THE WEIL PAIRING ON ELLIPTIC CURVES


 Peter Dalton
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1 THE WEIL PAIRING ON ELLIPTIC CURVES Background NonSingular Curves. Let k be a number field, that is, a finite extension of Q; denote Q as its separable algebraic closure. The absolute Galois group G k = GalQ/k = lim GalF/k is the K projective limit of Galois groups associated with finite, normal separable extensions F/k. Let I k[x 1, x 2,..., x n ] be an ideal, and define the sets { } XQ = P A n Q fp = 0 for all f I { } IX = f Q[x 1, x 2,..., x n ] fp = 0 for all P XQ I k Q. Since G F G k acts on Q, we define XF = XQ G F = XQ A n F, namely the F rational points, as the points fixed by this action. We think of X as a functor which takes fields F to algebraic sets XF, and say that X is an affine variety over k if IX Q[x 1, x 2,..., x n ] is a prime ideal. Proposition 1. Let X be an affine variety over k, and define the integral domain OX = Q[x 1, x 2,..., x n ]/IX. Then the map XQ mspec OX which sends P = a 1, a 2,..., a n to m P = x 1 a 1, x 2 a 2,..., x n a n is an isomorphism. Proof. The map is welldefined O/m P Q is a field. Conversely, let m be a maximal ideal of O. Fix a surjection O O/m Q, and denote a i Q as the image of x i O. It is easy to check that m = m P for P = a 1, a 2,..., a n. We define O = OX as the global sections of X or the coordinate ring of X. Often, we abuse notation and write X = Spec O. If we denote K = QX as its quotient field, we define the dimension of X as the transcendence degree of K over Q. We say that X is a curve if dimx = 1. Theorem 2. Let X be a curve over k, and write the ideal I = f 1, f 2,..., f m K[x 1, x 2,..., x n ] so that dimx = n m = 1. The following are equivalent: i. For each P XQ, the m n matrix f 1 f 1 f 1 P P P x 1 x 2 x n f 2 f 2 f 2 P P P Jac P X = x 1 x 2 x n f m f m f m P P P x 1 x 2 x n 1
2 yields an exact sequence: {0} T P X A n Q Jac P X A m Q {0}. That is, the Jacobian matrix Jac P X has rank m while the tangent space has dimension dim Q TP X = dimx. ii. The Zariski cotangent space has dimension dim Q m/m 2 = dimx for each maximal ideal m mspec O. iii. For each P XQ, denote O P as the localization of O at m P. Then m P O P is a principal ideal. iv. For each P XQ, O P is a discrete valuation ring. v. For each P XQ, O P is integrally closed. vi. O is a Dedekind Domain. This is essentially a a restatement of Proposition 9.2 on pages in AtiyahMacdonald. If any of these equivalent statements holds true, we say that X is a nonsingular curve. Proof. i ii. We have a perfect i.e., bilinear and nondegenerate pairing mp /m 2 P TP X Q defined by f, b1, b 2,..., b n n f P b i. x i Hence dim Q mp /m 2 P = dimq TP X = n m = dimx. ii iii. As m = m P is a maximal ideal, Nakayama s Lemma states that we can find ϖ m P where ϖ / m 2 P. Consider the injective map O/m P m P /m 2 P defined by x ϖ x. Clearly this is surjective if and only if m P O P = ϖ O P is principal. Recall now that dim Q O/mP = 1. iii = iv. Say that m P O P = ϖ O P as a principal ideal. In order to show that O P is a discrete valuation ring, it suffices to show that any nonzero x O P is in the form x = ϖ m y for some m Z and y O P. Consider the radical of the ideal generated by x: { } x = y O P y n x O P for some nonnegative integer n. As O P has a unique nonzero prime ideal, we must have x = m P O P. But then there is largest nonnegative integer m such that t m 1 / x O P yet ϖ m x O P. Hence y = x/ϖ m O P but y / m P. iv = v. Say that O P is a discrete valuation ring. Say that x K is a root of a polynomial equation x n + a 1 x n a n = 0 for some a i O P. Assume by way of contradiction that x / O P. Then v P x < 0, so that v P 1/x > 0, hence y = 1/x is an element of O P. Upon dividing by x n 1 we have the relation x = a 1 + a 2 y + + a n y n 1 O P. This contradiction shows that O P is indeed integrally closed. v = iii. Say that O P is integrally closed. We must construct an element ϖ O P such that m P O P = ϖ O P. Fix a nonzero x m P. By considering the radical x and noting that m P O P is a finitely generated O P module, we see that there exists some m Z such that m m P O P x O P yet m m 1 P O P x O P. Choose y m m 1 P such that y / x O P, and let ϖ = x/y be an element in K. Consider the module 1/ϖ m P O P O P ; we will show equality. As y / O P, we have 1/ϖ / O P, so that 1/ϖ is not integral over O P. Then 1/ϖ m P O P cannot be a finitely generated O P module, 2 i=1
3 we have 1/ϖ m P O P m P. As there is an element of 1/varpi m P O P which is not in m P, we must have equality: 1/ϖ m P O P = O P. Hence m P O P = ϖ O P as desired. v vi. A Dedekind domain is a Noetherian integral domain of dimension 1 that is integrally closed. But the localization O P is integrally closed for each maximal ideal m P if and only if O is integrally closed. Consult Theorem 5.13 on page 63 of AtiyahMacdonald. Examples. Choose {a 1, a 2, a 3, a 4, a 6 } k, and consider the polynomial fx, y = y 2 + a 1 x y + a 3 y x 3 + a 2 x 2 + a 4 x + a 6. Then X : fx, y = 0 is a curve over K. Define the Krational numbers b 2 = a a 2 b 4 = 2 a 4 + a 1 a 3 b 6 = a a 6 b 8 = a 2 1 a a 2 a 6 a 1 a 3 a 4 + a 2 a 2 3 a 2 4 Then X is nonsingular if and only if 0. c 4 = b b 4 c 6 = b b 2 b b 6 = b 2 2 b 8 8 b b b 2 b 4 b 6 Choose {a 0, a 1, a 2, a 3, a 4 } k, and consider the quartic polynomial fx = a 4 x 4 + a 3 x 3 + a 2 x 2 + a 1 x + a 0. Then X : y 2 = fx is a curve over k. If X has a krational point P = x 0, y 0, then it is birationally equivalent over k to the cubic curve v 2 = u 3 + A u + B in terms of A = a a 1 a 3 12 a 0 a 4 3 B = 2 a3 2 9 a 1 a 2 a a 0 a a2 1 a 4 72 a 0 a 2 a Then X is nonsingular if and only if 16 discf = 16 4 A B 2 = 0. The RiemannRoch Theorem Let X be a nonsingular curve over k = C. From now on, we will identity X with Xk, and embed X C. We ll explain how to choose such an embedding later. Meromorphic Functions. Let k = C denote the complex numbers. Let X C be a compact Riemann surface. We will denote O as the ring of holomorphic i.e., analytic functions on X, and K as the field of meromorphic functions on X. Let me explain. Say that f : U C is a function defined on an open subset U X. Using the embedding X R R which sends x + i y x, y, we say that f is smooth if fz = ux, y + i vx, y in terms of smooth functions u, v : U R, where z = x + i y. We may denote the set of all such by C U. By considering the identities f z = 1 f 2 x i f = 1 u y 2 x + v + i 1 v y 2 x u y f z = 1 f 2 x + i f = 1 u y 2 x v y 3 + i 1 v 2 x + u y
4 we see that the CauchyRiemann Equations imply that fz is holomorphic or antiholomorphic, respectively on U if and only if f/ z = 0 or f/ z = 0, respectively. Note that fz is holomorphic if and only if f z is antiholomorphic. Denote OU as the collection of such holomorphic functions on U. Since this is an integral domain, we may denote K U as its function field; this is the collection of meromorphic functions on U. The following diagram may be useful: We will denote O = OX and K = K X. {0} OU K U C U Meromorphic Differentials. Continue to let U X be an open subset. Denote Ω 0 C U, the collection of differential 0forms on U, as the set of smooth functions f on U. Similarly, denote Ω 1 C U, the collection of differential 1forms on U, as the set of sums ω = f dx + g dy = f i g 2 dz + f + i g 2 where f and g are smooth functions on U. Hence we have a canonical decomposition Ω 1 C U = Ω 1,0 C U Ω 0,1 C U as the direct sum of 1forms in the form ω = f dz or ω = f d z, respectively where f is a smooth function on U. In particular, ω Ω 1,0 C U or ω Ω 0,1 C U, respectively if and only if g = i f or g = i f, which happens if and only if ω z = i ωz. As complex conjugation acts on the set Ω 1 C U of differential 1forms via ωz ω z, we see that we may identify Ω 1 C U = Ω 1,0 C U and Ω 1 C U + = Ω 0,1 C U as the eigenspaces corresponding to the eigenvalues i, respectively. We have a differential map d : Ω 0 C U Ω 1 C U defined by f df = f f dz + z z d z. We say that a 1form ω is a holomorphic differential or antiholomorphic differential, respectively if ω = f dz or ω = f d z, respectively for some holomorphic or antiholomorphic, respectively function f on U. Denote ΩU as the collection of holomorphic differentials on U. Similarly, we say that a 1form ω is a meromorphic differential or antimeromorphic differential, respectively if ω = f/g dz or ω = f/g d z, respectively for some holomorphic or antiholomorphic, respectively functions f and g on U. Denote Ω K U as the collection of meromorphic differentials on U. The following diagram may be useful: {0} ΩU Ω K U Ω 1,0 C U Note that ΩX is the collection of holomorphic differentials on X. Homology Groups. Let H 1 X, Z denote the free abelian group of closed loops γ in X. It is wellknown that H 1 X, Z Z 2g for some nonnegative integer g; we call g the genus of X. Complex conjugation γ γ acts on these closed loops, so we may consider eigenspaces corresponding to the eigenvalues 1 either reversing or preserving direction generated by this involution: H 1 X, Z = H 1 X, Z H 1 X, Z + where H 1 X, Z Z g. Upon tensoring with C, we have the homology group H 1 X, C C 2g, with eigenspaces H 1 X, C C g. We have a nondegenerate, bilinear pairing H 1 X, C ΩX C, n i γ i, ω n i ω. i i γ i Note here that ω must be a holomorphic differential on X, so that each loop γ i H 1 X, Z. This implies the following results: 4 d z
5 Proposition 3. Let OX be the collection of such holomorphic functions on X, ΩX be the collection of holomorphic differentials on X, and H 1 X, Z Z 2g be the free abelian group of closed loops γ in X. ΩX Hom C H1 X, C, C C g. As the map O ΩX defined by f f dz is an isomorphism, we see that ΩX is an Omodule of rank 1, but a complex vector space of dimension g. Examples. The unit sphere is given by S 2 R = { } u, v, w R 3 u 2 + v 2 + w 2 = 1. Stereographic Projection is the map π : C S 2 R defined by 2 Rez πz = z 2 + 1, 2 Imz z 2 + 1, z 2 1 z 2 with inverse π 1 u, v, w = u + i v w. Of course, the inverse sends the north pole u, v, w = 0, 0, 1 to z =, so we actually find a birational equivalence between X = P 1 C = C { } and S 2 R. We consider X a compact Riemann surface although it cannot really be imbedded in the complex plane. Consider the differential 1form ω = dz. This is clearly a holomorphic differential on A 1 C = C, but upon making the substitution w = 1 z = ω = dz = dw w 2 we see that ω is not holomophic on X = P 1 C. In fact, X has no nonzero holomorphic differentials only meromorphic ones! so its genus must be g = 0. Fix complex numbers g 2, g 3 such that g g2 3. We define a meromorphic map : C C implicitly via the relation z = z dx 4 x 3 g 2 x g 3 = z 2 = 4 z 3 g 2 z g 3. This is the Weierstrass paefunction. Hence the map z z, z induces a short exact sequence {0} Λ C EC {0} in terms of a lattice Λ = Z[ω 1, ω 2 ], generated by integrating around the poles of the cubic polynomial, and the complex points on the elliptic curve E : y 2 = 4 x 3 g 2 x g 3. We have the compact Riemann surface { } X = z = m ω 1 + n ω 2 C 0 m 1 and 0 n 1 C Λ EC. The collection of meromorphic functions on X C is K = C z, z. Note that the differential ω = dz = d = dx y = 2 dy 12 x 2 g 2 5
6 is not only meromorphic on C, it is actually holomorphic. As this is the only such differential, we see that ΩX C consists of constant multiples of ω = dx/y. In particular, g = 1. Divisors. Denote DivX as the collection of divisors; these are formal sums a = P n P P over the points P X, where all but finitely many of the integers n P are zero. The degree of a divisor is the integer dega = P n P. There is a partial ordering on DivX: given another divisor b = P m P P, we say a b when n P m P for all points P. The map K /k DivX which sends f P ord P f P is injective. In fact, we have the following short exact sequence: {1} K /k DivX PicX {0}. Similarly, any nonzero memomorphic differential ω = f dz for some meromorphic function f O, so define divω = divf = P ord P f P. As ΩX O, we say c = divω 0 is a canonical divisor for any nonzero meromorphic differential ω 0. We have the following commutative diagram, where the rows and columns are exact: {1} {0} {0} {1} K /k div Div 0 X JacX {0} = {1} K /k div DivX PicX {0} deg deg {1} {1} div DivX/Div 0 X NSX {0} {1} {0} {0} The quotient group JacX = Div 0 X/Divk of degree 0 divisors modulo principal divisors is the Jacobian of X; the quotient group PicX = DivX/Divk of divisors modulo principal divisors is the Picard group or the divisor class group of X; and the quotient group N SX = PicX/JacX is the NéronSeveri group of X. RiemannRoch Theorem. For any divisor a = P n P P, we wish to consider the following two complex vector spaces: { } H 0 a = f k la = dim divf a {0} C H 0 a { } = dega = n P H 1 a = ω Ω K X {0} divω a P X {0} δa = dim C H 1 a Note the change in the signs for the ordering! The main question here concerns the relationship between H 0 a, H 1 a, and H 1 X, Z. We have the following results: Proposition 4. Any divisor a can be written as a difference a = b p for divisors such that b, p 0. Since a a+p = b, we have H 0 a H 0 b. One shows by induction 6 =
7 that la lb degb + 1. In particular, H 0 a is a finite dimensional complex vector space. For each canonical divisor c = divω 0, the map ω ω/ω 0 shows that H 1 a H 0 c a = δa = lc a. In particular H 1 a is also a finite dimensional complex vector space. Say a = 0 is the zero divisor. Then H 0 0 = C consists of the constant functions, while H 1 0 = ΩX consists of the holomorphic differentials. In particular, H 0 c H 1 0 C g. In the 1850 s, Bernhard Riemann proved the inequality la dega + 1 g. student, Gustav Roch, showed more precisely: In 1864, his Remarks. Theorem 5 RiemannRoch. for any canonical divisor c. la dega lc a = la dega δa = 1 g. The paper appears in Crelle s Journal as Über die Anzahl der willkürlichen Constanten in algebraischen Functionen. This is usually called the RiemannRoch Theorem. Sadly, both Riemann and Roch died two years later in Italy of tuberculosis: Riemann aged 39, and Roch aged 26. In 1874, Max Noether and Alexander von Brill gave a refinement of Roch s result, and were the first to call it the RiemannRoch Theorem. In 1929, F. K. Schmidt generalized the Roch s result to algebraic curves. Subsequent generalizations were given by Friedrich Hirzebruch, JeanPierre Serre, and Alexander Grothendieck. Classification via the Genus Let me give some applications. Now we can let k = Q be an algebraically closed field, O be a Dedekind domain, and K be its quotient field. We will let X = Spec O be our nonsingular curve. Recall that for any divisor a = P n P P we have the identity dim k H 0 a dega dim k H 0 c a = 1 g where H 0 a = { f K divf + a 0 }. We see two facts right away regarding a canonical divisor c = divω 0 : g = dim k H 0 c, which we see by choosing a = 0. degc = 2 g 2, which we see by choosing a = c. We will show that, in some cases, we can classify X depending on the genus g. 7
8 Genus 0. We show that g = 0 if and only if X P 1 k. Proposition 6. If X P 1 k, then JacX {0} whereas PicX NSX Z. Proof. Choose O = k[x] as the polynomial ring in one variable, so that its quotient field K = kx consists of those rational functions in one variable. Each nonzero prime ideal m P O is in the form m P = x a for some P = a k, so we have a onetoone correspondence mspec O k. We define A 1 k = Spec O as the affine line over k. In order to make this a projective line, we add in the point at infinity: P 1 k = A 1 k {P }. Fix a nonnegative integer d, and consider the divisor b = d P of the point at infinity. We show that H 0 b = { f K divf + b 0 } consists of those polynomials of degree at most d. As the divisor of x K is P 0 P we see that ord P f d for any polynomial f = d i=0 a i x i. Hence f H 0 b. Conversely, let f H 0 b. Write f = g/h for some polynomials g, h O. If h has degree greater than 0, then it contains a nontrivial zero in k, so that f has a pole at some point in k. Hence h must be a constant. If g has degree greater than d then ord P g < d. Hence g has degree at most d. This shows in particular the equality lb = degb + 1. We show that any divisor a = P n P P can be expressed as a sum a = b + divf. Since affine points P = x a for some a k, we may choose fx = a k x an P, so that divf = P ord P f P = P n P P P = a d P for d = dega. Proposition 7. g = 0 if and only if X P 1 k. Proof. Let b = 2 g P be the divisor of degree 2 g associated with the point at infinity. We have seen that dim k H 0 a = degb + 1 in this case, so the RiemannRoch Theorem states that g = dim k H 0 c b. But degc b = 2 so that H 0 c b = {0}, showing that g = 0. Conversely assume that g = 0. We will construct a birational map X P 1 k. Let b = P as the divisor of a point in X. Then degc b < 0 so that H 0 c b = {0}. The RiemannRoch Theorem states that lb = 2. Fix a nonconstant function f H 0 b. For each a k, we note that ord P f a 0 for P P and ord P f a 1, so divf a = P a P for some point P a in X. As O/P k, define a map f : X P 1 k which sends a prime ideal P to the projective point fp = f mod P : 1. Note that fp a = a : 1 and fp = 1 : 0. As this map is onetoone and onto, we see that X P 1 k. Base Points. Given a divisor a DivX, define a complete linear system as the set a { } = b DivX b 0 and a = b + divf for some f k. Note that deg a = degb is independent of the choice of b a. It is easy to see that this fits into the following exact sequence: {1} k H 0 a {0} a {0} = div {1} k A n k {0} P n 1 k {0} where n = la. This relates affine vector spaces with projective vector spaces. In particular, the complete linear system c P g 1 k has deg c = 2 g 1. We say that a point P X is a base point if b P for all b c. 8
9 Proposition 8. X P 1 k whenever X has a base point. If g 1, then X is base point free. Proof. Say that P is one such base point. If f H 0 c is a nonzero function, then div1/f + c = b P so that div1/f + c P 0. Hence H 0 c H 0 c P, so the RiemannRoch Theorem states that dim k H 0 P = 1 g + deg P + dim k H 0 c P 2. Let f H 0 P be a nonconstant function. Following the same argument as above, divf a = P a P, so that the map f : X P 1 k is the desired isomorphism. Genus 1. Assume that k has characteristic different from 2 or 3. Proposition 9. g = 1 if and only if X Ek for some E : y 2 = x 3 + A x + B with 4 A B 2 0. Proof. Assume that g = 1. Fix a positive integer d, and consider the divisor b = d P. Then degc b = d < 0, so that H 0 c b = {0}. The RiemannRoch Theorem states that dim k H 0 b = 1 g + degb + dim k H 0 c b = d. Let {1, u} and {1, u, v} be bases for H 0 2 P and H 0 3 P, respectively. Since the set {1, u, v, u 2, u v, v 2, u 3 } of seven functions is contained in a vector space H 0 6 P of dimension 6, we must have a linear combination in the form a 1 + a 2 u + a 3 v + a 4 u 2 + a 5 u v + a 6 v 2 + a 7 u 3 = 0 for some a i k. Note that {1, u, v, u 2, u v} is a basis for H 0 5 P so we must have a 6, a 7 0. Upon making the substitutions x = 3 a 2 5 4a 4 a 6 12 a 6 a 7 u y = 108 a 6 a 7 a3 + a 5 u + 2 a 6 v A = 27 a a 4 a 2 5 a 6 16 a 2 4 a a 3 a 5 a 6 a a 2 a 2 6 a 7 B = 54 a a 4 a 4 5 a a 2 4 a 2 5 a a 3 4 a a 3 a 3 5 a 6 a a 3 a 4 a 5 a 2 6 a 7 72 a 2 a 2 5 a 2 6 a a 2 a 4 a 3 6 a a 2 3 a 2 6 a a 1 a 3 6 a 2 7 we find the identity y 2 = x 3 + A x + B. Denote this curve by E. We construct a birational map X Ek. Choose a, b k satisfying b 2 = a 3 + A a + B. Since {1, x} and {1, x, y} are bases for H 0 2 P and H 0 3 P, respectively, we have divx a = P a,b + P a, b 2 P and divy b = P a,b + P a,b + P a,b 3 P. As O/P k, consider that map f : X P 2 k which sends a prime ideal P to the projective point fp = x mod P : y mod P : 1. Note that fp a,b = a : b : 1 and fp = 0 : 1 : 0. As this map is onetoone and onto, we see that X Ek. Elliptic Curves. As before, assume that k has characteristic different from 2 or 3. Fix A, B k such that 4 A B 2 0. Let X P 2 k denote the collection of krational points on y 2 = x 3 + A x + B. We say that X is an elliptic curve. We will show that X is an abelian group with respect to some operation. 9
10 Theorem 10. Assume that g = 1. Then X JacX. In particular, X is an abelian group. Proof. This is the content of Proposition 3.4 in Chapter III.3.5 in Silverman s The Arithmetic of Elliptic Curves : we will construct a birational map κ : X JacX. Fix a point P X and send κ : X JacX by P P P. To see why this map is surjective, choose a Div 0 X and set b = a + P. Since degc b < 0, the RiemannRoch Theorem states that dim k H 0 b = 1 g + degb + dim k H 0 c b = 1. Let f H 0 b be nonzero; as this space is 1dimensional we must have divf = P b for some unique point P. Hence a = P P divf for some unique P X. We explain how the group law on elliptic curves can be derived from the Riemann Roch Theorem. Fix a point P X and denote O = 0 : 1 : 0. Given two points P, Q X draw a line in P 2 k going through them. Rather explicitly, if P = p 1 : p 2 : p 0 and Q = q 1 : q 2 : q 0, then the line is in the form fx 1, x 2, x 0 = 0 in terms of the linear polynomial p 1 p 2 p 0 fx 1, x 2, x 0 = q 1 q 2 q 0 x 1 x 2 x 0. It is easy to see that divf = P + Q + P Q 3 O for some point P Q. Now consider the line going through P Q and P ; this is in the form gx 1, x 2, x 0 = 0 for some linear polynomial. Again, it is easy to see that divg = P Q + P Q + P 3 O for some point P Q. Hence we find that P Q P = P P + Q P divf/g. Hence the map X JacX defined by P P P yields an associative group law. Note that P is the identity, which we often choose as P = O. Theorem 11. Let X be an elliptic curve, and let D = m i=1 n i P i be a divisor on E. Then D = divf for some rational function f : X P 1 if and only if both m i=1 n i = 0 in Z and m i=1 [n i] P i = O in X. The notation [n]p = P P P is the sum of P a repeated n times in X. Proof. This is the content of Corollary 3.5 in Chapter III.3.5 in Silverman s The Arithmetic of Elliptic Curves : We have seen that the map κ : X JacX which sends P P O is an isomorphism. Assume that D = divf. Then i n i = deg D = deg divf = 0, and i [n i]p i = i [n i] P O = κ 1 D = κ 1 divf. Tate Pairing and Weil Pairing Group Law. Now let k be any number field, and choose {a 1, a 2, a 3, a 4, a 6 } k. The set E : fx, y = 0 in terms of the polynomial fx, y = y 2 + a 1 x y + a 3 y x 3 + a 2 x 2 + a 4 x + a 6 10
11 is a curve over k. Define the krational numbers b 2 = a a 2 b 4 = 2 a 4 + a 1 a 3 b 6 = a a 6 b 8 = a 2 1 a a 2 a 6 a 1 a 3 a 4 + a 2 a 2 3 a 2 4 c 4 = b b 4 c 6 = b b 2 b b 6 = b 2 2 b 8 8 b b b 2 b 4 b 6 Then E is nonsingular if and only if 0. In this case, E is an elliptic curve. We review the group law : Ek Ek Ek defined above: Given two points P = p 1 : p 2 : p 0 and Q = q 1 : q 2 : q 0 in Ek draw a line fx 1, x 2, x 0 = 0 in P 2 k going through them in terms of the linear polynomial p 1 p 2 p 0 fx 1, x 2, x 0 = q 1 q 2 q 0 = divf = P + Q + P Q 3 O. x 1 x 2 x 0 Now consider the line going through P Q and O; this is in the form gx 1, x 2, x 0 = 0 for some linear polynomial, where divg = P Q + P Q 2 O for some point P Q Ek. Isogenies. Let E and E be two elliptic curves defined over k. An isogeny is a rational map φ : EQ E Q defined over k such that φo = O. Since φ : E E induces a map φ : QE QE which sends f f φ, we define the degree of φ as the degree of the extension QE/φ QE. Theorem 12. Let φ : E E be an nonconstant isogeny of degree m between elliptic curves over k. φ is a group homomorphism, that is, φp Q = φp φq as a sum in E Q for any P, Q EQ. The map kerφ Gal QE/φ QE which sends T to the function τ T g : P gp T is an isomorphism. In particular, kerφ = m. There exists a unique dual isogeny φ : E E such that the composition φ φ = [m] : E E E sends P [m] P on E. Proof. This first statement the content of Theorem 4.8 in Chapter III.4 of Silverman s The Arithmetic of Elliptic Curves : It follows from a diagram chase. EQ φ E Q P Q φp Q = φp φq κ 1 JacE κ 2 1 φ JacE P Q O = P + Q 2 O φp Q O = φp + φq 2 O For the second statement, we begin by showing the map is welldefined. Each T kerφ maps to that automorphism τt which sends a function g QE to that function τ T g : P gp T. 11
12 If g φ QE, then g = f φ for some f QE, so that τt g is that function which sends P EQ to τt g P = f φp φt = f φp O = gp. Hence τt acts trivially on φ QE. Clearly the map T τt is a welldefined injection. Conversely, degφ = φ 1 Q for some Q E Q. Fix P φ 1 Q. Then the map τ P : φ 1 O φ 1 Q which sends T P T is a onetoone correspondence, so that Gal QE/φ QE = degφ = φ 1 Q = φ 1 O = kerφ. For the third statement, consider the extension QE/[m] QE with Galois group ker [m]. Since [m] T = O for any T kerφ by Lagrange s Theorem, we see that kerφ ker [m]. In particular, we have the following tower of fields: [m] QE φ QE QE This shows that the map [m] : E E is in the form [m] = φ φ for some rational map φ : E E. Note that we have the following diagram: E Q φ EQ Q [m] P κ 2 JacE φ κ 1 1 JacE Q O T kerφ P T T for any P EQ such that φp = Q. In particular, φ φ P = φq = [m] P so that φo = φ φo = [m] O = O. If φ is any other dual isogeny, then φ φ φ = [m] [m] = [0] on E, so that φ φ = [0] must be constant. This shows that φ is the unique rational map with φ φ = [m] and φo = O, so φ must be an isogeny. Examples. Consider an elliptic curve E : y 2 + a 1 x y + a 3 y = x 3 + a 2 x 2 + a 4 x + a 6 where a i k. Given a point P = x : y : 1 in Ek, we have [m] P = O if and only of ψ m P = 0 in terms of the division polynomials 1 for m = 1, 2 y + a 1 x + a 3 = 4 x 3 + b 2 x b 4 x + b 6 for m = 2, ψ m P = 3 x 4 + b 2 x b b 6 x + b 8 for m = 3, ψ 2 P [ 2 x 6 + b 2 x b 4 x b 6 x b 8 x 2 + b 2 b 8 b 4 b 6 x + b 4 b 8 b 2 6 ] for m = 4. Other division polynomials can be generated by the recursive relation ψ m+n P ψ m n P ψ 1 P 2 = ψ m+1 P ψ m 1 P ψ n P 2 ψ n+1 P ψ n 1 P ψ m P 2 12
13 for any integers m and n. In fact, the multiplicationbym map [m] : Ek Ek sends P to [m]p = φ m P /ψ m P 2 : ω m P /ψ m P 3 : 1 in terms of the polynomials x for m = 1, φ m P = x 4 b 4 x 2 2 b 6 x b 8 for m = 2, φ 1 P ψ m P 2 ψ m+1 P ψ m 1 P for m 2. y for m = 1, ω m P = a 1 φ 2 P ψ 2 P 2 a 3 ψ 2 P 4 + ψ 4 P 2 ψ 2 P a 1 φ m P ψ m P 2 + a 3 ψ m P ψ m 1P 2 ψ m+2 P + ψ m 2 P ψ m+1 P 2 2 ψ 2 P for m = 2, for m 2. In particular, deg ψ m P 2 = m 2 1, so that the multiplicationbym map is an isogeny of degree m 2. In fact, ker [m] Z m Z m and [m] = [m]. Consider the elliptic curves E : y 2 = x 3 + a x 2 + b x E : Y 2 = X 3 + A X 2 + B X where A = 2 a B = a 2 4 b where a, b, A, B k satisfy b B 0. It is easy to check that T = 0 : 0 : 1 is a krational point of order 2, that is, [2]T = O. Then we have a maps φ : E E and φ : E E which send φ : x 1 : x 2 : x 0 x 2 2 x 0 : x 2 b x 2 0 x2 1 : x2 1 x 0 φ : X 1 : X 2 : X 0 2 X2 2 X 0 : X 2 B X0 2 X2 1 : 8 X2 1 X 0 It is easy to check that kerφ = { 0 : 0 : 1, 0 : 1 : 0 } Z 2 and that φ φ = [2] is the multiplicationby2 map. Hence both φ and φ are 2isogenies. Let A EQ C/Λ be any finite subgroup such that G k acts trivially. Then we can find an isogeny φ : E E such that kerφ A. One can construct E explicitly using the cohomology group H 1 G k, A. Usually, one focuses on subgroups in the form A Z m Z m or A Z n, but we can certainly consider others such as A Z m Z n. Weil Pairing. For any isogeny φ : E E and its dual φ : E E, the kernels E[φ] = kerφ and E [ φ] = ker φ are intimately related. Theorem 13. Let φ : E E be a nonconstant isogeny of degree m between elliptic curves over k. Denote E[φ] = kerφ Ek and E [ φ] = ker φ E k as the kernels of the isogeny and its dual. Then there exists a pairing satisfying the following properties: e φ : kerφ ker φ µ m 13
14 Bilinearity: For all S kerφ and T ker φ, we have e φ S 1 S 2, T = e φ S 1, T e φ S 2, T e φ S, T 1 T 2 = e φ S, T 1 e φ S, T 2 NonDegenerate: e φ S, T = 1 for all S kerφ, then T = O. Galois Invariant: σ e φ S, T = e φ σs, σt for all σ Gk. Compatibility: If ψ : E E is another isogeny, then e ψ φ P, Q = e ψ φp, Q for all P kerψ φ and Q ker ψ. Proof. We follow Section III.8 on pages and Exercise 3.15 on page 108 of Joseph Silverman s The Arithmetic of Elliptic Curves. Let T ker φ E [m]. According to Theorem 11, there are functions f T QE and g T QE satisfying divf T = m T m O divg T = φ T O = T kerφ P T T where P φ 1 T E[m]. Since divg m T = divf T φ, we may assume without loss of generality that f T φ = g m T. For any S kerφ, consider the map EQ P 1 Q which sends X g T X S/g T X. Since g T X S m = f T φx φs = ft φx = gt X m, we see that this map takes on only finitely may values and hence must be constant. We define the Weil pairing e φ : kerφ ker φ µ m as the mth root of unity e φ S, T = g T X S/g T X. We show Bilinearity. For the first factor we have e φ S 1 S 2, T = g T X S 1 S 2 g T X = g T X S 1 S 2 g T X S 2 = e φ S 1, T e φ S 2, T. gt X S 2 g T X = g T X S 1 g T X gt X S 2 g T X For the second factor, fix T 1, T 2 ker φ. Using Theorem 11 again, we can find functions f 1, f 2, f 3 QE and g 1, g 2, g 3 QE satisfying divf 1 = m T 1 m O divf 2 = m T 2 m O divf 3 = m T 1 T 2 m O divg 1 = φ T 1 O divg 2 = φ T 2 O divg 3 = φ T 1 T 2 O = f 1 φ = g1 m f 2 φ = g2 m f 3 φ = g3 m Similarly, we can find a function h QE such that divh = T 1 T 2 T 1 T 2 + O, and so f3 f m 3 div = m divh = = h m g3 = = h φ m. f 1 f 2 f 1 f 2 g 1 g 2 14
15 Hence g 3 = c g 1 g 2 h φ for some constant c Q. This gives e φ S, T 1 T 2 = g 3X S g 3 X = g 1X S g 1 X = g 1X S g 1 X g2x S g 2 X = e φ S, T 1 e φ S, T 2. g2x S g 2 X h φx O h φx h φx φs h φx We show NonDegeneracy. Say that e φ S, T = 1 for all S kerφ. Then g T X S = g T X for all X EQ. Following the ideas in Theorem 12, we see that g T φ QE, so that g T = h T φ for some h T QE. Since h T φ m = g m T = f T φ, we find that f T = h m T, and so divh T = T O. According to Theorem 10, we must have T = O. Galois Invariance is clear. We show Compability using the following diagram: EQ φ E Q ψ E Q kerψ φ φ φ φ kerψ φ ψ ker φ ψ kerφ ker ψ Say that ψ : E E is an isogeny of degree n. For each Q ker ψ ker ψ φ, there are functions d Q, f Q QE, g Q QE, and h Q QE satisfying divd Q = n Q n O divf Q = m n Q m n O divg Q = ψ Q O divh Q = ψ φ Q O = d Q ψ = g n Q f Q ψ φ = h mn Q f Q = d m Q g Q φ = h Q We define the pairings e ψ : kerψ ker ψ µ mn and e ψ φ : kerψ φ ker ψ µ mn via e ψ S, Q = g Q X S/g Q X and e ψ φ P, Q = h Q Y P /h Q Y, respectively. If we write X = φy, then g Q X φp e ψ φp, Q = = g Q φy φp = h QY P = e ψ φ P, Q. g Q X g Q φy h Q Y This completes the proof. Examples. Consider the elliptic curve E : y 2 + a 1 x y + a 3 y = x 3 + a 2 x 2 + a 4 x + a 6 Say φ = [2] is the multiplicationby2 map. Recall that the 2division polynomial is ψ 2 x = 2 y + a 1 x + a 3 = 4 x 3 + b 2 x b 4 x + b 6. If we denote e as one of the roots of 15
16 this polynomial, then T = e : a 1 e a 3 : 1 as a point of order m = 2. We denote the functions f T P = x e g T P = 4 e2 + b 2 e + b e x 2 x 2 = ft [2] P = g T P y + a 1 x + a 3 Consider the elliptic curves E : y 2 = x 3 + a x 2 + b x E : Y 2 = X 3 + A X 2 + B X where A = 2 a B = a 2 4 b where a, b, A, B k satisfy b B 0. Then we have a maps φ : E E and φ : E E which send φ : x 1 : x 2 : x 0 x 2 2 x 0 : x 2 b x 2 0 x2 1 : x2 1 x 0 φ : X 1 : X 2 : X 0 2 X2 2 X 0 : X 2 B X0 2 X2 1 : 8 X2 1 X 0 where φ φ = [2] is the multiplicationby2 map. Note that kerφ = { T, O } is the kernel, there T = 0 : 0 : 1 is a krational point of order 2, that is, [2]T = O. We denote the functions f T Q = X g T P = y = ft φ P = g T P 2. x There is an easy way to interpret the Weil pairing. Consider the multiplicationbym map [m] : E E. Since E[m] Z m Z m over Q, we can choose a basis {T 1, T 2 }. Then define e m : E[m] E[m] µ m via S = [a]t 1 [b]t 2, and T = [c]t 1 [d]t 2 to ζm ad bc. The only downside to making this definition is one would have to prove that E[m] Z m Z m! Tate Pairing. We discuss how a specific example of an isogeny gives information about the elliptic curve. Theorem 14. Say that E is an elliptic curve over k as above. Denote the 2division polynomial as ψ 2 x = 2 y+a 1 x+a 3 = 4 x 3 + b 2 x b 4 x + b 6. This has distinct roots e 1, e 2, e 3 Q, and so E : Y 2 = X e 1 X e 2 X e 3. Moreover, E[2] = { T EQ [2] T = 0 } { } = e 1 : 0 : 1, e 2 : 0 : 1, e 3 : 0 : 1, 0 : 1 : 0 Z 2 Z 2. Assume that E[2] Ek. Consider the map defined by e 2 : Ek 2 Ek k E[2] k, P, T 2 { 1 if T = O, X e otherwise; where P = X : Y : 1 and T = e : 0 : 1. This is a perfect pairing i.e., 16
17 NonDegeneracy: If e 2 P, T = 1 for all T E[2] then P 2 Ek. Bilinearity: For all P, Q Ek and T E[2] we have e 2 P Q, T = e 2 P, T e 2 Q, T, e 2 P, T 1 T 2 = e 2 P, T 1 e 2 P, T 2. Proof. Choose P = p 1 : p 2 : p 0 Ek, and say that e 2 P, T = 1 for all T E[2]. To show P 2 Ek it suffices to exhibit P Ek such that P = [2]P. If P = O we may choose P = O as well, so assume p 0 0. Upon considering T = e : 0 : 1, we see that f i = p1 p 0 e i k for i = 1, 2, 3; we choose the signs so that p 2 p 0 = f 1 f 2 f 3. It is easy to check that the desired krational point is P e1 e 3 e 2 e 3 = f 1 f 3 f 2 f 3 + e e 1 e 2 e 1 e 3 e 2 e 3 3 : f 1 f 2 f 1 f 3 f 2 f 3 : 1. We show e 2 P Q, T = e 2 P, T e 2 Q, T. If T = O there is nothing to show since e 2 P, T = e 2 Q, T = e 2 P Q, T = 1 so assume that T = e : 0 : 1. Choose two points P = p 1 : p 2 : p 0 and Q = q 1 : q 2 : q 0 in Ek. Draw a line through them, say a x 1 + b x 2 + c x 0 = 0, and assume that it intersects E at a third point R = r 1 : r 2 : r 0. The projective curve E is defined by the homogeneous polynomial F x 1, x 2, x 0 = x 2 2 x 0 x 1 e 1 x 0 x 1 e 2 x 0 x 1 e 3 x 0 so the intersection with the line a x 1 + b x 2 + c x 0 = 0 admits the factorization p 0 q 0 r 0 F x 1, x 2, x 0 = p 1 x 0 p 0 x 1 q 1 x 0 q 0 x 1 r 1 x 0 r 0 x 1. When x 1 : x 2 : x 0 = b e : a e c : b is the point where the lines a x 0 + b x 1 + c x 0 = 0 and x 1 e x 0 = 0 intersect, we have the equality b 3 p1 q1 r1 e e e = F b e, a e c, b = a e + c 2 b. p 0 q 0 r 0 This implies the congruence e 2 P, T e 2 Q, T e 2 R, T 1 mod k 2. We conclude that e 2 P Q, T = e 2 P, T e 2 Q, T. We show e 2 P, T 1 T 2 = e 2 P, T 1 e 2 P, T 2. If T 1 = T 2 then e 2 P, T 1 T 2 = e 2 P, O = 1 e 2 P, T 1 2 = e 2 P, T 1 e 2 P, T 2. If T 1 T 2, we may assume T 1 = e 1 : 0 : 1 and T 2 = e 2 : 0 : 1. If either T 1 or T 2 is O there is nothing to show. Then T 1 T 2 = e 3 : 0 : 1. The identity 2 p1 p1 p1 p2 e 1 e 2 e 3 = p 0 p 0 p 0 p 0 implies the congruence e 2 P, T 1 e 2 P, T 2 e 2 P, T 1 T 2 1 mod k 2. e 2 P, T 1 T 2 = e 2 P, T 1 e 2 P, T 2. We conclude that Remarks. This sometimes called the Tate pairing. This is not quite a perfect pairing: nondegeneracy holds on the right, but not on the left. 17
18 Since e 2 P, T is bilinear, it is easy to compute its value when P E[2]. For example, write T i = e i : 0 : 1 so that we find: e 2 T i, T i 1 = e i e i 1, e 2 T i, T i+1 = e i e i+1 = e 2 T i, T i = e 2 T i, T i 1 e 2 T i, T i+1 = e i e i 1 e i e i+1. If k is a number field, the image in k / k 2 is actually finite. One uses this to conclude that Ek/2 Ek is finite as well. This was first shown for k = Q by Mordell. Say that we can write Ek Ek tors Z r for some finite group Ek tors Z m Z n and some nonnegative integer r; this nonnegative integer is called the rank of E over k. Then we can write Ek 2 Ek Ek tors 2 Ek tors Z r 2, Ek tors 2 Ek tors = {1} if m and n are odd, Z 2 if m is even but n is odd, Z 2 Z 2 if both m and n are even. The Theorem above concerns the case where m and n are both even. Hence we can determine the rank r if we can determine the image of this pairing. There is a more general construction for each positive integer m: e m : Ek k E[m] m Ek k m assuming E[m] Ek. This pairing is used quite often in cryptography, especially when k = F p is a finite field of order p 1 mod m Z so that E[m] Z m Z m. It is not a coincidence that the Tate pairing is defined via f T Q = X e. In general, say that φ : E E is a nonconstant isogeny of degree m. We have seen that for each T E [ φ], there are functions f T QE and g T QE such that divf T = m T m O divg T = φ T O = f T φ = g m T. You can actually choose f T and g T to have coefficients in k. This yields a perfect pairing E k φ Ek ker φ k k m, P, T ft P mod k m. One can derive this pairing from the Weil pairing. We will see in general that the Weil pairing e φ : kerφ ker φ µ m yields a cup product on Galois cohomology: H i G k, E[φ] H j G k, E [ φ] H i+j G k, µ m. Indeed, there is a short exact sequence {O} E[φ] EQ 18 φ E Q {O}
19 so Galois cohomology gives the diagram E k φ Ek δ ker φ k k m H 1 G k, E[φ] H 0 G k, E [ φ] H 1 G k, µ m 19
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