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Plane Conics in Algebraic Geometry

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Plane Conics in Algebraic Geometry PLANE CONICS IN ALGEBRAIC GEOMETRY ERIC YAO Abstract. We first examine points and lines within projective spaces. Then we classify affine conics based on the classification of projective conics. Based on the parametrization of conics, we also prove two easy cases of B´ezout's Theorem. In the end we turn to the discussion of the space of conics and the notion of pencils of conics. Contents 1. Introduction 1 2. 2 ; 1 , and projective transformations 2 PR PR 3. Classification of Conics in P2 and A2 6 4. Parametrization of Conics in 2 12 PR 5. Spaces of Conics 14 Acknowledgments 16 References 16 1. Introduction Example 1.1. We describe the parametrization of all integer solutions for X2 + Y 2 = Z2. Because the given equation is homogeneous (every term has the same degree), we can use factorization to simplify the task. When Z 6= 0, we parametrize the equation by factorizing out Z2. We then have X 2 Y 2 Z 2 X Y the equation ( Z ) + ( Z ) = ( Z ) . Now define x to be Z and y to be Z . The new equation x2 + y2 = 1 represents a circle in R2. Since X; Y; Z are integers, (x; y) represent all points on the circle whose coor- dinates are both rational numbers. We choose an arbitrary point O = (1; 0), and from O draw a line y = λ(x − 1), for every slope λ 2 Q. In other words, λ = l=m, where l; m 2 N. There are two variables, x and y, and two equations. Solving the equations, we have: λ2 − 1 X −2λ Y y x = = ; y = = ; λ = : λ2 + 1 Z λ2 + 1 Z x − 1 X = l2 − m2;Y = 2lm; Z = l2 + m2: Here l; m 2 N and l is co-prime with m. When l = m = 0, the above formula provides the solution (0; 0; 0), where Z = 0. So these formulas include all integer solutions for X2 + Y 2 = Z2. Date: AUGUST 29, 2014. 1 2 ERIC YAO This example illustrates the process of solving algebraic equations with geomet- ric methods. Dehomogenization reduces the number of variables in the equation, which enables us to find geometric representations of equations on a plane. These intersections between the set of lines with rational slopes starting from O and the circle generate a set of rational points. Multiplying coordinates of these points by Z results in integer solutions of X; Y for X2 + Y 2 = Z2. Similarly, dehomogenization and projection are also used in finding coordinates of points in projective spaces. 2. 2 ; 1 , and projective transformations PR PR By defining projective spaces, projective planes and projective lines, we can analyze the relationship between projective and affine objects. We will also discover how parallel lines intersect in projective spaces. And we will compare projective transformations with affine transformations. Definition 2.1. The projective space P(V ) is the set of lines passing through the origin of a vector space V. The projection of V = 3 is the projective plane 2 . R PR When V = 2, the projective space is the projective line 1 . R PR We can understand projective planes based on equivalence classes and homoge- neous coordinates. By definition, homogeneous coordinates are coordinates that are invariant up to scaling. For example: (x; y; z) = (2x; 2y; 2z): In R3, every line passing through origin can be expressed by the equation aX + bY + cZ = 0. If one point (x; y; z) lies on a line, then all of multiples of this point, α(x; y; z), also lie on the same line. So all points on one line form an equivalence class represented by (x; y; z). In a projective space, we denote this equivalence class by point (x; y; z). Here (x; y; z) is a homogeneous coordinate. So 2 is the set of homogeneous co- PR ordinates that represent all lines passing through the origin in R3. However, the origin in R3 is not included in projective space. 2 = ( 3 n 0)= ∼; PR R and each homogeneous coordinate [a:b:c] is a representative of its equivalence class, (a; b; c) ∼ (λa, λb, λc): We analyze the projective space 2 by dividing it into affine components. PR Proposition 2.2. 2 = 2 t 1 : PR A PR Proof. The set of all equivalence classes [X : Y : Z] in R3 consists of [x : y : 1] X Y (where x = Z ; y = Z ;Z 6= 0) and [X : Y : 0]. The first case contains a copy of R2. We call this affine plane A2 since it is invariant up to scaling (multiply by λ) and shearing (the shifting of the vector space while preserving a line; for projective plane, this is the line at infinity, which we will discuss later). The equivalence classes represented by [X : Y : 0] do not intersect with any points from [x : y : 1] as they do not share the same third coordinates. Thus these 2 two sets of elements in PR are disjoint. Because [x : y : 1] = ( 2 n 0)= ∼ and [X : Y : 0] = 1 ; A PR 2 = 2 t 1 : PR A PR PLANE CONICS IN ALGEBRAIC GEOMETRY 3 Proposition 2.3. 1 = 1 t 0: PR A A 2 X Proof. The set of all equivalence classes [X : Y ] in R consists of [ Y : 1] (where Y 6= 0) and [X : 0]. The first case contains a copy of R1. Thus it is bijective to A1. X 0 [X : 0] represents the case of Y=0. Factoring out Y generates [ 0 ; 0 ]. Since any number divided by 0 does not generate a number, [X : 0] is the point at infinity 0 that does not intersect with [X : Y ]. We write it as A . Proposition 2.4. n = n t n−1: PR A P Proof. The subspaces for n include [x ; x : : : x ; 1] and [x ; x : : : x ; 0]. The for- PR 1 2 n 1 2 n mer subspace consists of a copy of Rn+1, and it is bijective to An. The latter space n−1 n−1 consists of lines passing through the origin in R , so it is P : The above propositions explicate the structure for projective spaces. By induc- tion, n = n t n−1 t · · · t 0: PR A A A Proposition 2.5. Parallel lines in 2 meet at the line at infinity. PR Proof. We start with the formula of lines in the Euclidean Space R3; ax + by + c = 0; with slope k = -a/b: Since projective spaces only use homogeneous coordinates, we want to homoge- nize the general equation for lines so that each term has the same degree. To ho- mogenize, we multiply the equation with a variable Z and define X = xZ; Y = yZ. Thus we have aX + bY + cZ = 0: [X : Y : Z] represents the equivalence classes in 2 . So aX + bY + cZ = 0 is PR invariant up to scaling and we can use [a : b : c] to represent the equivalence classes of lines. If two lines l1; l2 are parallel in the projective plane, then [a1 : b2] ∼ [a2 : b2]: Therefore we can scale the 3-tuples of both lines to make the first two coordinates equal. aX + bY + c1Z = aX + bY + c2Z: Since l1,l2 are parallel, c1 6= c2. The solution is Z = 0, which means the inter- section lies on the line at infinity in 2 . PR This counter-intuitive result actually corresponds to real-life experiences of seeing two train tracks intersecting at the horizon. Every pair of non-identical lines in the projective space has exactly one intersection. Projective space is regarded as affine space with a line at infinity. After understanding the structure and property of projective spaces, we are able to study transformations within it. Definition 2.6. An affine change of coordinates in R2 is of the form T (x) = Ax + B; where x = (x,y) 2 R2 and A is a 2 × 2 invertible matrix. Definition 2.7. A projective transformation of 2 is of the form T(X) = MX, PR where M is an invertible 3 x 3 matrix. The projective transformation defines an isomorphic mapping from one space to another. It preserves the linearity of the space and it is well-defined. 4 ERIC YAO Definition 2.8. T ([X]) is the set of images of projective transformations from elements in the equivalence class [X]. T ([X]) = fMxjx 2 [X]g. Exercise 2.9. [T (X)] = T ([X]): 2 3 a11 a12 a13 Proof. Take M = 4a21 a22 a235 . a31 a32 a33 2 3 22 33 22 033 a11 a12 a13 x x 0 T ([X]) = 4a21 a22 a235 44y55 = 44y 55 : 0 a31 a32 a33 z z 22λx033 22x033 [T (x : y : z)] = T ([x : y : z]) as 44λy055 = λ 44y055 : Therefore the projective λz0 z0 transformation is well defined. A projective transformation on an affine space 2 ⊂ 2 is the fractional-linear R PR az+b 2 2 transformation (a rational function of the form f(z) = cz+d ) mapping R to R x x a a b 7! (A + B)=cx + dy + e (A = 11 12 ;B = 1 ); y y a21 a22 b2 where 2 3 a11 a12 b1 M = 4a21 a22 b25 : c d e Proposition 2.10. There is a unique projective transformation in P1 that maps a triplet (0; 1; 1) to (P; Q; R) 2 P1. Proof. Since 1 ⊂ 1 , (0; 1; 1) ⊂ 1 : Respectively, their coordinates in 1 are A PR AR PR 0 7−! [0 : 1]; 1 7−! [1 : 1]; 1 7−! [1 : 0]: We want to show there exists a unique M2×2 such that M[0; 1; 1] = [P; R; Q] with P = [p1 : p2];Q = [q1 : q2];R = [r1 : r2]; where P; Q; R are distinct.
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