# such that (E, C)∼(EC,E[n]C

## Full text

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AND RELATED ALGORITHMS

Communicated by Alexandru Zaharescu

In this paper, we provide a new description of the fixed points of the action of Fricke’s involutionwnon the open modular curvesY0(n) and the corresponding algorithm for computing them. We develop first an algorithm in order to classify the complex elliptic curvesEfor which there exist cyclic subgroupsC(E,+) of ordern such that the elliptic curvesE and E/Care isomorphic, where n is a positive integer, and implement it in Magma. Next, we answer the following question: given a complex elliptic curveE, when can one find a cyclic subgroup C of ordern of (E,+) such that (E, C)(EC,E[n]C ), E[n] being then-torsion subgroup ofE, classifying in this manner the fixed points of the action ofwnon Y0(n). Several interesting examples are given in the third and fifth sections.

AMS 2010 Subject Classification: Primary 11G07, 11G15, 11Y16; Secondary 14D22.

Key words:elliptic curve, algorithm, cyclic subgroup, modular curve.

1. INTRODUCTION

Let E be an elliptic curve defined over the field of complex numbers C and C be a subgroup (not necessarily cyclic) of order n <∞ of (E,+). This means that C is a subgroup of order n of E[n] where by E[n] one denotes the n-torsion subgroup of E i.e. the set of points of order n in E: E[n] = {P ∈ E : [n]P = O}. Let also π : E → E/C be the natural projection.

SinceC acts effectively and properly discontinuous onE, the groupE/C has a structure of Riemann variety, compatible with the morphism π and moreover π is unramified of degreen: degπ=|π−1(O)|=|C|=n. It is known thatE/C is a complex elliptic curve and that if C is cyclic, one hasE[n]/C ∼=Z/nZ.

Denote by H the upper half plane i.e. {z ∈C|Im(z)>0}. One defines the open modular curves Y0(n) as the quotient space Γ0(n)/H that is the set of orbits {Γ0(n)τ : τ ∈ H}, where Γ0(n) is the “Nebentypus” congruence subgroup of levelnof SL2(Z), acting on Hfrom the left:

Γ0(n) ={

a b c d

∈SL2(Z)|c≡0(modn)}.

MATH. REPORTS16(66),4(2014), 477–502

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Following the notations of [5], an enhanced elliptic curve for Γ0(n) is by definition an ordered pair (E, C) where E is a complex elliptic curve and C a cyclic subgroup ofE of order n. Two pairs (E, C) and (E0, C0) are said to be equivalent if some isomorphism E∼=E0 takes C to C0. One denotes the set of equivalence classes with

S0(n) :={enhanced elliptic curves for Γ0(n)}/∼.

Furthermore, an element of S0(n) is an equivalence class [E, C]. S0(n) is a moduli space of isomorphism classes of complex elliptic curves and n-torsion data.

Denote now by Λτ the lattice Z+Zτ,τ ∈ Hand byEτ the elliptic curve C/Λτ. Then one has the following bijection:

S0(n)∼=Y0(n) given by [C/Λτ,h1/n+ Λτi]7→Γ0(n)τ (see [5], Theorem 1.5.1 for details.)

In the second section of this paper, we elementarily study the complex elliptic curves E for which there exist cyclic subgroups C ≤ (E,+) of order n such that the elliptic curves E and E/C are isomorphic, where n is a posi- tive integer. We point out that one can also study these curves by using the modular polynomial Φn(X, Y)∈ Z[X, Y]. Following [12], Theorem 5, Section 5, Φn(j, j0) = 0 if and only if there exists an elliptic curve E of j-invariant j, n-isogeneous to an elliptic curveE0 ofj-invariantj0. Consequently, finding the complex elliptic curvesE for which there exist cyclic subgroupsC ≤(E,+) of order n such that the elliptic curves E and E/C are isomorphic is equivalent to finding the j0ssuch that Φn(j, j) = 0.

We observe that E (and consequently E/C) are CM curves, obviously have isomorphic endomorphism rings and hence, these points (E, C) with E ∼=E/C are a special class of Heegner points on the modular curve X0(n) = Y0(n)∪ {cusps}. We recall that the Heegner points ofY0(n) as defined by Birch in [2] are pairs (E, E0) ofn-isogenous curves with the same endomorphism ring.

Heegner was the one introducing them in [8] while working on the class num- ber problem for imaginary quadratic fields and their importance is extensively described in the survey [3].

After elementarily studying the above mentioned class of Heegner points, upon imposing certain conditions (see Theorem 2.3 and Proposition 2.4), given a complex elliptic curve E, we answer the question of when there exists C ≤ (E,+) cyclic subgroup of order nof E such that (E, C)∼(EC,E[n]C ), studying in this manner the fixed points of the action of the Fricke involution

wn:=

0 −1

n 0

∈GL2(Q+)

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on the open modular curves Y0(n).

Throughout the third section, we work out several interesting examples by mean of Theorems 2.1 and 2.3 for the above class of Heegner points as well as for the fixed points of the action of wn on Y0(n), for the cases 2 ≤n ≤ 7 and n= 11.

In the fourth section we develop our first algorithm which classifies the complex elliptic curvesE for which there exist cyclic subgroupsC ≤(E,+) of order n such that E and E/C are isomorphic, where n is a positive integer.

Furthermore, by mean of the second algorithm, which we derive from the first one, we compute the fixed points of the action of wn on the open modular curves Y0(n) and obtain correct results. The number of fixed points of the action of Fricke’s involution on Y0(n) was a priori computed by Ogg (see [14], Proposition 3) and Kenku (see [10], Theorem 2) and this number, forn >3, is ν(n) = h(−n) +h(−4n) if n≡3(mod4) andν(n) =h(−4n) otherwise, where h(−n) is the class number of primitive quadratic forms of discriminant−nand ν(2) =ν(3) = 2. By using our second algorithm the reader will also obtain this number which as one can observe is not based on Riemann-Hurwitz’s formula nor on factoring Fricke’s involution as in [14] and [10].

Though the fixed points ofwnare known, in this paper we provide a new description of them.

It is easy to see that wn normalizes the group Γ0(n) and hence, gives an automorphism Γ0(n)z7→Γ0(n)wn(z) onY0(n). It is also easy to check that this automorphism is an involution. The following proposition is a known result which we will use throughout the paper and we skip its proof (the reader may consult [4], section IV), 4.4 or [9], Theorem 2.4 and Remark 5.5 for details):

Proposition1.1. The action ofwnon the moduli spaceS0(n)is given by:

[E, C]7→[E/C, E[n]/C].

The reader can find a survey of the known algorithms for computing isogenies between elliptic curves in [1]. These algorithms are mainly based on power series expansions of the Weierstrass ℘-function and computing its coefficients or by making use of the differential equations satisfied by certain functions related to the Weierstrass℘-function (see [1] for details) whereas in our algorithms we use elliptic curves as one dimensional tori i.e. in the form C/Λ with Λ complex lattice and not in the Weierstrass form given the fact that it suffices for the main goal of our paper namely to obtain the fixed points of the Fricke involution based on our first algorithm. We would like to point out that in [16], V´elu showed how for any given fieldK, given an elliptic curveE/K and the kernel of an isogeny havingE as domain, to compute the codomain of the isogeny as well as to establish the rational map of the isogeny. Kohel provided

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in [11] a new approach and instead of using a list of points in the kernel he computed the isogeny from the kernel polynomial. One can not determine all complex elliptic curves Ewhich admit cyclic subgroupsC such thatE/C ∼=E by using Kohel and V´elu’s methods. Starting with an elliptic curve E and a point generating the kernel might end up with the same curve in very few cases after factorization. One can get the same results as the ones obtained upon running our first algorithm by using the algorithms of Stark and Elkies, their approaches being different than our’s: in the algorithm of Stark (see [15]) one computes the kernel polynomial given the domain, codomain and degree of an isogeny. In Elkies algorithm (see [6]), by using the modular polynomial, given an elliptic curveE one computes curves which are isogeneous to it.

Throughout the last section of our paper, we provide interesting compu- tational examples by mean of our codes together with the CPU time it took to complete the corresponding calculations and the total memory usage.

2. MAIN RESULTS

With the background from the previous section we are ready to elementar- ily classify the complex elliptic curvesEfor which there exist cyclic subgroups C ≤(E,+) of ordernsuch that the elliptic curvesE andE/C are isomorphic:

Theorem2.1. Let E be a complex elliptic curve determined by the lattice h1, τi, τ ∈ H. Then:

i) ∃C≤(E,+) finite cyclic subgroup such that EC 'E⇔ ∃u, v∈Q such thatτ2 =uτ+vwith ∆ =u2+ 4v <0 (i.e. E admits complex multiplication);

ii) If τ satisfies the conditions of i) and u = uu1

2, v = vv1

2, u2 6= 0, v2 6=

0, u1, u2, v1, v2 ∈Z,(u1, u2) = (v1, v2) = 1, d2 = (u2, v2), then:

∃C ≤ (E,+) cyclic subgroup of order n which satisfies EC ' E ⇐⇒

∃(a, b0)∈Z2 with (a, b0) = 1 such that n= detM, where M is the matrix M =

a A b B

and (a, A, b, B) = a,u2dv1

2 b0,u2dv2

2 b0, a+ud1v2

2 b0

;

iii) The subgroup C from ii) is C=hu11+un21τi, whereu11, u21 are obtained in the following way:

Since detM = n and gcd(a, A, b, B) = 1 (one deduces easily this), the matrix M is arithmetically equivalent to the matrix:

M ∼

1 0 0 n

,

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hence,

∃U, V ∈GL2(Z) such that M =U ·

1 0 0 n

·V.

The elements u11, u21 are the first column of the matrix U =

u11 u12

u21 u22

.

Proof. We have that L=Z+Zτ andC =h¯gi ≤E= CL cyclic subgroup of order n hence, C = L+ZgL and n¯g = ¯0. It follows that ng ∈ L and hence, we can write g∈C as

(1) g= α12τ

n , α1, α2∈Z.

Since ord(¯g) =nit follows easily that gcd(α1, α2, n) = 1.

We have that EC ' L+CLZg

L

' L+C

Zg = LC0, where we denoted byL0 =L+Zg= Z+Zτ +Zg.

It is known that CL ' LC0 ⇔ ∃λ ∈ C such that λL = L0. Consequently E ' ECCL ' LC0 ⇔ ∃λ∈C such that

(2) λ(Z+Zτ) =Z+Zτ+Zg.

The relation (2) can be studied in the following way:

“⊆” Sinceg is given by (1), from the inclusion “⊆” one shows that (note that this is not an equivalence): ∃a, b, A, B ∈Zsuch that

(3)

(λ= a+bτn λτ = A+Bτn .

“⊇” The inclusion “⊇” is equivalent to

(4) ∃α, β, u, v, s, t∈Z such that





1 =αλ+βλτ τ =uλ+vλτ g=sλ+tλτ.

By replacing (3) and (1) in (4) we obtain (5)





1 =αa+bτnA+Bτn τ =ua+bτn +vA+Bτn

α12τ

n =sa+bτn +tA+Bτn

⇐⇒













a A b B

!

· α u β v

!

=nI2

a A b B

!

· s t

!

= α1

α2

!

.

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Note that the relation (2) is not yet equivalent to (3) and (5). In order to obtain the equivalent conditions for (2) let us look to the inclusion “⊆”.

The inclusion “⊆” means that there exista1, b1, c1, a2, b2, c2 ∈Zsuch that (6)

(λ=a1+b1τ +c1g

λτ =a2+b2τ +c2g ⇐⇒

(a+bτ

n =a1+b1τ+c1α12τ n A+Bτ

n =a2+b2τ +c2α12τ n

⇐⇒

(a=na1+c1α1

b=nb1+c1α2 and

(A=na2+c2α1

B =nb2+c2α2 ⇔ (7)

(aˆ= ˆc1αˆ1

ˆb= ˆc1αˆ2 and

(Aˆ= ˆc2αˆ1

Bˆ= ˆc2αˆ2 in Z nZ.

We obtain that the inclusion “⊆” is equivalent to the existence ofc1, c2∈ Z which both satisfy (6),i.e. (2) is equivalent to (5) and (6).

We have obtained that the hypothesis E ' EC withC cyclic subgroup of order nis equivalent to the relations (2) and (α1, α2, n) = 1,i.e. is equivalent to the relations (5), (6), (α1, α2, n) = 1 and (a+bτ)τ =A+Bτ. Assume now that these four conditions are satisfied and let us replace them with conditions which are easier to be verified.

Denote by M the matrix M =

a A b B

hence, Mˆ =

ˆa Aˆ ˆb Bˆ

=

1αˆ12αˆ1 ˆ

c1αˆ22αˆ2

in Z nZ =⇒ det ˆM = ˆ0 in Z

nZ =⇒ det M...n.

By using now (5) we obtain det(M)|n2 and so det(M) =nk, k|n. From (5) we get α = detMnB = Bk hence,k|B. Similarly one obtains k|b, k|A, k|a and hence, from the second equality of (5) we obtain that k|α1, α2. Since k|n and (α1, α2, n) = 1 we obtain k=±1, consequently det(M) =±n.

We will prove that the relations (5), gcd(α1, α2, n) = 1 and det(M) =±n imply (6):

In did, from (α1, α2, n) = 1 it follows that ∃µ1, µ2, µ3 ∈ Z such that µ1α12α23n = 1. Choose ˆc11\a+µ2b. We then have ˆc1αˆ1 = (µ1\a+µ2b) ˆα1 = a·µ[1α1 +b·µ[2α1 = a(1−µ\2α2−µ3n) +b·µ[2α1 = ˆa+ ˆ

µ2(bα\1−aα2).

From (5) we get that t = detM2−bα1 = 2±n−bα1 =⇒ aα2−bα1...n. Conse- quently ˆc1αˆ1 = ˆa,q.e.d.

Similarly ˆc1αˆ2 = (µ1\a+µ2b) ˆα2 = a·µ[1α2 +b ·µ[2α2 = a·µ[1α2 + b(1−µ\1α1−µ3n) = ˆb+ ˆµ1(aα\2−bα1) = ˆb,q.e.d.

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Similarly, by choosing ˆc21A\+µ2B one obtains the expressions of (6) regarding A, B.

We have obtained that (6) follows from the relations gcd(α1, α2, n) = 1, (5) and det(M) =±n. Moreover, we can renounce to the first equality of (5) since it can be deduced from the condition det(M) =±n:

In did, if det(M) =±nwe have M·M = det(M)·I2 = (±n)I2, hence, the existence of the elementsα, β, u, vfollows from the existence of the adjoint matrixM.

In conclusion, the hypothesis E ' EC with C cyclic subgroup of order n is equivalent to the following four relations: the second equality from (5), (α1, α2, n) = 1, det(M) =±nand (a+bτ)τ =A+Bτ.

We prove now that we can renounce at the second equality of (5) and at gcd(α1, α2, n) = 1, by replacing them in the above equivalence with the condition gcd(a, A, b, B) = 1 :

Let d = gcd(a, A, b, B). From the second equality from (5) we obtain d|α1, d|α2 and since d|detM = ±n, it follows that d|gcd(α1, α2, n) = 1, i.e.

gcd(a, A, b, B) = 1. We have that M is arithmetically equivalent to:

M =

a A b B

1 0 0 n

, i.e. ∃U, V ∈SL2(Z) such that a A

b B

=U·

1 0 0 n

·V.

The second condition of (5) becomes:

a A b B

· s

t

= α1

α2

⇐⇒U·

1 0 0 n

·V · s

t

= α1

α2

.

We denote by V · s

t

= s0

t0

and obtain the equivalent equation α1

α2

=U· s0

nt0

, U ∈SL2(Z).

Viceversa, remark that the second condition of (5) follows from det(M) =

±nand from the relation gcd(a, A, b, B) = 1 as it follows: chooses0, t0 ∈Zsuch that (s0, n) = 1. Consider the matrix (α1 α2) = (s0 nt0)·Ut, where U is an invertible matrix from the arithmetically equivalent decomposition of M. We have that gcd(α1, α2, n) = gcd((α1, α2), n) = gcd((s0, nt0), n) = gcd(s0, n) = 1 and the second equality of (5) follows from the previous construction,q.e.d.

Let us analyze now the condition (a+bτ)τ =A+Bτ. This means that τ is algebraic overQand satisfies the second degree equation

2 = (B−a)τ+A (∗)

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Let µτ = X2 −uX −v ∈ Q[X] be the minimal polynomial of τ, with u= uu1

2 ∈Q, v = vv1

2 ∈Q, u1, u2, v1, v2∈Z,gcd(u1, u2) = gcd(v1, v2) = 1 and let d2 := gcd(u2, v2). We identify now the coefficients of the equation (∗) and of the minimal polynomial ofτ and obtain:

(A

b =v= vv1

B−a 2

b =u= uu1

2

hence,

(v2A=v1b u2(B−a) =u1b.

Since

(v2A=v1b

gcd(v1, v2) = 1 we get that (v2|b

v1|A.

From

(u2(B−a) =u1b

gcd(u1, u2) = 1 it follows that



 u2|b u1|B−a

we already have v2|b hence, lcm[u2, v2]|b.

Consequently ∃b0 ∈Z such that





b= [u2, v2]b0 = ud2v2

2 b0 v2A=v1b

u2(B−a) =u1b

hence,





b= ud2v2

2 b0 A= ud2v1

2 b0 B=a+ud1v2

2 b0,

where d2 = gcd(u2, v2) andτ2 =uτ+v.

We prove now the equivalence:

gcd(a, A, b, B) = 1⇐⇒gcd(a, b0) = 1.

“=⇒” Denote by d0 = gcd(a, b0). It follows thatd0|gcd(a, A, b, B) hence, d0|1, i.e. d0 = 1.

“⇐= ” Let d= gcd(a, A, b, B). Then d|a and since gcd(a, b0) = 1 we get gcd(d, b0) = 1. Moreover d2 = (u2, v2) so let u2 = d2u02 and v2 = d2v02 with gcd(u02, v02) = 1. Since d|aand d|B we have that d|B−aand obtain:





d|b= ud2v2

2 b0 d|A= ud2v1

2 b0=u02v1b0 d|B−a= ud1v2

2 b0 =v20u1b0

hence,



 d|ud2v2

2 =u02v2 =u2v02 d|u02v1

d|v02u1.

If d|u02, since gcd(u02, v20) = 1 one obtains d|u1. But gcd(u1, u2) = 1 and consequently d= 1.

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If d|v2, using the fact that gcd(v1, v2) = 1 we get that d|u02. Since d|u02 and gcd(u1, u2) = 1 we have gcd(d, u1) = 1. Consequently d|v02 and since we already have d|u02, we obtaind= 1.

We prove now that n= det(M)>0:

In order to simplify the notations, denote by d := d2 and note that (a, A, b, B) =

a,u2dv1b0,u2dv2b0, a+u1dv2b0

. We obtain that det(M) =aB−bA=a

a+u1dv2b0

u2v2

d b0 u2v1

d b0

= a2+u1dv2ab0v1vd22u22b02=

a+u2d1v2b0 2

u4d21v222b02v1vd22u22b02 =

a+u2d1v2b0 2

u21v22+4v1v2u22

4d2 b02 =

a+u2d1v2b02

u22v22

hu1 u2

2

+4vv1

2

i

4d2 b02=

a+u2d1v2b02

u224dv222b02. Since ∆<0 and u2 6= 0, v2 6= 0 it follows easily that det(M)>0.

In order to describe the subgroup C we deduce that:

α1

α2

=U · s0

nt0

⇐⇒

α1

α2

=

u11 u12

u21 u22

· s0

nt0

i.e.

1 =u11s0+u12nt0 α2 =u21s0+u22nt0 . We obtain that:

g= α12τ

n = u11s0+u12nt0+ (u21s0+u22nt0

n =

u11s0+u21s0τ +n(u12t0+u22t0τ)

n = s0(u11+u21τ)

n +t0(u12+u22τ) =g0+t0(u12+u22τ), where we denoted by g0= s0(u11+u21τ)

n .

Consequently,C= L+ZgL = L+ZgL 0 =hg0i. But gcd(s0, n) = gcd(α1, α2, n) = 1 hence, there exist a, b ∈ Z such that as0 +bn = 1. It follows that ag0 = as0(u11+un 21τ) = (1−bn)(un11+u21τ) = u11+un21τ −b(u11+u21τ). We have obtained that C= L+ZgL 0 =hg0i=hu11+un21τi.

Remark 2.2. Fora, b0 fixed integers such that gcd(a, b0) = 1 the subgroup C is uniquely determined hence, we can denote C=Ca,b0.

Recall now that for E, E0 complex elliptic curves andC ≤(E,+), C0 ≤ (E0,+) cyclic subgroups of order n of E and E0 respectively, one has that (E, C)∼(E0, C0)⇐⇒ ∃u:E −→E0 isomorphism such thatu(C) =C0. The question we want to answer is the following:

“Let E be a complex elliptic curve. When ∃C ≤(E,+) cyclic subgroup of ordernofE such that (E, C)∼(EC,E[n]C ), whereE[n] ={x∈E :nx= 0}?”

In order to answer this question we will give the following criterion:

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Theorem 2.3. Let E be an elliptic curve defined over C satisfying the conditions of 2.1 i). Then the following are equivalent:

{∃C ≤(E,+) cyclic subgroup of order nof E such that (E, C)∼(EC,E[n]C )} ⇐⇒

{∃(a, b0)∈Z2 withgcd(a, b0) = 1 such that det(M) =n andn|Tr(M)} ⇐⇒

{∃(a, b0)∈Z2, with gcd(a, b0) = 1 such that det(M) =nand M2≡O2(modn)},

where M is the matrix from 2.1 ii) and Tr(M) the trace of M. Proof. We use the notations of Theorem 2.1.

Let E = CL with L = Z +Zτ ⊂ C lattice and C = h¯gi ≤ E = CL subgroup cyclic of ordern, consequentlyC= L+LZg. We have that EC ' L+CLZg

L

'

L+CZg = LC0, where we denoted by L0 = L+Zg = Z+Zτ +Zg. Recall that

CL ' LC0 ⇔ ∃λ∈Csuch thatλL=L0.

Letube defined in the following way: E= CL −→ EC = LC0 = L+ZgC isomor- phism, u(¯z) =λz. Thenc u(C) =u L+ZgL

=λ·L+ZgL0 = λL+ZλgL0 . SinceλL=L0 we have that u(C) = L0+LZ0λg = hcλg,+i. But E[n] =

1 nL

L and consequently

E[n]

C =

1n L L

C =

1n L L L+Zg

L

=

1 nL L+Zg =

1 nL

L0 . We obtain that:

(8) u(C) = E[n]

C ⇐⇒L0+Zλg = 1 n·L.

The relation (8) is equivalent toλL+Zλg=1n·L⇐⇒λ(L+Zg) =n1·L⇐⇒

λL0 = 1n·L⇐⇒λ2L= 1n·L⇐⇒nλ2L=L.Moreover, (9) L=nλ2L⇐⇒ ∃M1 ∈SL2(Z) such that

22τ

=M1· 1

τ

.

We denote byM1=

a1 A1 b1 B1

and by using the relations (3):

(λ=a+bτn λτ=A+Bτn we get:

22τ

=

a1 A1 b1 B1

· 1

τ

⇐⇒

(nλ2=a1+A1τ

2τ =b1+B1τ ⇐⇒

(a1+A1τ = (a+bτn )2 b1+B1τ =λ·(A+Bτ)

⇐⇒

(a1+A1τ = (a+bτ)n 2 b1+B1τ = (a+bτ)(A+Bτ)n .

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Note that A+Bτ =τ·(a+bτ), which implies that bτ2 = (B−a)τ+A.

By using this equation we obtain:

a1+A1τ = (a+bτn )2 ⇐⇒na1+nA1τ =a2+b2τ2+ 2abτ = a2+ 2abτ+b[(B−a)τ+A] = (a2+bA) +b(a+B)τ ⇐⇒

(a1 = a2+bAn A1= b(a+B)n . b1+B1τ = (a+bτ)(A+Bτ)

n = τ(a+bτ)n 2 ⇐⇒nb1+nB1τ =τ[(a2+bA)+b(a+B)τ] = (a2+bA)τ + (a+B)((B−a)τ+A) = (a+B)A+ (B2+bA)τ ⇐⇒

(b1= (a+B)An B1 = B2+bAn . We have that:

M2 =

a A b B

·

a A b B

=

a2+bA A(a+B) b(a+B) B2+bA

=

na1 nb1

nA1 nB1

.

It is easy to see that the relation (8) is equivalent to M2 ≡O2(modn):

“=⇒” We proved above that M2=n·M1t≡O2(modn).

“⇐= ” Since M2 ≡ O2(modn) we obtain that there exist a1, b1, A1, B1 ∈ Z such that

M2 =

na1 nb1 nA1 nB1

=n·M1t, where we denote by M1 =

a1 A1 b1 B1

.

Since M2 =n·M1t and det(M) = n we obtain that n2 =n2·det(M1), hence, det(M1) = 1, i.e. M1 ∈ SL2(Z). From the previous proof we obtain that the relation (8) is satisfied,q.e.d.

Moreover, from Hamilton-Cayley’s theorem we know that M2−(T r(M))·M+ det(M)I2 = 02.

Since det(M) = n,M ∈ SL2(Z) and gcd(a, A, b, B) = 1 it follows easily that

M2 ≡O2(modn)⇐⇒n|Tr(M) which completes the proof.

Proposition 2.4. Let E be a complex elliptic curve satisfying the condi- tion of Theorem 2.1 i), τ2=uτ+v, u, v ∈Q,∆ =u2+ 4v <0 (i.e. E admits complex multiplication).

∃C≤(E,+)cyclic subgroup of order nof E such that (E, C)∼(EC,E[n]C )

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only when n ∈ {1,2,3,−u4d22v222,−u22dv222} and moreover, this is happening only when the following conditions are satisfied:

a) The casen= −u4d22v222 occurs ⇐⇒2d|u1v2. This case is realized for b0 =±1 anda=−u2d1v2b0. b) The casen= −u22dv222 occurs ⇐⇒2d6 |u1v2. This case is realized for b0 =±2 anda=−u2d1v2b0. c) The casen= 2 occurs ⇐⇒ −u4d22v222 = 1 and 2d|u1v2. This case is realized for b0 =±1 anda=±1− u2d1v2b0. d) The casen= 3 occurs ⇐⇒ −ud22v222 = 3 and 2d6 |u1v2. This case is realized for b0 =±1 anda=±32u2d1v2b0.

Remark 2.5.For each complex elliptic curve satisfying the condition of Theorem 2.1 i) there exists a unique n ≥ 2 for which ∃C ≤ (E,+) cyclic subgroup of ordernsuch that (E, C)∼(EC,E[n]C ). In other words, if for a given n there exist cyclic subgroups of order n of E such that (E, C) ∼ (EC,E[n]C ) then for a different integermthere are no cyclic subgroups of the sameE such that (E, C)∼(EC,E[m]C ).

Proof.From the proof of Theorem 2.1 we have that (a, A, b, B) =

a,u2dv1b0,

u2v2

d b0, a+ u1dv2b0

and n=

a+ u2d1v2b02

u22v22

4d2 b02.

From Theorem 2.3 we have that there exist C ≤(E,+) cyclic subgroup of order n of E such that (E, C) ∼ (EC,E[n]C ) if and only if n = detM and n|Tr(M) =a+B. The relation n|Tr(M) =a+B is equivalent to

(10) n=

a+u1v2

2d b0 2

−u22v22

4d2 b02 | 2

a+u1v2

2d b0

.

By denoting x=a+u12dv2b0, (10) becomes

(11) x2− u22v22

4d2 b02 | 2x.

If x >2 then x2 >2x and consequently the relation (11) is impossible.

Ifx <−2 thenx2 >−2xand consequently the relation (11) is impossible.

We obtain that x∈[−2,2],x= m2, m∈Z.

If x= 2, by using the fact that ∆<0, (11) becomes 4−u22v22

4d2 b02 | 4, i.e. b0 = 0. Consequently x = a = 2, n = 4 and (a, b0) = (2,0). But gcd(a, b0) = 1, contradiction.

If x = −2 we obtain similarly that b0 = 0 and x = a = −2, hence, gcd(a, b0) = 2, contradiction.

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If x= 0 we have thata=−u2d1v2b0. We distinguish two cases:

I) If 2d|u1v2 it follows that b0|a. Since gcd(a, b0) = 1 it follows that b0 =±1.

Since b0 =±1 we have that n=x2u224dv222b02 =−u224dv222. This case is realized for b0 =±1 anda=−u2d1v2b0.

II) If 2d-u1v2, by using the fact that a=−u2d1v2b0 ∈Z, we have 2|b0 =⇒

∃b00∈Zsuch thatb0= 2b00. Consequentlya=−u1dv2b00and since gcd(a, b00) = 1 we get that b00=±1 hence, b0 =±2 .

Fromb0 =±2 we obtain thatn=x2u224dv222b02 =−u22dv222. This case is realized for b0 =±2 anda=−u2d1v2b0.

If x=±1 then n|2x=a+B =±2, consequentlyn∈ {1,2}.

If n= 1 thenC is trivial subgroup.

Ifn= 2 thenn=x2u224dv222b02= 2⇐⇒ −u224dv222b02 = 1⇐⇒u21v22+4v4d21v2u22b02

=−1.(*)

If 2|b0 then we have the equation

b0 2

2

·h u21

v2

d

2

+v1v2

u2

d

2i

= −1, hence, b20 =±1,i.e. b0 =±2.

The equation (*) becomes u21v22 + 4v1v2u22 = −d2 and, by using the fact thatu2 =du02 and v2=d·(v20) we obtain that u21(v02)2+ 4v1v2(u02)2=−1. But a perfect square is congruent to either 0(mod4) or 1(mod4), hence, from the previous equation−1≡0(mod4) or−1≡1(mod4) respectively, contradiction.

It remains that 2 6 |b0. Since a= ±1−u2d1v2b0 ∈ Z we get 2d|u1v2. The equation (∗) becomes n = b02·h

u1v2

2d

2

+ +v1v2

u2

d

2i

= −1, consequently b02= 1. In conclusion n= 2 implies 2d|u1v2 and u224dv222 =−1.

This case is realized for b0 =±1 anda=±1−u2d1v2b0.

If x=±32 we have thatn|2x=a+B=±3, hence, n∈ {1,3}.

If n= 1 thenC is trivial subgroup.

If n= 3 we have that n =x2u224dv222b02 = 94u224dv222b02|2x = ±3. The divisibility occurs if and only if u224dv222b02 = −34 ⇐⇒ b02· u21v22+4vd21v2u22 = −3.

It follows that either b02 = 1 or b02 = 3. The case b02 = 3 is impossible.

Consequently, it remains that b02 = 1 and u224dv222 =−3.

Sincex=±32 we obtain thata=±32u2d1v2b0∈Z, consequently 2d-u1v2. This case is realized for b0=±1 and a=±32u2d1v2b0.

If x=±12 thenn|2x=a+B=±1, consequently C is trivial subgroup.

Vice-versa, if 2d|u1v2 or 2d-u1v2 one can easily check that there exists nwith the properties specified in the statement of the Proposition.

The fact that such ann >1 is unique is also easy to check:

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If 2d|u1v2 and −u4d22v222 >1, there existsC cyclic of ordern= −u4d22v222. If 2d|u1v2 and −u4d22v222 = 1, there existsC cyclic of order 2.

If 2d-u1v2andn= −u22dv222 >1, there existsCcyclic of ordern= −u22dv222. If 2d-u1v2 and n= −ud22v222 = 1 we have that u21v22+4vd21v2u22 = −1. Since d|u2 and d|v2 it follows that there exist u02, v20 ∈ Z such that u2 = du02 and v2 = dv20. The equality u21v22+4vd21v2u22 = −1 means u21(v20)2 + 4v1v2(u02)2 =

−1, i.e. (u1(v02))2 ≡ −1(mod4), contradiction. Consequently, this case is not possible.

3. EXAMPLES. THE CASES 2n7 ANDn= 11

In this section, we classify (up to an isomorphism) the elliptic curves E which admit a subgroupC ≤(E,+) of order 2≤n≤7 and n= 11 such that

E

C ' E. We recall that complex elliptic curves are of the form CL for some L = Z+Zτ ⊂C where τ ∈ G =n

z = x+iy ∈C :−12 ≤ x < 12 and either

|z| ≥1 if x≤0 or |z|>1 if x >0o .

Let E be an elliptic curve satisfying the condition of Theorem 2.1,i).

E is isomorphic to an elliptic curveE0 = CL, whereL=Z+Zτ andτ ∈G.

Since an isomorphismu:E −→E0 is of the typeu(z) =A·z, A∈SL2(Z) one easily obtains that E0 satisfies the condition of Theorem 2.1,i). Hence, we can assume (up to an isomorphism) thatE is of the form CL withL=Z+Zτ ⊂C and τ ∈ G. Moreover, we observe that if τ = x+iy ∈ G, from |x| ≤ 12 and

|z|=x2+y2 ≥1 it follows that y234.

Let τ2 −uτ −v = 0, u, v ∈ Q,∆ = u2 + 4v < 0 and τ ∈ G. Then τ = i

|∆|

2 and, since τ ∈G, we get that−1≤u <1 and |∆| ≥3.

Since ∆ =u2+ 4v <0 we have that v <0. Without loss of generality, we can assume that v2 >0, v1 <0 and u2 >0. By using Theorem 2.1 and these restrictions imposed toτ we obtain the following results:

Proposition 3.1. There are exactly 3 elliptic curves E (up to an iso- morphism) which admit at least one subgroup C≤(E,+) of order 2 such that

E C 'E.

If we denote by L=Z+Zτ, they are:

a) E= CL, τ2 =−1;

b) E= CL, τ2 =−2;

c) E= CL, τ2 =−τ −2.

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