et +les cinq points $P_1,\ldots,P_5$) ; on pourra remarquer que $d(x-c) = +dx$. En déduire que le diviseur canonique de $\omega := \frac{dx}{y}$ +vaut $2[\infty]$, c'est-à-dire que $\ord_Q(\omega) = 0$ en tout point +$Q$ sauf $\ord_\infty(\omega) = 2$. Le théorème de Riemann-Roch +prédit plus exactement $\ell(n[\infty]) - \ell((2-n)[\infty])) = n + 1 +- g$ pour tout $n \in \mathbb{Z}$ : vérifier directement cette +affirmation à l'aide du résultat calculé à la question (5). + +\smallbreak + +(7) Aux questions (2) et (3), on a calculé exactement $\ord_Q(f)$ +(pour $f$ quelconque, écrit sous la forme $g_0 + g_1 y$) si $Q$ est +l'un des six points $P_1,\ldots,P_5,\infty$, en calculant séparément +$\ord_Q(g)$ si $g\in k(x)$ et $\ord_Q(y)$. À la question (4), on a +étudié $\ord_Q$ pour un quelconque autre point, on a calculé +$\ord_Q(g)$ et $\ord_Q(y - y_Q)$. Ceci permet-il de calculer +$\ord_Q(f)$ en général ? Si non, donner un exemple de fonction $f \in +K$ dont le calcul ne découle pas de ces valeurs. + + + +% +% +% +\end{document} diff --git a/controle-20230412.tex b/controle-20230412.tex new file mode 100644 index 0000000..65a37c0 --- /dev/null +++ b/controle-20230412.tex @@ -0,0 +1,660 @@ +%% 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{\val}{\operatorname{val}} +% +\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{2023-04-12} +\maketitle + +\pretolerance=8000 +\tolerance=50000 + +\vskip1truein\relax + +\noindent\textbf{Instructions.} + +This exam consists of a single lengthy problem. Although the +questions depend on each other, they have been worded in such a way +that the necessary information for subsequent questions is given in +the text. Thus, failure to answer one question should not make it +impossible to proceed to later questions. + +\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 7 pages (cover page included). +\else +This exam has 4 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 + + +% +% +% + +\textit{The goal of this problem is to study a representation of lines + in $\mathbb{P}^3$.} + +\smallskip + +We fix a field $k$. Recall that \emph{points} in $\mathbb{P}^3(k)$ +are given by quadruplets $(x_0{:}x_1{:}x_2{:}x_3)$ of “homogeneous +coordinates” in $k$, not all zero, defined up to a common +multiplicative constant, and that \emph{planes} in $\mathbb{P}^3(k)$ +are of the form $\{(x_0{:}x_1{:}x_2{:}x_3) \in \mathbb{P}^3(k) : u_0 +x_0 + \cdots + u_3 x_3 = 0\}$ (for some $u_0,\ldots,u_3$, not all +zero, defined up to a common multiplicative constant) which we can +denote as $[u_0{:}u_1{:}u_2{:}u_3]$ (a point of the +“dual” $\mathbb{P}^3$). Our goal is to find a representation for +lines. + +{\footnotesize It may be convenient, if so desired, to call $\langle + w\rangle$ the point in projective space $\mathbb{P}^{m-1}(k)$ + defined by a vector $w\neq 0$ in $k^m$ (i.e., if $w = + (w_0,\ldots,w_m)$ then $\langle w \rangle = (w_0{:}\cdots{:}w_m)$), + that is, the class of $w$ under collinearity. \par} + +\bigskip + +\textbf{(1)} Given $x := (x_0,\ldots,x_3) \in k^4$ and $y := +(y_0,\ldots,y_3) \in k^4$, let us define $x\wedge y := (w_{0,1}, +w_{0,2}, w_{0,3}, w_{1,2}, w_{1,3}, w_{2,3}) \in k^6$ where $w_{i,j} +:= x_i y_j - x_j y_i$. What is $(\lambda x)\wedge(\mu y)$ in relation +to $x\wedge y$? Under what necessary and sufficient condition do we +have $x\wedge y = 0$? What is $x\wedge(\lambda x+\mu y)$ in relation +to $x\wedge y$? + +\begin{answer} +The $w_{i,j}$ are bilinear in $x,y$ (they are $2\times 2$ +determinants) so $(\lambda x)\wedge(\mu y) = \lambda\mu(x\wedge y)$. +Vanishing of $w_{i,j}$ means $(x_i,x_j)$ is proportional to +$(y_i,y_j)$ so vanishing of all the $w_{i,j}$ means precisely that +$x$ or $y$ is zero or that $x$ and $y$ are collinear. Again by +bilinearity, we have $x\wedge(\lambda x+\mu y) = \lambda(x\wedge x) + +\mu(x\wedge y)$ which is just $\mu(x\wedge y)$ since $x\wedge x$ +is $0$. +\end{answer} + +\textbf{(2)} Show that if $V \subseteq k^4$ is a $2$-dimensional +vector subspace, then the set of $x\wedge y$ for $x,y\in V$ is a +$1$-dimensional subspace of $k^6$. + +\begin{answer} +Consider $u,v$ a basis of $V$: then $u\wedge v$ is nonzero, and any +element of $V$ can be written $\lambda u + \mu v$, and then $(\lambda +u + \mu v) \wedge (\lambda' u + \mu' v) = (\lambda \mu' - \lambda' +\mu)(u \wedge v)$ by (1) (or by composition of determinants). In +other words, any $x\wedge y$ with $x,y\in V$ is collinear to $u\wedge +v$, and the latter is nonzero, and since $\lambda \mu' - \lambda' \mu$ +can obviously take every value in $k$, we see that $\{x\wedge y : +x,y\in V\}$ is the line spanned by $u\wedge v$. +\end{answer} + +\textbf{(3)} Deduce from (2) that if $L \subseteq \mathbb{P}^3(k)$ is +a line, then $(w_{0,1}{:}w_{0,2}{:}w_{0,3} {:} \penalty0 +w_{1,2}{:}w_{1,3}{:}w_{2,3})$, where $w_{i,j} := x_i y_j - x_j y_i$ as +above, and where $(x_0{:}x_1{:}x_2{:}x_3)$ and +$(y_0{:}y_1{:}y_2{:}y_3)$ are two distinct points on $L$, is a +well-defined point in $\mathbb{P}^5(k)$, not depending on the chosen +points on $L$ nor on the homogeneous coordinates representing them. + +\begin{answer} +Calling $\langle w\rangle$ the point in projective space +$\mathbb{P}^{m-1}(k)$ defined by a vector $w\neq 0$ in $k^m$, if $L$ +is a line in $\mathbb{P}^3(k)$ we can see it as $\{\langle v\rangle : +v\in V\}$ for a $2$-dimensional vector subspace $V \subseteq k^4$, and +we have seen that $\langle x\wedge y\rangle \in \mathbb{P}^4(k)$ +exists when $x$ and $y$ are not collinear (so that $x\wedge y \neq +0$), i.e., when $\langle x\rangle \neq \langle y\rangle$, and does not +depend on the $x,y \in V$. +\end{answer} + +\bigskip + +The $w_{i,j}$ in question are known as the \textbf{Plücker + coordinates} of $L$. + +\textbf{(4)} Show that any point $(z_0{:}z_1{:}z_2{:}z_3)$ on the +line $L$ as above satisfies +\[ +w_{0,1} z_2 - w_{0,2} z_1 + w_{1,2} z_0 = 0 +\tag{$*$} +\] +— and analogously by replacing $0,1,2$ by three distinct coordinates +in $\{0,1,2,3\}$. + +\begin{answer} +Expanding the $3\times 3$ determinant +\[ +\left| +\begin{matrix} +x_0&y_0&z_0\\ +x_1&y_1&z_1\\ +x_2&y_2&z_2\\ +\end{matrix} +\right| +\] +gives $w_{0,1} z_2 - w_{0,2} z_1 + w_{1,2} z_0$. If $\langle +z\rangle$ is in the line through $\langle x\rangle$ and $\langle +y\rangle$, meaning the three vectors $x,y,z$ are linearly dependent, +then this determinant is zero. The same holds, of course, for any +other choice of coordinates instead of $0,1,2$. +\end{answer} + +\textbf{(5)} Deduce from (4) that the $w_{i,j}$ satisfy the following +relation: +\[ +w_{0,1} w_{2,3} - w_{0,2} w_{1,3} + w_{0,3} w_{1,2} = 0 +\tag{$\dagger$} +\] + +\begin{answer} +By (4), we have $w_{0,1} x_2 - w_{0,2} x_1 + w_{1,2} x_0 = 0$ and +$w_{0,1} y_2 - w_{0,2} y_1 + w_{1,2} y_0 = 0$. Adding $y_3$ times the +first to $-x_3$ times the second gives the stated +relation ($\dagger$). +\end{answer} + +\bigskip + +The projective algebraic variety defined by ($\dagger$) in +$\mathbb{P}^5$ is known as the \textbf{Plücker quadric}. In other +words, we have shown above how to associate to any line $L$ in +$\mathbb{P}^3(k)$ a $k$-point $(w_{0,1}:\cdots:w_{2,3})$ on the +Plücker quadric. We now consider the converse. + +\textbf{(6)} Assuming $(w_{0,1}{:}w_{0,2}{:}w_{0,3} {:} \penalty0 +w_{1,2}{:}w_{1,3}{:}w_{2,3})$ in $\mathbb{P}^5(k)$ +satisfies ($\dagger$) (viꝫ. belongs to the Plücker quadric), and +assuming also that $w_{0,3} \neq 0$, show that the two points +$(w_{0,3}{:}w_{1,3}{:}w_{2,3}{:}0)$ and +$(0{:}w_{0,1}{:}w_{0,2}{:}w_{0,3})$ in $\mathbb{P}^3(k)$ are +meaningful and distinct, and that the line joining them has the +Plücker coordinates $(w_{0,1}:\cdots:w_{2,3})$ that were given. +(\emph{Hint:} \underline{first} compute $(w_{0,3},w_{1,3},w_{2,3},0) +\wedge (0,w_{0,1},w_{0,2},w_{0,3})$ and then use the result, with the +Plücker relation and the fact that $w_{0,3} \neq 0$ to conclude.) + +\begin{answer} +We straightforwardly compute $(w_{0,3},w_{1,3},w_{2,3},0) \wedge +(0,w_{0,1},w_{0,2},w_{0,3}) = (w_{0,3} w_{0,1}, w_{0,3} w_{0,2}, +w_{0,3}^2, \penalty0 w_{1,3} w_{0,2} - w_{2,3} w_{0,1}, w_{1,3} +w_{0,3}, w_{2,3} w_{0,3})$. By Plücker's relation ($\dagger$), +$w_{1,3} w_{0,2} - w_{2,3} w_{0,1} = w_{1,2} w_{0,3}$, so we get +$w_{0,3}$ times $(w_{0,1}, w_{0,2}, w_{0,3}, \penalty0 w_{1,2}, +w_{1,3}, w_{2,3})$. Assuming $w_{0,3} \neq 0$, this is a nonzero +vector, which implies (by question (1)) that +$(w_{0,3},w_{1,3},w_{2,3},0)$ and $(0,w_{0,1},w_{0,2},w_{0,3})$ are +nonzero and non-collinear, so that the two points +$(w_{0,3}{:}w_{1,3}{:}w_{2,3}{:}0)$ and +$(0{:}w_{0,1}{:}w_{0,2}{:}w_{0,3})$ are meaningful and distinct, and +by the computation we have just done, the Plücker coordinates of the +line joining them is the set of coordinates $(w_{0,1}:\cdots:w_{2,3})$ +that were given. +\end{answer} + +\textbf{(7)} Deduce from (6) that any $(w_{0,1}{:}w_{0,2}{:}w_{0,3}{:} +\penalty0 w_{1,2}{:}w_{1,3}{:}w_{2,3})$ in $\mathbb{P}^5(k)$ +satisfying ($\dagger$) (viꝫ. belonging to the Plücker quadric) is the +set of Plücker coordinates of a (clearly unique) line in +$\mathbb{P}^3(k)$. (\emph{Hint:} what needs to be proved is that the +assumption $w_{0,3} \neq 0$ in (6) is harmless: explain how it can be +arranged by a judicious permutation of coordinates.) + +\begin{answer} +We have seen in (6) that when $w_{0,3} \neq 0$ then +$(w_{0,1}{:}w_{0,2}{:}w_{0,3}{:} \penalty0 +w_{1,2}{:}w_{1,3}{:}w_{2,3})$ are the Plücker coordinates of a line +in $\mathbb{P}^3(k)$. But all $w_{i,j}$ cannot be zero (as they are +given in $\mathbb{P}^5$), and we can always permute coordinates in +such a way that any $w_{i,j} \neq 0$, which is sure to exist, becomes +$w_{0,3}$, and the formula $w_{0,1} w_{2,3} - w_{0,2} w_{1,3} + +w_{0,3} w_{1,2} = 0$ is invariant under any permutation of coordinates +(for this is is enough to check a cyclic permutation and a +transposition; keep in mind that $w_{j,i} = -w_{i,j}$ when rewriting +so as $i<j$), so we have confirmed the result in all cases. +\end{answer} + +\bigskip + +At this point, we have established a bijection between the set of +lines $L$ in $\mathbb{P}^3(k)$ and the set of $k$-points in the +Plücker quadric defined by ($\dagger$) in $\mathbb{P}^5$; we know how +to compute Plücker coordinates from two distinct points lying on $L$ +(by definition). We now wish to compute Plücker coordinates for a +line that is described as the the intersection of two planes. + +\textbf{(8)} Rephrase (4) to deduce that, if $L$ is a line with +Plücker coordinates $(w_{0,1}:\cdots:w_{2,3})$, then the planes +$[w_{1,2} : {-w_{0,2}} : w_{0,1} : 0]$ and $[0 : w_{2,3} : {-w_{1,3}} + : w_{1,2}]$ both contain $L$. Now consider these as points in the +dual $\mathbb{P}^3$ and show that the Plücker coordinates of the line +$L^*$ joining the two points in question are: $[w_{2,3} : {-w_{1,3}} : + w_{1,2} : w_{0,3} : {-w_{0,2}} : w_{0,1}]$, provided $w_{1,2} \neq +0$. + +\begin{answer} +The relation ($*$) of (4), namely $w_{1,2} z_0 - w_{0,2} z_1 + w_{0,1} +z_2 = 0$, means precisely that $(z_0{:}z_1{:}z_2{:}z_3)$ is on the +plane $[w_{1,2} : {-w_{0,2}} : w_{0,1} : 0]$. Shifting coordinates +cyclically, it is also on the plane $[0 : w_{2,3} : {-w_{1,3}} : + w_{1,2}]$. Computing the Plücker coordinates as defined in +questions (1) to (3) for the line through these two (dual) points +gives $[w_{1,2} w_{2,3} : {-w_{1,2} w_{1,3}} : w_{1,2}^2 : {w_{0,2} + w_{1,3} - w_{0,1} w_{2,3}} : {-w_{0,2} w_{1,2}} : w_{0,1} + w_{1,2}]$ provided not all are zero. By Plücker's +relation ($\dagger$), $w_{0,2} w_{1,3} - w_{0,1} w_{2,3} = w_{0,3} +w_{1,2}$, and now we can divide all coordinates by $w_{1,2}$ if it is +nonzero, giving $[w_{2,3} : {-w_{1,3}} : w_{1,2} : w_{0,3} : + {-w_{0,2}} : w_{0,1}]$. +\end{answer} + +\textbf{(9)} Deduce from (8) that if $L$ is a line in +$\mathbb{P}^3(k)$ with Plücker coordinates $(w_{0,1} : w_{0,2} : +w_{0,3} : \penalty0 w_{1,2} : w_{1,3} : w_{2,3})$, and if $L^*$ +denotes the “dual” line in the dual $\mathbb{P}^3$, that is, the line +consisting of all points corresponding to planes containing $L$, then +$L^*$ has (dual) Plücker coordinates $[w_{2,3} : {-w_{1,3}} : w_{1,2} + : w_{0,3} : {-w_{0,2}} : w_{0,1}]$. (\emph{Hint:} the only thing +that needs to be proved is that the assumption $w_{1,2} \neq 0$ in (8) +is harmless: explain how it can be arranged by a judicious permutation +of coordinates.) + +\begin{answer} +We have seen in (8) that when $w_{1,2} \neq 0$ then the line $L^*$ +dual to $L$ is given by $[w_{2,3} : {-w_{1,3}} : w_{1,2} : w_{0,3} : + {-w_{0,2}} : w_{0,1}]$. But all $w_{i,j}$ cannot be zero +(question (3)), and we can always permute coordinates in such a way +that any $w_{i,j} \neq 0$, which is sure to exist, becomes $w_{1,2}$, +and the formula $(w_{0,1} : w_{0,2} : w_{0,3} : \penalty0 w_{1,2} : +w_{1,3} : w_{2,3}) \mapsto [w_{2,3} : {-w_{1,3}} : w_{1,2} : w_{0,3} : + {-w_{0,2}} : w_{0,1}]$ is covariant under any permutation of +coordinates (for this is is enough to check a cyclic permutation and a +transposition), so we have confirmed the result in all cases. +\end{answer} + +\textbf{(10)} If $[u_0{:}u_1{:}u_2{:}u_3]$ and +$[v_0{:}v_1{:}v_2{:}v_3]$ are distinct planes in $\mathbb{P}^3(k)$, +how can we compute the Plücker coordinates of their line of +intersection? (\emph{Hint:} first describe the Plücker coordinates of +the line $L^*$ joining the corresponding points in the “dual” +$\mathbb{P}^3$, and then apply the result of (9), together with +projective duality.) + +\begin{answer} +Le line $L^*$ joining the points $[u_0{:}u_1{:}u_2{:}u_3]$ and +$[v_0{:}v_1{:}v_2{:}v_3]$ of the dual $\mathbb{P}^3$ is given by the +Plücker coordinates $w_{i,j} = u_i v_j - u_j v_i$. We then obtain the +Plücker coordinates of $L$ as $(w_{2,3} : {-w_{1,3}} : w_{1,2} : +w_{0,3} : {-w_{0,2}} : w_{0,1})$ (the formula given in (9) for +computing the Plücker coordinates of $L^*$ from those of $L$: it is +involutive and projective duality ensures that it also computes the +Plücker coordinates of $L$ from those of $L^*$), in other words $(u_2 +v_3 - u_3 v_2 : {-u_1 v_3 + u_3 v_1} : u_1 v_2 - u_2 v_1 : u_0 v_3 - +u_3 v_0 : {-u_0 v_2 + u_2 v_0} : u_0 v_1 - u_1 v_0)$ +\end{answer} + +\bigskip + +\textbf{(11)} Explain how, by expanding a $4\times 4$ determinant +expressing the fact that four points $(x_0{:}x_1{:}x_2{:}x_3)$, +$(y_0{:}y_1{:}y_2{:}y_3)$, $(z_0{:}z_1{:}z_2{:}z_3)$ and +$(p_0{:}p_1{:}p_2{:}p_3)$ in $\mathbb{P}^3$ are coplanar, we can +obtain a formula for the plane through a line $L$ defined by its +Plücker coordinates and a point $(z_0{:}z_1{:}z_2{:}z_3)$ not situated +on $L$. (It is not required to go through the full computations: just +explain how it would work.) + +\begin{answer} +Consider the $4\times 4$ determinant +\[ +\left| +\begin{matrix} +x_0&y_0&z_0&p_0\\ +x_1&y_1&z_1&p_1\\ +x_2&y_2&z_2&p_2\\ +x_3&y_3&z_3&p_3\\ +\end{matrix} +\right| +\] +whose vanishing expresses the fact that $x,y,z,p$ are in a common +hyperplane in $k^4$, i.e., that the points $\langle x\rangle$, +$\langle y\rangle$, $\langle z\rangle$ and $\langle p\rangle$ are +coplanar. Expanding it with respect to the final column $p$ gives a +linear condition for $p$ to be in the plane $P$ spanned by $\langle +x\rangle$, $\langle y\rangle$, $\langle z\rangle$, whose coefficients +are $3\times 3$ determinants, which are the coordinates of the +plane $P$ in the dual $\mathbb{P}^3$. Expanding those $3\times 3$ +determinants with respect to the final column $z$ writes them in terms +of the Plücker coordinates of the line $L$ through $\langle x\rangle$, +and $\langle y\rangle$, and the point $\langle z\rangle$. + +To be precise (although this was not asked), we get: +\[ +\begin{aligned} +\relax [\; +& - w_{1,2} z_3 + w_{1,3} z_2 - w_{2,3} z_1 \\ +:\; +& w_{2,3} z_0 + w_{0,2} z_3 - w_{0,3} z_2 \\ +:\; +& w_{0,3} z_1 - w_{1,3} z_0 - w_{0,1} z_3 \\ +:\; +& w_{0,1} z_2 - w_{0,2} z_1 + w_{1,2} z_0 +\;] +\end{aligned} +\] +(the last coordinate being precisely given by ($*$)). +\end{answer} + +For your additional information: if $L$ and $L'$ are two lines in +$\mathbb{P}^3(k)$ with Plücker coordinates $(w_{0,1}:\cdots:w_{2,3})$ +and $(w'_{0,1}:\cdots:w'_{2,3})$ respectively, then $L$ and $L'$ meet +in a common point (or equivalently, belong to a common plane) +iff\footnote{This relation (the “polarization” of the quadratic + relation ($\dagger$)) can be interpreted by saying that the line + in $\mathbb{P}^5$ joining the two points $(w_{0,1}:\cdots:w_{2,3})$ + and $(w'_{0,1}:\cdots:w'_{2,3})$ on the Plücker quadric is entirely + contained in said quadric.} +\[ +w_{0,1} w'_{2,3} - w_{0,2} w'_{1,3} + w_{0,3} w'_{1,2} ++ w_{2,3} w'_{0,1} - w_{1,3} w'_{0,2} + w_{1,2} w'_{0,3} = 0 +\tag{$\ddagger$} +\] +(it is not required to prove this). + +\bigskip + +\textbf{(12)} Briefly summarize all of the above, emphasizing how we +have obtained formulæ allowing algorithmic computation of all possible +geometric constructions between points, lines and planes +in $\mathbb{P}^3$. + +\begin{answer} +For projective subspaces in $\mathbb{P}^3$ we can represent: +\begin{itemize} +\item \textbf{points} by their homogeneous coordinates + $(x_0{:}x_1{:}x_2{:}x_3)$ (which are arbitrary not all zero, defined + up to a common multiplicative constant), +\item \textbf{lines} by their Plücker coordinates + $(w_{0,1} : w_{0,2} : w_{0,3} : \penalty0 + w_{1,2} : w_{1,3} : w_{2,3})$ (which are not all zero, and subject + to the sole condition ($*$) of belonging to the Plücker quadric, + defined up to a common multiplicative constant), or equivalently by + their dual Plücker coordinates which are the same up to a + permutation and some changes of sign, $[w_{2,3} : {-w_{1,3}} : + w_{1,2} : w_{0,3} : {-w_{0,2}} : w_{0,1}]$, +\item \textbf{planes} by their dual coordinates + $[u_0{:}u_1{:}u_2{:}u_3]$ (which are arbitrary not all zero, defined + up to a common multiplicative constant). +\end{itemize} + +We can then compute: +\begin{itemize} +\item whether a point lies on a line by checking the relation ($*$) + and all its permutations of coordinates (cyclic permutations + suffice), +\item whether a point lies on a plane by the relation $u_0 x_0 + + \cdots + u_3 x_3 = 0$, +\item whether a line lies in a plane by checking whether the dual + point of the plane lies on the dual line (this gives $w_{2,3} u_2 + + w_{1,3} u_1 + w_{0,3} u_0 = 0$ and cyclic permutations thereof), +\item the line joining two distinct points by computing the Plücker + coordinates as $2\times 2$ determinants as defined in question (1), +\item the plane though a line and a point not lying on it by the + formula found as explained in question (11), +\item the intersection line of two distinct planes by computing dual + Plücker coordinates as explained in question (10), +\item the point of intersection of a line and a plane by taking the + dual of the formula for the plane through a line and a point, +\item a sample point on a line as $(w_{0,3} : w_{1,3} : w_{2,3} : 0)$ + or some coordinate permutation thereof (as shown in question (6)), +\item a sample plane through a line dually to the previous point, + namely $[w_{1,2} : {-w_{0,2}} : w_{0,1} : 0]$ (as shown in + question (8)), +\item whether two lines meet, or equivalently, belong to a common + plane, by using the criterion stated before (12), in which case + their point of intersection can be computed by intersecting one of + the lines with a sample plane through the other (as explained in the + previous items), and their common plane can be computed dually. +\end{itemize} +\end{answer} + +\bigskip + +\centerline{\hbox to3truecm{\hrulefill}} + +\medskip + +\textbf{(13)} Independently of all of the above, compute the number of +lines in $\mathbb{P}^3(\mathbb{F}_q)$ (for example by counting the +number of pairs of distinct points in $\mathbb{P}^3(\mathbb{F}_q)$ and +the number of pairs of distinct points in a given line +$\mathbb{P}^1(\mathbb{F}_q)$), where $q$ is a prime power and +$\mathbb{F}_q$ denotes the finite field with $q$ elements. + +\begin{answer} +There are $\frac{q^4-1}{q-1} = q^3 + q^2 + q + 1$ points in +$\mathbb{P}^3(\mathbb{F}_q)$. There are thus $(q^3 + q^2 + q + 1)(q^3 ++ q^2 + q) = q (q^3 + q^2 + q + 1) (q^2 + q + 1)$ pairs of distinct +points in $\mathbb{P}^3(\mathbb{F}_q)$, each defining a line. There +are $\frac{q^2-1}{q-1} = q + 1$ points in +$\mathbb{P}^2(\mathbb{F}_q)$. There are thus $q(q + 1)$ pairs of +distinct points defining any given line. Thus, there are $(q (q^3 + +q^2 + q + 1) (q^2 + q + 1)) / (q (q+1)) = (q^2 + q + 1)(q^2 + 1)$ +lines in $\mathbb{P}^3(\mathbb{F}_q)$ (that is, $q^4 + q^3 + 2q^2 + q ++ 1$). +\end{answer} + +\textbf{(14)} Deduce from (13) and (1)–(7) the number of +$\mathbb{F}_q$-points on the hypersurface of degree $2$ (“quadric”) +$\{X_0 X_3 + X_1 X_4 + X_2 X_5 = 0\}$ in $\mathbb{P}^5(\mathbb{F}_q)$ +with coordinates $(X_0:\cdots:X_5)$. Assuming $q \equiv 1 \pmod{4}$, +deduce the number of $\mathbb{F}_q$-points on the hypersurface of +degree $2$ (“quadric”) $\{Z_0^2 + \cdots + Z_5^2 = 0\}$ in +$\mathbb{P}^5(\mathbb{F}_q)$ with coordinates $(Z_0:\cdots:Z_5)$ +(\emph{hint:} $q \equiv 1 \pmod{4}$ means $-1$ is a square in +$\mathbb{F}_q$, so we can factor $Z^2 + Z^{\prime2}$). + +\begin{answer} +We have seen in (1)–(7) that $\mathbb{F}_q$-points on the Plücker +quadric are in bijection with lines in $\mathbb{P}^3(\mathbb{F}_q)$, +and in (13) that there are $(q^2 + q + 1)(q^2 + 1) = q^4 + q^3 + 2q^2 ++ q + 1$ of them. Thus, there are that many points on the Plücker +quadric. The equation of the latter can be written in the form $X_0 +X_3 + X_1 X_4 + X_2 X_5 = 0$ by a simple linear coordinate change, +i.e., projective transformation ($X_0 = w_{0,1}$, $X_1 = w_{0,2}$, +$X_2 = w_{0,3}$, $X_3 = w_{2,3}$, $X_4 = -w_{1,3}$ and $X_5 = +w_{1,2}$) which certainly does not change the number of points. + +As for $Z_0^2 + \cdots + Z_5^2 = 0$, if we call $\sqrt{-1}$ some fixed +square root of $-1$ (which exists when $q \equiv 1 \pmod{4}$ as this +means that the Legendre symbol $(\frac{-1}{q})$ is $1$), we can write +$Z_0^2 + Z_1^2 = (Z_0+\sqrt{-1}\,Z_1) (Z_0-\sqrt{-1}\,Z_1)$ and +similarly for $Z_2^2 + Z_3^2$ and $Z_4^2 + Z_5^2$, and since the +linear transformation $X_0 = Z_0+\sqrt{-1}\,Z_1$, $X_1 = +Z_2+\sqrt{-1}\,Z_3$, $X_2 = Z_4+\sqrt{-1}\,Z_5$, $X_3 = +Z_0-\sqrt{-1}\,Z_1$, $X_4 = Z_2-\sqrt{-1}\,Z_3$, $X_5 = +Z_4-\sqrt{-1}\,Z_5$ is invertible, there are still the same number of +points. +\end{answer} + +\bigskip + +\centerline{\hbox to3truecm{\hrulefill}} + +\medskip + +(This question is more difficult; it is independent of (13)\&(14).) + +\textbf{(15)} Let $h \in k[t_0,t_1,t_2,t_3]$ be a homogeneous +polynomial, so that it defines a Zariski closed set (hypersurface) $X +:= \{h(x_0,x_1,x_2,x_3) = 0\}$ in $\mathbb{P}^3$. Show that the of +lines contained in $X$ defines a Zariski closed subset $Y$ of the +Plücker quadric in $\mathbb{P}^5$. (To be completely clear, this +means\footnote{Here $k^{\alg}$ denotes the algebraic closure of $k$, + but feel free to assume that $k$ is algebraically closed ($k = + k^{\alg}$) in this question.}: there is a Zariski closed set $Y$ in +$\mathbb{P}^5$, defined over $k$ and contained in the Plücker quadric +$Q$ (defined by $\dagger$), such that, for $w \in Q(k^{\alg})$, we +have $w \in Y(k^{\alg})$ if and only if $L_w \subseteq X(k^{\alg})$, +where $L_w$ denotes the line in $\mathbb{P}^3(k^{\alg})$ having +Plücker coordinates $w$.) + +The important part of this question is: how can we compute equations +for $Y$ given the equation $h=0$ of $X$? + +\begin{answer} +We have seen in question (6) that, so long as $w_{0,3} \neq 0$, the +line $L_w$ defined by $(w_{0,1}{:}w_{0,2}{:}w_{0,3} {:} \penalty0 +w_{1,2}{:}w_{1,3}{:}w_{2,3})$ in $Q$ is the line through +$(w_{0,3}{:}w_{1,3}{:}w_{2,3}{:}0)$ and +$(0{:}w_{0,1}{:}w_{0,2}{:}w_{0,3})$. In other words, $\lambda,\mu$ it +is the line consisting of points $(\lambda w_{0,3} : \lambda w_{1,3} + +\mu w_{0,1} : \lambda w_{2,3} + \mu w_{0,2} : \mu w_{0,3})$. This +line is included in $X$ iff $h(\lambda w_{0,3}, \lambda w_{1,3} + \mu +w_{0,1}, \lambda w_{2,3} + \mu w_{0,2}, \mu w_{0,3}) = 0$ for all +$\lambda,\mu$ in $k^{\alg}$, which just means that, seen as a +polynomial in $\lambda,\mu$ now seen as two \emph{indeterminates}, +this is identically zero. But this is a homogeneous polynomial in +$\lambda,\mu$ whose coefficients are homogeneous polynomials in the +$w_{i,j}$, so equating all these coefficients to $0$ gives homogeneous +equations in the $w_{i,j}$ (coordinates in $\mathbb{P}^5$) for $L_w$ +to be included in $X$. This only holds so long as $w_{0,3} \neq 0$ +(when it is zero, some of the resulting equation will be trivial); +however, at least one $w_{i,j}$ must be zero in any case, so writing +the corresponding equations for all permutations of coordinates, +together with the equation ($\dagger$) of the Plücker quadric itself +(to ensure that the $w_{i,j}$ do correspond to a line in +$\mathbb{P}^3$) gives us the equations of the desired $Y$. + +To illustrate that this is actually algorithmic, the following Sage +code computes the equations for the set $Y$ of lines inside the +“diagonal cubic surface” $X := \{x_0^3 + x_1^3 + x_2^3 + x_3^3 = 0\}$: +{\fontsize{8}{10}\relax +\begin{verbatim} +sage: R.<x0,x1,x2,x3,w01,w02,w03,w12,w13,w23,u,v> = PolynomialRing(QQ,12) +sage: xvars = [x0,x1,x2,x3] +sage: wvars = [[0,w01,w02,w03],[-w01,0,w12,w13],[-w02,-w12,0,w23],[-w03,-w13,-w23,0]] +sage: # Plücker equation: +sage: plucker = w01*w23 - w02*w13 + w03*w12 +sage: # Equation of the surface X: +sage: h = x0^3 + x1^3 + x2^3 + x3^3 +sage: deg = h.degree() +sage: # All possible permutations of (0,1,2,3): +sage: perm4 = [(j0,j1,j2,j3) for j0 in range(4) for j1 in range(4) for j2 in range(4) +....: for j3 in range(4) if len(set([j0,j1,j2,j3]))==4] +sage: # Generate the ideal I of the set Y of lines in X, as above: +sage: I = R.ideal([plucker] + [h.subs(dict([(xvars[j0],wvars[j0][j3]*u), (xvars[j1],w +....: vars[j1][j3]*u+wvars[j0][j1]*v), (xvars[j2],wvars[j2][j3]*u+wvars[j0][j2]*v), ( +....: xvars[j3],wvars[j0][j3]*v)])).coefficient({u:deg-k,v:k}) for k in range(deg) fo +....: r (j0,j1,j2,j3) in perm4]) +sage: # Compute its radical: +sage: I0 = I.radical() +sage: # This really means Y is 0-dimensional in projective space: +sage: I0.dimension() +7 +sage: # This computes its number of geometric points (i.e., geometric lines on X): +sage: hp = I0.hilbert_polynomial() ; hp.leading_coefficient()*factorial(hp.degree()) +27 +\end{verbatim} +\par}\noindent (notation is as above except that $\lambda,\mu$ have +been called \texttt{u},\texttt{v}); the above code proves that thare +are $27$ geometric lines in the surface $\{x_0^3 + x_1^3 + x_2^3 + +x_3^3 = 0\} \subseteq \mathbb{P}^3$ (over $\mathbb{Q}$). +\end{answer} + + + + +% +% +% +\end{document} 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} diff --git a/exercices.tex b/exercices.tex index 6d8ff3e..ec90648 100644 --- a/exercices.tex +++ b/exercices.tex @@ -33,6 +33,9 @@ % \newcommand{\id}{\operatorname{id}} \newcommand{\alg}{\operatorname{alg}} +\newcommand{\ord}{\operatorname{ord}} +\newcommand{\val}{\operatorname{val}} +\newcommand{\divis}{\operatorname{div}} % \DeclareUnicodeCharacter{00A0}{~} % @@ -751,4 +754,369 @@ $(A,B;C,D)$ (sur $m$). % % % + +\exercice + +Dans cet exercice, on se place sur un corps $k$ de caractéristique +différente de $2$ et $3$. + +Soit $C := \{ y^2 = x^3 - x \}$ la variété algébrique affine dans +$\mathbb{A}^2$ définie par l'annulation du polynôme $h := y^2 - x^3 + +x \in k[x,y]$. + +(1) Donner l'équation de la complétée projective $C^+$ de $C$ +dans $\mathbb{P}^2$ (c'est-à-dire, l'adhérence de Zariski de $C$ +dans $\mathbb{P}^2$) dont les coordonnées seront notées $(T{:}X{:}Y)$ +(en identifiant le point $(x,y)$ de $\mathbb{A}^2$ avec $(1{:}x{:}y)$ +de $\mathbb{P}^2$). Quels sont ses points « à l'infini » +(c'est-à-dire situés sur $C^+$ mais non sur $C$) ? + +\begin{corrige} +La complétée projective de $Z(h) \subseteq \mathbb{A}^2$ est donnée +par l'équation $T^{\deg h}\, h(\frac{X}{T}, \frac{Y}{T})$, +c'est-à-dire $T Y^2 = X^3 - T^2 X$. Les points à l'infini sont donnés +en mettant $T=0$ (équation de la droite à l'infini) avec cette +équation, ce qui donne $X^3=0$ soit $X=0$, si bien que le seul point +est $(0{:}0{:}1)$ (point à l'infini dans la direction « verticale »). +\end{corrige} + +(2) Montrer que $C^+$ est lisse. + +\begin{corrige} +Si $h^+\in k[T,X,Z]$ est le polynôme homogène $T Y^2 - X^3 + T^2 X$, +il s'agit de vérifier que $\frac{\partial h^+}{\partial T} = Y^2 + 2T +X$ et $\frac{\partial h^+}{\partial X} = -3X^2 + T^2$ et +$\frac{\partial h^+}{\partial Y} = 2 T Y$ n'ont pas de zéro commun +(autre que $T=X=Y=0$ qui ne définit pas un point de $\mathbb{P}^2$) +sur la clôture algébrique $k^{\alg}$ de $k$. Or $2TY=0$ implique soit +$T=0$ soit $Y=0$ (car le corps est de caractéristique $\neq 2$), dans +le premier cas les deux autres équations donnent $Y^2=0$ donc $Y=0$ et +$-3X^2=0$ donc (comme le corps est de caractéristique $\neq 3$) que +$X=0$ ; et dans le second cas, on a $2TX=0$, le cas $T=0$ a déjà été +traité, reste à regarder $X=0$, mais on a alors $T^2=0$ par une autre +équation, donc $T=0$ ; donc dans tous les cas $T=X=Y=0$. + +(On pouvait aussi trouver les relations $X^3 = \frac{1}{6} T \, +\frac{\partial h^+}{\partial T} - \frac{1}{3} X \, \frac{\partial + h^+}{\partial X} - \frac{1}{12} Y \, \frac{\partial h^+}{\partial + Y}$ et $Y^3 = Y \, \frac{\partial h^+}{\partial T} - X \, +\frac{\partial h^+}{\partial Y}$ et $T^4 = \frac{3}{2} T X \, +\frac{\partial h^+}{\partial T} + T^2 \, \frac{\partial h^+}{\partial + X} - \frac{3}{4} X Y \, \frac{\partial h^+}{\partial Y}$ qui +collectivement montrent que $\frac{\partial h^+}{\partial T}$ et +$\frac{\partial h^+}{\partial X}$ et $\frac{\partial h^+}{\partial Y}$ +engendrent un idéal irrelevant puisque contenant $X^3,Y^3,T^4$ (donc +ne peuvent pas toutes s'annuler simultanément dans $\mathbb{P}^2$). +\end{corrige} + +(3) On considère maintenant $h := y^2 - x^3 + x$ comme élément de +$k(x)[y]$, c'est-à-dire comme polynôme en l'indéterminée $y$ sur le +corps $k(x)$ des fractions rationnelles en l'indéterminée $x$. +Montrer qu'il est irréductible (on pourra pour cela vérifier que +l'élément $x^3 - x$ de $k(x)$ n'est pas le carré d'un élément de +$k[x]$ et en déduire qu'il n'est pas le carré d'un élément de $k(x)$). +En déduire que le quotient $K := k(x)[y]/(h)$ est un corps. En +déduire que $K := k(x)[y]/(h)$ est le corps des fonctions rationnelles +aussi bien de $C$ que de $C^+$ (on pourra remarquer que $k(x)[y]/(y^2 +- x^3 + x)$ contient $k[x,y]/(y^2 - x^3 + x)$). + +\begin{corrige} +Remarquons d'abord que $x^3 - x$ n'est pas le carré d'un élément +de $k[x]$ : c'est clair car le carré d'un polynôme sur un corps est de +degré pair. On en déduit que ce n'est pas non plus le carré d'un +élément de $k(x)$, car si on écrivait un tel élément $u/v$ avec $u,v$ +polynômes sans facteur commun (ce qui a un sens car $k[x]$ est un +anneau factoriel — c'est-à-dire qu'il admet une décomposition unique +en éléments irréductibles), son carré serait $u^2/v^2$ avec $u^2,v^2$ +également sans facteur commun, donc on doit avoir $v$ constant pour ne +pas avoir de dénominateur et finalement $u$ est un polynôme. + +Le fait que $x^3 - x$ ne soit pas un carré dans $k(x)$ signifie que $h +:= y^2 - x^3 + x \in k(x)[y]$ n'a pas de racine. Mais il est de +degré $2$, donc sa seule factorisation non-triviale possible serait en +deux facteurs de degré $1$, ce qui implique qu'il aurait deux racines, +et on vient de voir qu'il n'y en a pas. Ainsi, $h$ est irréductible +(en tant qu'élément de $k(x)[y]$). + +Le quotient $K := k(x)[y]/(h)$ est donc un corps car il est de la +forme $K = E[y]/(h)$ avec $E$ un corps et $h$ un polynôme irréductible +en une seule variable sur $E$. (Rappels : $K$ est un anneau intègre +puisque $uv=0$ dans $K$ signifie que $u,v$ relevés à $E[y]$, sont +multiples de $h$, mais comme $h$ est irréductible, l'un des deux doit +être multiple de $h$, donc nul dans $K$ ; et $K$ est alors un anneau +intègre de dimension finie sur $E$, donc un corps car la +multiplication $K \to K, z \mapsto az$ par un élément $a$ non nul est +injective donc bijective car entre espaces vectoriels de même +dimension finie.) + +Comme $C = Z(y^2 - x^3 + x)$ est affine, l'anneau $\mathcal{O}(C)$ des +fonctions \emph{régulières} sur $C$ est $k[x,y] / (y^2 - x^3 + x)$. +Cet anneau est contenu dans $K$ (au sens où le morphisme évident +$\mathcal{O}(C) \to K$, défini en envoyant chacun de $x$ et $y$ sur +l'élément du même nom, et qui passe au quotient par $y^2 - x^3 + x$, +est injectif puisque tout multiple de $y^2 - x^3 + x$ dans $k(x)[y]$ +qui est dans $k[x,y]$ est déjà multiple de $y^2 - x^3 + x$ dans +$k[x,y]$). Puisque le corps $K$ contient $\mathcal{O}(C)$, il +contient son corps des fractions, qui est le corps $k(C)$ des +fonctions rationnelles de $C$ ; mais réciproquement, comme $k(C)$, vu +dans $K$, contient à la fois $x$ et $y$, il doit contenir d'abord le +corps engendré par $x$, soit $k(x)$, et ensuite l'anneau engendré par +$y$ au-dessus de ce corps, qui est justement $K$. + +Ceci montre que $K = k(C)$. Comme $C$ est un ouvert de Zariski (non +vide, donc dense) de $C^+$ (précisément, c'est l'ouvert $T \neq 0$), +ils ont le même corps des fonctions rationnelles, donc $K = k(C^+)$ +aussi. +\end{corrige} + +(4) Expliquer pourquoi tout élément de $K := k(x)[y]/(h)$ possède une +représentation unique sous la forme $g_0 + g_1\, y$ où $g_0$ et $g_1$ +sont des fractions rationnelles en l'indéterminée $x$ (et où on a noté +abusivement $y$ pour la classe de $y$ modulo $h$). Expliquer comment +on calcule les sommes et les produits dans $K$ sur cette écriture. +Expliquer comment la connaissance d'une relation de Bézout $ug + vh = +1 \in k(x)[y]$ permet de calculer l'inverse d'un élément $g = g_0 + +g_1\, y$ de $K$. À titre d'exemple, calculer l'inverse de $y$ +dans $K$ (on pourra observer ce que vaut $y^2$ dans $K$). + +\begin{corrige} +Par division euclidienne dans $E[y]$ où $E = k(x)$ (noter que $E$ est +un corps), tout élément de $E[y]$ s'écrit de façon unique sous la +forme $q h + g$ où $\deg g < \deg h = 2$. C'est-à-dire que tout +élément de $E[y]$ est congru modulo $h$ à un unique élément $g \in +E[y]$ de degré $<2$, qu'on peut alors écrire sous la forme $g_0 + +g_1\, y$ où $g_0,g_1 \in E = k(x)$. + +Pour ajouter deux éléments écrits sous cette forme, on ajoute +simplement les $g_0,g_1$ correspondants. Pour les multiplier, on +effectue le produit dans $E[y]$ et on effectue une division +euclidienne par $h$ pour se ramener à un degré $<2$, ce qui, en +l'espèce, revient simplement à remplacer $y^2$ par $x^3 - x$. + +Une relation de Bézout $ug + vh = 1$ dans $E[y]$ se traduit en $ug = +1$ dans $E[y]/(h) =: K$, ce qui signifie que $u$ est l'inverse de $g$. +Or on sait qu'on peut (par l'algorithme d'Euclide étendu dans $E[y]$) +calculer une telle relation de Bézout dès lors que $g$ et $h$ ont pour +pgcd $1$ (c'est-à-dire que $g$ n'est pas multiple de $h$, i.e., pas +nul dans $K$). À titre d'exemple, comme $y^2 = x^3 - x$ dans $K$, on +a $\frac{1}{y} = \frac{y}{x^3-x}$. +\end{corrige} + +On rappelle le fait suivant : pour chaque point $P$ de la courbe +$C^+$, il existe une et une seule fonction $\ord_P\colon K\to +\mathbb{Z}\cup\{\infty\}$ qui vérifie les propriétés suivantes : +\textbf{(o)} $\ord_P(g) = \infty$ si et seulement si $g=0$,\quad +\textbf{(k)} $\ord_P(c) = 0$ si $c\in k$,\quad +\textbf{(i)} $\ord_P(g_1 + g_2) \geq \min(\ord_P(g_1), \ord_P(g_2))$ +(avec automatiquement l'égalité lorsque $\ord_P(g_1) \neq +\ord_P(g_2)$),\quad \textbf{(ii)} $\ord_P(g_1 g_2) = \ord_P(g_1) + +\ord_P(g_2)$,\quad \textbf{(n)} $1$ est atteint par $\ord_P$, et enfin +\quad \textbf{(r)} $\ord_P(g) \geq 0$ si $g$ est définie en $P$ (avec +automatiquement $\ord_P(g) > 0$ si $g$ s'annule en $P$). + +(5) On va chercher à mieux comprendre la fonction $\ord_O$ lorsque $O$ +est le point $(0,0)$ de la courbe $C$.\quad (a) Posons $e := +\ord_O(x)$ : pourquoi a-t-on $e \geq 1$ ?\quad (b) Cherchons à +comprendre ce que vaut $\ord_O$ sur le sous-corps $k(x)$ de $K$ ne +faisant pas intervenir $y$. Montrer que $\ord_O(g) = e\cdot +\val_0(g)$ si $g \in k[x]$, où $\val_0(g)$ désigne l'ordre du zéro de +$g$ à l'origine en tant que polynôme en une seule variable $x$ +(c'est-à-dire le plus grand $r$ tel que $x^r$ divise $g$). En déduire +que $\ord_O(g) = e\cdot \val_0 (g)$ pour tout $g\in k(x)$, où +$\val_0(g)$ désigne l'ordre du zéro de $g$ à l'origine en tant que +fonction rationnelle en une seule variable $x$ (c'est-à-dire $\val_0$ +de son numérateur moins $\val_0$ de son dénominateur).\quad +(c) Calculer $\ord_O(y^2)$ et en déduire $\ord_O(y)$ (en faisant +intervenir le nombre $e$).\quad (d) En déduire comment calculer +$\ord_O(g_0 + g_1\, y)$ pour $g_0,g_1\in k(x)$ (toujours en faisant +intervenir le nombre $e$).\quad (e) En faisant intervenir la propriété +(n) (de normalisation de $\ord_O$), en déduire la valeur de $e$ et +finalement la valeur de $\ord_O(g_0 + g_1\, y)$ pour $g_0,g_1 \in +k(x)$. + +\begin{corrige} +(a) On a $e := \ord_O(x) \geq 1$ car $x$ s'annule en $O$ (par la + propriété (r)). + +(b) De $\ord_O(x) = e$ on déduit $\ord_O(x^i) = e\cdot i$ par la + propriété (ii), donc $\ord_O(c x^i) = e\cdot i$ si $c\in k^\times$ + par la propriété (k), et donc, par la propriété (i), que $\ord_O(c_r + x^r + \cdots + c_n x^n) = e\cdot r$ si $r\leq n$ et $c_r,\ldots,c_n + \in k$ et $c_r \neq 0$, ce qui signifie précisément $\ord_O(g) = + e\cdot\val_0(g)$ si $g\in k[x]$. Si $g = u/v \in k(x)$ avec $u,v\in + k[x]$, on a $\val_0(g) = \val_0(u) - \val_0(v)$ et $\ord_O(g) = + \ord_O(u) - \ord_O(v)$ (par la propriété (ii)), donc toujours + $\ord_O(g) = e\cdot\val_0(g)$. + +(c) Comme $y^2 = x^3 - x$ dans $K$, on a $\ord_O(y^2) = e\cdot + \val_0(x^3 - x) = e$. On en déduit $\ord_O(y) = \frac{1}{2}e$ + (propriété (ii)). + +(d) On a vu $\ord_O(g_0) = e\cdot\val_0(g_0)$ en (b), et $\ord_O(g_1\, + y) = e\cdot(\val_0(g_1)+\frac{1}{2})$ puisque $\ord_O(y) = + \frac{1}{2}e$. Comme $\val_0(g_0)$ et $\val_0(g_1)+\frac{1}{2}$ ne + peuvent pas être égaux, la propriété (i) donne $\ord_O(g_0 + g_1\, + y) = e\cdot\min(\val_0(g_0), \val_0(g_1)+\frac{1}{2})$ quels que + soient $g_0,g_1\in k(x)$. + +(e) On vient de voir $\ord_O(g_0 + g_1\, y) = e\cdot\min(\val_0(g_0), + \val_0(g_1)+\frac{1}{2})$ : pour que ceci ne prenne que des valeurs + entières, $e$ doit être pair ; mais pour que ceci puisse prendre la + valeur $1$ (donc tous les entiers), $e$ doit être exactement égal + à $2$. Finalement, on a donc $\ord_O(g_0 + g_1\, y) = + \min(2\val_0(g_0), 2\val_0(g_1)+1)$. +\end{corrige} + +(6) En notant $(\infty)$ le point à l'infini de $C^+$, montrer que le +diviseur $\divis(x)$ de $x$ (vu comme fonction rationnelle sur $C^+$) +vaut $2[O] - 2[\infty]$. En déduire $\ord_\infty(y)$, et en déduire +$\divis(y) = [O] + [P] + [Q] - 3[\infty]$ où $P=(1,0)$ et $Q=(-1,0)$. + +\begin{corrige} +À la question (5), on a calculé $\ord_O(x) = 2$. En tout autre point +de $C^+$ non situé à l'infini (c'est-à-dire, situé sur $C$), la +fonction $x$ n'a ni zéro ni pôle (elle n'a pas de pôle car $x$ est une +fonction régulière sur $\mathbb{A}^2$ et notamment sur $C$, et elle +n'a pas de zéro car le seul point de $C$ où $x$ s'annule vérifie +aussi $y=0$ d'après l'équation $y^2 = x^3 - x$, donc est $(0,0) =: +O$). Comme le degré total du diviseur de $x$ doit être $0$, l'ordre +en $\infty$ doit forcément être $-2$, autrement dit $\divis(x) = 2[O] +- 2[\infty]$. + +Comme $\ord_\infty(x) = -2$, on a $\ord_\infty(x^3) = -6$ et +$\ord_\infty(y^2) = \ord_\infty(x^3 - x) = -6$, donc $\ord_\infty(y) = +-3$. Comme la fonction $y$ est régulière sur $\mathbb{A}^2$ et +notamment sur $C$, elle n'a pas d'autre pôle que $\infty$, et elle +s'annule en les points $(x,y)$ de $C$ où $y=0$ et $x^3-x=0$ +c'est-à-dire $x(x-1)(x+1)=0$, qui sont donc $O,P,Q$ (qui sont +distincts car $k$ est de caractéristique $\neq 2$). Comme le degré +total de $\divis(y)$ doit être $0$, l'ordre en chacun de $O,P,Q$ doit +forcément être $1$ (puisque ce sont trois entiers strictement positifs +de somme $3$), autrement dit $\divis(y) = [O] + [P] + [Q] - +3[\infty]$. +\end{corrige} + +(7) Toujours en notant $(\infty)$ le point à l'infini de $C^+$, +montrer que $\ord_\infty(g_0 + g_1\, y) = \min(2\val_\infty(g_0), +2\val_\infty(g_1)-3)$ pour $g_0,g_1 \in k(x)$, où $\val_\infty(g)$ +désigne la valuation usuelle de $g$ à l'infini en tant que fonction +rationnelle en une seule variable $x$, c'est-à-dire le degré du +dénominateur moins le degré du numérateur. + +\begin{corrige} +On a vu $\ord_\infty(x) = -2 = 2\val_\infty(x)$ : on en déduit que +$\ord_\infty(g) = 2\val_\infty(g)$ pour tout $g\in k[x]$ puis pour +tout $g\in k(x)$ (comme en (5)(b)). Comme $\ord_\infty(y) = -3$, on a +$\ord_\infty(g_0 + g_1\, y) = \min(2\val_\infty(g_0), +2\val_\infty(g_1)-3)$ (en utilisant la propriété (i) et en remarquant +que $2\val_\infty(g_0)$ et $2\val_\infty(g_1)-3$ ne peuvent pas être +égaux). +\end{corrige} + +(8) Pour $n\in \mathbb{N}$, soit $\mathscr{L}(n[\infty]) := \{0\} \cup +\{f\in k(C)^\times : \divis(f) +n[\infty] \geq 0\}$ l'espace vectoriel +sur $k$ des fonctions rationnelles sur $C^+$ ayant au pire un pôle +d'ordre $n$ en $\infty$ (et aucun pôle ailleurs). Décrire +explicitement $\mathscr{L}(n[\infty])$ comme l'ensemble des $g_0 + +g_1\,y$ avec $g_0,g_1\in k[x]$ vérifiant certaines contraintes sur +leur degré : en déduire la valeur de $\ell (n[\infty]) := \dim_k +\mathscr{L}(n[\infty])$ et notamment le fait que $\ell (n[\infty]) = +n$ si $n$ est assez grand. + +\begin{corrige} +Dire que $g_0 + g_1\, y$ appartient à $\mathscr{L}(n[\infty])$ +signifie qu'elle est régulière partout sauf peut-être en $\infty$, et +que par ailleurs $\ord_\infty(g_0 + g_1\, y) \geq -n$. La première +condition signifie $g_0,g_1\in k[x]$ ; la seconde signifie +$\min(2\val_\infty(g_0), 2\val_\infty(g_1)-3) \geq -n$, c'est-à-dire +$\max(2\deg(g_0), 2\deg(g_1)+3) \leq n$, autrement dit, +$\mathscr{L}(n[\infty])$ est l'ensemble des $g_0 + g_1\,y$ avec +$g_0,g_1\in k[x]$ vérifiant $\deg(g_0) \leq \frac{1}{2}n$ et +$\deg(g_1) \leq \frac{1}{2}(n-3)$. Comme la dimension de l'espace des +polynômes vérifiant $\deg(g) \leq B$ pour $B\in \mathbb{R}$ vaut +$\max(0,\lfloor B\rfloor + 1)$, on trouve finalement $\ell (n[\infty]) += \max(0,\lfloor \frac{1}{2}n\rfloor + 1) + \max(0,\lfloor +\frac{1}{2}(n-3)\rfloor + 1)$, ce qui, après examen des quelques +premiers cas, donne +\[ +\ell (n[\infty]) = \left\{ +\begin{array}{ll} +0&\hbox{~si $n<0$}\\ +1&\hbox{~si $n=0$ (ou $n=1$)}\\ +n&\hbox{~si $n\geq 1$}\\ +\end{array} +\right. +\] +pour dimension de l'espace des fonctions rationnelles sur $C^+$ ayant +au pire un pôle d'ordre $n$ en $\infty$ (et aucun pôle ailleurs). +\end{corrige} + +(9) En comparant la valeur trouvée pour $\ell (n[\infty])$ avec la +formule de Riemann-Roch, calculer le genre de $C^+$. + +\begin{corrige} +La formule de Riemann-Roch prédit $\ell(D) - \ell(K-D) = \deg(D) + 1 - +g$ où $g$ est le genre de la courbe (à déterminer), et notamment +$\ell(D) = \deg(D) + 1 - g$ si $\deg(D) > 2g-2$. Or on vient de voir +que pour $D := n[\infty]$ avec $n$ assez grand (et notamment +$\deg(n[\infty]) = n$ plus grand que ce qu'on voudra), on a +$\ell(n[\infty]) = n = \deg(n[\infty])$. On en déduit $g = 1$. +\end{corrige} + +(10) Montrer que $\omega := \frac{dx}{y} = \frac{2\,dy}{3x^2-1} \in +\Omega^1(C)$ et montrer que cette différentielle est d'ordre $0$ +en $\infty$. Expliquer pourquoi $dx$ et $dy$ sont régulières (i.e., +d'ordre $\geq 0$) sur $C$ et ne peuvent pas s'annuler (i.e., avoir un +ordre $>0$) simultanément en un point de $C$. (Pour la seconde +affirmation, on pourra noter que si $f$ est régulière sur $C$ alors +$df = f'_x\,dx + f'_y\,dy$ avec $f'_x,f'_y$ régulières sur $C$, et +qu'il n'est pas possible que tous les $df$ s'annulent en $M$.) En +remarquant que $y$ et $3x^2-1$ ne s'annulent jamais simultanément +sur $C$, montrer que $\omega$ est d'ordre $0$ en tout point (elle n'a +ni zéro ni pôle), i.e., $\divis(\omega) = 0$. En déduire un calcul du +genre de $g$ indépendant des questions (8) et (9). + +\begin{corrige} +En dérivant $y^2 = x^3 - x$ on trouve $2y\,dy = (3x^2-1)\,dx$ dans le +$k(C)$-espace vectoriel $\Omega^1(C)$, d'où $\frac{dx}{y} = +\frac{2\,dy}{3x^2-1}$ comme annoncé. + +En $\infty$, on a $\ord_\infty(dx) = \ord_\infty(x) - 1 = -3$ (car +$\ord_\infty(x) = -2 \neq 0$), et $\ord_\infty(y) = -3$, donc +$\ord_\infty(\omega) = 0$. + +En tout point (géométrique) $M$ de $C$ on a $\ord_M(x) \geq 0$ donc +$\ord_M(dx) \geq 0$, et $\ord_M(y) \geq 0$ donc $\ord_M(dy) \geq 0$. +Par ailleurs, si $f$ est une fonction rationnelle sur $C$, on a $df = +f'_x\,dx + f'_y\,dy$ où $f'_x,f'_y$ sont les dérivées partielles (au +sens usuel) de $f$ par rapport à $x,y$ respectivement (si $f$ est +écrite, disons, de la forme $g_0 + g_1\,y$ avec $g_0,g_1\in k(x)$, +alors $f'_x = g'_0 + g'_1\,y$ et $f'_y = g_1$) ; en choisissant une +fonction régulière sur $C$ (donc dans $k[x,y]/(y^2-x^3+x)$) qui est +une uniformisante en $M$, on a $f'_x,f'_y$ régulières et $\ord_M(df) = +0$, ce qui interdit qu'on ait simultanément $\ord_M(dx) > 0$ et +$\ord_M(dy) > 0$. + +En tout point (géométrique) $M$ de $C$ où $y$ ne s'annule pas, on a +$\ord_M(\omega) \geq 0$ car $\ord_M(dx) \geq 0$ et $\ord_M(y) = 0$ ; +de même, en tout point où $3x^2-1$ ne s'annule pas, on a +$\ord_M(\omega) \geq 0$ car $\ord_M(dy) \geq 0$ et $\ord_M(3x^2-1) = +0$. Comme $y$ et $3x^2-1$ ne s'annulent jamais simultanément sur $C$ +(car $y$ ne s'annule qu'en $O=(0,0)$, $P=(1,0)$ et $Q=(-1,0)$, et +$3x^2-1$ n'est nul en aucun de ces points), ceci montre +$\ord_M(\omega) \geq 0$ en tout $M$. Mais si on avait $\ord_M(\omega) +> 0$ en un point $M$, les écritures $dx = y\,\omega$ et $dy = +\frac{3x^2-1}{2}\,\omega$ montreraient que $dx,dy$ s'annulent toutes +les deux en $M$, ce qui n'est pas possible. Donc $\ord_M(\omega) = 0$ +en tout $M$ de $C$, et on a déjà montré par ailleurs que +$\ord_\infty(\omega) = 0$. Ceci prouve $\divis(\omega) = 0$. + +Comme $\deg(\divis(\omega)) = 2g-2$ pour n'importe quelle $\omega \in +\Omega^1(C)$, ceci montre $2g-2 = 0$ soit $g=1$. +\end{corrige} + + +% +% +% \end{document} |