Theorem 6.1 The addition defined above makes the points of E into an abelian group with O as the identity element. Proof. Let s assume that K is

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1 6 Elliptic curves Elliptic curves are not ellipses. The name comes from the elliptic functions arising from the integrals used to calculate the arc length of ellipses. Elliptic curves can be parametrised by elliptic functions in a similar way as circles can be parametrised by sine and cosine. Elliptic curves have the very special property that their points also have a natural commutative group structure. Elliptic curves have been studied intensively in number theory, they played a crucial part in Wiles s proof of Fermat s Last Theorem. Elliptic curves over finite fields have a great practical importance, too, their groups of points can be used for a public key cryptography algorithm similar to RS, but as the group structure is more complicated, smaller primes can be used to achieve the same level of security. Definition. n elliptic curve is a projective variety isomorphic to a nonsingular curve of degree 3 in P 2 together with a distinguished point O E. Definition. (Chord and tangent process) Let E be a non-singular curve of degree 3 in P 2 and fix O E. For, B E define + B E as follows: let Q be the third intersection point of the line B with E, and then + B is the third intersection point of the line OQ with E. B Q Q B O If = B or O = Q, then the line B or the line OQ is taken to be the tangent line at or O, resp. If any line in this construction is tangent to E, the third intersection point is defined using intersection multiplicities. 75

2 The picture below shows the calculation of 2. We take the tangent line at, let Q be the third intersection point of this tangent line with E, and then 2 is the third intersection point of the line OQ with E. Q 2 Q O To calculate the negative of a point in this group structure, first take the tangent line at O and let M be its 3rd point of intersection with E. To find, just take the line through and M, its 3rd point of intersection with E is. O M 76

3 Worked examples of calculations on elliptic curves can be found in the handout pdf. lthough elliptic curves are projective, calculations are usually carried out in affine coordinates. If the affine equation has the special form y 2 = x 3 + ax 2 + bx + c, the so-called Weierstraß form, then its projective closure has equation Y 2 Z = X 3 +ax 2 Z +bxz 2 +Z 3. The only point of intersection with the line Z = 0 is (0 : 1 : 0), often called simply the point at. By convention, if the equation is in this form, O is taken to be (0 : 1 : 0) even if it is not mentioned explicitly. If a, b, c R, then x 3 + ax 2 + bx + c = 0 can have 1 or 3 real roots, in the first case the elliptic curve y 2 = x 3 + ax 2 + bx + c has one real component like the curve below on the left, in the second case the elliptic curve has two real components like the curve below on the right The lines passing through (0 : 1 : 0) are the lines parallel to the y-axis in affine terms. If the point Q in the addition process has co-ordinates (u, v), the line through O and Q is the line x = u, and since as the curve is symmetric about the x-axis, its 3rd intersection point with E is (u, v) = + B. The line Z = 0 and E have intersection multiplicity 3 at (0 : 1 : 0), therefore the point M in the construction of is also O, so if = (s, t), then = (s, t). The calculations are much simpler if the equation is in Weierstraß form, one of our goals will be to transform elliptic curves into this form. The diagrams below illustrate the addition of two points, doubling a point and taking the negative of a point on an elliptic curve is Weierstraß form. 77

4 Q 4 2 B Q x x x 2 Q 2 2 Q B Theorem 6.1 The addition defined above makes the points of E into an abelian group with O as the identity element. Proof. Let s assume that K is algebraically closed, otherwise we can replace it by its algebraic closure. It follows directly from the addition process that O + = for every E, + B = B + for every, B E and that the point constructed above satisfies + ( ) = 0. The substantial part of the proof is to prove associativity. Let, B, C E. The diagram below shows the calculation of ( + B) + C and + (B + C), except for the last step. S is the 3rd intersection point of the line C Q with E, so ( + B) + C is the 3rd intersection point of the line OS with E. S is the 3rd intersection point of the line R with E, so + (B + C) is the 3rd intersection point of the line OS with E. Therefore in order to prove the equality ( + B) + C = + (B + C) it is sufficient to show that S = S. Let s assume that O,, B, C, Q, Q, R, R are all distinct and they are not equal to S or S. 78

5 O C Q B B R R B C Q S S' Let L 1 = 0 be the equation of the line BQ, let L 2 = 0 be the equation of the line OR R and let L 3 = 0 be the equation of the line C QS. Let F be the cubic curve defined by L 1 L 2 L 3 = 0, consisting of the union of the green lines in the diagram. Let M 1 = 0 be the equation of the line BCR, let M 2 = 0 be the equation of the line OQ Q and let M 3 = 0 be the equation of the line RS. Let G be the cubic curve defined by M 1 M 2 M 3 = 0, consisting of the union of the red lines in the diagram. E F {O,, B, C, Q, Q, R, R, S}. s E is irreducible, it has no common component with F, therefore the intersection must consist of exactly these 9 points by Bézout s Theorem. Then by Theorem 5.3, any cubic curve that passes through O,, B, C, Q, Q, R, R also passes through S. Similarly, E G = {O,, B, C, Q, Q, R, R, S }. Since G contains O,, B, C, Q, Q, R, and R, as we noted above it must also contain S. Since S is different from the other 8 points, this implies S = S as required. Now let s consider the assumption that O,, B, C, Q, Q, R, R are all distinct and they are not equal to S or S. This implies O, then given, the points O,, B, Q and Q will be all distinct for all but finitely many choices of B. Similarly, for given and B, the points O,, B, C, Q, Q, R, R will be all distinct and not equal to S or S for all but finitely many choices of C. Therefore ( + B) + C = + (B + C) holds for (, B, C) in a non-empty subset of E E E. Over R or C we can use continuity to prove ( + B) + C = + (B + C) for all, B and C. Over 79

6 any field, we can make E E E into a variety (this is not as simple as in the affine case because P m P n = P m+n, E E E will be a projective variety in P 26 ), and then we have two morphisms Φ 1, Φ 2 : E E E E, Φ 1 (, B, C) = ( + B) + C, Φ 2 (, B, C) = + (B + C), which agree outside a proper subvariety of E E E E. s E E E E is irreducible, this implies Φ 1 (, B, C) = Φ 2 (, B, C) for all, B, C. We noted earlier that if the equation of the elliptic curve is of the form Y 2 Z = X 3 + ax 2 Z + bxz 2 + Z 3, then the line Z = 0 only intersects the curve at (0 : 1 : 0), so by Bézout s Theorem, the intersection multiplicity at that point must be 3. This implies that the tangent line to the elliptic curve at (0 : 1 : 0) is the line Z = 0 and furthermore, that (0 : 1 : 0) is a special point, since in general a curve and the tangent line to it at a point only have intersection multiplicity 2. Definition non-singular point P of a plane curve C (affine or projective) is called an inflection point if and only if the tangent line to C at P has intersection multiplicity at least 3 with C at P. Warning: Inflection points are preserved under affine or projective equivalence, but not necessarily under more general isomorphisms of varieties. Proposition 6.2 Let E be a non-singular curve of degree 3 in P 2 and assume that O E is an inflection point. (i) point P O has order 2 in the group structure on E if and only if the tangent line at P passes through O. (ii) point P O has order 3 in the group structure on E if and only if it is an inflection point. Proof. (i) 2P = O is equivalent to P = P. s O is an inflection point, P is simply the 3rd intersection point of the line OP with E as shown on the diagram below on the left. It is equal to P if and only if the line OP is tangent to E at P. 80

7 O O P P Q 2P P Q 2P (ii) 3P = O is equivalent to P = 2P. Let Q be the 3rd intersection point of E with the tangent line to E at P. Then 2P = Q is the 3rd intersection point of the line OQ with E as shown on the diagram above on the right. By the method of calculating the negative of a point, Q = Q = 2P. P = 2P if and only if P = Q, i. e., if and only if P is an inflection point. In particular, (i) implies that if the equation of the curve is in the form y 2 = x 3 + ax 2 + bx + c, the points of order 2 are the points where the tangent line is parallel to the y-axis, which are the points whose y co-ordinate 0 and whose x co-ordinate is one of the roots of x 3 + ax 2 + bx + c. Example: Find the points of order 2 on the elliptic curve with affine equation y 2 = x 3 2x 4 over C. The points of order 2 are the points of the form (α, 0), where α is a root x 3 2x 4. x 3 2x 4 = (x 2)(x 2 + 2x + 2), so the roots are 2 and 1 ± i, hence the points of order 2 are (2, 0), ( 1 + i, 0) and ( 1 i, 0). It follows from part (ii) of the above proposition that the 3rd point of intersection of a line through two inflection points of an elliptic curve is also an inflection point. Definition. Let f(x, y) K[x, y]. The Hessian of f is the determinant f xx f xy f x f xy f yy f y f x f y 0, where f x denotes the partial derivative of f with respect to x, etc. Let F (X, Y, Z) K[X, Y, Z] be a homogeneous polynomial. The Hessian of F is the determinant F XX F XY F XZ F XY F Y Y F Y Z F XZ F Y Z F ZZ. 81

8 Proposition 6.3 Let C be a curve in 2 or in P 2, defined by an irreducible polynomial f(x, y) K[x, y] or by an irreducible homogeneous polynomial F (X, Y, Z) K[X, Y, Z]. Let H be the Hessian of f or F and let D be the curve H = 0. In the affine case the inflection points of C are the elements of C D which are non-singular points of C, in the projective case the same holds under the additional assumption that if the field K has finite characteristic, the characteristic does not divide deg F 1. Proof. In the affine case the slope of the tangent line to the curve f(x, y) = 0 is f x /f y by implicit differentiation. Inflection points are critical points of f x /f y on C, which occur where the gradients of f x /f y and f are parallel, including the possibility that the former is (0, 0). ( fxx f y f xy f x ( f x /f y ) = f 2 x, f xyf y f yy f x f 2 x so the condition for parallelarity with f = (f x, f y ) is f x (f xy f y f yy f x ) f y (f xx f y f xy f x ) = 0. The expression on the left-hand side is exactly the Hessian of f. point of C is singular if and only if f x = f y = 0 at that point, if this holds, then H = 0 at that point automatically, but if f x, f y are not both 0, i. e., the point is non-singular, then the above equation implies that ( f x /f y ) is a scalar multiple of f, so the slope of the tangent line has a critical point there and the point is an inflection point. In the projective case let f K[x, y], f(x, y) = F (x, y, 1) be the dehomogenisation of F with respect to Z, then F (X, Y, Z) = Z d f(x/z, Y/Z), where d = deg F. By using the chain rule and the product rule, we can express the partial derivatives of F in terms of those of f and after some calculation it turns out that F XX F XY F XZ ( f xx f xy f x ) F XY F Y Y F Y Z F XZ F Y Z F ZZ = (d 1)2 Z 3(d 2) f xy f yy f y f x f y 0 + d d 1 (f xxf yy (f xy ) 2 )f. Therefore the intersection points of C and D which do not lie on the line Z = 0 correspond to the intersection points of the affine curve f = 0 and the curve defined by the Hessian of f, so if they are non-singular points of C, they are inflection points. The definition of the Hessian for homogeneous polynomials is symmetric in X, Y and Z, therefore we can also use this argument with X and Y to deduce that all elements of C D are inflection points of C if they are non-singular points of C. 82 ),

9 The diagram below shows a cubic curve (blue, thicker), its Hessian (red, thinner), the inflection points, and the tangent lines to the original cubic curve at the inflection points. There are some special phenomena for cubic curves, which do not happen for curves of higher degree. The tangent lines are also tangent to the Hessian curve, and it is also true that the inflection points of the Hessian curve are the same as those of the original curve, although this cannot be seen in this diagram. s the projective Hessian has degree 3(d 2), the number of inflection points is at most 3d(d 2) and for general curves over an algebraically closed field this number is achieved. If d = 3, the number of inflection points is 9 (if the characteristic of the field is not 3). If the coefficients of the elliptic curve E are real, then at least one inflection point has to be real, since non-real complex solutions come in conjugate pairs. In fact, an elliptic curve over R always has 3 real inflection points. Example: Find the real inflection points of the curve y 2 = x 3 + 4x 2 + 3x 1. Let f(x, y) = x 3 + 4x 2 + 3x 1 y 2. Its Hessian is 6x x 2 + 8x + 3 H = 0 2 2y 3x 2 + 8x + 3 2y 0 = 2(3x2 + 8x + 3) 2 4y 2 (6x + 8). To find the solutions f = H = 0, we consider H 4(6x + 8)f = 2(3x 2 + 8x + 3) 2 4(6x + 8)(x 3 + 4x 2 + 3x 1) = 6x 4 32x 3 36x x + 50, 83

10 which only involves x. x = 1 is a root, the corresponding values of y are y = ± 7. These are the real inflection points, together with the point at infinity, (0 : 1 : 0). 6x 4 32x 3 36x x + 50 = 0 has another real root x 3.473, but the corresponding values of y are imaginary. The diagram below show this elliptic curve, its Hessian curve, the two inflection points and the tangent lines there x 2 4 Since if O is an inflection point, then the inflection points are exactly the points P satisfying 3P = 0, so over an algebraically closed field of characteristic other than 3, there are 9 such points. From Proposition 6.2 (i) and the remarks after the proof it follows that if the equation of the elliptic curve is in Weierstraß form and the field K is algebraically closed, then then there are 4 points satisfying 2P = O, O itself and the 3 points of order 2. These are special cases of a more general phenomenon. The number of points P on an elliptic curve such that np = O is n 2 if the field K is algebraically closed and its characteristic does not divide n, in particular this holds over C. If the characteristic of K is a prime p, then the number of points such that p k P = 0 may be p k or just 1, in the latter case the curve is called supersingular (not related to the definition of singular points, the elliptic curve is always a non-singular variety). 84

11 Examples: 1. Let K be an algebraically closed field of characteristic 2. The elliptic curve defined by the affine equation y 2 + y = x 3 + ax 2 + bx + c and O = (0 : 1 : 0) is supersingular for any a, b, c K. It has no point of order 2, because any line passing through O has affine equation x = α for some α K. y 2 + y = α 3 + aα 2 + bα + c has two distinct solutions for any α K, they are of the form β, β + 1 for some β K, so the line x = α is never tangent to the curve y 2 + y = y 2 + y = x 3 + ax 2 + bx + c. 2. Let K be an algebraically closed field of characteristic 3. The elliptic curve defined by the affine equation y 2 = x 3 + bx + c and O = (0 : 1 : 0) is supersingular for any b K \ {0}, c K. In this case the Hessian turns out to be 2b 2, a non-0 constant, so there are no inflection points other than O = (0 : 1 : 0) and there are no points of order 3. Theorem 6.4 If K is algebraically closed and its characteristic is not 2 or 3, any non-singular cubic curve in P 2 is projectively equivalent to one with an equation of the form Y 2 Z = X 3 + pxz 2 + qz 3 (y 2 = x 3 + px + q in affine form). Proof. Step 1. Choose an inflection point of E and then do a projective transformation that maps that inflection point to (0 : 1 : 0) and the tangent line at the inflection point to the line Z = 0. This step eliminates the Y 3, XY 2 and X 2 Y terms from the equation. fter this step the equation can be converted to affine form and the remaining steps can be carried out in affine form, if preferred. Step 2. Write the equation with the terms containing Y on one side and the other terms, only containing X and Z on the other side. Let α and β be the coefficients of Y 2 Z and X 3, resp. Multiply the whole equation by β 2 /α 3 and then use βx/α and βy/α as new variables, this will make the coefficients of Y 2 Z and X 3 equal to 1. (In the affine form the equation will look like y 2 + a 1 xy + a 3 y = x 3 + a 2 x 2 + a 4 x + a 6, this is often called a Weierstraß form, too.) Step 3. Complete the square with respect to Y to eliminate XY Z and Y Z 2 terms (xy and y in the affine form). (This is where we need that the characteristic is not 2.) Step 4. Complete the cube with respect to X to eliminate the X 2 Z term (x 2 in the affine form). (This is where we need that the characteristic is not 3.) worked example of this procedure form can be found at 85

12 maths.manchester.ac.uk/~gm/teaching/mth32062/ellipticequation.pdf. s we noted previously, a cubic curve defined by an equation with real coefficients always has real inflection points, so the 1st step can be carried out over R and we obtain an equation with p, q real. However, if the coefficients are rational, there is no guarantee that there exists an inflection point with rational coefficients, so p, q may not be rational. Remark. p and q are not unique, we can multiply the equation y 2 = x 3 +px+q by α 6 for some α K \ {0} and then we can rewrite it as (α 3 y) 2 = (α 2 x) 3 + (α 4 p)(α 2 x) + α 6 q, so the parameters α 4 p and α 6 q determine an isomorphic curve. Definition. The j-invariant of the curve y 2 = x 3 4p 3 +px+q is j = p q. 2 The rest of this chapter is not examinable. nother common form of the equation of the elliptic curve is the Legendre form y 2 = x(x 1)(x λ), where λ 0, 1. j can be expressed in terms of λ as j = 256 (λ2 λ + 1) 3 λ 2 (λ 1) 2. (*) λ is not uniquely defined, depending on which two roots of the cubic are chosen to be mapped to 0 and 1, it can take 6 values, λ, 1 λ, 1/λ, 1/(1 λ), λ/(λ 1) and (λ 1)/λ, but they all give the same value of j. It is not clear from either of these definitions that j is really an invariant of the curve E, since there are choices in various steps of the process of transforming the equation to the Weierstraß or Legendre form. If E is a non-singular curve in P 2, then from any point E one can draw four tangent lines to E. Let λ be the cross ratio of the slopes of these lines, then the formula (*) gives the j-invariant. This implies that the j invariant is invariant under projective transformations. It requires some more sophisticated tools to show that if two cubic curves in P 2 are isomorphic, then they are projectively equivalent. Theorem 6.5 j is indeed an invariant of the elliptic curve, i. e., all possible Weierstraß and Legendre forms of the same curve give the same value for j. If the field K is algebraically closed, two elliptic curves over K are isomorphic if and only if they have the same j-invariant. Non-singular cubic curves with different j invariant are not just not isomorphic, but they are not birationally equivalent either. Non-singular cubic curves are not rational, i. e., they are not birationally equivalent to P 1. The study of points with rational co-ordinates is a very active area of number 86

13 theoretical research. Mordell s Theorem states that the group of rational points is finitely generated, i. e., as an abstract group it is isomorphic to T Z r, where T is a finite group and r 0 is a non-negative integer, called the rank of the elliptic curve. Mazur s Theorem states that the group T has at most 12 elements. It is not known whether r can be arbitrarily large, but there exists an example with r