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diff --git a/controle-20260415.tex b/controle-20260415.tex new file mode 100644 index 0000000..99d9223 --- /dev/null +++ b/controle-20260415.tex @@ -0,0 +1,342 @@ +%% 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} +\let\exercice=\exercise +\renewcommand{\qedsymbol}{\smiley} +\renewcommand{\thefootnote}{\fnsymbol{footnote}} +% +\newcommand{\id}{\operatorname{id}} +\newcommand{\alg}{\operatorname{alg}} +\newcommand{\ord}{\operatorname{ord}} +\newcommand{\norm}{\operatorname{N}} +% +\DeclareUnicodeCharacter{00A0}{~} +\DeclareUnicodeCharacter{A76B}{z} +% +\DeclareMathSymbol{\tiret}{\mathord}{operators}{"7C} +\DeclareMathSymbol{\traitdunion}{\mathord}{operators}{"2D} +% +\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{FMA-4AC05-TP / ACCQ205\\Final exam — answer key\\{\normalsize Algebraic curves}} +\else +\title{FMA-4AC05-TP / ACCQ205\\Final exam\\{\normalsize Algebraic curves}} +\fi +\author{} +\date{2026-04-15} +\maketitle + +\pretolerance=8000 +\tolerance=50000 + +\vskip1truein\relax + +\noindent\textbf{Instructions.} + +This exam consists of \textcolor{red}{XXX} 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 permitted. + +Use of electronic devices of any kind is prohibited. + +\medbreak + +Duration: 2 hours + +\ifcorrige +This answer key has \textcolor{red}{XXX} pages (this cover page included). +\else +This exam has \textcolor{red}{XXX} 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 seven distinct points $p_1,\ldots,p_7$ in the +projective plane $\mathbb{P}^2$ over a field $k$ is a \textbf{Fano + configuration} when the points satisfy the alignment conditions +depicted in the following figure: + +\begin{center} +\vskip-2ex\leavevmode +\begin{tikzpicture} +\coordinate (P1) at (-2cm,0); +\coordinate (P2) at (2cm,0); +\coordinate (P3) at (0,0); +\coordinate (P4) at (0,3.464cm); +\coordinate (P5) at (-1cm,1.732cm); +\coordinate (P6) at (1cm,1.732cm); +\coordinate (P7) at (0cm,1.155cm); +\draw (P2)--(P4); +\draw (P1)--(P4); +\draw (P3)--(P4); +\draw (P1)--(P2); +\draw (P2)--(P5); +\draw (P1)--(P6); +\draw (P3) to[out=180,in=240] (P5) to[out=60,in=120] (P6) to[out=300,in=0] (P3); +\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=north east] at (P1) {$p_1$}; +\node[anchor=north west] at (P2) {$p_2$}; +\node[anchor=north] at (P3) {$p_3$}; +\node[anchor=south] at (P4) {$p_4$}; +\node[anchor=south east] at (P5) {$p_5$}; +\node[anchor=south west] at (P6) {$p_6$}; +\node[anchor=north west] at (P7) {$p_7$}; +\end{tikzpicture} +\vskip-5ex\leavevmode +\end{center} + +This means: the seven points are distinct; all the following sets of +points are aligned: $\{p_2, p_4, p_6\}$, $\{p_1, p_4, p_5\}$, $\{p_3, +p_4, p_7\}$, $\{p_1, p_2, p_3\}$, $\{p_2, p_5, p_7\}$, $\{p_1, p_6, +p_7\}$ and $\{p_3, p_5, p_6\}$; and no other set of three of the $p_i$ +are aligned. + +The goal of this exercise is to determine over which fields $k$ a Fano +configuration exists, and compute the coordinates of its points. + +We fix an arbitrary 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 denote by $(x{:}y{:}z)$ the (homogeneous) coordinates of a +point, and write $[u{:}v{:}w]$ for 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 to compute coordinates, +%% however!) + +\emph{We assume for questions (1)–(3) below that $p_1,\ldots,p_7$ is a +Fano 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 $p_4=(1{:}0{:}0)$ and $p_2=(0{:}1{:}0)$ and $p_1=(0{:}0{:}1)$ and +$p_7=(1{:}1{:}1)$. \emph{We shall now do so.} + +\begin{answer} +No three of the four points $p_4,p_2,p_1,p_7$ 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 $p_4=(1{:}0{:}0)$ and $p_2=(0{:}1{:}0)$ +and $p_1=(0{:}0{:}1)$ and $p_7=(1{:}1{:}1)$. +\end{answer} + +\textbf{(2)} Compute the coordinates (i.e., equations) of the lines +$\ell_{123}$ and $\ell_{347}$, and deduce the coordinates of the point +$p_3$. Analogously compute the coordiantes of $p_5$ and $p_6$. + +\begin{answer} +Denoting $p\vee q$ the line through distinct points $p$ and $q$, we +get $\ell_{123} = p_1 \vee p_2 = [1{:}0{:}0]$ and $\ell_{347} = +p_4\vee p_7 = [0{:}-1{:}1]$. Denoting by $\ell\wedge m$ the point of +intersection of distinct lines $\ell$ and $m$, we get $p_3 = +\ell_{123} \wedge \ell_{347} = (0{:}1{:}1)$. + +Similar computations (or using the cyclic symmetry $p_1 \mapsto p_2 +\mapsto p_4 \mapsto p_1$ which corresponds to a cyclic permutation of +coordinates) gives $p_5 = (1{:}0{:}1)$ and $p_6 = (1{:}1{:}0)$. +\end{answer} + +\textbf{(3)} Using the last alignment condition that hasn't yet been +used, give a \emph{necessary} condition for a Fano configuration to +exist in $\mathbb{P}^2(k)$. + +\begin{answer} +The remaining condition is the alignment of $p_3,p_5,p_6$. This is +expressed by the vanishing of the determinant of their coordinates, +or, equivalently, by computing $p_3 \vee p_5 = [1{:}1{:}-1]$ and +expressing the fact that $p_6$ lies on it. The necessary condition we +get is: $2=0$ in $k$, in other words, the field $k$ is of +characteristic $2$. + +Thus, we have shown that a Fano configuration does not exist in a +field of characteristic $\neq 2$. +\end{answer} + +\textbf{(4)} Conversely, use the previously computed coordinates to +explain why this necessary condition on $k$ is also sufficient for a +Fano configuration to exist. + +\begin{answer} +If $k$ is any field, then setting $p_4=(1{:}0{:}0)$ and +$p_2=(0{:}1{:}0)$ and $p_1=(0{:}0{:}1)$ and $p_7=(1{:}1{:}1)$ and $p_3 += (0{:}1{:}1)$ and $p_5 = (1{:}0{:}1)$ and $p_6 = (1{:}1{:}0)$ ensures +six of the seven required alignments. And if $k$ is of +characteristic $2$ then $p_3,p_5,p_6$ are also aligned for the reasons +explained in the previous question. But furthermore, this gives an +identification of the $7$ points with the points of +$\mathbb{P}^2(\mathbb{F}_2)$ (where $\mathbb{F}_2$ is seen as a +subfield of $k$), and since $\mathbb{P}^2(\mathbb{F}_2)$ has +$7$ lines, there are no other alignments than the prescribed ones. +\end{answer} + +\medskip + +\textbf{(5)} \emph{Independently of all previous questions,} show that +the number of labeled projective bases in $\mathbb{P}^2(\mathbb{F}_q)$ +(in other words, $4$-uples $(a,b,c,d)$ of points such that no $3$ are +aligned) is given by the formula: $q^3\, (q-1)^2\, (q+1)\, (q^2+q+1)$. + +\emph{Hint:} One possible approach is to count the number of possible +choices for point $a$, then $b$, then $c$, then $d$; another possible +approach is to count matrices in $\mathit{GL}_3(\mathbb{F}_q)$ by +counting the possibilities for the first, then second, then third +columns, and deduce the cardinality of $\mathit{PGL}_3(\mathbb{F}_q)$. +Both approaches give the same formula (although in a slightly +different way). + +\begin{answer} +First aproach: there are $q^2+q+1$ possibilities for the point $a$, +because that is the cardinality of $\mathbb{P}^2(\mathbb{F}_q)$. For +the point $b$, since it needs to be different from $a$, we are left +with $q^2+q$ possibilities. For the point $c$, since it cannot belong +to the line $ab$, which has $q+1$ points, we are left with $q^2$ +possibilities. Finally, for the last point $d$, there are three lines +to be ruled out ($ab$, $ac$ and $bc$), each one having $q+1$ points, +but as they meet pairwise in a single point, they have $3(q+1)-3 = 3q$ +point together, and we are left with $(q^2+q+1)-3q = q^2-2q+1 = +(q-1)^2$ possibilities for $d$. This means there are $(q^2+q+1)\, +(q^2+q)\, q^2\, (q-1)^2 = q^3\, (q-1)^2\, (q+1)\, (q^2+q+1)$ labeled +projective bases. + +Second approach: to construct a matrix in +$\mathit{GL}_3(\mathbb{F}_q)$, we first choose its first column, which +can be any nonzero vector, giving us $q^3-1$ possibilities; then we +choose its second, which can be any vector not collinear with the +first, giving us $q^3-q$ possibilities; then we choose the third, +which can be any vector not in the vector space spanned by the first +two, leaving us $q^3-q^2$ possibilities. Thus, there are +$(q^3-1)\,(q^3-q)\,(q^3-q^2)$ elements in +$\mathit{GL}_3(\mathbb{F}_q)$. But since +$\mathit{PGL}_3(\mathbb{F}_q)$ is its quotient by the subgroup of +order $q-1$ consisting of homotheties (multiplication by a nonzero +constant), there are $\frac{(q^3-1)\,(q^3-q)\,(q^3-q^2)}{q-1} = +(q^2+q+1)\,(q^3-q)\,(q^3-q^2) = q^3\, (q-1)^2\, (q+1)\, (q^2+q+1)$ +elements of $\mathit{PGL}_3(\mathbb{F}_q)$. This is also the number +of labeled projective bases because $\mathit{PGL}_3(\mathbb{F}_q)$ +acts simply transitively on such. +\end{answer} + +\medskip + +Let us now say that a \textbf{labeled Fano configuration}\footnote{In +French: “configuration de Fano étiquetée”.} is a $7$-tuple of points +$(p_1,\ldots,p_7)$ satisfying the same conditions as previously. (In +other words, the difference is that the unlabeled Fano configuration +is just the set $\{p_1,\ldots,p_7\}$ of seven points, whereas the +labeled Fano configuration has the information of which is $p_1$, +which is $p_2$, etc.) + +\smallskip + +\textbf{(6)} How many labeled Fano configurations are there in +$\mathbb{P}^2(\mathbb{F}_{2^d})$? Compute this number of $d=1$ and +$d=2$ (that is, in $\mathbb{P}^2(\mathbb{F}_2)$ and +$\mathbb{P}^2(\mathbb{F}_4)$). + +\emph{Note:} You can write numbers as products, there is no need to +fully compute the multiplications by hand. + +\begin{answer} +We have seen in questions (1)–(4) that, over a field of +characteristic $2$, a labeled Fano configuration is constructed in a +unique way from a labeled projective basis (which serves as +$p_4,p_2,p_1,p_7$). Thus, the number of Fano configurations in +$\mathbb{P}^2(\mathbb{F}_{2^d})$ equals the number of labeled +projective bases, which equals $2^{3d} (2^d-1)^2 (2^d+1) +(2^{2d}+2^d+1)$. For $d=1$ this gives $8\times 3\times 7 = 168$; and +for $d=2$ this gives $64\times 9 \times 5 \times 21 = 60\,480$. +\end{answer} + +\textbf{(7)} Deduce the number of \emph{unlabeled} Fano configurations +in $\mathbb{P}^2(\mathbb{F}_4)$. (\emph{Hint:} The previous question +provides a way to count the number of labeled Fano configurations for +each unlabeled one!) + +\begin{answer} +The number of ways to label a given Fano configuration, i.e., the +number of labeled Fano configurations for each unlabeled one, equals +the number of Fano configurations in $\mathbb{P}^2(\mathbb{F}_2)$, +which we have seen is $168$. So we are left with $60\,480 \, / \, 168 += 360$ (unlabeled) Fano configurations in $\mathbb{P}^2(\mathbb{F}_4)$. +\end{answer} + +% +% +% +\end{document} |