%% 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}