CONIC AND CUBIC

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 7 9. Parametrized Cubic Plane Curves 7 10. The (y2 = x(x 1)(x )) 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 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 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 that intersects this at (0, 1) with positive slope = where p, q Z are q 2 coprime; the equation of this line is y = x+1. Solve the system of for the 2 1 2 line and circle to find that the point of intersection besides (0, 1) is ( 2+1 , 2+1 ). Now, it is obvious that X = 2, Y =1 2 and Z = 2 + 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 = x +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 2 1 2 is of the form ( 2+1 , 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 , 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 ane spaces. Fur- thermore, there6 is a stratification,6 6 in some sense, by anes 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 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 ane conic. Restricting the homogenization to ane 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 diculty. ⇤ Remark 3.5. There are three fewer cases in 2 than in .In 2 , , , PR R PR and 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 : 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. Thus, a nonempty, nondegenerate conic C P2 is projectively equivalent to ⇢ k (XZ = Y 2).

5. Homogenous Forms in 2 Variables Definition 5.1. Let F (U, V ) be a nonzero form, or homogenous polynomial, of degree d in U, V ,withcoecientsink.Write d d 1 i d i d F (U, V )=adU + ad 1U V + + aiU V + + a0V . ··· ··· Now, associate with F an inhomogenous polynomial in 1 variable d d 1 i f(u)=adu + ad 1u + + aiu + + a0. ··· ··· So for ↵ k, f(↵) = 0 if and only if u ↵ divides f(u). Equivalently, this implies and2 is implied by (U ↵V )dividingF (U, V ) which is itself equivalent to F (↵, 1) = 0. Thus, zeros of F on P1 correspond to zeros of f except when ↵ = , k 1 represented by (1, 0). However, F (1, 0) = 0 if and only if ad = 0, which happens if and only if deg f

Proposition 5.3. Let F (U, V ) be a nonzero form of degree d in U, V . Then F has 1 at most d zeros on Pk. Proof. Let m be the name for the multiplicity of the zero of F at the point (1, 0). By definition,1 this means that d m is the degree of f, the single variable inhomogenous polynomial defined above. A1 polynomial in one variable will have at most deg f roots, so the proposition is proved. ⇤ Remark 5.4. If k is algebraically closed and the zeros are counted using the above 2 definition of multiplicity, then F will have exactly d zeros on Pk.

6. Two Simple Cases of Bezout’s´ Theorem B´ezout’s theorem says that two plane curves C, D with deg C = m, deg D = n intersect at a total of mn points. However, there are three preconditions that must be satisfied: (1) the field must be algebraically closed (2) points of intersection must be counted with the correct multiplicity 2 (3) the curves must be considered in Pk in order to include intersections at infinity. Here, B´ezout’s theorem will be proved for the two simple cases when one of the curves is either a line or conic.

Theorem 6.1. Let L P2 be a line, C P2 be a nondegenerate conic and D P2 ⇢ k ⇢ k ⇢ k be a curve defined by D :(Gd(X, Y, Z) = 0), where G is a form of degree d in X, Y, Z.AssumethatL D and C D, then the number of points in L D d and the number of points6⇢ in C D6⇢ 2d. When multiplicities are counted,{ \ } the inequality will still hold, and if{ k\is algebraically} closed, equality will hold.

2 Proof. First, consider a line L Pk. It is given by an equation = 0 for a linear form . Parametrically, it can⇢ be given as X = a(U, V ),Y = b(U, V ),Z = c(U, V ) where a, b, c are linear forms in U, V . 2 Next, consider a nondegenerate conic in Pk. Parametrically, it can be expressed as X = a(U, V ),Y = b(U, V ),Z= c(U, V )wherea, b, c are quadratic forms in U, V be- cause C is a projective transformation of XZ = Y 2, which in turn is parametrically (X, Y, Z)=(U 2,UV,V2). Thus, C is given by X U 2 (6.2) Y = M UV 0Z 1 0 V 2 1 for M, a nonsingular 3 3 matrix.@ A @ A So the intersections of L⇥and C with D are given by by finding the values of ratios (U : V ) that satisfy

F (U, V )=Gd(a(U, V ),b(U, V ),c(U, V )) = 0. For the intersection with L, F is a form of degree d and for the intersection with C, F is a form of degree 2d in U, V . So by Proposition 5.3, the theorem holds. ⇤ Next, we will prove a corollary, but before this, we must consider a quick propo- sition that will be used in the proof that follows. 6 JAMES W. KISELIK

Proposition 6.3. Let k be a field with at least four elements, and 2 2 C :(XZ = Y ) P ; ⇢ k if Q(X, Y, Z) is a quadratic form which vanishes on C, then Q = (XZ Y 2). 2 Proof. The equation for C is Q0 = XZ Y , so you can substitute Q0 XZ for 2 Y .ThusQ = Q0 + A(X, Z)+YB(X, Z)where is a linear form. Parametrize C so X = U 2,Y= UV, Z = V 2. This means that Q = 0 if and only if A(U 2,V2)+ UVB(U 2,V2)=0 k[U, V ]. And this is true if and only if A(X, Z)=B(X, Z)=0 because A(U 2,V2)2 only contains terms with even exponents and UVB(U 2,V2) only 2 2 2 2 contains terms with odd exponents in the form A(U ,V )+UVB(U ,V ). ⇤ 2 Corollary 6.4. If P1,...,P5 P are distinct points and no four points are 2 R collinear, there exists at most one conic through P1,...,P5.

Proof. Suppose for the sake of contradiction that C1 and C2 are conics such that C = C and P ,...,P C C . 1 6 2 { 1 5}⇢ 1 \ 2 C1 is nonempty, so first consider it if it were nondegenerate. Then C1 is pro- 2 2 1 jectively equivalent to the parametrized curve C1 = (U ,UV,V ) (U, V ) P . { | 2 2 2} By B´ezout’s theorem, C1 C2.IfQ2 is the equation of C2,thenQ2(U ,UV,V ) ⇢ 1 is equivalent to 0 for all (U, V ) P . The above proposition shows that Q2 is a multiple of (XZ Y 2), so this contradicts2 the fact that C = C . 1 6 2 If C1 is degenerate, then it’s a line or line pair. So C1 = L0 L1, and C2 = L0 L2 for distinct lines L and L .ThenC C = L (L L [). However, this would[ 1 2 1 \ 2 0 [ 1 \ 2 mean that four points are collinear, which is a contradiction. ⇤ 7. Space of All Conics 3 Definition 7.1. The space of all conics is defined as S2 = quadratic forms on R = { } 3 3 symmetric matrixes = R6. { ⇥ } ⇠ If Q S , express it as 2 2 Q = aX2 +2bXY + + fZ2. ··· For 2 P0 =(X0,Y0,Z0) P , 2 R consider the relation P C :(Q = 0). This will take the form 0 2 Q(X ,Y ,Z )=aX2 +2bX Y + + fZ2 =0. 0 0 0 0 0 0 ··· 0 For a fixed P0, this will be a linear equation in (a, b, . . . , f). So then, 5 6 S2(P0)= Q S2 Q(P0)=0 = R S2 = R { 2 | } ⇠ ⇢ and S (P ) is a 5-dimensional hyperplane. For P ,...,P 2 , define in the same 2 0 1 n PR manner 2 S (P ,...,P )= Q S Q(P ) = 0 for i =1,...,n . 2 1 n { 2 2| i } Then there are n linear equations in the 6 coecients (a, b, . . . , f) of Q. So we can state the following proposition. Proposition 7.2. dim S (P ,...,P ) 6 n. 2 1 n And in certain circumstances, equality will hold.

Corollary 7.3. If n 5 and no four points of P1,...,Pn are collinear, then dim S (P ,...,P )=6 n. 2 1 n CONICANDCUBICPLANECURVES 7

Proof. Corollary 6.4 implies that if n = 5, dim S2(P1,...,P5) 1. If n 4, then add points until there are 5 in total preserving the condition that no four points are collinear. Each added point imposes at most one linear condition, so 1 = dim S (P ,...,P ) dim S (P ,...,P ) (5 n) 1. Equality comes from 2 1 5 2 1 n proposition 7.2 and the first line of this proof. ⇤ Remark 7.4. If 6 points P ,...,P 2 are given, then they might or might not 1 6 PR lie on a conic. 2 Remark 7.5. B´ezout’s theorem states that two conics will intersect in four points. 2 Corollary 7.3 states that if there are four points P1,...,P4 P ,thenifthese 2 R points are not collinear, S2(P1,...,P4) has a dimension of 2. If you choose Q1,Q2 as a basis for S2(P1,...,P4), then there are two conics C1,C2 such that C1 C2 = P ,...,P . \ { 1 4} 8. Degenerate Conics in a Pencil

Definition 8.1. A pencil of conics is a family of the form C(,µ) :(Q1 +µQ2 = 0). Each element is a that depends linearly on the parameters (, µ), 1 which can be thought of as points of Pk. C(,µ) is degenerate if and only if det(Q1 + µQ2) = 0. This condition can also be expressed as abd ghj (8.2) F (, µ)=det bce+ µ hik =0 0 1 0 1 def jkl @ A @ A Because F (, µ) is a homogenous cubic form in , µ, proposition 5.3 can be applied to deduce a further proposition. 2 Proposition 8.3. Suppose C(,µ) is a pencil of conics of Pk such that F (, µ) is not identically zero (so it will have at least one nondegenerate conic). Then the pencil has at most 3 degenerate conics. If k = R, then the pencil as at least one . Proof. A cubic form has three zeros or fewer. Specifically for R, a cubic form must have at least one zero. ⇤ With that proposition, we will turn our attention to plane cubic curves.

9. Parametrized Cubic Plane Curves Definition 9.1. A is the curve defined by a homogenous form of degree 3, F (X, Y, Z) = 0 for projective coordinates X : Y : Z . { } Just like we were able to parametrize conic curves, some cubic plane curves can be parametrized. Proposition 9.2. There is a map that is bijective on points from one dimensional ane space to the nodal cubic C :(y2 = x3 + x2) R2. ⇢ Proof. Recalling the geometric argument used at the beginning of the paper, sub- stitute y = tx, a line through (0, 0) with slope t, into the equation and solve for x to determine the intersections of these lines and the curve. The equation becomes x3 + x2 =(tx)2, which simplifies to x3 +( t2 + 1)x2 = 0. Solutions for x are 0, 8 JAMES W. KISELIK with multiplicity of 2, and also t2 1. This means points of intersection of the conic and y = tx are (0, 0) with multiplicity of two and (t2 1,t3 t), so the curve is the image of the map : R1 R2 given by t (t2 1,t3 t). ! ! ⇤ Proposition 9.3. There is a map that is bijective on points from one dimensional ane space to the cuspidal curve C :(y2 = x3) R2. ⇢ Proof. Using the same method as above, we substitute into the equation and get x3 t2x2 = 0. The solutions are 0 with a multiplicity of two and t2, so points of intersection are (0, 0) with multiplicity of two and also (t2,t3). So the curve is the image of : R R2 where t (t2,t3). ! ! ⇤ However, not all cubic curves can be rationally parametrized.

10. The Curve (y2 = x(x 1)(x )) Has No Rational Parametrization Theorem 10.1. Let k be a field that is not of characteristic 2, and let k with =0, 1; let f,g k(t) be rational functions such that f 2 = g(g 1)(g 2). Then f,g6 k. 2 2 Proof. Because k[t] is a unique factorization domain, you can write p f = with p, q k[t] and coprime, q 2 r g = with r, s k[t] and coprime. s 2

Once denominators are cleared, the equation becomes p2s3 = q2r(r s)(r s). Because p, q are coprime, q2 divides s3 and because r, s are coprime, s3 divides q2. Therefore, q2 = as3 where a is a unit of k[t], so a k.Then,as =(q )2 is a 2 s square in k[t]. Furthermore, p2 = ar(r s)(r s), so there are nonzero constants b, c, d k such that br, c(r s), and d(r s) are all squares in k[t]. If r, s are shown2 to be constants, thenp, q are also constants, which proves the theorem. In order to prove r, s are constants, consider K, the algebraic closure of k, and notice that r, s K[t] satisfy the conditions of the following lemma, Fermat’s method of infinite descent.2 Lemma 10.2. Let K be an algebraically closed field. If r, s K[t] are coprime elements and 4 distinct linear combinations are squares in K[t]2, then r, s K. 2 Proof. Neither the hypothesis nor conclusion of the lemma is a↵ected if r, s are replaced with r0 = ar + bs, s0 = cr + ds,witha, b, c, d K and ad bc = 0. So the 4 given squares can be assumed to be r, r s, r s,2 s.Sor = u2,s=6 v2 with u, v K[t] coprime, and with max (deg u, deg v) < max (deg r, deg s). By contradiction,2 assume that max (deg r, deg s) > 0 and is the minimum amongst all r, s that satisfy the conditions of this lemma. Then both r s = u2 v2 = (u v)(u + v) and r s = u2 v2 =(u pv)(u + pv) are squares in K[t]. By the coprimeness of u, v, it can be concluded that u v, u + v, u pv, u + pv are squares. This is a contradiction because max (degr, deg s) was assumed to be minimal. ⇤ CONICANDCUBICPLANECURVES 9

11. Linear Systems The next goal of this paper is to lay the groundwork for proving Cayley-Bacharach Theorem.

Definition 11.1. Sd = forms of degree d in (X, Y, Z) . Any F Sd can be { i j k } 2 expressed uniquely as F =⌃aijkX Y Z with aijk k and the sum is taken over all i, j, k 0withi + j + k = d. 2 d+2 2 So Sd is a vector space such that dim Sd = 2 . For points P1,...,Pn P , let 2 S (P ,...,P )= F S F (P ) = 0 for i =1,...,n , d 1 n { 2 d | i } which is a subset of Sd. For Pi =(Xi : Yi : Zi), each F (Pi)=F (Xi,Yi,Zi)=0is one linear condition on F such that Sd(P1,...,Pn) is a vector space of dimension d+2 n. n Lemma 11.2. Suppose that k is an infinite field, and let F S . 2 d (1) Let L P2 be a line. If F 0 on L, then F is divisible in k[X, Y, Z] by ⇢ k ⌘ the equation of L. In other words, F = H F 0, where H is the equation of · L and F 0 Sd 1. 2 (2) Let C P2 be a nonempty nondegenerate conic. If F 0 on C, then F ⇢ k ⌘ is divisible in k[X, Y, Z] by the equation of C. In other words, F = Q F 0, · where Q is the equation of C and F 0 Sd 2. 2 Proof. For the first part of the lemma, change coordinates so that H = X. For any F Sd, there is a unique expression F = X Fd0 1 + G(Y,Z). All monomials involving2 X are put in the first term of the sum· and the remainder must be a polynomial only in terms of Y,Z.SoF 0 on L if and only if G 0 on L if and only if G(Y,Z) = 0 because of proposition⌘ 5.3. ⌘ The second part of the lemma is very similar to proposition 6.3. First, change coordinates so that Q = XZ Y 2.SubstituteXZ Q for Y 2 where it occurs in F and the remaining terms have degree less than or equal to 1 in Y .Thusthese terms are expressed A(X, Z)+YB(X, Z). C is the parametrized conic given by X = U 2,Y = UV,Z = V 2,soF 0 on C if and only if ⌘ A(U 2,V2)+UVB(U 2,V2) 0 ⌘ on C, which is equivalent to A(U 2,V2)+UVB(U 2,V2)=0 k[U, V ]. 2 By considering terms of even and odd degrees in the form A(U 2,V2)+UVB(U 2,V2), we can conclude that A(X, Z)=B(X, Z) = 0. ⇤ 2 2 Corollary 11.3. Let L :(H = 0) Pk be a line and let C :(Q = 0) Pk be a ⇢ 2 ⇢ nondegenerate conic. If points P1,...,Pn Pk are given, consider Sd(P1,...,Pn) for some fixed d, then 2 (1) If P ,...,P L, P ,...,P L,anda>d, then 1 a 2 a+1 n 62 Sd(P1,...,Pn)=H Sd 1(Pa+1,...,Pn). · (2) If P ,...,P C, P ,...,P C,anda>2d, then 1 a 2 a+1 n 62 Sd(P1,...,Pn)=Q Sd 2(Pa+1,...,Pn). · 10 JAMES W. KISELIK

Proof. Both proofs are e↵ectively identical, so I will just prove the first. If F is homogenous and of degree d, and the curve D :(F = 0) meets L in points P1,...,Pa with a>d, we can use B´ezout’s Theorem. Because L D, by lemma 11.2, F = H ⇢ · F 0. Because Pa+1,...,Pn L, then we conclude that F 0 Sd 1(Pa+1,...,Pn). ⇤ 62 2 12. Cayley-Bacharach Theorem Definition 12.1. A set of points is called conconic if they all lie on a nondegenerate conic.

2 Proposition 12.2. Let k be an infinite field, and points P1,...,P8 P are dis- 2 k tinct. If no 4 of P1,...,P8 are collinear, and no 7 lie on a nondegenerate conic, then dim S3(P1,...,P8)=2. Proof. There are three cases. First, consider the case in which no 3 points are collinear and no 6 are conconic. Suppose for the sake of contradiction that dim S (P ,...,P ) 3, and let P ,P 3 1 8 9 10 be distinct points that lie on the line L = P1P2.Then dim S (P ,...,P ) dim S (P ,...,P ) 2 1. 3 1 10 3 1 8 So there exists F =0withF S (P ,...,P ). But by corollary 11.3, F = H Q 6 2 3 1 10 · with Q S2(P3,...,P8). So there is a contradiction because Q can be either non- degenerate,2 but have 6 conconic points, or it can be a line pair (or double line) with at least 3 collinear points.

The next two cases are degenerate. The first is where P ,P ,P L are collinear 1 2 3 2 and L :(H = 0). Let P9 be a fourth point on L. By corollary 11.3, S (P ,...,P )=H S (P ,...,P ). 3 1 9 · 2 4 8 Because no 4 points of P4,...,P8 are collinear, by corollary 7.3, we have that dim S2(P4,...,P8) = 1, and so dim S3(P1,...,P9) = 1. This implies, then, that dim S (P ,...,P ) 2. 3 1 8 

Finally, let P1,...,P6 C be conconic with C :(Q = 0), a nondegenerate conic. Choose P Q to be distinct2 from P ,...,P . By corollary 11.3, 9 2 1 6 S (P ,...,P )=Q S (P ,P ). 3 1 9 · 1 7 8 Because the line L = P7P8 is unique, S3(P1,...,P9) is the one dimensional space that is spanned by QL. Therefore, dim S (P ,...,P ) 2. 3 1 8  ⇤ Now, the Cayley-Bacharach Theorem follows almost immediately from this propo- sition.

Corollary 12.3. Let C1,C2 be cubic curves whose intersection consists of nine distinct points, C C = P ,...,P . If a cubic D passes through P ,...,P , 1 \ 2 { 1 9} 1 8 then it also passes through P9.

Proof. If any 4 points of P1,...,P8 were on a line L,thenC1 and C2 would both meet L in 4 points, and thus would contain L. This contradicts the assumption that the intersection is nine distinct points. Similarly, no 7 points can be conconic. This means that the assumptions of 12.2 are satisfied, so dim S3(P1,...,P8) = 2. Thus, the equations F1,F2 of the cubics C1,C2 form a basis of S3(P1,...,P8). This CONICANDCUBICPLANECURVES 11 means D :(G = 0) where G = F1 + µF2. Both F1,F2 vanish at P9,soG does as well. ⇤ 13. Pascal’s Theorem A simple and fun application of Cayley-Bacharach theorem is Pascal’s theorem, 2 or ’the mystic hexagon.’ The theorem concerns a hexagon ABCDEF in Pk with its pairs of opposite sides extended until they meet at points P, Q, R. Of note, the fact that the hexagon is in projective space addresses the issue that will arise in, for instance, a hexagon where the opposite sides will meet at infinity. Theorem 13.1. For a hexagon as described above, the points A, B, C, D, E, F are conconic if and only if P, Q, R are collinear. Proof. If lines through AF and CD meet at P , through AB and DE meet at Q, and through BC and EF meet at R, then we can consider two triples of lines

L1 : AF P, L2 : DEQ, L3 : BCR and M1 : CDP, M2 : ABQ, M3 : EFR.

So we can define two cubics C1 = L1 +L2 +L3 and C2 = M1 +M2 +M3. It’s obvious that C1 C2 = A, B, C, D, E, F, P, Q, R so now we can apply Cayley-Bacharach theorem.\ { } ( )First,ifPQR are collinear with L = PQR, then define to be the conic through( A, B, C, D, E, which exists and is unique by corollary 7.3. Thus L +isa cubic though A, B, C, D, E, P, Q, R, so by Cayley-Bacharach theorem it must also contain F . F cannot be on the line L,soF . This proves that the six points A, B, C, D, E, F are conconic. 2 ( ) Conversely, if ABCDEF are on a conic , and L = PQ then L +isa cubic) passing through A, B, C, D, E, F, P, Q, so by Cayley-Bacharach theorem, it also passes through R. R cannot be on or else is a line pair, so R L, and 2 PQR are collinear. ⇤ Acknowledgments. I would like to thank my mentor, Daniel Le, for proposing a topic I enjoyed studying, patiently explaining concepts, and asking questions that helped me to better understand. I would also like to thank Peter May for organizing the REU.

References [1] Miles Reid Undergraduate University of Warwick. 2013.