CONIC AND CUBIC PLANE CURVES JAMES W. KISELIK Abstract. In this paper, I will introduce some basic notions of conic and cubic plane curves in 2 including their definitions, parametrizations, and PR some basic theorems associated with each. These theorems include two cases of B´ezout’stheorem, Cayley-Bacharach theorem, and Pascal’s theorem. Contents 1. An Introductory Example 1 2. Defining Conics in 2 2 PR 3. Classification of Conics in 2 3 PR 4. Parametrization of Conics 4 5. Homogenous Forms in 2 Variables 4 6. Two Simple Cases of B´ezout’s Theorem 5 7. Space of All Conics 6 8. Degenerate Conics in a Pencil 7 9. Parametrized Cubic Plane Curves 7 10. The Curve (y2 = x(x 1)(x λ)) Has No Rational Parametrization 8 11. Linear Systems− − 8 12. Cayley-Bacharach Theorem 10 13. Pascal’s Theorem 11 Acknowledgments 11 References 11 1. An Introductory Example We begin with an example that asks a fundamentally algebraic question, but we answer this question using a very geometric proof. The use of geometry in algebraic proofs and algebra in geometric ones will be very important throughout the remainder of the paper. In this example, the question is if we can find a way to represent all rational solutions to the Pythagorean theorem. The Pythagorean theorem says that if a right triangle has legs of lengths X and Y and hypotenuse of length Z,thenX2 + Y 2 = Z2.WhenX, Y , and Z are integers, they are collectively referred to as a Pythagorean triple. Familiar examples of triples include (3, 4, 5), (5, 12, 13), and (8, 15, 17), but it is possible to construct a method for finding all solutions. Proposition 1.1. There is a method for representing the general form of Pythagorean triples. Date:28August2014. 1 2 JAMES W. KISELIK X2 Y 2 X Proof. First, reformulate the equation as Z2 + Z2 = 1. Next, define x = Z and Y 2 2 y = Z , so the formula can be temporarily rewritten as x + y = 1. Consider a p line that intersects this circle at (0, 1) with positive slope λ = where p, q Z are q 2 coprime; the equation of this line is y = λx+1. Solve the system of equations for the 2λ 1 λ2 line and circle to find that the point of intersection besides (0, 1) is ( λ−2+1 , λ−2+1 ). Now, it is obvious that X = 2λ, Y =1 λ2 and Z = λ2 + 1. Because the equation is homogenous, scales− of a solution− will also be solutions to the equation. Choosing a scale of kq2 with k Z will give general forms of a triple as X = 2kpq, Y = k(q2 p2) and Z = k(2p2 + q2). ⌥ ± − ± ⇤ The picture for one of the lines with rational slope looks like this: A 6 y = λx +1 x2 + y2 =1 '$ r - B r &% A is (1, 0) and B is the other point of intersection that will vary for each line, but 2λ 1 λ2 is of the form ( λ−2+1 , λ−2+1 ). Thus, the algebraic problem has become a very easily solvable problem of finding the other point where a line and a conic intersect. 2. Defining Conics in 2 PR Our first goal is to define conics in the real projective plane, 2 . In order to PR achieve this goal, we will first recall how conics are defined in R2 and we will define 2 . Then we will combine these notions in order to define conics in 2 . PR PR Definition 2.1. In R2, conics are plane curves defined by the familiar equation (2.2) q(x, y)=ax2 + bxy + cy2 + dx + ey + f =0. Definition 2.3. The real projective plane, 2 , is defined as lines of 3 through the origin . PR R Equivalently, it can also be thought of as ratios X : Y : Z{ where X, Y, Z are not } all 0. { } Proposition 2.4. The two above definitions of 2 are, indeed, equivalent. Fur- PR thermore, P2 contains a copy of R2 as well as P1 P0 . R R [ R Proof. In cases when Z = 0, choose a unique representation of each ratio from 6 X Y its equivalence class by setting x = Z and y = Z , then consider ratios of x : y :1. Thus, we can conveniently define all the possible ratios in terms{ of only two variables.} This conception obviously makes sense with respect to the other definition 2 = lines of 3 through the origin . Simply think of each ratio as PR R giving a unique ({x, y) coordinate pair on the plane} (Z = 1), then there is a line that passes through the origin and that point. Because there is a line representing each point of 2, we can say that 2 contains a copy of 2. And from each line R PR R passing through the origin, the coordinates of any point on the line besides 0 will give the corresponding ratio. X However, there are instances when Z = 0. In these cases, if Y = 0, set x = Y and consider ratios of x :1. Each ratio gives a unique coordinate6 where a line through the origin{ on the plane} (Z = 0) passes through the line y = 1. Finally, CONICANDCUBICPLANECURVES 3 there is a single ratio when Z = 0 and Y = 0, but X = 0. This is the line along the x-axis. These cases with Z = 0 are referred to as the6 line at infinity, which is a copy of P1 = R . R [{1} ⇤ Therefore, Z =0,Y =0,X = 0 give an open covering by affine spaces. Fur- thermore, there6 is a stratification,6 6 in some sense, by affines spaces of decreasing dimension. For the real projective plane, this can be seen when considering that 2 2 1 2 P = R P = R R . R [ R [ [{1} By this logic, in general, the projective space for Rn+1 is n n n 1 n n 1 P = R P − = R R − ... R . R [ R [ [ [ [{1} Now that the real projective plane has been defined, we can almost define conics in 2 . Before that, we must create a bijection between (2.2), the inhomogenous PR quadratic used to define conics in R2, and the homogenous quadratic (2.5) Q(X, Y, Z)=aX2 + bXY + cY 2 + dXZ + eY Z + fZ2 =0. The bijection q Q is defined by q(x, y)=Q(x, y, 1) and Q(X, Y, Z)=Z2q(x, y) X $ Y with x = Z and y = Z . This is the projective closure of the affine conic. Restricting the homogenization to affine space will give the original equation. Definition 2.6. In 2 , conics are curves defined by Q(X, Y, Z) = 0. PR Because the equation is homogenous, Q(X, Y, Z) = 0 is well-defined for each equivalence class. 3. Classification of Conics in 2 PR Conics defined in R2 can be classified with relative ease simply by varying the constants a, b, c, d, e, f. However, it is often easier to work in 2 , so it is our next PR goal to find a way to classify conics here. Let k be any field such that its characteristic is not 2. This is a requirement that will hold for the entirety of the paper whenever we work in a field that is more general than R. Proposition 3.1. For a homogenous quadratic polynomial, there is a bijection between it and a symmetric bilinear form on k3 given by the formula abd (3.2) aX2 +2bXY + cY 2 +2dXZ +2eY Z + fZ2 bce . $ 0 def1 Theorem 3.3. Let V be a vector space over k and Q : V@ k a quadraticA form, then there exists a basis of V such that Q = ✏ x2 + ✏ x2 + ! + ✏ x2 ,with✏ k 1 1 2 2 ··· n n i 2 where ✏i can be varied by squares. Proof. This follows from the Gram-Schmidt orthonormalization process. ⇤ Corollary 3.4. If a suitable system of coordinates is chosen, any conic in 2 can PR be categorized as one of the following: (1) nondegenerate conic, C :(X2 + Y 2 Z2 = 0); (2) empty set, C :(X2 + Y 2 + Z2 = 0);− (3) pair of lines, C :(X2 Y 2 = 0); − 4 JAMES W. KISELIK (4) the point (0, 0, 1), C :(X2 + Y 2 = 0); (5) double line, C :(X2 = 0); (6) 2 , C : (0 = 0). PR Proof. Any ✏ R can only be either 0 or a nonzero square. Thus, consider 2 ± the above theorem only with ✏i = 0 or 1. By systematically going through each possible value of ✏ for each variable, the± possibilities can be enumerated without difficulty. ⇤ Remark 3.5. There are three fewer cases in 2 than in .In 2 , ellipses, hyperbolas, PR R PR and parabolas all fall under case (1), and both intersecting and parallel pairs of lines fall under case (3). 4. Parametrization of Conics For a nondegenerate, nonempty conic C in 2 , take new coordinates (X + PR Z, Y, Y Z). Thus, C is projectively equivalent to (XZ = Y 2). The curve can be parametrized− so that Φ: 1 C 2 by (U : V ) (U 2 : UV : V 2). The inverse PR PR map : C 1 is defined by! (X⇢: Y : Z) (X :7!Y )=(Y : Z)whereX, Z = 0. PR It is well-defined! because X, Y, Z are not all7! 0. 6 Proposition 4.1. Let k be any field of characteristic =2,andV a 3-dimensional k-vector space; let 6 Q : V V k ⌦ ! be a nondegenerate quadratic form on V .If0 = e V satisfies Q(e )=0, then 6 1 2 1 V has a basis e1,e2,e3 such that 2 Q(x1e1 + x2e2 + x3e3)=x1x3 + ax2.