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+%% This is a LaTeX document.  Hey, Emacs, -*- latex -*- , get it?
+\documentclass[12pt,a4paper]{article}
+\usepackage[a4paper,margin=2.5cm]{geometry}
+\usepackage[english]{babel}
+\usepackage[utf8]{inputenc}
+\usepackage[T1]{fontenc}
+%\usepackage{ucs}
+\usepackage{times}
+% A tribute to the worthy AMS:
+\usepackage{amsmath}
+\usepackage{amsfonts}
+\usepackage{amssymb}
+\usepackage{amsthm}
+%
+\usepackage{mathrsfs}
+\usepackage{wasysym}
+\usepackage{url}
+%
+\usepackage{graphics}
+\usepackage[usenames,dvipsnames]{xcolor}
+\usepackage{tikz}
+\usetikzlibrary{matrix,calc}
+\usepackage{hyperref}
+%
+%\externaldocument{notes-accq205}[notes-accq205.pdf]
+%
+\theoremstyle{definition}
+\newtheorem{comcnt}{Whatever}
+\newcommand\thingy{%
+\refstepcounter{comcnt}\smallskip\noindent\textbf{\thecomcnt.} }
+\newcommand\exercise{%
+\refstepcounter{comcnt}\bigskip\noindent\textbf{Exercise~\thecomcnt.}\par\nobreak}
+\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}