Super Congruences. Master s Thesis Mathematical Sciences

Transcription

1 Suer Congruences Master s Thesis Mathematical Sciences Deartment of Mathematics Author: Thomas Attema Suervisor: Prof. Dr. Frits Beukers Second Reader: Prof. Dr. Gunther L.M. Cornelissen

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3 Abstract In 011 the Chinese mathematician Zhi-Wei Sun ublished an article containing 100 oen conjectures about congruences involving hyergeometric sums. It haens to be the case that we can relate many of these congruences to the number of oints on ellitic curves with comlex multilication. The comlex multilication roerty of these ellitic curves will give us a way to find the number of F -oints on these ellitic curves and by means of these cardinalities certain congruences are imlied. i

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5 Preface Number theory has always had my interest. This is why I asked Frits Beukers to be my suervisor. He first suggested another subject but later on came with the idea to write a thesis on suer congruences. The starting oint of this thesis was an article [Sun11c] of the Chinese mathematician Zhi-Wei Sun. In this article Sun collected 100 unsolved conjectures about suer congruences. This thesis studies a new aroach to solve some of these conjectures. At first ellitic curves were merely a tool to solve certain congruences. But since the results on this subject were very interesting a reasonable amount of this thesis is dedicated to ellitic curves. After a little work the congruences follow from the results on ellitic curves. Acknowledgements I would like to thank Frits Beukers for his active suervision the ast year. I am very grateful for the many talks we had and for all of the useful suggestions he made. I would also like to thank Sander Dahmen, he heled solve some of the roblems I encountered. Furthermore I would like to exress my gratitude to the other staff members at the Deartment of Mathematics in Utrecht. They have taught me many courses and were always oen to questions. These staff members have made the ast five years at the University of Utrecht into a very leasant and informative exerience. In articular I would like to thank Gunther Cornelissen for taking the time to read my thesis. Furthermore I would like to thank Kaser Dokter for reading my thesis and suggesting some imrovements. Thomas Attema 4 June 013 iii

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7 Contents Abstract Preface i iii Introduction 1 1 Ellitic Curves The grou E/K Singular Cubic Curves Isogenies Frobenius Endomorhism Rationality Cardinality Formulas 19.1 Multilication by d Minimal Models Gamma Function Gauss Sums Suer Congruences The cubic curve C 1 t : y + xy + x 3 + t = The cubic curve C t : y + xy ty + x 3 = The cubic curve C 3 t : y + xy + x 3 + tx = The cubic curve C 4 t : y = xx 1x 16t Conjectures Rational t-values Quadratic t-values Other extensions of Q Primes inert in L Other j-invariants v

8 Bibliograhy 90 vi

9 Introduction Congruences aear very often in number theory and other fields of mathematics. Therefore congruences have been extensively studied. In this thesis we will study congruences modulo rimes. In articular we will be interested in suer congruences. Suer congruences are congruences that hold not only modulo, but also modulo higher owers of. These suer congruences are not easily solved and require new tools. The congruences we will study in this thesis are all related to hyergeometric sums. A hyergeometric sum is a sum k c k, such that the ratio of successive terms c k+1 /c k is a rational function of k. We will be interested in finite hyergeometric sums involving binomial coefficients. In articular sums of the form reduced modulo owers of a rime, where a k is a roduct of binomial coefficients. 1 k=0 a k m k The first one to study these truncated hyergeometric sums was Fernando Rodriguez-Villegas. In [RV03] Rodriguez-Villegas observed some congruences associated to truncated hyergeometric sums. He also observed the connection to certain manifolds and the number of their F -oints. Later Zhi-Wei Sun collected various conjectures on suer congruences associated to hyergeometric sums in his aer [Sun11c]. One of the conjectures Zhi-Wei Sun ublished in his aer is the following one. Conjecture [Sun11c, A9]. Let > 3 be a rime. Then 1 4m! m! 4 56 m 4x mod if = 1 and = x + y, 0 mod if = 1, i.e. if 5, 7 mod 8. Here we see a hyergeometric sum on the left hand side of the congruence. The right hand side can be related to the number of F -oints on a secific ellitic curve. So if we are able to comute the number of F -oints on this ellitic curves we will be able to evaluate this hyergeometric sum modulo. The key roerty that this ellitic curve ossesses is comlex multilication. An ellitic curve has comlex multilication when its endomorhism ring is larger then the ring of integers Z. The CM -roerty imlies extra symmetries on the ellitic curve. These extra symmetries will enable us to find the cardinality of such ellitic curves over finite fields. The two cases of the congruence in the revious conjecture already show the relation with the endomorhism ring of the ellitic curve. The first case namely comes from the rimes that slit in the endomorhism ring and the second case comes from the rimes that are inert in the endomorhism ring. 1

10 In chater 1 we gather the necessary theory about ellitic curves. This enables us, in chater, to comute the number of F -oints on ellitic curves E/Q with comlex multilication. In chater 4 we will relate these cardinalities to the hyergeometric sums as mentioned above. Before we can do this we need some extra theory about Gauss-sums and the -adic Gamma function. This necessary theory is catured in chater 3. Using the gathered theorems and roositions we are able to rove some congruences, some of which are suer congruences. The study of these congruences has led to conjectures about stronger results. In chater 5 we state some conjectures on congruences modulo higher owers of rimes. Some of these conjectures can be roven using the theorems in the aer [CVH91] of Coster and Van Hamme.

11 Chater 1 Ellitic Curves An ellitic curve is a non-singular rojective curve of genus one with a secified oint O. Now let E be an ellitic curve over a field K. If the characteristic of K is not equal to and 3 then the affine art of E/K can be written down exlicitly by a short Weierstrass equation E : y = fx = x 3 + Ax + B, 1.1 where A, B K. From now on we will assume that chark, 3 unless secified otherwise. The ellitic curve E/K consists of all oints x, y K satisfying this equation and the unique oint O at infinity. The ellitic curve E/K is actually an algebraic grou defined over K. The addition formulas make the set of oints EK into an additive abelian grou with identity O. If we consider ellitic curves in short Weierstrass form the only change of variables reserving this form of the equation is given by x = u x, y = u 3 y, u 4 A = A, u 6 B = B, 1. for some u K, where K is an algebraic closure of K. Here x, y satisfies the equation E : y = x 3 + Ax + B and x, y satisfies the equation E : y = x 3 + A x + B. So we see there are not many choices for isomorhisms between ellitic curves in short Weierstrass form. The discriminant of an ellitic curve E/K in short Weierstrass form is given by E = 164A 3 + 7B. So if we consider the change of variables as in equation 1. we see E = u 1 E. The non-singularity of E/K is equivalent to E 0. Hence if u K then E is non-singular if and only if E is non-singular. The j-invariant of an ellitic curve is given by 4A 3 je = 178 4A 3 + 7B. Now let E and E be isomorhic ellitic curves. Then there exists an u K such that the 3

12 isomorhism between E and E is given by a change of variables as in equation 1.. Hence je = 4A A 3 + 7B = 4u 4 A u 4 A 3 + 7u 6 B 4A 3 = 178 4A 3 + 7B = je. Therefore two isomorhic ellitic curves have the same j-invariant. We actually have the following roosition see also [Sil09,.45]. Proosition 1.1. Two ellitic curves are isomorhic over K if and only if they have the same j-invariant. Proof. We already saw that isomorhic curves have the same j-invariant. So let E : y = x 3 + Ax + B, E : y = x 3 + A x + B be ellitic curves with the same j-invariant. Hence which imlies so 4A A 3 + 7B = 178 4A 3 4A 3 + 7B, A 3 4A 3 + 7B = A 3 4A 3 + 7B, A 3 B = A 3 B. 1.3 First suose AB 0. Then je = je 0, 178 and equation 1.3 shows that A B = 0 imlies A = B = 0. But this can not haen since E is non-singular and thus E 0. So A B 0. Now equation 1.3 can be rewritten as Hence if we take u = A 3 B A = B. 1 1 A 4 B 6 = K, A B we obtain an isomorhism of the form as in equation 1.. Now suose A = 0. Then je = je = 0, hence A = 0. Moreover B 0 and B 0, since E and E are non-singular. So we obtain an isomorhism of the form as in equation 1. by taking u = 1 B 6 B K. Finally suose B = 0 then je = 178 and we find A, A 0 and B = 0. So we can take u = 1 A 4 A K to find an isomorhism. We say two ellitic curves E 1 /K and E /K are isomorhic over a field L containing K if there exists an isomorhism from E 1 to E defined over L. 4

13 Let E : y = x 3 + Ax + B be an ellitic curve over Q. We then call any other ellitic curve over Q with the same j-invariant a twist of E. Note that any twist of E is isomorhic to E over Q but not necessarily over Q itself. If now je 0, 178, i.e. A, B 0, then we see that any twist of E is given by y = x 3 + Au 4 x + Bu 6 and since this curve has to be defined over Q we see u 4 Q and u 6 Q, therefore u Q. Hence if je 0, 178 then any twist of E is given by E D : y = x 3 + AD x + BD 3 for some D Q. These ellitic curves are called quadratic-twists since E and E D are isomorhic over Q D. If je = 0 or je = 178 then also higher order twists are ossible. An isomorhism between two quadratic twists is given by φ : E E D, x, y Dx, D Dy. 1.1 The grou E/K As mentioned before the set of oints E/K is a grou with identity element O at infinity. The addition formulas can be written down as quotients of olynomials. If we namely take two oints x 1, y 1 and x, y on the ellitic curve E : y = x 3 + Ax + B defined over K such that x 1 x then we can define λ = y 1 y, x 1 x and we have x 1, y 1 + x, y = x 3, y 3 with x 3 = λ x 1 x and y 3 = λx 1 λx 3 y 1. It is an easy comutation to check that these formulas define a oint on E. Now if we want to double a oint x, y on the ellitic curve E then x, y = x, y + x, y = x, y with x = x4 Ax 8Bx + A 4y and y = x3 Ax + B 3x x + Ax. y The final case consists of two distinct oints P and Q on E with the same x-coordinates. We must then have P = x, y and Q = x, y. These oints are inverses of each other so P +Q = O and P = Q. These formulas define a grou action on the ellitic curve E. Moreover all addition formulas are defined over K. So if two oints P and Q are defined over K then so are their sum and inverses. Therefore the grou E/K has a subgrou EK of all K-rational oints. Theorem 1.. Mordell-Weil Let K be a number field. Then the grou EK is finitely generated. Proof. See [Sil09,.07] By the Mordell-Weil theorem we see that the grou EK has the form EK = EK tors Z r, where r is the rank of the grou EK and EK tors is the torsion subgrou of EK i.e. all oints of finite order. Moreover by the Mordell-Weil theorem it follows that EK tors is finite. 5

14 From the addition formulas it follows that a oint P = x, y E has order if and only if y = 0. The oints P for which P = O form a subgrou of E. We call this subgrou the -torsion subgrou of E and denote it by E[]. Now since E/K is an additive grou we can multily oints by integers. Namely for m Z mp = P + P P. }{{} m times Hence we can look at the m-torsion subgrou of E for any m Z, E[m] = {P E : mp = O}. The formulas for the multilication by m will also be quotients of corime olynomials in K[x, y]. From these olynomials the grou E[m] is easily deduced by setting the denominators equal to zero. Hence comuting the grou E[m] comes down to solving olynomial equations. Note that we have E tors = m Z E[m] and EK tors = m Z EK[m]. The olynomial whose roots are the x-coordinates of E[m] \ {O} is called the m th -division olynomial of E/K and is denoted by ψ m. These olynomials can be comuted by using the addition formulas of the ellitic curve E/K. The division olynomials will be a useful tool in determining the torsion subgrous of ellitic curves. We will now consider the case K = Q. By erforming a variable change as in 1., if necessary, any ellitic curve over Q can be written as E : y = x 3 + Ax + B for some A, B Z. So from now on we will assume that an ellitic curve over Q is given this way. Note that this also gives E Z. Let E/Q be an ellitic curve and let be a rime. Then we can reduce A and B modulo to obtain a curve E : y = x 3 + Ax + B over the finite field F, where we use the fact that we can find a Weierstrass model with A, B Z. Note that E has discriminant E = E F, which might be zero. So E/F may not be an ellitic curve since it can be singular. In articular we have that E is non-singular if and only if E. We say E has good reduction modulo if the ellitic curve E/Q has a model which reduces to a non-singular curve over F. We say E has bad reduction modulo otherwise. For the grou of oints of E/F with coordinates in F we simly write EF. Note that EF is a finite grou since F is finite. We can now count the number of oints in EF the following way EF = 1 + x 3 + Ax + B 1 + = x 3 + Ax + B, x F x F where we just count the number of solutions of y = x 3 + Ax + B over F. Here x is the Legendre symbol defined for all x Z by x = 1 if x is a quadratic residue modulo, 1 if x is a quadratic nonresidue modulo, 0 if x. 6

15 From number theory we know that Moreover we see, by the above formula, that x x 1 mod. a := x 3 + Ax + B = + 1 EF. x F By the Hasse-Weil theorem see for examle [ST9,.110] we have the following estimate for all rimes. a <, Also note that from the above formulas we find, for any quadratic twist E D of E, a ED = x 3 + AD x + BD 3 x F = Dx 3 + AD Dx + BD 3 Dx F D 3 x 3 + Ax + B = Dx F D = a E. 1. Singular Cubic Curves Let C be a cubic curve over K given by a short Weierstrass equation y x 3 Ax B = 0 for some A, B K. We say C is singular if there exists a oint P C at which the artial derivatives of y x 3 Ax B vanish simultaneously. This means that if a cubic curve is nonsingular it has a well-defined tangent line at every oint. Now if these artial derivatives vanish at the oint x 0, y 0, then y 0 = 0 and x 0 is a double root of fx = x 3 + Ax + B. Conversely if fx has a double root at x 0, then x 0, 0 is a singular oint of C. There are two ossible tyes of singularity deending on whether fx = x 3 + Ax + B has a double root or a trile root. When fx has a double root at x 0, we say the singular oint x 0, 0 is a node. In this case we can aly a coordinate change to find the following tyical equation for a singular cubic curve with a node, C : y = x x + 1. This curve has two distinct tangent directions in its singular oint 0, 0, as can be seen in the following figure. This figure dislays the grah of the affine art of the singular cubic curve C : y = x x + 1 over Q. 7

16 x Figure 1.1: C : y = x x + 1 The other ossible tye of singularity haens when fx has a trile root. After a change of coordinates we obtain the equation C : y = x 3. This cubic curve has a singularity in 0, 0, which we call a cus. The following figure shows the affine art of the curve C : y = x 3 over Q. y x Figure 1.: C : y = x 3 We have already mentioned that a cubic curve is singular if and only if its discriminant equals 0. The next lemma enables us to determine the tye of singularity by using the short Weierstrass equation. Lemma 1.3. Let C : y = x 3 + Ax + B be a cubic curve over a field K with chark, 3. Then a C is non-singular if and only if C 0, b C has a node if and only if C = 0 and A 0, c C has a cus if and only if C = A = 0. 8

17 Proof. See roosition 1.4 of [Sil09,.45]. Remark. All quantities we define for cubic curves in short Weierstrass form are also defined for general cubic curves. The formulas for these quantities are more comlicated, but the theorems and lemmas still hold in this more general case. We will however not bother the reader with these tedious formulas. Singular cubic curves behave in a very different way from non-singular cubic curves. Any line through a singular oint intersects this oint with multilicity at least two, therefore this line has at most one more intersection with the cubic curve. This means that we are able to arameterize singular cubic curves in a nice way. Lets start with the first tye of singularity. We assume the cubic curve is given by C : y = x x + 1 over K. If we then take t = y x, we find x, y = t 1, t 3 t. In articular we find all oints of CK\{O} by substituting different values of t K. Moreover t 1, t 3 t = s 1, s 3 s imlies either t = s or t = s = ±1. We thus find a bijection between CK \ {O} and K \ {1}, therefore CK = K. If we now consider the singular cubic curve C : y = x 3 over K, we find an even simler arametrization by taking x, y = t, t 3. Moreover t, t 3 = s, s 3 imlies s = t. We thus find a bijection between CK \ {O} and K. Any singular cubic curve over Q with a node is isomorhic over Q to a twist of C : y = x x + 1. Therefore we see that a general cubic curve C over Q with a node C is isomorhic to a curve of the form y = x x + D for some D Z. Hence C F = + 1 Since the cubic curve C : y = x 3 has no twists other than itself we see that any cubic curve with a cus is isomorhic to C over Q. The following theorem now follows. Theorem 1.4. Let C be a singular cubic curve defined over Q. Then C has a short Weierstrass model with coefficients in Z. In articular C : y = x 3 in which case D CF = + 1, or C : y = x x + D for some D Z in which case 1.3 Isogenies CF = + 1. D. We will now review some theory about mas between ellitic curves. Definition 1.1. Let E/K and E /K be ellitic curves. An isogeny from E to E is a morhism φ : E E such that φo E = O E, defined over the algebraic closure K. Two ellitic curves E and E are called isogenous if there is an isogeny φ from E to E such that φe {O E }. 9

18 We will see that isogenies have some nice roerties. One of these roerties is that isogenies are grou homomorhisms. This statement is roven in theorem 4.8 of [Sil09,.71]. Examle 1.1. Let E/K be an ellitic curve. As E is an abelian grou we can multily oints by integers. Since the addition formulas are quotients of olynomials in K[x, y] we see that the multilication by m ma is a morhism of varieties with mo = O. So for every m Z there exists an isogeny [m] : E/K E/K, P mp. Notice that the kernel of this ma is exactly the m-torsion subgrou E[m] of E. A morhism between two curves is either constant or surjective see [Sil09,.0]. So an isogeny is either constant or surjective. Any morhism between varieties also has a degree d Z. The geometric interretation of the degree d of a morhism is that any searable morhism of degree d has d elements in its kernel. Moreover for any isogeny φ from E to E of degree d there exists a unique dual isogeny ˆφ from E to E such that ˆφ φ from E to E is given by multilication by d see [Sil09, III.6.1]. Hence the relation of being isogenous is actually an equivalence relation. Examle 1.. Let E 1 : y = x ax + ax + b be an ellitic curve over Q with a, b Z. And let E : y = x + ax ax 7a 4b be another ellitic curve. Then we have the following isogeny φ : E 1 E, x, y x + a + b x a, 1 a + b x a y. It is now a simle calculation to check that this ma is well-defined and indeed an isogeny. For two ellitic curves E and E we can now use the fact that these curves are abelian grous. We define the grou of all isogenies from E to E by HomE, E := {isogenies from E to E }. Since E is a grou, for any two isogenies φ, ψ there exists a new isogeny φ + ψ defined by φ + ψp = φp + ψp, which imlies HomE, E is actually a grou with as identity the trivial isogeny. If now E = E we can also comose isogenies, hence EndE := HomE, E is a ring. We call this ring the endomorhism ring of E. The multilication on EndE is given by the comosition of morhisms. We omit here the full roof of the fact that φ + ψ is an isogeny and also the roof that EndE really is a ring. For these roofs we refer to [Sil09]. As we have seen in examle 1.1 there exists an endomorhism [m] for every m Z and every ellitic curve E. So for every ellitic curve E we have Z EndE. However there may also exist other endomorhisms as the next examle shows. Examle 1.3. Let E : y = x 3 x be an ellitic curve over Q. endomorhism [i] : E E, x, y x, iy. This endomorhism is well defined since iy = y = x 3 + x = x 3 + x. Then we can define the Moreover the kernel of this endomorhism is {O}, hence the isogeny [i] is not given by a multilication by m ma. So EndE Z. 10

19 Proosition 1.5. The endomorhism ring of an ellitic curve E/K is either Z, an order in an imaginary quadratic field or an order in a quaternion algebra. If chark = 0, then only the first two are ossible and if chark 0 then only the last two are ossible. Proof. See corollary 9.4 of [Sil09,.10] and theorem 3.1 of [Sil09,.144]. There exist ellitic curves over a field with characteristic 0 such that the endomorhism ring is isomorhic to Z. For chark = 0 we say E/K has comlex multilication or simly E/K is a CM-curve if the endomorhism ring is strictly larger then Z. It is actually the case that for each of the 9 imaginary quadratic fields with class number 1 there is at least one order in that field aearing as endomorhism ring of an ellitic curve E/Q see [Cre97,.103]. Moreover all endomorhism rings of ellitic curves over Q are given as an order in such a field. If E/K is an ellitic curve with comlex multilication where the endomorhism ring is an order R in the quadratic imaginary field K we say E has comlex multilication by K or E has comlex multilication by R. Also note that two ellitic curves over Q with the same j-invariant are isomorhic over Q. From this it follows that their endomorhism rings are isomorhic. We can now create table 1.3 in which we find a CM-curve for every ossible endomorhism ring of ellitic curves over Q. Ellitic Curve Endomorhism Ring j-invariant E 1 : y = x 3 x Z[i] 178 E : y = x 3 11x 14 Z[i] 66 3 E 3 : y = x Z[ 1+ 3 ] 0 E 4 : y = x 3 15x + Z[ 3] 30 3 E 5 : y = x 3 70x 151 Z[ ] 0 3 E 6 : y = x 3 35x 98 Z[ 1+ 7 ] 15 3 E 7 : y = x 3 595x 5586 Z[ 7] 55 3 E 8 : y = x 3 480x E 9 : y = x x E 10 : y = x 3 608x E 11 : y = x x E 1 : y = x x E 13 : y = x x Z[ ] Z[ ] Z[ ] Z[ ] Z[ ] Z[ ] Table 1.1: Ellitic Curves with Comlex Multilication Note that these equations are not uniquely determined by their j-invariant. We could also have taken twists. These curves and their twist are the only ellitic curves over Q with comlex multilication. Moreover all these endomorhism rings have a field of fractions with class number 1, which imlies that the integral closures of these rings are rincial ideal domains. In articular the integral closures are unique factorization domains, which will be a useful roerty later on. 11

20 If the endomorhism ring of an ellitic curve E/K is an order in a quaternion algebra then we say E/K is suersingular. Notice that by roosition 1.5 we must have chark > 0 for any suersingular ellitic curve E/K. Moreover if an ellitic curve is suersingular it does not mean it is singular. For a CM-curve E with endomorhism ring Z[α] the dual of an endomorhism π Z[α] is given by its comlex conjugate π see [Sil94,.97]. Hence the degree of an endomorhism is given by ππ and since a CM-curve is defined over a field of characteristic zero we see that any morhism is searable, hence for any π Z[α] In articular E[m] = m for any m Z. kerπ = ππ. By roosition 1.5 we also see that the endomorhism ring of an arbitrary ellitic curve is a ring extension of Z. Moreover we see that EndE is free of rank 1, or 4 as a Z-algebra. Hence there exist x 1,..., x n EndE that form a Z-basis for EndE for some n {1,, 4} EndE = Zx 1 Zx... Zx n. Now for any x EndE we consider the Z-linear multilication by x ma, M x : EndE EndE, a xa. With a Z-basis as above we can describe this ma by an n n matrix with coefficients in Z. We define the norm and trace of an element x EndE by Nx := det M x and Trx := trace M x. Both the norm and the trace are indeendent of the choice of basis. We can also construct the characteristic olynomial f x of x EndE, namely f x = detx Id EndE M x. This is a monic olynomial of degree n and it has coefficients in Z. Moreover f x x = 0 for all x EndE. Notice also that the constant coefficient of f x equals 1 n Nx and that the second highest coefficient equals Trx. 1.4 Frobenius Endomorhism Let E : y = x 3 +Ax+B be an ellitic curve over F q for some rime ower q = n. Because this ellitic curve is defined over a finite field we can define the so called Frobenius endomorhism Fr q : E E, x, y x q, y q. This ma is well-defined since for any x, y F q, we have x q = x and x + y q = x q + y q + q 1 q i=1 i x i y q i = x q + y q, because q i for all 1 i q 1 and charfq =. Hence for x, y E we have 0 = y x 3 Ax B = y x 3 Ax B q = y q x q 3 A q x q B q = y q x q 3 Ax q B, where the last equality follows since A, B F q. So x q, y q E. 1

21 Theorem 1.6. Let E/F q be an ellitic curve, let be the Frobenius endomorhism and let Fr q : E E, x, y x q, y q, aq = q + 1 EF q. a Let π and π be the roots of the olynomial X aqx + q. Then π and π are comlex conjugates satisfying π = π = q, and for every n 1, b The Frobenius endomorhism satisfies Proof. See theorem.3.1 of [Sil09,.14]. EF q n = q n + 1 π n π n. Fr q aq Fr q +q = 0. We thus see that the trace and norm of the Frobenius endomorhism are resectively given by TrFr q = q + 1 EF q and NFr q = q. And the characteristic olynomial of Fr q in EndE is given by P X = X aqx + q. So Fr q is a zero π EndE of this olynomial. Moreover aq = π + π and ππ = q. Note that the olynomial P X = X aqx + q has discriminant aq 4q. But by the Hasse-Weil bound we have aq < q for all q. Hence the discriminant of P X is negative and therefore is P X irreducible in Z[X]. Also note that we have q = ππ and the number of oints in EF q is determined by π. So we can comute EF q by finding the correct factorization of q in EndE. For ellitic curves that arise as reductions of CM-curves we have the following theorem: Theorem 1.7. Let E/Q be an ellitic curve with comlex multilication. Let be a rime such that slits in K, where K is the field of fractions of End Q E and such that E has good reduction modulo. Then there exists a rime π End Q E such that = ππ and Furthermore End F E = End Q E. EF = + 1 π π. Sketch of the roof. We will not give the full roof of this theorem, since it requires some tools beyond the scoe of this thesis. The first imortant ste of the roof is that the reduction modulo induces a natural ma End Q E End F E. This ma reserves the degrees of the endomorhisms. Now if the conditions of the theorem are fulfilled then this ma is an isomorhism. So there exists a π End Q E which mas to Fr. Hence π has the same degree as Fr, which is. But over the comlex numbers the degree 13

22 of π is just its norm. So Nπ = ππ = in End Q E. The following fact we will use is that for searable endomorhisms the degree equals the cardinality of the kernel. Therefore we see EF = ker1 Fr = deg1 Fr. If we now again use the fact that the reduction ma reserves degrees, we find deg1 Fr = deg1 π = 1 π1 π = + 1 π π. This theorem gives a nice way to comute the number of oints EF of an ellitic curve E/Q with comlex multilication. At least for rimes of good reduction that slit in EndE. So what haens if a rime ramifies or is inert in EndE? 1.5 Rationality In this section we will only consider ellitic curves over fields with characteristic zero. By roosition 1.5 this means that the endomorhism rings of these ellitic curves are orders in imaginary quadratic fields. Let E be an ellitic curve over a field K with chark = 0 and let R C such that EndE = R. Then there exists a unique isomorhism [ ] : R EndE, such that for any invariant differential ω Ω E see [Sil09] [α] ω = αω for all α R. For a roof of this statement see roosition 1.1 of [Sil94,.97]. From this isomorhism it follows that we can denote every endomorhism by [α] for some unique α R. First we will rove the following useful theorem. Theorem 1.8. Let E 1 and E be two ellitic curves over a finite field F q. If there exists an isogeny from E 1 to E defined over F q then E 1 F q n = E F q n for any n 1. Proof. Let φ : E 1 E be an isogeny defined over F q. And let Fr q be the Frobenius endomorhism which acts on E 1 and on E since both are defined over F q. Let X ax + q be the characteristic olynomial of Fr q in EndE 1. Since φ is an isogeny defined over F q it is given as the quotient of two olynomials in F q, hence φ commutes with Fr q. Moreover the multilication by m ma also commutes with φ since φ is a homomorhism of grous. So on the on hand we have φfr q a Fr q +qp = φ0p = φo 1 = O, for any P E 1. But on the other hand we have φfr q a Fr q +qp = Fr q φp a Fr q φp + qφp = Fr q a Fr q +qφp. So Fr q a Fr q +qφp = 0 for any P E 1. The surjectivity of isogenies now imlies Fr q a Fr q +q = 0 in EndE. So the characteristic olynomials of Frobenius are the same on both curves. Then by theorem 1.6 it now follows that E 1 F q n = E F q n for any n 1. 14

23 Remark. The converse of this theorem holds as well. In theorem 1 of [Tat66] Tate roves the following statement from which theorem 1.8 follows. Theorem 1.9. Let E 1 and E be two ellitic curves over a finite field k. Then the following two statements are equivalent: a There exists an isogeny from E 1 to E defined over k. b E 1 and E have the same number of oints defined over k for every finite extension k of k. Let E 1 and E be two ellitic curves defined over Q such that there exists an isogeny from E 1 tot E that is defined over Q. For rimes such that E 1 and E have good reduction modulo we see that the isogeny reduces to an isogeny from E 1 F to E 1 F defined over F. Therefore E 1 F = E F for all rimes such that E 1 and E have good reduction. Examle 1.4. Let E 1 : y = x ax + ax + b be an ellitic curve over Q with a, b Z and let E : y = x + ax ax 7a 4b be another ellitic curve see also examle 1.1. Then we have the following isogeny φ : E 1 E, x, y x + a + b x a, 1 a + b x a y. For rimes such that E 1 and E have good reduction modulo we see by theorem 1.8 that E 1 F = E F. Theorem 1.8 already shows the imortance of some rationality questions which arise when trying to find the cardinality of an ellitic curve over a finite field. We will be interested in whether certain isogenies are defined over Q. The following theorem gives a lot of information about the field of definition of isogenies and endomorhisms. Here E σ is the ellitic curve obtained by alying σ AutQ to the coefficients of E and [α] σ E is the endomorhism obtained by alying σ to the coordinate functions. Theorem R Q. Then a Let E/Q be an ellitic curve with comlex multilication by the ring [α] σ E = [ασ ] E σ for all α R and σ AutQ. b Let E be an ellitic curve defined over a field L C with comlex multilication by the quadratic imaginary field K C. Then every endomorhism of E is defined over the comositum LK. c Let E 1 /L and E /L be ellitic curves defined over a field L C. Then there is a finite extension L /L such that every isogeny from E 1 to E is defined over L. Proof. See theorem. of [Sil94,.105]. Let now E/Q be an ellitic curve with comlex multilication such that [ d ] EndE. Then for any σ AutC, E σ = E since E is defined over Q. Moreover [ d ] σ [ d σ ] [ = = ± ] d, 15

24 where the sign deends on whether σ fixes d or not. We can now twist the curve E with d to obtain the quadratic twist E d. An isomorhism between these curves is given by φ : E E d, x, y dx, d dy. This isomorhism is defined over Q d and since both E d and E are defined over Q we find another isomorhism φ σ between these curves for every σ AutC. By the exlicit formulas given for φ we see that φ σ = ±φ, where the sign deends on whether σ fixes d or not. But this means that the isogeny [ d ] φ : E E d, is fixed by AutC. Hence the isogeny φ [ d ] is defined over Q. Since the isogeny φ is an isomorhism, this roves the following roosition. Proosition Let E/Q be an ellitic curve with comlex multilication such that [ d ] EndE. Then there exists an isogeny defined over Q from E to the quadratic twist E d. Moreover the kernel of this isogeny is equal to the kernel of the endomorhism [ d ]. We have already found a method to determine the cardinality of EF when slits in EndE. Of course we are also interested in the cardinality of EF for rimes that are inert in EndE. The following lemma decides whether a rime slits, ramifies or is inert in a quadratic field extension K of Q. Lemma 1.1. Let d be a square free integer, K = Q d and let O K be the ring of integers of K. Then for an odd rime Z we have: a if = 1 then is inert in K, b if = 1 then is slit in K, b if = 0 then is ramified in K, where is the discriminant of K. Proof. First suose is an odd rime that is not inert in K. Then = 1 for rime ideals 1 and = O K / 1 = Z/Z = a Z : a d mod 1 = a d mod 1 d d = a d mod = = 1 or = 0. Now since = d if d 1 mod 4 and = 4d if d, 3 mod 4 we see square free integers d. Hence a follows from the above. d Now suose = = 1 for an odd rime. Then a Z/Z \ {0} : a d mod = a d = a da + d =, a d, a + d. 16 d = for all

25 Suose, a d =, a + d, then a = a d + a + d, a d Z = Z. But this contradicts the fact that a. Hence, a d and, a + d are two distinct rimes lying above, and since K is a quadratic extension their roduct must equal. So, a d, a + d = and is slit in K. d Now suose = 0 for an odd rime. Then d so, d =, d, d. But since d is square free we have gcdd, =, hence, d =. Therefore it follows that is d ramified in K if = 0. We are now ready to rove the following theorem. Theorem Let E/Q be an ellitic curve with comlex multilication given by the quadratic imaginary field K C. Let be an odd rime such that is inert in K and such that E has good reduction modulo. Then EF = + 1. Proof. We may assume that K = Q d for some square free integer d > 0. Then there exists an a Z such that a d EndE. By roosition 1.11 this imlies there exists an isogeny defined over Q from E to the quadratic twist E da. Hence by theorem 1.8 EF = E da F, for all rimes of good reduction. Let now a be such that Then In articular EF = + 1 a. E da F = + 1 d a = a, for all rimes of good reduction. But by lemma 1.1 d = 1, d a. if is inert in K. Hence a = 0 if is inert in K and the theorem follows. 17

26 18

27 Chater Cardinality Formulas In this section we will use theorem 1.7 and theorem 1.13 to find the cardinality EF for some ellitic curves E/Q. We take an ellitic curve E/Q with comlex multilication by the quadratic imaginary field K such that E has good reduction modulo a rime. By theorem 1.7 we then have to factor in EndE to find the cardinality of EF. In articular theorem 1.7 shows us that if a rational rime slits in K, then there exists a π K such that = ππ and Moreover theorem 1.13 shows us that EF = + 1 π π. EF = + 1 if is inert in K. This factorization is however not uniquely determined by and the endomorhism ring, since we may for examle multily π with 1 to find another factorization of. So counting the number of oints comes down to finding the correct factorization of the rime. Lemma 1.1 gave us conditions to determine whether a rime slits or is inert in a quadratic imaginary field K. So for an ellitic curve E/Q with comlex multilication given by K = Q d and any rime of good reduction we have d EF = + 1 if = 1..1 Proosition.1. Let E 1 : y = x 3 x be the ellitic curve over Q with comlex multilication given by Z[i]. Then for any odd rime + 1 if 1 mod 4, E 1 F = + 1 x if 1 mod 4 and = x + y, with x 1 mod 4. Proof. First of all E 1 has discriminant E 1 = 64 = 6, therefore E 1 has good reduction modulo all odd rimes. By lemma 1.1 we see that an odd rime Z slits in Z[i] if = 1 and is inert if = 1. But = 1 1 mod 4 and = 1 1 mod 4. 19

28 So the first case follows from theorem We now assume that 1 mod 4. Hence slits in Z[i]. This means that we can write = ππ for some π Z[i]. Then π = a + bi with a, b Z and since Z[i] is an unique factorization domain any other factorization of will be given by multilying π with a unit. Since Z[i] = {1, 1, i, i} we are left with 4 candidates for the correct factorization of, as in theorem 1.7. Next we comute the 4 th -division olynomial ψ 4 x of E 1 whose roots are the x-coordinates of the oints of E 1 [4] \ {O}. We find ψ 4 x = 8xx 1x + 1x + 1x x 1x x 1, with roots {0, ±1, ±i, ±1 ± }. Moreover if we take fx = x 3 x we find fi = i = 1 i, f i = i = 1 + i, f1 + = = +, f1 = 6 4 =, f 1 + = = i, f 1 = 6 4 = i +, and since f0 = f1 = f 1 = 0 we have E 1 [4] = {O, 0, 0, 1, 0, 1, 0, i, ±1 i, i, ±1 + i, +, ± +,, ±, +, ±i,, ±i + }. Now notice that since 1 mod 4 we have i = 1 F, which means there exists a square root of 1 in F. Hence {O, 0, 0, 1, 0, 1, 0, i, ±1 i, i, ±1 + i} E 1 F [4]. And since E 1 F [4] is a subgrou of E 1 [4] we see that E 1 F [4] {8, 16}, hence 8 divides the order of the subgrou E 1 F [4]. But this means that 8 divides the order of the grou E 1 F. By theorem 1.7 it now follows for 1 mod 4 that E 1 F = + 1 π + π = + 1 a 0 mod 8. Hence if 1 mod 8 then a 1 mod 4 and if 5 mod 8 then a 1 mod 4. Now notice that since = ππ = a + b 1 mod 4, either a or b has to be odd and they can not be both odd. By the above we see that a has to be odd. Altogether we find + 1 if 1 mod 4, E 1 F = + 1 x if 1 mod 4 and = x + y, with x 1 mod 4. Here we have used the following identity { 1 if ±1 mod 8, = 1 if ±3 mod 8. 0

29 Proosition.. Let E : y = x 3 11x 14 be the ellitic curve over Q with comlex multilication given by Z[i]. Then for any odd rime + 1 if 1 mod 4, E F = + 1 x if 1 mod 4 and = x + y, with x 1 mod 4. Proof. The discriminant of this ellitic curves is given by E = 9, hence E has good reduction modulo all odd rimes. Moreover we can write E 1 : y = x 1x + x, E : y = x + x x 7. Therefore by examle 1.4 we have the following isogeny φ : E 1 E, x, y x + x 1, which is defined over Q. It now follows by theorem 1.8 that E 1 F = E F, for all odd rimes, which comletes the roof. 1 x 1 y, Proosition.3. Let E 3 : y = x be the ellitic curve over Q with comlex multilication given by Z[ω], where ω = Then for any rime > if 1 mod 3, E 3 F = + 1 x 3 x if 1 mod 3 and = x + 3y. Proof. First of all the discriminant of this curve is given by E 3 = Hence E 3 has good reduction modulo for all rimes unequal to and 3. Moreover by lemma 1.1 we see that is inert in Z[ω] if 3 = 1 and slits if 3 = 1. By the quadratic recirocity law we now have 3 = = = 1. Hence 3 = { 1 if 1 mod 3, = 3 1 if 1 mod 3. Which roves the roosition for the case 1 mod 3. So now assume 1 mod 3. Hence slits in Qω and by theorem 1.7 we can write = ππ for some π Z[ω]. Let π = a + bω be as in theorem 1.7. Using the Nagel-Lutz theorem see [ST9] we can calculate the comlete torsion subgrou of EQ, consisting of all oints of finite order defined over Q. We find that the torsion of EQ is given by { 1, 0, 0, 1, 0, 1,, 3,, 3, O}. Therefore EQ contains a oint of order and a oint of order 3. We see that by reducing the coordinates modulo, for > 3 we have 1

30 that EF contains a oint of order and a oint of order 3. Hence 6 EF for all rimes > 3. Moreover 1 mod 3 is an odd rime, hence 1 mod 6. So altogether we find 0 EF + 1 π + π a b mod 6.. However Z[ω] is a unique factorization domain, hence any other factorization of is given by multilying π with a unit. We have Z[ω] = {1, 1, ω, ω, ω, ω }, which gives us 6 ossible candidates for the correct factorization of. If we can eliminate 5 of them we have determined π u to conjugation. By equation. we have π + π = a + b mod 6. So b has to be even which imlies that a has to be odd since = ππ = a + ab + b is an odd rime. These congruences give us some restrictions on the factorization of in the endomorhism ring. All other factorizations of are found by multilying π with a unit. Comuting the other candidates gives us π π π mod 6, ωπ = aω + bω = b + a bω, ωπ = b + b aω, ω π = b a aω, ω π = a b + aω. Since a is odd and b is even we see that the condition π + π mod 6 determines π u to conjugation. Moreover since b is even we have π = a + bω = a + b + b 3 Z[ 3]. Hence = ππ = x + 3y for some x, y Z. But x = π + π mod 6 which imlies x 1 mod 3. Altogether we thus find the following formula, which holds for all rimes > if 1 mod 3, E 3 F = x + 1 x if 1 mod 3 and = x + 3y. 3 Proosition.4. Let E 4 : y = x 3 15x + be the ellitic curve over Q with comlex multilication given by Z[ 3]. Then for any rime > if 1 mod 3, E 4 F = x + 1 x if 1 mod 3 and = x + 3y. 3 Proof. The discriminant of E 4 is given by E 4 = 8 3 3, hence E 4 has good reduction for all rimes > 3. We can now write E 3 : y = x + 1x x + 1, E 4 : y = x x + x 11.

31 Therefore by examle 1.4 we have the following isogeny φ : E 3 E 4, x, y x + 3 x + 1, which is defined over Q. It now follows by theorem 1.8 that E 3 F = E 4 F, for all rimes, 3, which comletes the roof. 1 3 x + 1 y,.1 Multilication by d For other CM-curves over Q we need some extra tools to find formulas as the ones in the revious section. So let E/Q be an ellitic curve with comlex multilication. Suose d Z >0 such that [ d] EndE. We will study the kernel of this endomorhism, which we denote by E[ d]. Since charq = 0 it follows that E[ d] = deg[ d d] = d = d. Now notice that for P E[ d] we have [d]p = [ d][ d]p = O. So E[ d] is a subgrou of E[d]. Therefore the x-coordinates of the oints in E[ d] \ {O} are zeros of the d th -division olynomial ψ d. Now let E : y = x 3 + Ax + B be a short Weierstrass equation of E for some A, B Q and let σ GalQ/Q. Then for any x, y E we have 0 = σ0 = σy x 3 Ax B = σy σx 3 σaσx σb = σy σx 3 Aσx B, where the last inequality follows since A, B Q. Hence for any x, y EQ we see σx, y = σx, σy E, so σ mas EQ to EQ. Since the addition formulas are defined over Q we see that σ defines a grou isomorhism and therefore σe[m] = E[m]. Lemma.5. Let E be an ellitic curve over Q. Then the grou E[ d] is fixed by GalQ/Q. Proof. By roosition 1.11 we see that E[ d] is the kernel of an Q-isogeny from E to its twist E d. Therefore the lemma immediately follows. For any oint x, y E we have x, y E. Moreover these oints are distinct when y 0, i.e. when x, y / E[]. If we take d odd, then E[] E[d] = {O}. Therefore any zero of the d th -division olynomial ψ d gives us two oints of E[d]. Moreover GalQ/Q fixes E[ d]. Hence if we factor ψ d over Q and we find a unique factor of degree d 1, then the zeros of this factor give us a unique set of d 1 oints in E[d] \ {O}, for which the x-coordinates are fixed by GalQ/Q. This means that these oints together with O form the set E[ d]. E[ d] is a subgrou with d elements. Hence if d is a rime E[ d] = Z/dZ. In articular the field Q E [ d ], obtained by adding all coordinates of E [ d ] to Q, has [ d ] Gal Q E /Q Z/dZ. Hence Q E [ d ] is an abelian extension of Q. 3

32 Theorem.6 Kronecker-Weber. Let K be an abelian extension of Q. Then there is a ositive integer n such that K Q ζ n, ζ n = e πi/n. Proof. See theorem 4.1 of [Was97,.319]. When we study the roof of the Kronecker-Weber theorem, as stated in [Was97], we find an even stronger result. This roof starts with an abelian extension K/Q and it constructs the cyclotomic field Q ζ n, where n = e. ramifies The roduct is taken over all rational rimes that ramify in the extension K/Q. The next and most difficult ste of the roof consists of roving that K Q ζ n. We have thus seen that n is only divisible by rimes that ramify in K/Q. Moreover a rime ramifies in Qζ n /Q if and only if it divides n. It follows that, if we take n to be minimal, the rimes ramifying in Qζ n are exactly the rimes ramifying in K. We will now assume that we always take n to be minimal when alying the Kronecker-Weber theorem. Let now d be a rime such that [ d ] EndE. Then Q E [ d ] /Q is an abelian extension and the Kronecker-Weber theorem imlies the existence of a rimitive root of unity ζ n such that Q E [ d ] Qζ n. But the Galois grou of Qζ n is simly Z/nZ and any σ GalQζ n /Q is given by σζ n = ζ a n for some a Z/nZ. Furthermore we see that for a rime of good reduction σ sending ζ n to ζ n acts exactly as the Frobenius endomorhism Fr after reduction modulo. Hence if we are able to determine the rimitive root of unity ζ n such that Q E [ d ] Qζ n, we will be able to determine the action of Frobenius on EF [ d] by exlicit comutations in characteristic 0. This means that we will be able to determine residue class of the Frobenius endomorhism modulo d. Altogether it follows that if there exists an element π EndE such that σ : E[ d] E[ d], P πp we find Fr π mod d and since a = TrFr = Fr +Fr we will be able to determine a modulo d with this information. Proosition.7. Let E 5 : y = x 3 70x 151 be the ellitic curve over Q with comlex multilication given by Z[ ]. Then for any rime > if 1, 3 mod 8, E 5 F = + 1 x if 1, 3 mod 16 and = x + y such that x 1 mod 4, + 1 x if 9, 11 mod 16 and = x + y such that x 3 mod 4. Proof. Firstly the discriminant of this curve is equal to E 5 = = So E 5 has good reduction modulo all rimes > 3. Moreover 1 if 1, 3 mod 8 = 0 if = 1 otherwise. So the first case now follows from equation.1. 4

33 Next we consider the case 1, 3 mod 8. Then slits in Q, hence = ππ for some π Z[ ]. Now since Z[ ] is a unique factorization domain and Z[ ] = {±1} we see that the factorization is unique u to the sign of π and π. We have π = a + b for some a, b Z and = ππ = a + b. Hence a is odd and we only have to determine the sign of a to find a = TrFr = π + π = a. Therefore it is enough to determine the residue class of a modulo 4. To determine this residue class we will consider the endomorhism [ ] : E 5 E 5. For a oint P E 5 [ ] we see 4P = [ ][ ]P = O, hence E 5 [ ] E 5 [4]. Moreover E 5 [] E 5 [ ]. But we can comute the x-coordinates of E 5 [] by solving We find x 3 70x 151 = x + 1x 1x 16 = 0. E 5 [] = {O, 1, 0, 6 9e 4πi/16 + 9e 1πi/16, 0, 6 + 9e 4πi/16 9e 1πi/16, 0}. We also know that the degree of the endomorhism [ ] is 8, hence E 5 [ ] = 8. So there are 4 other oints in this subgrou, but these oints only give us two new x-coordinates since x, y x, y for all x, y E 5 \ E 5 []. We comute the 4 th -division olynomial ψ 4 x = 8x + 1x 1x 16x + 4x 18x 4 4x 3 756x 158x Since the grou E 5 [ ] is stable under GalQ/Q we see that the x-coordinates we need are zeros of x + 4x 18. Hence altogether we find E 5 [ ] \ E 5 [] = { 1 9e 4πi/16 + 9e 1πi/16, 54e 6πi/ e 10πi/16, 1 9e 4πi/16 + 9e 1πi/16, 54e 6πi/16 54e 10πi/16, 1 + 9e 4πi/16 9e 1πi/16, 54e πi/ e 14πi/16, 1 + 9e 4πi/16 9e 1πi/16, 54e πi/16 54e 14πi/16 } In articular we see that all coordinates of E[ ] are elements of the field Qe πi/16. For any rime we now have σ GalQe πi/16 /Q such that σ : E 5 [ ] E 5 [ ], e πi/16 e πi/16. Hence σ is determined by the residue class of modulo 16. And since σ gets maed to Fr after reduction modulo we see that acting on E 5 F [ ] the Frobenius endomorhism only deends on the residue class of modulo 16. Therefore it follows that for two rimes and q we have q mod 16 = Fr Fr q mod = 1 a 1 aq mod = 1 a 1 aq mod 4, = a aq mod 8, where the second to last imlication follows since a, aq Z. This means that we only have to determine a for one in each residue class modulo 16. By simly counting the number of oints on the reduction of E 5 modulo we find a a17 = + 1 E 5 F 17 = 18 4 mod 8, if 1 mod 16, a a19 = + 1 E 5 F 19 = 0 18 mod 8, if 3 mod 16, a a41 = + 1 E 5 F 41 = mod 8, if 9 mod 16, a a11 = + 1 E 5 F 11 = mod 8, if 11 mod 16. 5

34 Therefore we have { + 1 x if 1, 3 mod 16 and = x + y such that x 1 mod 4, E 5 F = + 1 x if 9, 11 mod 16 and = x + y such that x 3 mod 4. Remark. The roof above shows that if QE[ d] Qζ n for some n th -root of unity ζ n, then we only have to determine a modulo d for one rime in each residue class of Z/nZ. Since for two rimes 1 mod n that slit in QEndE, we have Fr 1 Fr mod d. Note that these endomorhism act on different ellitic curves since Fr 1 acts on the reduction of E modulo 1 and Fr acts on the reduction of E modulo. However we do have End F1 E = End Q E = End F E. So the different Frobenius endomorhisms we are considering in the roof above all lie in isomorhic endomorhism rings. Proosition.8. Let E [ 6 : y = ] x 3 35x 98 be the ellitic curve over Q with comlex multilication given by Z. Then for any rime, if 1,, 4 mod 7, E 6 F = x + 1 x if 1,, 4 mod 7 and = x + 7y. 7 Proof. The discriminant of E 6 is given by E 6 = 1 7 3, hence E 6 has good reduction modulo all rimes, 7. Moreover 7 = So the first case now follows from equation.1. Now let be a rime that slits in Z 1 if 1,, 4 mod 7 0 if = 7 1 otherwise. [ 1+ 7 ], i.e. 1,, 4 mod 7. So we have π = c + d 7, for some c, d Z of the same arity such that ππ =. Moreover the Frobenius endomorhism is given by multilication by π and therefore a = + 1 E 6 F = π + π = c. But since x 3 35x 98 = x 7x + 7x + 14 we see that 7, 0 E 6 []. Therefore for any rime of good reduction there exists a oint of order in EF. By grou theory it now follows that EF, hence a = c is even. This means we can write π = a + b 7, [ for some a, b Z and = a + 7b ]. Moreover Z 1+ 7 = {±1}, hence a and b are unique u to their sign. 6