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diff --git a/controle-20260415.tex b/controle-20260415.tex new file mode 100644 index 0000000..49a7c5a --- /dev/null +++ b/controle-20260415.tex @@ -0,0 +1,811 @@ +%% 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 five 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. + +\medbreak + +Indicative and approximate grading scheme: 1 point per question, for a +final score out of 20 (so it will not be necessary to answer all +questions to get a perfect 20/20 score). + +\medbreak + +\ifcorrige +This answer key has 9 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 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 $k$ (or “$k$-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 until question (4).} + +\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 coordinates 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 elements of $\mathit{GL}_3(\mathbb{F}_q)$ 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 approach: 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 \textbf{unlabeled} Fano +configuration is just the set $\{p_1,\ldots,p_7\}$ of seven points, +whereas the labeled Fano configuration is the tuple: it 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 for $d=1$ 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} + +% +% +% + +\exercise + +In this exercise, we consider the affine plane $\mathbb{A}^2$ with +coordinates $(x,y)$ as a subset of the projective plane $\mathbb{P}^2$ +with coordinates $(T{:}X{:}Y)$ by $(x,y) \mapsto (1{:}x{:}y)$. We +work over the field $\mathbb{R}$ of real numbers, but we will also +consider some complex points (i.e., $\mathbb{C}$-points). + +\textit{Definitions:} A \textbf{translation} of $\mathbb{A}^2$ is a +map $(x,y) \mapsto (x,y) + (a,b)$ for certain (fixed) $(a,b) \in +\mathbb{R}^2$. A \textbf{vector homothety} is a map $(x,y) \mapsto (c +x, cy)$ for certain (fixed) $c \in \mathbb{R}^\times := +\mathbb{R}\setminus\{0\}$. A \textbf{vector rotation} is a map $(x,y) +\mapsto (ux + vy,\, -vx + uy)$ for certain (fixed) $(u,v) \in +\mathbb{R}^2$ satisfying $u^2+v^2 = 1$. An \textbf{affine similitude} +is an element of the group generated by translations, vector +homotheties and vector rotations (this is a subgroup of the group of +all affine transformations). + +\textbf{(1)} Describe the matrices, in $\mathit{PGL}_3(\mathbb{R})$ of +the extensions to $\mathbb{P}^2$ of the three kinds of transformations +we just defined (translations, vector homotheties, and vector +rotations). + +\begin{answer} +The matrices in question are, with the same notations as in the +definition: +\[ +\begin{pmatrix}1&0&0\\a&1&0\\b&0&1\\\end{pmatrix}\quad,\quad +\begin{pmatrix}1&0&0\\0&c&0\\0&0&c\\\end{pmatrix}\quad,\quad +\begin{pmatrix}1&0&0\\0&u&v\\0&-v&u\\\end{pmatrix} +\] +(each one being defined, of course, only up to multiplication by a +constant). +\end{answer} + +\medskip + +We define the complex points $I := (0 : 1 : \sqrt{-1})$ and $J := (0 : +1 : -\sqrt{-1})$ in $\mathbb{P}^2(\mathbb{C})$, also known as the +\textbf{cyclic points}. + +\textbf{(2)} Show that $I$ and $J$ are fixed under every affine +similitude (extended to $\mathbb{P}^2$). + +\begin{answer} +We compute the product of the matrices found in (1) with the column +vectors giving the coordinates of $I$ and $J$. Translations and +homotheties fix $I$ and $J$ simply because they are on the line at +infinity ($T=0$). For a rotation acting on $I$ we find, with the same +notations as previously, $(0 : u + v \sqrt{-1} : -v + u \sqrt{-1})$, +and we observe that these are the coordinates of $I$ multiplied by $u ++ v \sqrt{-1}$, so it is the same point; the same argument works +for $J$. +\end{answer} + +\textbf{(3)} Conversely, show that every real projective +transformation of $\mathbb{P}^2$ (i.e., every element of +$\mathit{PGL}_3(\mathbb{R})$) which fixes $I$ and $J$ is an affine +similitude. (\emph{Hint:} You may want to first observe that it +stabilizes the line $\ell_\infty := IJ$ and conclude that it is an +affine transformation.) + +\begin{answer} +The line $IJ$ is the line at infinity $\{T=0\}$. A real projective +transformation fixing $I$ and $J$ must stabilize the line through them +(because projective transformations preserve alignment). This means +that it is, in fact, an affine transformation, or equivalently, given +by a matrix of the form: +\[ +\begin{pmatrix}1&0&0\\a&m&n\\b&p&q\\\end{pmatrix} +\] +(with $m,n,p,q\in\mathbb{R}$). Now fixing $I$ imposes the condition +that $(0 : m + n \sqrt{-1} : p + q \sqrt{-1})$ is $(0 : 1 : +\sqrt{-1})$, which means that $p + q \sqrt{-1} = \sqrt{-1}\,(m + n +\sqrt{-1}) = -n + m \sqrt{-1}$, and by identifying real and imaginary +parts we conclude $p = -n$ and $q = m$. So our matrix is now of the +form: +\[ +\begin{pmatrix}1&0&0\\a&m&n\\b&-n&m\\\end{pmatrix} +\] +and if we let $c = \sqrt{m^2+n^2}$ (a real number) and $u := m/c$ and +$v = n/c$ (which satisfy $u^2 + v^2 = 1$), the above matrix is the +product (i.e., the composition) +\[ +\begin{pmatrix}1&0&0\\a&1&0\\b&0&1\\\end{pmatrix}\, +\begin{pmatrix}1&0&0\\0&c&0\\0&0&c\\\end{pmatrix}\, +\begin{pmatrix}1&0&0\\0&u&v\\0&-v&u\\\end{pmatrix} +\] +as required. +\end{answer} + +\textbf{(4)} In this question, let $A := (0,0)$ and $B := (1,0)$ in +$\mathbb{A}^2$. Compute the equation of the line $(AI\wedge BJ) \vee +(AJ\wedge BI)$. (Here, $\ell\wedge m$ denotes the intersection point +of the lines $\ell,m$, and $P\vee Q$ or just $PQ$ denotes the line +through $P,Q$.) + +\begin{answer} +Let us write as usual $[\lambda{:}\mu{:}\nu]$ for the line $\{\lambda +x+\mu y+\nu z = 0\}$. We find $AI = [0 : 1 : \sqrt{-1}]$ and $BJ = [1 + : -1 : \sqrt{-1}]$, so $AI\wedge BJ = (2 : 1 : \sqrt{-1})$. The +formulas for $AJ$, $BI$ and $AJ\wedge BI$ are obtained by exchanging +$\sqrt{-1}$ with $-\sqrt{-1}$ so it is not necesary to recompute them. +Finally, the sought-after line $(AI\wedge BJ) \vee (AJ\wedge BI)$ is +obtained by joining $(2 : 1 : \sqrt{-1})$ with $(2 : 1 : -\sqrt{-1})$, +and this gives $[1 : -2 : 0]$. So it is the line $\{2X=T\}$, or (the +projective extension of) $\{x = \frac{1}{2}\}$. +\end{answer} + +\textbf{(5)} Show that for any two distinct points $A,B$ in +$\mathbb{A}^2(\mathbb{R})$ there is an affine similitude taking +$(0,0)$ to $A$ and $(1,0)$ to $B$. (\emph{Hint:} You can use simple +arguments of standard elementary Euclidean plane geometry for this +question, independently of all previous questions. Alternatively, you +can use the previous questions and a fact from projective geometry +seen in the course.) + +\begin{answer} +By Euclidean geometry arguments: using a translation we can place any +point of $\mathbb{A}^2(\mathbb{R})$ in any given place, so we can +assume without loss of generality that $A = (0,0)$; using a homothety +we multiply distances by a constant $c\neq 0$, so we can assume +without loss of generality that the distance $AB$ is $1$, and $B$ is +now a point on the unit circle; finally, we can rotate around the +origin to get $B$ in $(1,0)$. This provides the required affine +similitude. + +By a projective geometry argument: for any two distinct $A,B$ in +$\mathbb{A}^2(\mathbb{R})$, the points $A,B,I,J$ form a projective +basis of $\mathbb{P}^2(\mathbb{C})$ (in detail: the points $A,B$ are +not aligned with $I,J$ because they are not on the line $\ell_\infty$ +at infinity, and the points $I,J$ cannot be on the line $AB$ because +$AB$ is a \emph{real} line and therefore also $AB\wedge \ell_\infty$ +is a real point). So if we let $A_0 := (0,0)$ and $B_0 := (1,0)$, +there is a unique complex projective transformation taking +$A_0,B_0,I,J$ to $A,B,I,J$; now that complex transformation is real +because its complex conjugate takes $A_0,B_0,J,I$ to $A,B,J,I$, so it +is the same. But we have seen in question (3) that a real projective +transformation of $\mathbb{P}^2$ which fixes $I$ and $J$ is an affine +similitude, so we have answered the question. + +Alternatively, it is also possible to answer the question by a direct +computation of the coefficients of the matrix. +\end{answer} + +\textbf{(6)} Conclude that, for any two distinct points $A,B$ in +$\mathbb{A}^2(\mathbb{R})$, the perpendicular bisector\footnote{In +French: “la médiatrice”. The perpendicular bisector of $[AB]$ is the +line of points at equal distance from $A$ and $B$ in Euclidean +geometry. In this context, it is also the perpendicular line to $AB$ +through the midpoint of $[AB]$.} of $[AB]$ can be constructed as the +line $(AI\wedge BJ) \vee (AJ\wedge BI)$. + +\begin{answer} +We have seen in question (4) that the construction $(AI\wedge BJ) \vee +(AJ\wedge BI)$ gives the perpendicular bisector $x = \frac{1}{2}$ of +the two points $A = (0,0)$ and $B = (1,0)$. Since we have seen in +question (5) that any two distinct points in +$\mathbb{A}^2(\mathbb{R})$ can be brought to this position by an +affine similitude, and since affine similitudes preserve perpendicular +bisectors (because each one of translations, vector homotheties and +vector rotations preserve angles and midpoints), the construction +works for any two distinct points $A,B$. +\end{answer} + +% +% +% + +\exercise + +\textit{Reminder:} A morphism $\varphi \colon \mathbb{A}^2_k \to +\mathbb{A}^2_k$ (over a field $k$) is simply given by two polynomials +$P,Q \in k[x,y]$; the morphism takes a geometric point $(x,y) \in +\mathbb{A}^2(k^{\alg})$ to $(P(x,y), \, Q(x,y))$. + +Explain why a $\varphi$ as above takes values in $\mathbb{A}^2 +\setminus\{(0,0)\}$ (i.e., $\varphi(x,y) \neq (0,0)$ for every +geometric point $(x,y)$) \emph{if and only if} there exist $U,V \in +k[x,y]$ such that $U P + V Q = 1$. + +\begin{answer} +The condition that $\varphi(x,y) \neq (0,0)$ for all geometric points +$(x,y)$ means, by definition, that the affine algebraic variety +$Z(P,Q)$ defined by the equations $P=0$ and $Q=0$ has no geometric +points. By Hilbert's Nullstellensatz, this is equivalent to demanding +that $P,Q$ generate the unit ideal in $k^{\alg}[x,y]$. But we have +seen in the course that this is also equivalent to demanding that they +generate the unit ideal in $k[x,y]$ (because the latter is the +intersection of the former with $k[x,y]$): that is, that there exist +$U,V \in k[x,y]$ such that $U P + V Q = 1$. +\end{answer} + +% +% +% + +\exercise + +In the affine plane $\mathbb{A}^2_k$ with coordinates $(x,y)$ over a +field $k$ of characteristic $\neq 2$, we consider the affine algebraic +variety $C$ with equation $y^2 = x^3 + x^2$. + +\textbf{(0)} Draw an approximate picture of $C$ for $k = \mathbb{R}$. +(You may wish to answer the next question first.) + +\begin{answer} +Over the real numbers, the not-necessarily-smooth affine curve $C$ +looks like this: +\begin{center} +\begin{tikzpicture}[scale=2] +%% \draw[step=.2cm,help lines] (-1.25,-1.25) grid (1.25,1.25); +\draw[->] (-1.15,0) -- (1.15,0); \draw[->] (0,-1.15) -- (0,1.15); +\draw (0.777778,-1.037037) .. controls (0.481481,-0.555556) and (0.222222,-0.222222) .. (0,0) ; % t from -4/3 to -1 +\draw (0,0) .. controls (-0.666667,0.666667) and (-1,0.333333) .. (-1,0); % t from -1 to 0 +\draw (-1,0) .. controls (-1,-0.333333) and (-0.666667,-0.666667) .. (0,0); % t from 0 to 1 +\draw (0,0) .. controls (0.222222,0.222222) and (0.481481,0.555556) .. (0.777778,1.037037); % t from 1 to 4/3 +\end{tikzpicture} +\end{center} + +It intersects the $x$ axis in two points, $(-1,0)$ and $(0,0)$. At +the former, it has a vertical tangent as one checks by computing +partial differentials. At the latter, the partial differentials +vanish (the point is not smooth): one can get an idea of what it looks +like by considering the lowest order terms $y^2 \approx x^2$, +suggesting the two lines $y = \pm x$; or by taking the derivative +around $-1$ and $1$ of the parametric equation $\psi$ found in the +next question. +\end{answer} + +\textbf{(1)} For $t \neq 1,-1$ (in $k^{\alg}$), show that the line $y += t\,x$ intersects $C$ in a unique point $\psi(t)$ different from $O +:= (0,0)$, compute its coordinates, and explain why this formula +defines a morphism $\psi \colon \mathbb{A}^1 \to C$. What are +$\psi(-1)$ and $\psi(1)$? + +\begin{answer} +For $t \neq 1,-1$, a point other than $(0,0)$ satisfying $y = t\, x$ +and $y^2 = x^3 + x^2$ satisfies $t^2 x^2 = x^3 + x^2$ so $t^2 = x + +1$, so $x = t^2 - 1$ and $y = t^3 - t$, and conversely these +coordinates define a point on $C$ and on the line $y = tx$ for any $t$ +(even $t = 1,-1$). Since $t^2 - 1$ and $t^3 - t$ are polynomials, +this defines a morphism $\psi \colon \mathbb{A}^1 \to C$. Both +$\psi(-1)$ and $\psi(1)$ equal $O$. +\end{answer} + +\textbf{(2)} Given $(x,y)$ a (geometric) point of $C$ different +from $O$, how can we compute $t$ such that $(x,y) = \psi(t)$ (remember +how $\psi$ was constructed!)? Deduce that there is a morphism $\tau +\colon C\setminus\{O\} \to \mathbb{A}^1\setminus\{\pm 1\}$ such that +$\tau \circ (\psi|_{\mathbb{A}^1\setminus\{\pm 1\}})$ is the identity +on $\mathbb{A}^1\setminus\{\pm 1\}$ and $\psi \circ \tau$ is the +identity on $C\setminus\{O\}$. Conclude that $C\setminus\{O\}$ is +isomorphic to $\mathbb{A}^1\setminus\{\pm 1\}$. On the other hand, is +$\psi$ itself an isomorphism (why or why not)? + +\begin{answer} +The construction of $\psi$ was to take the point other than $O$ in the +intersection of $C$ and line $y = t\, x$: so we recover $t$ as $y/x$. +This defines a morphism $\tau \colon C\setminus\{O\} \to +\mathbb{A}^1\setminus\{\pm 1\}$, namely $(x,y) \mapsto y/x$: it makes +sense because $x\neq 0$ on $C\setminus\{O\}$ (clearly, $(0,0)$ is the +only point of $C$ with $x=0$, since the equation then implies $y=0$ as +well); and it is $\neq 1,-1$ because the only solution to $y^2 = x^3 + +x^2$ with $y = \pm x$ is $(0,0)$ (which we ruled out). The fact that +$\tau \circ (\psi|_{\mathbb{A}^1\setminus\{\pm 1\}})$ is the identity +amounts to checking the obvious fact $(t^3-t)/(t^2-1) = t$ (for $t\neq +-1,1$); and the fact that $\psi \circ \tau$ is the identity follows +from our computations in question (1). So we have morphisms $\tau +\colon C\setminus\{O\} \to \mathbb{A}^1\setminus\{\pm 1\}$ and +$\psi|_{\mathbb{A}^1\setminus\{\pm 1\}} \colon +\mathbb{A}^1\setminus\{\pm 1\} \to C\setminus\{O\}$ which are inverse +of one another, letting us conclude that $C\setminus\{O\}$ is +isomorphic to $\mathbb{A}^1\setminus\{\pm 1\}$. + +On the other hand, $\psi$ itself is \emph{not} an isomorphism because +both $-1$ and $1$ get mapped to $O$. + +(Intuitively, $C$ is obtained by taking $\mathbb{A}^1$ and identifying +the two points $-1$ and $1$.) +\end{answer} + +\smallskip + +We now see $\mathbb{A}^2_k$ as a subset of the projective plane +$\mathbb{P}^2_k$ with coordinates $(W{:}X{:}Y)$ by $(x,y) \mapsto +(1{:}x{:}y)$. We call $\bar C$ the projective completion of $C$, in +other words, the Zariski closure of $C$ inside $\mathbb{P}^2$. + +\textbf{(3)} What is the equation of $\bar C$? What are its points at +infinity (in other words, the points of $\bar C$ on the line $W=0$)? +Where (in which direction) should we imagine these points to be in the +drawing of question (0)? + +\begin{answer} +The equation of $\bar C$ is obtained by homogeneizing the equation +$y^2 - x^3 - x^2 = 0$ of $C$ (substituting $x = X/W$ and $y = Y/W$ and +multiplying by $W^3$ to get rid of denominators): this gives us $W Y^2 +- X^3 - W X^2$. Intersecting with the line at infinity $W=0$ gives +$X=0$, so we have a unique point $(0{:}0{:}1)$ (remember that +$(0{:}0{:}Y)$ defines a \emph{unique} point in $\mathbb{P}^2$). This +point is the point at infinity in the vertical direction (because it +lies, say, on the line $X=0$ which is the $y$ axis): this is +intuitively in accordance with the fact that the curve seems to go to +infinity “more vertically”. +\end{answer} + +\textbf{(4)} For each point at infinity $P$ found in question (3), +determine whether $C$ is smooth at $P$ and, if it is, compute the +tangent line $T_P C$ to $C$ at $P$. (\emph{Hint:} You will need to +switch to a different $\mathbb{A}^2$ so that the point $P$ is no +longer at infinity.) Where should we imagine this tangent line $T_P +C$ to be in the drawing of question (0)? + +\begin{answer} +We have found only one point at infinity, $P = (0{:}0{:}1)$. We +introduce new affine coordinates where it is not at infinity by taking +$Y = 0$ to be our line at infinity, in other words we put $w' = W/Y$ +and $x' = X/Y$. In the complement $\mathbb{A}^{2\prime} := +\mathbb{P}^2 \setminus \{Y=0\}$ of this line at infinity, the equation +of the curve $\bar C \cap \mathbb{A}^{2\prime}$ is given by +dehomogeneizing the equation of $\bar C$: it is $w' - x^{\prime 3} - +w' x^{\prime 2} = 0$. Our point $P$ is now given by $(w',x') = +(0,0)$. Differentiating $g := w' - x^{\prime 3} - w' x^{\prime 2}$ +with respect to $w'$ and $x'$ at $(0,0)$ gives $\left. \frac{\partial + g}{\partial w'} \right|_{(0,0)} = 1$ and $\left. \frac{\partial + g}{\partial x'} \right|_{(0,0)} = 0$, so the curve $\bar C$ is +smooth at $P$ and has tangent line $w' = 0$ in $\mathbb{A}^{2\prime}$, +that is $W = 0$ in $\mathbb{P}^2$. This tangent line is the line at +infinity of our original affine chart $\mathbb{A}^2$: in the drawing +of question (0), one should imagine that it is at infinity in every +direction. +\end{answer} + +\textbf{(5)} Show that $\psi$ extends to a morphism $\bar\psi \colon +\mathbb{P}^1 \to \bar C$. Why is it surjective (that is, surjective +on the geometric points)? + +\begin{answer} +We can extend $\psi$ either by explicitly describing an equation for +it: if $(U{:}V)$ are coordinates on $\mathbb{P}^1$ with $t = V/U$ +being the coordinate formerly used on $\mathbb{A}^1$, then we can +homogeneize the formulas for $\psi$ as follows: +\[ +(U{:}V) \; \mapsto \; (U^3 : UV^2 - U^3 : V^3 - U^2V) +\] +— the coordinates being given in order $(W{:}X{:}Y)$. Clearly, $U^3$, +$UV^2 - U^3$ and $V^3 - U^2V$ are homogeneous of the same degree; and +they never vanish simultaneously, because if $U^3$ and $V^3 - U^2 V$ +vanish then $U=0$ and $V=0$ which is meaningless for homogeneous +coordinates on $\mathbb{P}^1$; furthermore, these satisfy $W Y^2 - X^3 +- W X^2$, so we have our morphism $\bar\psi \colon \mathbb{P}^1 \to +\bar C$. + +It is surjective because we have already seen in (1) that every point +other than the point $P$ at infinity is in the image of $\psi$ (for +points $M$ other than $P$ and $O$, take the slope of the line $OM$; +and for $O$ we have seen that it is attained twice); as for the point +$P = (0{:}0{:}1)$, it is $\bar\psi((0{:}1))$. +\end{answer} + +% +% +% + +\exercise + +In this exercise, we let $k$ be the field $\mathbb{R}$ or +$\mathbb{C}$, and we consider the curve $\bar C$ which is the +projective completion (in other words, the Zariski closure inside +$\mathbb{P}^2_k$) of $C := \{x^2 + y^2 + 1 = 0\} \subseteq +\mathbb{A}^2_k$. + +\textbf{(1)} Recall why $\bar C$ is a (smooth) curve and why the field +$K := k(\bar C)$ of rational functions on $\bar C$ is +$k(x)[y]/(x^2+y^2+1)$ (which can also be written, suggestively, as +$k(x)(\sqrt{-1-x^2})$). A short answer will suffice. + +\begin{answer} +The projective completion $\bar C$ of $C$ has homogeneous equation +$X^2 + Y^2 + Z^2 = 0$ given by homogeneizing that of $C$. It is +smooth because the partial differentials, $2X$, $2Y$, $2Z$ of this +equation never vanish simultaneously in $\mathbb{P}^2$ (they generate +the irrelevant ideal $(X,Y,Z)$). We have seen in the course that, +under such circumstances, $k(\bar C) = k(x)[y]/(x^2+y^2+1)$. +\end{answer} + +\textbf{(2)} Briefly recall why we can express elements of $K$ in the +form $f_0 + f_1 y$ with $f_0,f_1 \in k(x)$, and how sums and products +are computed. As an example of how to compute divisions, compute +$\frac{1}{y}$ and $\frac{1}{y-1}$ in this form. + +\begin{answer} +By performing Euclidean division by $x^2+y^2+1$ in $k(x)[y]$ and +keeping only the remainder, we express elements of $K = +k(x)[y]/(x^2+y^2+1)$ in as polynomials of degree $<2$ in $y$, which is +exactly of the form $f_0 + f_1 y$ with $f_0,f_1 \in k(x)$. Sums are +computed termwise ($(f_0 + f_1 y) + (g_0 + g_1 y) = (f_0+g_0) + +(f_1+g_1)y$), and products are computed by taking the Euclidean +division, which amounts to replacing the possible $y^2$ term by +$-1-x^2$: in other words, $(f_0 + f_1 y) \cdot (g_0 + g_1 y) = +(f_0 g_0 - (x^2+1) f_1 g_1) + (f_1 g_0 + f_0 g_1)y$. + +To compute $\frac{1}{y}$, we search for a Bézout relation $u\,y + +v\cdot (x^2+y^2+1) = 1$ in $k(x)[y]$, which is easy because $-y^2 + +(x^2+y^2+1) = x^2+1$ is an element of $k(x)$, so $-\frac{y}{x^2+1}\,y ++ \frac{1}{x^2+1}\,(x^2+y^2+1) = 1$, and this shows that +$-\frac{1}{x^2+1}\,y$ is the inverse of $y$ in $K$. + +Similarly, to compute $\frac{1}{y-1}$, we search for a Bézout relation +$u\,(y-1) + v\cdot (x^2+y^2+1) = 1$ in $k(x)[y]$, which is again easy +because $-(y+1)(y-1) + (x^2+y^2+1) = x^2+2$ is an element of $k(x)$, +so $-\frac{y+1}{x^2+2}\,(y-1) + \frac{1}{x^2+2}\,(x^2+y^2+1) = 1$, and +this shows that $-\frac{1}{x^2+2} -\frac{1}{x^2+2}\,y$ is the +inverse of $y-1$ in $K$. +\end{answer} + +\textbf{(3)} Explain why, when $k = \mathbb{C}$, then $K$ is +isomorphic to $k(t)$ (the field $\operatorname{Frac}(k[t])$ of +rational fractions in one indeterminate $t$). Describe an explicit +isomorphism. (\textit{Hint:} If we let $x' := \sqrt{-1}\,x$ and $y' +:= \sqrt{-1}\,y$, what does $C$ become ?) + +\begin{answer} +Letting $x' := \sqrt{-1}\,x$ and $y' := \sqrt{-1}\,y$ (which describes +an invertible linear transformation), the equation of $C$ becomes +$x^{\prime 2} + y^{\prime 2} = 1$. But we have seen in the course +that this is rational with an explicit isomorphism $k(t) \to +k(x')[y']/(x^{\prime 2} + y^{\prime 2} - 1)$ being given by $t \mapsto +\frac{y'}{x'+1}$ and conversely $x' \mapsto \frac{1-t^2}{1+t^2}$ and +$y' \mapsto \frac{2t}{1+t^2}$ (rational parametrization of the +circle). This shows that $K$ for $k=\mathbb{C}$ is isomorphic to +$k(t)$ by $t \mapsto \frac{y}{x-\sqrt{-1}}$ and conversely $x \mapsto +-\sqrt{-1}\,\frac{1-t^2}{1+t^2}$ and $y \mapsto +-\sqrt{-1}\,\frac{2t}{1+t^2}$. +\end{answer} + +\textbf{(4)} Show that $K$ is \emph{not} isomorphic to $k(t)$ when $k += \mathbb{R}$. (\textit{Hint:} The equation $u^2 + v^2 + 1 = 0$ has +no real solution, so none in $k(t)$, but it has an obvious one +in $K$.) + +\begin{answer} +In the field $k = \mathbb{R}$ the equation $u^2 + v^2 + 1 = 0$ has no +solutions because $u^2 + v^2 + 1 > 0$. Consequently, it also has no +solution in $k(t)$ because if $u^2 + v^2 + 1 = 0$ in $k(t)$, finding +some $\theta$ at which $u,v$ do not have a pole (which is possible as +a rational fraction has only finitely many poles), we can evaluate +$u,v$ at $\theta$, giving $u(\theta)^2 + v(\theta)^2 + 1 = 0$ in +$\mathbb{R}$, again a contradiction. + +On the other hand, $x^2 + y^2 + 1 = 0$ in $K$ by definition of $K$. +So the equation $u^2 + v^2 + 1 = 0$ has no solution in $k(t)$ and has +one in $K$, showing that $k(t)$ and $K$ cannot be isomorphic. +\end{answer} + +% +% +% +\end{document} |