AND RELATED ALGORITHMS

BOGDAN CANEPA and RADU GABA

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)∼(^{E}_{C},^{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 Y_{0}(n) as the quotient space Γ_{0}(n)/H that is the set
of orbits {Γ_{0}(n)τ : τ ∈ H}, where Γ_{0}(n) is the “Nebentypus” congruence
subgroup of levelnof SL2(Z), acting on Hfrom the left:

Γ_{0}(n) ={

a b c d

∈SL2(Z)|c≡0(modn)}.

MATH. REPORTS16(66),4(2014), 477–502

Following the notations of [5], an enhanced elliptic curve for Γ_{0}(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 (E^{0}, C^{0}) are said to be
equivalent if some isomorphism E∼=E^{0} takes C to C^{0}. One denotes the set of
equivalence classes with

S0(n) :={enhanced elliptic curves for Γ_{0}(n)}/∼.

Furthermore, an element of S_{0}(n) is an equivalence class [E, C]. S_{0}(n) is
a moduli space of isomorphism classes of complex elliptic curves and n-torsion
data.

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

S_{0}(n)∼=Y_{0}(n) given by [C/Λ_{τ},h1/n+ Λ_{τ}i]7→Γ_{0}(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 Φ_{n}(X, Y)∈ Z[X, Y]. Following [12], Theorem 5, Section
5, Φn(j, j^{0}) = 0 if and only if there exists an elliptic curve E of j-invariant j,
n-isogeneous to an elliptic curveE^{0} ofj-invariantj^{0}. 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 j^{0}ssuch 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) =
Y_{0}(n)∪ {cusps}. We recall that the Heegner points ofY_{0}(n) as defined by Birch
in [2] are pairs (E, E^{0}) 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)∼(^{E}_{C},^{E[n]}_{C} ), studying
in this manner the fixed points of the action of the Fricke involution

w_{n}:=

0 −1

n 0

∈GL_{2}(Q^{+})

on the open modular curves Y_{0}(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 w_{n} on Y_{0}(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 w_{n} on the open modular
curves Y_{0}(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 ofw_{n}on the moduli spaceS_{0}(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

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 ^{E}_{C} 'E⇔ ∃u, v∈Q such
thatτ^{2} =uτ+vwith ∆ =u^{2}+ 4v <0 (i.e. E admits complex multiplication);

ii) If τ satisfies the conditions of i) and u = ^{u}_{u}^{1}

2, v = ^{v}_{v}^{1}

2, u2 6= 0, v_{2} 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 ^{E}_{C} ' E ⇐⇒

∃(a, b^{0})∈Z^{2} with (a, b^{0}) = 1 such that n= detM, where M is the matrix
M =

a A b B

and (a, A, b, B) =
a,^{u}^{2}_{d}^{v}^{1}

2 b^{0},^{u}^{2}_{d}^{v}^{2}

2 b^{0}, a+^{u}_{d}^{1}^{v}^{2}

2 b^{0}

;

iii) The subgroup C from ii) is C=h^{u}^{11}^{+u}_{n}^{21}^{τ}i, whereu_{11}, u_{21} 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

,

hence,

∃U, V ∈GL_{2}(Z) such that M =U ·

1 0 0 n

·V.

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

u11 u12

u_{21} u_{22}

.

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

(1) g= α_{1}+α_{2}τ

n , α_{1}, α_{2}∈Z.

Since ord(¯g) =nit follows easily that gcd(α_{1}, α_{2}, n) = 1.

We have that ^{E}_{C} ' _{L+}^{C}^{L}_{Z}_{g}

L

' _{L+}^{C}

Zg = _{L}^{C}0, where we denoted byL^{0} =L+Zg=
Z+Zτ +Zg.

It is known that ^{C}_{L} ' _{L}^{C}0 ⇔ ∃λ ∈ C such that λL = L^{0}. Consequently
E ' ^{E}_{C} ⇔ ^{C}_{L} ' _{L}^{C}0 ⇔ ∃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+bτ}_{n}
λτ = ^{A+Bτ}_{n} .

“⊇” 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 =α^{a+bτ}_{n} +β^{A+Bτ}_{n}
τ =u^{a+bτ}_{n} +v^{A+Bτ}_{n}

α1+α2τ

n =s^{a+bτ}_{n} +t^{A+Bτ}_{n}

⇐⇒

a A b B

!

· α u β v

!

=nI2

a A b B

!

· s t

!

= α1

α_{2}

!

.

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+c1α1+α2τ n A+Bτ

n =a2+b2τ +c2α1+α2τ n

⇐⇒

(a=na1+c1α1

b=nb_{1}+c_{1}α_{2} and

(A=na2+c2α1

B =nb_{2}+c_{2}α_{2} ⇔
(7)

(aˆ= ˆc1αˆ1

ˆb= ˆc_{1}αˆ_{2} and

(Aˆ= ˆc2αˆ1

Bˆ= ˆc_{2}αˆ_{2} in Z
nZ.

We obtain that the inclusion “⊆” is equivalent to the existence ofc_{1}, c_{2}∈
Z which both satisfy (6),i.e. (2) is equivalent to (5) and (6).

We have obtained that the hypothesis E ' ^{E}_{C} 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

=

cˆ_{1}αˆ_{1} cˆ_{2}αˆ_{1}
ˆ

c_{1}αˆ_{2} cˆ_{2}αˆ_{2}

in Z nZ =⇒ det ˆM = ˆ0 in Z

nZ =⇒ det M...n.

By using now (5) we obtain det(M)|n^{2} and so det(M) =nk, k|n. From
(5) we get α = _{detM}^{nB} = ^{B}_{k} hence,k|B. Similarly one obtains k|b, k|A, k|a and
hence, from the second equality of (5) we obtain that k|α_{1}, α_{2}. Since k|n and
(α1, α2, n) = 1 we obtain k=±1, consequently det(M) =±n.

We will prove that the relations (5), gcd(α_{1}, α_{2}, n) = 1 and det(M) =±n
imply (6):

In did, from (α_{1}, α_{2}, n) = 1 it follows that ∃µ_{1}, µ_{2}, µ_{3} ∈ Z such that
µ_{1}α_{1}+µ_{2}α_{2}+µ_{3}n = 1. Choose ˆc_{1} =µ_{1}\a+µ_{2}b. We then have ˆc_{1}αˆ_{1} =
(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 = ^{aα}_{detM}^{2}^{−bα}^{1} = ^{aα}^{2}_{±n}^{−bα}^{1} =⇒ aα_{2}−bα_{1}...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\_{1}α_{1}−µ_{3}n) = ˆb+ ˆµ_{1}(aα\_{2}−bα_{1}) = ˆb,q.e.d.

Similarly, by choosing ˆc2 =µ1A\+µ2B one obtains the expressions of (6) regarding A, B.

We have obtained that (6) follows from the relations gcd(α_{1}, α_{2}, 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)·I_{2} = (±n)I_{2}, hence,
the existence of the elementsα, β, u, vfollows from the existence of the adjoint
matrixM^{∗}.

In conclusion, the hypothesis E ' ^{E}_{C} with C cyclic subgroup of order
n is equivalent to the following four relations: the second equality from (5),
(α_{1}, α_{2}, 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|α_{1}, d|α_{2} and since d|detM = ±n, it follows that d|gcd(α_{1}, 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 ∈SL_{2}(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

=
s^{0}

t^{0}

and obtain the equivalent equation
α_{1}

α2

=U·
s^{0}

nt^{0}

, 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: chooses^{0}, t^{0} ∈Zsuch
that (s^{0}, n) = 1. Consider the matrix (α_{1} α_{2}) = (s^{0} nt^{0})·U^{t}, where U is an
invertible matrix from the arithmetically equivalent decomposition of M. We
have that gcd(α_{1}, α_{2}, n) = gcd((α_{1}, α_{2}), n) = gcd((s^{0}, nt^{0}), n) = gcd(s^{0}, 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

bτ^{2} = (B−a)τ+A (∗)

Let µ_{τ} = X^{2} −uX −v ∈ Q[X] be the minimal polynomial of τ, with
u= ^{u}_{u}^{1}

2 ∈Q, v = ^{v}_{v}^{1}

2 ∈Q, u_{1}, u_{2}, v_{1}, v_{2}∈Z,gcd(u_{1}, u_{2}) = gcd(v_{1}, v_{2}) = 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= ^{v}_{v}^{1}

B−a 2

b =u= ^{u}_{u}^{1}

2

hence,

(v_{2}A=v_{1}b
u_{2}(B−a) =u_{1}b.

Since

(v_{2}A=v_{1}b

gcd(v_{1}, v_{2}) = 1 we get that
(v_{2}|b

v_{1}|A.

From

(u_{2}(B−a) =u_{1}b

gcd(u1, u2) = 1 it follows that

u_{2}|b
u_{1}|B−a

we already have v2|b
hence, lcm[u_{2}, v_{2}]|b.

Consequently ∃b^{0} ∈Z such that

b= [u2, v2]b^{0} = ^{u}_{d}^{2}^{v}^{2}

2 b^{0}
v_{2}A=v_{1}b

u2(B−a) =u1b

hence,

b= ^{u}_{d}^{2}^{v}^{2}

2 b^{0}
A= ^{u}_{d}^{2}^{v}^{1}

2 b^{0}
B=a+^{u}_{d}^{1}^{v}^{2}

2 b^{0},

where d_{2} = gcd(u_{2}, v_{2}) andτ^{2} =uτ+v.

We prove now the equivalence:

gcd(a, A, b, B) = 1⇐⇒gcd(a, b^{0}) = 1.

“=⇒” Denote by d^{0} = gcd(a, b^{0}). It follows thatd^{0}|gcd(a, A, b, B) hence,
d^{0}|1, i.e. d^{0} = 1.

“⇐= ” Let d= gcd(a, A, b, B). Then d|a and since gcd(a, b^{0}) = 1 we get
gcd(d, b^{0}) = 1. Moreover d_{2} = (u_{2}, v_{2}) so let u_{2} = d_{2}u^{0}_{2} and v_{2} = d_{2}v^{0}_{2} with
gcd(u^{0}_{2}, v^{0}_{2}) = 1. Since d|aand d|B we have that d|B−aand obtain:

d|b= ^{u}_{d}^{2}^{v}^{2}

2 b^{0}
d|A= ^{u}_{d}^{2}^{v}^{1}

2 b^{0}=u^{0}_{2}v1b^{0}
d|B−a= ^{u}_{d}^{1}^{v}^{2}

2 b^{0} =v_{2}^{0}u_{1}b^{0}

hence,

d|^{u}_{d}^{2}^{v}^{2}

2 =u^{0}_{2}v_{2} =u_{2}v^{0}_{2}
d|u^{0}_{2}v1

d|v^{0}_{2}u_{1}.

If d|u^{0}_{2}, since gcd(u^{0}_{2}, v_{2}^{0}) = 1 one obtains d|u_{1}. But gcd(u_{1}, u_{2}) = 1 and
consequently d= 1.

If d|v_{2}, using the fact that gcd(v_{1}, v_{2}) = 1 we get that d|u^{0}_{2}. Since d|u^{0}_{2}
and gcd(u_{1}, u_{2}) = 1 we have gcd(d, u_{1}) = 1. Consequently d|v^{0}_{2} and since we
already have d|u^{0}_{2}, we obtaind= 1.

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

In order to simplify the notations, denote by d := d_{2} and note that
(a, A, b, B) =

a,^{u}^{2}_{d}^{v}^{1}b^{0},^{u}^{2}_{d}^{v}^{2}b^{0}, a+^{u}^{1}_{d}^{v}^{2}b^{0}

. We obtain that det(M) =aB−bA=a

a+^{u}^{1}_{d}^{v}^{2}b^{0}

−

u2v2

d b^{0}
u2v1

d b^{0}

=
a^{2}+^{u}^{1}_{d}^{v}^{2}ab^{0}−^{v}^{1}^{v}_{d}^{2}_{2}^{u}^{2}^{2}b^{02}=

a+^{u}_{2d}^{1}^{v}^{2}b^{0}
2

−^{u}_{4d}^{2}^{1}^{v}_{2}^{2}^{2}b^{02}−^{v}^{1}^{v}_{d}^{2}_{2}^{u}^{2}^{2}b^{02} =

a+^{u}_{2d}^{1}^{v}^{2}b^{0}
2

−

u^{2}_{1}v^{2}_{2}+4v1v2u^{2}_{2}

4d^{2} b^{02} =

a+^{u}_{2d}^{1}^{v}^{2}b^{0}2

−

u^{2}_{2}v^{2}_{2}

hu1 u2

2

+4^{v}_{v}^{1}

2

i

4d^{2} b^{02}=

a+^{u}_{2d}^{1}^{v}^{2}b^{0}2

−^{u}^{2}^{2}_{4d}^{v}^{2}^{2}2^{∆}b^{02}.
Since ∆<0 and u2 6= 0, v_{2} 6= 0 it follows easily that det(M)>0.

In order to describe the subgroup C we deduce that:

α1

α_{2}

=U ·
s^{0}

nt^{0}

⇐⇒

α1

α_{2}

=

u11 u12

u_{21} u_{22}

·
s^{0}

nt^{0}

i.e.

(α_{1} =u_{11}s^{0}+u_{12}nt^{0}
α_{2} =u_{21}s^{0}+u_{22}nt^{0} .
We obtain that:

g= α1+α2τ

n = u11s^{0}+u12nt^{0}+ (u21s^{0}+u22nt^{0})τ

n =

u11s^{0}+u21s^{0}τ +n(u12t^{0}+u22t^{0}τ)

n = s^{0}(u11+u21τ)

n +t^{0}(u12+u22τ) =g^{0}+t^{0}(u12+u22τ),
where we denoted by g^{0}= s^{0}(u_{11}+u_{21}τ)

n .

Consequently,C= ^{L+Zg}_{L} = ^{L+Zg}_{L} ^{0} =hg^{0}i. But gcd(s^{0}, n) = gcd(α_{1}, α_{2}, n) =
1 hence, there exist a, b ∈ Z such that as^{0} +bn = 1. It follows that ag^{0} =
a^{s}^{0}^{(u}^{11}^{+u}_{n} ^{21}^{τ)} = ^{(1−bn)(u}_{n}^{11}^{+u}^{21}^{τ)} = ^{u}^{11}^{+u}_{n}^{21}^{τ} −b(u_{11}+u_{21}τ). We have obtained
that C= ^{L+Zg}_{L} ^{0} =hg^{0}i=h^{u}^{11}^{+u}_{n}^{21}^{τ}i.

Remark 2.2. Fora, b^{0} fixed integers such that gcd(a, b^{0}) = 1 the subgroup
C is uniquely determined hence, we can denote C=C_{a,b}^{0}.

Recall now that for E, E^{0} complex elliptic curves andC ≤(E,+), C^{0} ≤
(E^{0},+) cyclic subgroups of order n of E and E^{0} respectively, one has that
(E, C)∼(E^{0}, C^{0})⇐⇒ ∃u:E −→E^{0} isomorphism such thatu(C) =C^{0}.
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)∼(^{E}_{C},^{E[n]}_{C} ), whereE[n] ={x∈E :nx= 0}?”

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

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)∼(^{E}_{C},^{E[n]}_{C} )} ⇐⇒

{∃(a, b^{0})∈Z^{2} withgcd(a, b^{0}) = 1 such that det(M) =n andn|Tr(M)} ⇐⇒

{∃(a, b^{0})∈Z^{2}, with gcd(a, b^{0}) = 1 such that det(M) =nand
M^{2}≡O_{2}(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 = ^{C}_{L} with L = Z +Zτ ⊂ C lattice and C = h¯gi ≤ E = ^{C}_{L}
subgroup cyclic of ordern, consequentlyC= ^{L+}_{L}^{Z}^{g}. We have that ^{E}_{C} ' _{L+}^{C}^{L}_{Z}_{g}

L

'

L+CZg = _{L}^{C}0, where we denoted by L^{0} = L+Zg = Z+Zτ +Zg. Recall that

CL ' _{L}^{C}0 ⇔ ∃λ∈Csuch thatλL=L^{0}.

Letube defined in the following way: E= ^{C}_{L} −→ ^{E}_{C} = _{L}^{C}0 = _{L+Zg}^{C} isomor-
phism, u(¯z) =λz. Thenc u(C) =u ^{L+Zg}_{L}

=λ·^{L+Zg}_{L}0 = ^{λL+Zλg}_{L}0 . SinceλL=L^{0}
we have that u(C) = ^{L}^{0}^{+}_{L}^{Z}0^{λg} = 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

L^{0} . We obtain that:

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

C ⇐⇒L^{0}+Zλg = 1
n·L.

The relation (8) is equivalent toλL+Zλg=^{1}_{n}·L⇐⇒λ(L+Zg) =_{n}^{1}·L⇐⇒

λL^{0} = ^{1}_{n}·L⇐⇒λ^{2}L= ^{1}_{n}·L⇐⇒nλ^{2}L=L.Moreover,
(9) L=nλ^{2}L⇐⇒ ∃M_{1} ∈SL_{2}(Z) such that

nλ^{2}
nλ^{2}τ

=M_{1}·
1

τ

.

We denote byM_{1}=

a_{1} A_{1}
b1 B1

and by using the relations (3):

(λ=^{a+bτ}_{n}
λτ=^{A+Bτ}_{n}
we get:

nλ^{2}
nλ^{2}τ

=

a_{1} A_{1}
b1 B1

· 1

τ

⇐⇒

(nλ^{2}=a1+A1τ

nλ^{2}τ =b_{1}+B_{1}τ ⇐⇒

(a_{1}+A_{1}τ = ^{(a+bτ}_{n} ^{)}^{2}
b1+B1τ =λ·(A+Bτ)

⇐⇒

(a1+A1τ = ^{(a+bτ)}_{n} ^{2}
b_{1}+B_{1}τ = ^{(a+bτ}^{)(A+Bτ)}_{n} .

Note that A+Bτ =τ·(a+bτ), which implies that bτ^{2} = (B−a)τ+A.

By using this equation we obtain:

a1+A1τ = ^{(a+bτ}_{n} ^{)}^{2} ⇐⇒na1+nA1τ =a^{2}+b^{2}τ^{2}+ 2abτ =
a^{2}+ 2abτ+b[(B−a)τ+A] = (a^{2}+bA) +b(a+B)τ ⇐⇒

(a_{1} = ^{a}^{2}^{+bA}_{n}
A_{1}= ^{b(a+B)}_{n} .
b1+B1τ = (a+bτ)(A+Bτ)

n = ^{τ(a+bτ)}_{n} ^{2} ⇐⇒nb1+nB1τ =τ[(a^{2}+bA)+b(a+B)τ] =
(a^{2}+bA)τ + (a+B)((B−a)τ+A) = (a+B)A+ (B^{2}+bA)τ ⇐⇒

(b_{1}= ^{(a+B)A}_{n}
B1 = ^{B}^{2}^{+bA}_{n} .
We have that:

M^{2} =

a A b B

·

a A b B

=

a^{2}+bA A(a+B)
b(a+B) B^{2}+bA

=

na1 nb1

nA_{1} nB_{1}

.

It is easy to see that the relation (8) is equivalent to M^{2} ≡O_{2}(modn):

“=⇒” We proved above that M^{2}=n·M_{1}^{t}≡O_{2}(modn).

“⇐= ” Since M^{2} ≡ O2(modn) we obtain that there exist a1, b1, A1, B1 ∈ Z
such that

M^{2} =

na_{1} nb_{1}
nA_{1} nB_{1}

=n·M_{1}^{t}, where we denote by M_{1} =

a_{1} A_{1}
b_{1} B_{1}

.

Since M^{2} =n·M_{1}^{t} and det(M) = n we obtain that n^{2} =n^{2}·det(M_{1}),
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
M^{2}−(T r(M))·M+ det(M)I_{2} = 0_{2}.

Since det(M) = n,M ∈ SL_{2}(Z) and gcd(a, A, b, B) = 1 it follows easily
that

M^{2} ≡O_{2}(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), τ^{2}=uτ+v, u, v ∈Q,∆ =u^{2}+ 4v <0 (i.e. E admits
complex multiplication).

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

only when n ∈ {1,2,3,^{−u}_{4d}^{2}^{2}^{v}2^{2}^{2}^{∆},^{−u}^{2}^{2}_{d}^{v}2^{2}^{2}^{∆}} and moreover, this is happening only
when the following conditions are satisfied:

a) The casen= ^{−u}_{4d}^{2}^{2}^{v}2^{2}^{2}^{∆} occurs ⇐⇒2d|u_{1}v2.
This case is realized for b^{0} =±1 anda=−^{u}_{2d}^{1}^{v}^{2}b^{0}.
b) The casen= ^{−u}^{2}^{2}_{d}^{v}2^{2}^{2}^{∆} occurs ⇐⇒2d6 |u_{1}v2.
This case is realized for b^{0} =±2 anda=−^{u}_{2d}^{1}^{v}^{2}b^{0}.
c) The casen= 2 occurs ⇐⇒ ^{−u}_{4d}^{2}^{2}^{v}_{2}^{2}^{2}^{∆} = 1 and 2d|u_{1}v_{2}.
This case is realized for b^{0} =±1 anda=±1− ^{u}_{2d}^{1}^{v}^{2}b^{0}.
d) The casen= 3 occurs ⇐⇒ ^{−u}_{d}^{2}^{2}^{v}_{2}^{2}^{2}^{∆} = 3 and 2d6 |u_{1}v_{2}.
This case is realized for b^{0} =±1 anda=±^{3}_{2} −^{u}_{2d}^{1}^{v}^{2}b^{0}.

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)∼(^{E}_{C},^{E[n]}_{C} ). In other words, if for a given
n there exist cyclic subgroups of order n of E such that (E, C) ∼ (^{E}_{C},^{E[n]}_{C} )
then for a different integermthere are no cyclic subgroups of the sameE such
that (E, C)∼(^{E}_{C},^{E[m]}_{C} ).

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

a,^{u}^{2}_{d}^{v}^{1}b^{0},

u2v2

d b^{0}, a+ ^{u}^{1}_{d}^{v}^{2}b^{0}

and n=

a+ ^{u}_{2d}^{1}^{v}^{2}b^{0}2

−^{u}^{2}^{2}^{v}^{2}^{2}^{∆}

4d^{2} b^{02}.

From Theorem 2.3 we have that there exist C ≤(E,+) cyclic subgroup
of order n of E such that (E, C) ∼ (^{E}_{C},^{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 b^{0}
2

−u^{2}_{2}v^{2}_{2}∆

4d^{2} b^{02} | 2

a+u1v2

2d b^{0}

.

By denoting x=a+^{u}^{1}_{2d}^{v}^{2}b^{0}, (10) becomes

(11) x^{2}− u^{2}_{2}v_{2}^{2}∆

4d^{2} b^{02} | 2x.

If x >2 then x^{2} >2x and consequently the relation (11) is impossible.

Ifx <−2 thenx^{2} >−2xand consequently the relation (11) is impossible.

We obtain that x∈[−2,2],x= ^{m}_{2}, m∈Z.

If x= 2, by using the fact that ∆<0, (11) becomes 4−^{u}^{2}^{2}^{v}^{2}^{2}^{∆}

4d^{2} b^{02} | 4,
i.e. b^{0} = 0. Consequently x = a = 2, n = 4 and (a, b^{0}) = (2,0). But
gcd(a, b^{0}) = 1, contradiction.

If x = −2 we obtain similarly that b^{0} = 0 and x = a = −2, hence,
gcd(a, b^{0}) = 2, contradiction.

If x= 0 we have thata=−^{u}_{2d}^{1}^{v}^{2}b^{0}. We distinguish two cases:

I) If 2d|u_{1}v_{2} it follows that b^{0}|a. Since gcd(a, b^{0}) = 1 it follows that
b^{0} =±1.

Since b^{0} =±1 we have that n=x^{2}−^{u}^{2}^{2}_{4d}^{v}^{2}^{2}_{2}^{∆}b^{02} =−^{u}^{2}^{2}_{4d}^{v}^{2}^{2}_{2}^{∆}.
This case is realized for b^{0} =±1 anda=−^{u}_{2d}^{1}^{v}^{2}b^{0}.

II) If 2d-u1v2, by using the fact that a=−^{u}_{2d}^{1}^{v}^{2}b^{0} ∈Z, we have 2|b^{0} =⇒

∃b^{00}∈Zsuch thatb^{0}= 2b^{00}. Consequentlya=−^{u}^{1}_{d}^{v}^{2}b^{00}and since gcd(a, b^{00}) = 1
we get that b^{00}=±1 hence, b^{0} =±2 .

Fromb^{0} =±2 we obtain thatn=x^{2}−^{u}^{2}^{2}_{4d}^{v}^{2}^{2}2^{∆}b^{02} =−^{u}^{2}^{2}_{d}^{v}2^{2}^{2}^{∆}.
This case is realized for b^{0} =±2 anda=−^{u}_{2d}^{1}^{v}^{2}b^{0}.

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

If n= 1 thenC is trivial subgroup.

Ifn= 2 thenn=x^{2}−^{u}^{2}^{2}_{4d}^{v}^{2}^{2}2^{∆}b^{02}= 2⇐⇒ −^{u}^{2}^{2}_{4d}^{v}^{2}^{2}2^{∆}b^{02} = 1⇐⇒^{u}^{2}^{1}^{v}^{2}^{2}^{+4v}_{4d}2^{1}^{v}^{2}^{u}^{2}^{2}b^{02}

=−1.(*)

If 2|b^{0} then we have the equation

b^{0}
2

2

·h
u^{2}_{1}

v2

d

2

+v_{1}v_{2}

u2

d

2i

= −1,
hence, ^{b}_{2}^{0} =±1,i.e. b^{0} =±2.

The equation (*) becomes u^{2}_{1}v^{2}_{2} + 4v_{1}v_{2}u^{2}_{2} = −d^{2} and, by using the fact
thatu_{2} =du^{0}_{2} and v_{2}=d·(v_{2}^{0}) we obtain that u^{2}_{1}(v^{0}_{2})^{2}+ 4v_{1}v_{2}(u^{0}_{2})^{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 |b^{0}. Since a= ±1−^{u}_{2d}^{1}^{v}^{2}b^{0} ∈ Z we get 2d|u_{1}v_{2}. The
equation (∗) becomes n = b^{02}·h

u1v2

2d

2

+ +v_{1}v_{2}

u2

d

2i

= −1, consequently
b^{02}= 1. In conclusion n= 2 implies 2d|u_{1}v_{2} and ^{u}^{2}^{2}_{4d}^{v}^{2}^{2}2^{∆} =−1.

This case is realized for b^{0} =±1 anda=±1−^{u}_{2d}^{1}^{v}^{2}b^{0}.

If x=±^{3}_{2} 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 =x^{2} −^{u}^{2}^{2}_{4d}^{v}^{2}^{2}_{2}^{∆}b^{02} = ^{9}_{4} − ^{u}^{2}^{2}_{4d}^{v}^{2}^{2}_{2}^{∆}b^{02}|2x = ±3. The
divisibility occurs if and only if ^{u}^{2}^{2}_{4d}^{v}^{2}^{2}2^{∆}b^{02} = −^{3}_{4} ⇐⇒ b^{02}· ^{u}^{2}^{1}^{v}^{2}^{2}^{+4v}_{d}_{2}^{1}^{v}^{2}^{u}^{2}^{2} = −3.

It follows that either b^{02} = 1 or b^{02} = 3. The case b^{02} = 3 is impossible.

Consequently, it remains that b^{02} = 1 and ^{u}^{2}^{2}_{4d}^{v}^{2}^{2}2^{∆} =−3.

Sincex=±^{3}_{2} we obtain thata=±^{3}_{2}−^{u}_{2d}^{1}^{v}^{2}b^{0}∈Z, consequently 2d-u_{1}v_{2}.
This case is realized for b^{0}=±1 and a=±^{3}_{2} −^{u}_{2d}^{1}^{v}^{2}b^{0}.

If x=±^{1}_{2} thenn|2x=a+B=±1, consequently C is trivial subgroup.

Vice-versa, if 2d|u_{1}v2 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:

If 2d|u_{1}v_{2} and ^{−u}_{4d}^{2}^{2}^{v}2^{2}^{2}^{∆} >1, there existsC cyclic of ordern= ^{−u}_{4d}^{2}^{2}^{v}2^{2}^{2}^{∆}.
If 2d|u_{1}v_{2} and ^{−u}_{4d}^{2}^{2}^{v}2^{2}^{2}^{∆} = 1, there existsC cyclic of order 2.

If 2d-u_{1}v_{2}andn= ^{−u}^{2}^{2}_{d}^{v}2^{2}^{2}^{∆} >1, there existsCcyclic of ordern= ^{−u}^{2}^{2}_{d}^{v}2^{2}^{2}^{∆}.
If 2d-u_{1}v_{2} and n= ^{−u}_{d}^{2}^{2}^{v}2^{2}^{2}^{∆} = 1 we have that ^{u}^{2}^{1}^{v}^{2}^{2}^{+4v}_{d}2^{1}^{v}^{2}^{u}^{2}^{2} = −1. Since
d|u_{2} and d|v_{2} it follows that there exist u^{0}_{2}, v_{2}^{0} ∈ Z such that u_{2} = du^{0}_{2} and
v_{2} = dv_{2}^{0}. The equality ^{u}^{2}^{1}^{v}^{2}^{2}^{+4v}_{d}2^{1}^{v}^{2}^{u}^{2}^{2} = −1 means u^{2}_{1}(v_{2}^{0})^{2} + 4v_{1}v_{2}(u^{0}_{2})^{2} =

−1, i.e. (u_{1}(v^{0}_{2}))^{2} ≡ −1(mod4), contradiction. Consequently, this case is not
possible.

3. EXAMPLES. THE CASES 2≤n≤7 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 ^{C}_{L} for some
L = Z+Zτ ⊂C where τ ∈ G =n

z = x+iy ∈C :−^{1}_{2} ≤ x < ^{1}_{2} 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 curveE^{0} = ^{C}_{L}, whereL=Z+Zτ andτ ∈G.

Since an isomorphismu:E −→E^{0} is of the typeu(z) =A·z, A∈SL2(Z) one
easily obtains that E^{0} satisfies the condition of Theorem 2.1,i). Hence, we can
assume (up to an isomorphism) thatE is of the form ^{C}_{L} withL=Z+Zτ ⊂C
and τ ∈ G. Moreover, we observe that if τ = x+iy ∈ G, from |x| ≤ ^{1}_{2} and

|z|=x^{2}+y^{2} ≥1 it follows that y^{2} ≥ ^{3}_{4}.

Let τ^{2} −uτ −v = 0, u, v ∈ Q,∆ = u^{2} + 4v < 0 and τ ∈ G. Then
τ = ^{u±}^{i}

√|∆|

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

Since ∆ =u^{2}+ 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= ^{C}_{L}, τ^{2} =−1;

b) E= ^{C}_{L}, τ^{2} =−2;

c) E= ^{C}_{L}, τ^{2} =−τ −2.