\documentclass[11pt]{article} \usepackage[noBBpl]{mathpazo} %\documentclass[12pt]{article} %\usepackage{mathptmx} \usepackage{amsfonts} \usepackage{amsmath} \usepackage{amssymb} \usepackage{amsthm} \usepackage{cancel} \pdfpagewidth 8.5in \pdfpageheight 11in \topmargin 0in \headheight 0in \headsep 0in \textheight 9in \oddsidemargin 0in \evensidemargin 0in \textwidth 6.5in \theoremstyle{plain} \newtheorem*{theorem}{Theorem} \newtheorem*{lemma}{Lemma} \newtheorem*{prop}{Proposition} \newtheorem*{cor}{Corollary} \theoremstyle{definition} \newtheorem*{defn}{Definition} \newtheorem*{prob}{Problem} \newtheorem*{ex}{Example} \theoremstyle{remark} \newtheorem*{remark}{Remark} \newtheorem*{note}{Note} \newtheorem*{claim}{Claim} \newtheorem*{case}{Case} \newtheorem*{conclusion}{Conclusion} \newcommand{\N}{\mathbb N} \newcommand{\Z}{\mathbb Z} \newcommand{\Q}{\mathbb Q} \newcommand{\R}{\mathbb R} \newcommand{\C}{\mathbb C} \newcommand{\PP}{\mathbb P} \newcommand{\abs}[1]{\lvert#1\rvert} \newcommand{\Abs}[1]{\left\lvert#1\right\rvert} \newcommand{\norm}[1]{\lVert#1\rVert} \newcommand{\Norm}[1]{\left\lVert#1\right\rVert} \newcommand{\inner}[1]{\langle#1\rangle} \newcommand{\pd}[2]{\frac{\partial#1}{\partial#2}} \newcommand{\unit}[1]{\:\mathrm{#1}} \newcommand{\e}[1]{\times10^{#1}} \DeclareMathOperator{\Real}{Re} \DeclareMathOperator{\Imag}{Im} \DeclareMathOperator{\rank}{rank} \begin{document} \begin{center} \section*{Math 145 -- Algebraic Geometry} \subsection*{Final Exam} \subsection*{David Philipson} \end{center} \textbf{1.} \nopagebreak\textbf{i.} By the algebraic identity $b^2+b = (-1-b)^2-(-1-b)$, if $(a,b)$ is on the cubic then $(a,-1-b)$ is on the cubic as well. Furthermore, the points $[a:b:1]$, $[a:-1-b:1]$ and $[0:1:0]$ are collinear, hence their sum equals zero. Therefore, $-(a,b) = (a,-1-b)$. For the next part, first suppose $a \neq 0$. Then the line through $(0,0)$ and $(a,b)$ is given by $y = mx$, where $m = b/a$. We seek the third point of intersection between this line and the cubic by substituting into the cubic equation, and find $m^2x^2 + mx = x^3 - x$. Cancelling $x$ from both sides (we already have the $x=0$ solution), we find \[ x^2 - m^2x^2 -(m+1) = 0 \] The product of the roots is thus $-(m+1)$, and we know one root is $a$. Therefore, \[ x = \frac{-(m+1)}{a} = \frac{-(a+b)}{a^2},\quad\text{hence}\quad y = mx = \frac{-b(a+b)}{a^3}, \] but we want the inverse of this point. By the formula for inverses found above, \[ P + Q = \left(\frac{a+b}{a^2}, \frac{ba+b^2-a^3}{a^3}\right). \] Next suppose $a = 0$ but $b \neq 0$. Then the line through $P$ and $Q$ is given by $x = 0$. >From the equation for the cubic we get $0 = b^2 + b = b(b+1)$, and since by assumption $b \neq 0$ it must be $-1$. But $(0,-1)$ is the inverse of $(0,0)$ by the formula found above, hence $P + Q$ is the zero element; that is, the point at infinity $[0:1:0]$. Finally, if $P = Q$, then writing the equation for the cubic $f = y^2+y-x^3+x = 0$ we find that \[ \pd fx = \pd fy = 1 \] hence the tangent line is $y = -x$. Substituting into the equation for the cubic, we get $x^2-x = x^3-x$, hence $0 = x^3-x^2 = x^2(x-1)$. Hence the third point of intersection is $(1,-1)$. Finding its inverse by the formula above, we conclude that $P+Q = (1,0)$. \textbf{ii.} Using the results from above, we compute: \[ \begin{array}{ccccc} P = (0,0) & \qquad & 2P = (1,0) & \qquad & 3P = (-1,-1) \\ 4P = (2,-3) & \qquad & 5P = (1/4,-5/8) & \qquad & 6P = (6,14) \end{array} \] Each point is rational.\\ \\ \textbf{2.} \nopagebreak\textbf{i.} Since $A$ is symmetric, by the spectral theorem there exists an orthonormal basis of eigenvectors. Let $C$ be the matrix whose columns are this basis, and note $C^{-1} = C^T$. Then $C^TAC$ is diagonal, so make the change of coordinates $X^{\mathrm{new}} = C^T X$. Under these coordinates, we have $Q([x:y:z]) = (x\:y\:z) C^T A C (x\:y\:z)^T$; that is, \[ Q = \lambda_1x^2 + \lambda_2y^2 + \lambda_3 z^2, \] where the $\lambda_i$ are the eigenvalues of $A$. Then a point $[x:y:z]$ is singular if and only if the three partial derivatives of $Q$ vanish, that is \[ 2\lambda_1 x = 2\lambda_2 y = 2\lambda_3z = 0. \] If such a point exists, then at least one of $x,y,z$ is nonzero; WLOG let $x \neq 0$. Then the last equality gives $\lambda_1 = 0$, hence $\det A = \lambda_1\lambda_2\lambda_3 = 0$. Conversely, if $\det A = 0$, then one of the $\lambda_i$ must be zero; WLOG let $\lambda_1 = 0$. Then $[1:0:0]$ is a singularity. \textbf{ii.} If $(a_{ij})$ and $(b_{ij})$ are respectively the matrices associated to $Q_1$, and $Q_2$, then $A_{\lambda,\mu} = (\lambda a_{ij} + \mu b_{ij})$. In calculating the determinant, each term is the product of three such linear terms in $\lambda$ and $\mu$, hence the determinant is a cubic polynomial in $\lambda$ and $\mu$. $Q_{\lambda,\mu}$ is singular if and only if $\det A_{\lambda,\mu} = 0$. Furthermore, for $\lambda \neq 0$ we have $\det A_{\lambda,0} = \lambda^3 \det Q_1 \neq 0$ since $Q_1$ is nonsingular and by (i). That is, $\det A_{\lambda,\mu}$ contains a $\lambda^3$ term, and similarly it contains a $\mu^3$ term. Hence, if $\det A_{\lambda,\mu}$ equals zero and eitehr of $\lambda$ or $\mu$ is zero, then the other must be as well, which yields no cubic. Thus, we may assume $\lambda,\mu$ are nonzero, and we may further assume $\mu = 1$ (since $Q_{\lambda,\mu} = Q_{c\lambda,c\mu}$). Since $\det A_{\lambda,1} = 0$ is a cubic equation, it has at most three solutions, and therefore any pencil of conics contains at most three singular conics. \textbf{iii.} Change coordinates such that \[ p_1 = [1:0:0] \qquad p_2 = [0:1:0] \qquad p_3 = [0:0:1] \qquad p_4 = [1:1:1] \] Let $Q$ be the equation for a conic passing through these points. $Q$ cannot contain an $x^2$ term, as then $Q([1:0:0])$ would not equal zero, and likewise $Q$ cannot contain an $y^2$ or $z^2$ term. Hence, $Q$ must appear $a\,xy + b\,yz + c\,xz$, which passes through the first three points for any $a,b,c$. The conic passes through $p_4$ as well if and only if $a+b+c = 0$. The subspace $\{(a,b,c) : a+b+c = 0\}$ has dimension two (by rank-nullity, since $\rank(1\:1\:1) = 1$). One basis is $\{(1,1,-2),(1,-2,1)\}$. Therefore, all such $Q$ appear \[ \lambda(xy+yz-2xz) + \mu(xy-2yz+xz) = 0 \] for some $\lambda,\mu$. To show that this gives a pencil, we must check that the conics $f = xy + yz - 2xz$ and $g = xy-2yz+xz$ are nonsingular. Taking derivatives, \[ \pd fx = y-2x\qquad \pd fy = x+z \qquad \pd fz = y-2z \] If all are to vanish, then from the $x$- and $z$-partials we conclude $x = z$, but then the $y$-partial does not vanish unless $x = y = z = 0$, which is impossible in $\PP^2$. Therefore, $f$ gives a nonsingular conic, and similarly $g$ does as well. >From (ii), there are at most three nonsingular conics in the pencil. Three such conics are $xy - yz$, $xy-xz$, and $yz-xz$, which have singularities at respectively $[1:0:1]$, $[0:1:1]$, and $[1:1:0]$. Therefore, they are the only ones.\\ \\ \textbf{3.} \nopagebreak\textbf{i.} If $p$ is an inflection point, then the tangent line at $p$ intersects the cubic only at that point (with multiplicity three). Hence, the three points of intersection are all $p$, and so $p+p+p = 0$. The converse holds similarly. The third intersection of the line through $p,q$ with the cubic is $-(p+q)$. Since the group law on $C$ is commutative, $3\cdot -(p+q) = -(3p + 3q) = 0$, hence $-(p+q)$ is an inflection point as well. \textbf{ii.} Let $\tilde x = x-\frac13(1+\lambda)z$, and note that $p_0 = [0:1:0]$ remained unchanged. Then the equation becomes \[ y^2z = \left(\tilde x + \frac{1+\lambda}3\right) \left(\tilde x +\left(\frac{1+\lambda}3 - 1\right)z\right)\left(\tilde x + \left(\frac{1+\lambda}3 - \lambda\right)z\right), \] and when the right-hand side is viewed as a polynomial in $x$, the sum of the roots is zero. Hence, it expands to $y^2z = \tilde x^3 + a\tilde xz^2 + bz^3$ as desired. \textbf{iii.} Let $f = x^3+axz^2+bz^3-y^2z$ be the equation for the cubic. If $z = 0$, then the equation becomes $x^3 = 0$, and so the only point on the cubic is $[0:1:0]$. Furthermore, this point is an inflection point since all second derivatives vanish except for $f_{yz} = -2$ (it is also the zero element). Otherwise, let $z = 1$ and let $f = x^3+ax+b-y^2$. Suppose $(u,v)$ is an inflection point on $y^2 = x^3+ax+b$. >From (i), $(u,v)$ has order 3 and so $-(u,v) = 2\cdot(u,v)$. As usual, $-(u,v) = (u,-v)$. To compute $2\cdot(u,v)$, the tangent line through $(u,v)$ is given by $2v(y-v) = (3u^2+a)(x-u)$. Solving for $y$ and squaring both sides, we find \[ y^2 = \frac{(3u^2+a)^2}{4v^2}x^2 + (\text{lower order terms}) = x^3+bx+c \] To solve for $x$ (and thus find the other point of intersection), we observe that after moving everything to the right-hand side we have a cubic for which the coefficient of the $x^2$ term is $-(3u^2+a)^2/(4v^2)$, and so the negative of this must be the sum of the roots. We know that two of the roots are $u$, hence the third root is $(3u^2+a)^2/(4v^2)-2u$. This is the $x$-coordinate of $2\cdot(u,v)$. Since $-(u,v) = 2\cdot(u,v)$, we therefore have \begin{gather*} u = \frac{(3u^2+a)^2}{4v^2} - 2u \\ 12uv^2 = (3u^2+a)^2 \\ 12u(u^3+bu+c) = (3u^2+a)^2 \end{gather*} >From here on, we write $x$ in place of $u$ (we used $u$ before to distinguish from the variables $x,y$ while describing the line and cubic). We show that $g = 12x(x^3+bx+c) - (3x^2+a)^2 = 0$ has exactly four solutions. Suppose on the contrary that $x$ is a double root of the quartic. Then the derivative equals zero at $x$; that is, \[ \frac{dg}{dx} = 48x^3+24ax+12b-(36x^3+12ax) = 12(x^3+ax+b) = 0. \] Hence, $x^3+ax+b = 0$, which has two consequences. First, since this is the expression for $y^2$, we have $y = 0$. Secondly, $g$ reduces to $-(3x^2+a)^2 = 0$, hence $\pd fx = 3x^2+a = 0$. We therefore have that \[ f(x,0) = 0,\qquad\pd fx(x,0) = 0,\quad \text{and} \quad \pd fy(x,0) = 0 \] and so $(x,0)$ is a singularity of the original conic, a contradiction because it was nonsingular. Therefore, $g$ has no double roots, so there are exactly four values of $x$ satisfying it. Furthermore, for no such value of $x$ does $x^3+ax+b = 0$ (or we arrive at the same contradiction), and so for each $x$ there are exactly two values of $y$ satisfying $y^2 = x^3+ax+b$. This gives a total of nine inflection points: $[0:1:0]$, and then the eight pairs by choosing one of four values for $x$ and two for $y$. \textbf{iv.} >From (i) and (iii), the inflection points form a subgroup with nine elements and each element has order 3. Therefore, the subgroup is isomorphic to $\Z/3\Z \times \Z/3\Z$. \textbf{v.} We first seek such points with $z = 0$. As in (iii), the only such point on the cubic is $p_0$, and trivially the tangent line through $p_0$ passes through $p_0$. Next suppose $z = 1$. Then for any point $p$ whose tangent line passes through $p_0$, we have $p+p+p_0 = 0$, hence $p$ has order 2. Conversely, any point of order $2$ is such a point. For such points, $p = -p$. Since $-(x,y) = (x,-y)$, such points are those of the form $(x,0)$. There are three such values of $x$, the solutions to $x^3+ax+b = 0$ (we know there are no repeated roots by the reasoning in (iii)). Therefore, there are a total of four such points, each having order $2$. Since we argued that they are the only points of order 2, they must form a subgroup (since in an abelian group, the sum of two elements with order 2 also has order 2). Hence, they form a subgroup isomoprhic to $\Z/2\Z \times \Z/2\Z$.\\ \\ \textbf{4.} \textbf{i.} As suggested, assume the singularity is at $[0:0:1]$ and let $f$ be the polynomial giving the cubic. Then since $\pd fx = \pd fy = 0$, $f$ may not contain an $xz^2$ or $yz^2$ term. Furthermore, since $\pd fz = 0$, $f$ may not contain a $z^3$ term. That is, $z$ does not appear to a power higher than 1 in $f$, and so we may write the cubic as \[ (\text{quadratic polynomial in }x,y)\cdot z = Q(x,y) \] where $Q$ is a cubic polynomial in $x,y$. The quadratic polynomial is either the square of a linear term or the product of two distinct linear terms in $x$ and $y$. In the first case, let $y^{\mathrm{new}}$ be that linear term, or in the second let $x^{\mathrm{new}}$ and $y^{\mathrm{new}}$ be the two linear terms. We obtain \[ y^2z = Q(x,y)\quad \text{or} \quad xyz = Q(x,y) \] First consider the first case, $y^2z = Q(x,y)$. We note first that $Q$ must contain an $x^3$ term, as otherwise both sides are divisible by $y$ contradicting that the conic is irreducible. First make a scaling change of coordinates of $x$ so that this $x^3$ term has coefficient $1$. We seek to ``complete the cube'' on the right-hand side. Given the coefficient of $x^2y$ in $Q$, there are specific coefficients for $xy^2$ and $y^3$ such that the right-hand side is a perfect cube. By making the change of coordinates $z \mapsto \lambda x + \mu y + z$, with the appropriate choices of $\lambda$ and $\mu$, we produce the terms $\lambda xy^2$ and $\mu y^3$, which can be used to complete the cube. This produces $y^2z = (x+by)^3$. With the final change of coordinates $x^{\mathrm{new}} = x+by$, we end with $y^2z = x^3$ as desired. Now consider the second case, $xyz = Q(x,y)$. As before, $Q$ must contain an $x^3$ and a $y^3$ term, or else it would violate irreducibility. First make scaling change of coordinates of $x$ and $y$ so that the coefficients of $x^3$ and $y^3$ are $1$. Then by making the change of coordinates $z \mapsto \lambda x + \mu y + z$, we once again complete the cube on the right-hand side (this time filling in the terms $x^2y$ and $xy^2$). This produces $xyz = (x+y)^3$. Now make the change of coordinates \[ \left\{ \begin{aligned} x' &= x+y\\ y' &= x-y\\ z' &= -z/4 \end{aligned}\right. \qquad \text{with inverse} \qquad \left\{ \begin{aligned} x &= (x'+y')/2 \\ y &= (x'-y')/2 \\ z &= -4z' \end{aligned}\right. \] Under these change of coordinates, the equation becomes $(y^2-x^2)z = x^3$, or equivalently $y^2z = x^2(x+z)$ as desired. \textbf{ii.} Consider the case $f = x^2(x+z) - y^2z = 0$. Then $\pd fx = 3x^3+2xz$, $\pd fy = -2yz$, and $\pd fz = x^2-y^2$. If the second is to vanish, then $y = 0$ or $z = 0$. If $y = 0$, then from the $z$-partial we get $x = 0$, so we find the point $[0:0:1]$, which is indeed singular. If $z = 0$, then from the $x$-partial we get $x = 0$, then from the $z$-partial we get $z = 0$. But then $x = y = z = 0$, which is impossible. Therefore, the only singularity is $[0:0:1]$. In the case $f = x^3-y^2z = 0$, the partials are $\pd fx = 3x^2$, $\pd fy = -2yz$, and $\pd fz = -y^2$. If all are to vanish, then we must have $x = y = 0$, so we get the unique singular point $[0:0:1]$. The reducible cubic given by $xyz = 0$ has the three singularities $[1:0:0]$, $[0:1:0]$, and $[0:0:1]$.\\ \\ \textbf{5.} \nopagebreak\textbf{i.} Since $C$ and $D$ do not share irreducible components, they intersect in at most $de$ points. Hence, there are at most $\binom{de}{2}$ lines through two points of $C \cap D$. Draw a curve which does not share an irreducible component with $C$, $D$, or any of these lines. This curve intersects each of the above in finitely many places, and so there are points on the curve which are not in $C$, $D$, or any of the lines. Choose one of these points and make a change of coordinates to place it at $[1:0:0]$. This is possible, as we have already proved the stronger result that change of coordinates may be used to move four points with no three collinear to $[1:0:0]$, $[0:1:0]$, etc. \textbf{ii.} If $p = [a:b:c] \in C\cap D$, then there exists a common solution for $x$ in each of the equations $P(x:b:c)$ and $Q(x:b:c)$ (that solution is $a$). Hence, $R_{P,Q}(b,c) = 0$, and so $bz-cy$ must be a linear factor of $R_{P,Q}(x,y)$. Therefore, $\mu_p(C,D) \geq 1$. The converse is confusing: since the definition of $\mu_p(C,D)$ does not make any reference to $a$, it seems as though $\mu_p(C,D) \geq 1$ allows the first coordinate of $p$ to take on any value (while holding $b$ and $c$ fixed), and certainly not all such points are in $C\cap D$. One way out is to insist that intersection multiplicity be defined only for points in $C\cap D$, as is implied in the paragraph at the start of the problem. In this case, the converse holds by definition. One thing that can be said for certain is that $\mu_p(C,D) \geq 1$ implies that there exists some $a'$ such that $[a':b:c]$ is in $C\cap D$, and because of the choice of coordinates this $a'$ is unique. \textbf{iii.} We first note that given coordinates $y,z$, there is at most one value of $x$ such that $[x:y:z] \in C\cap D$. Indeed, if there were two such values, then the point $[1:0:0]$ would be on the line between them, contradicting the choice of coordinates. Factor $R_{P,Q}(y,z)$ into $de$ (not necessarily distinct) linear terms of the form $b_iz-c_iy$. Suppose $b_iz-c_iy$ appears to the $k$th power in this factorization. Then $R_{P,Q}(b_i,c_i) = 0$, and so there exists a point $a_i$ which is a common root of $P(x:b_i:c_i)$ and $Q(x:b_i:c_i)$. Hence, $[a_i:b_i:c_i] \in C \cap D$, and by the discussion above there is a unique such $a_i$. Let $p_i = [a_i:b_i:c_i]$, and we have $\mu_{p_i}(C,D) = k$. By (ii), all points in $C\cap D$ will be found through this process. Hence, when summing $\mu_p(C,D)$ over all points in $C\cap D$, we find one point for each distinct factor in $R_{P,Q}(y,z)$, and that point contributes to the sum the power to which that factor is raised. Since the sum of the powers of all factors is $de$, we conclude that this sum equals $de$ as desired. \textbf{iv.} Call the curves $C$ and $D$ respectively, with equations $F$ and $G$. We first find the points in $C\cap D$; from $F = 0$, we must have $x + y = 0$, from $G = 0$ we have $x = 0$ or $x+z = 0$. In the first case, we get the point $[0:0:1]$ while in the second we get $[-1:1:1]$. We can now observe directly that these equations satisfy the criteria on the coordinate system given in (i); first, $[1:0:0]$ lies on neither $C$ nor $D$, and secondly $[1:0:0]$ does not lie on the line through $[0:0:1]$ and $[-1:1:1]$. In general, $R_{F,G}^{(x)} = \prod (\mu_j-\lambda_i)$, where the $\lambda_j$ are the roots of $F$ when viewed as a polynomial in $x$ and the $\mu_i$ are likewise the roots of $G$. Here, the roots of $F$ are $-y$ repeated $d$ times, while the roots of $G$ are $0$ and also $-z$ repeated $e-1$ times. Therefore, \[ R_{F,G}(y,z) = (y-z)^{d(e-1)}y^d \] The factor $y-z$, viewed as $bz-cy$, gives $b = c = -1$, which then gives the intersection point $[-1:1:1]$. Likewise, the factor $y$, viewed as $bz-cy$, gives $b = 0, c = -1$, which then gives the intersection point $[0:0:1]$. Therefore, $[-1:1:1]$ has intersection multiplicity $d(e-1)$, and $[0:0:1]$ has intersection multiplicity $d$. This satisfies Bezout's theorem, since $d(e-1) + d = de$. \end{document}