%% 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}