Advances in Pure Mathematics, 2013, 3, 304-308
http://dx.doi.org/10.4236/apm.2013.32043 Published Online March 2013 (http://www.scirp.org/journal/apm)
On the Torsion Subgroups of Certain Elliptic Curves
over * 
E
Yoon Kyung Park
School of Mathematics, Korea Institute for Advanced Study, Seoul, Republic of Korea
Received October 2, 2012; revised December 3, 2012; accepted December 15, 2012
ABSTRACT
Let be an elliptic curve over a given number field
K
. By Mordell’s Theorem, the torsion subgroup of defined
over is a finite group. Using Lutz-Nagell Theorem, we explicitly calculate the torsion subgroup for cer-
tain elliptic curves depending on their coefficients.
E


tors
E
K
223
246
,
Keywords: Elliptic Curve; Rational Point
1. Introduction
A
is fourth-power free and , then
A cubic curve over the field in Weierstrass form is
given by projectively
223
13
y
w axyw aywxaxw a xwa w
with coefficients in
K
. Then there is a unique
K
rational point on the line at infinite
. If the above is an elliptic curve, then is
a nonsingular point and we deal with the curve by
working with the affine form

0,1,0w

0,1,0
2
2 46
,,xy
0w
23
13 .
y
axy ayxaxax a (1)
Hereafter assume that
K
is a number field. Since the
field characteristic of
K
is , we can study 0
23
y
xAxB 
32
427
(2)
instead of (1.1). When the discriminant
E
A
BE

EK

tors
EK

E
 is nonzero, is a nonsingular curve.
By Mordell’s theorem, is a finitely generated
abelian group and its torsion subgroup is a
finite abelian group. Mazur proved that of an
elliptic curve over the rational numbers must be iso-
morphic to one of the following 15 types [1]:
E
,1
22
NN
 
10,12
,14.NN


This paper is focused on knowing how the coefficients
A
and of (1.2) determine tors . For the earlier
work, we see the cases
B

E
A
or is zero in [2]: B
Theorem 1. Let be the elliptic curve E
23
y
1) If 0B

22,ifisasquarein,
4,if 4,
2, otherwise.
tors
A
EA







 



B0A 2) If is sixth-power free and , then


43
6,if 1,
3,if4322 3 ,orifissquarenot1,
2,ifiscubicnot 1,
0, otherwise.
tors
E
B
BB
B


 










xAxB with
A
and in . B
It is too hard to determine the group

tors with-
out any relation between the coefficients. Hence we
consider the elliptic curve as follows:
E


23
yxfkxgk (3)






f
with ,kgk k. Then Theorem 1 yields the
case when one of




f
k and
g
k is zero and




 
max ,1 deg deg
kk
fkgk. In this paper, we deal


 
max ,2 deg deg
kk
fkgk. with the case
Theorem 2. Let



23 2
:63362Ey xkxkk

 
kk

359 4kk
(4)
be the elliptic curve with in . Suppose that is
an integer such that

and there is no
integer satisfying h

2
433 1khhh or





2
41331hhh

. Then
*This work was supported by NRF 2012-0006901.
C
opyright © 2013 SciRes. APM

Y. K. PARK 305











2
22
2
22
4,20or34mod 35,suchthat31
satisfying32and665
2,20or34mod 35,suchthat31
satisfying32and 665
2,is congruentto one
tors
kl
mmlll
kl
Emmll
k

 

 
 

 




2
2
of theelementsof theset
andsuch that31,
0, otherwise.
K
lkll










 





and
2issquare,
and
2issquare,
kll
lm
kll
llm



modulo 35
where

5, 27, 29,32Kx x 

235:4, 7,12,15, 22, 2.
2. The Proof of Theorem 2
We use the Lutz-Nagell Theorem and we have to cal-
culate
p
p
EE
E

if has a good reduction at the prime
.
pTheorem 3. (Lutz-Nagell) Let be an elliptic cur-
ve (1.1) with coefficients in and
p
E
Ep
be a obtained
curve by reducing coefficients of modulo . And let
E
E
 
,,1PyP


be the discriminant of .
1) If 1 a and if is in
, then
0Px
tors
E

x
P and


y
P are integers;
2) For any 1, if is in a

,PxPy
 
,1P


tors
E,
then

4
x
P and

8
y
P are integers;
3) If is an odd prime such that
p
E
p

, then the
restriction to of the reduction homomorphism

Etors


:
p
p is one-to-one. The same conclu-
sion is valid for if
p
E
2p
rE
2
E


and ;
1
4) If , so that is given by
0a
2
0E
23 ,
13
aaa
y
xAxB 

,1

E
P

0yP
and if is in tors , then either
( and has order 2) or else
 
,PxP yP

0yP

and
divides .

2
yP 32 27B
x Ax 
4dA 
23
:Ey
Proof. See [2]. □
Lemma 4. Let be the elliptic
curve over B
p
 and be a point in

,Pxy


p
E

,Pxy

which is not a point at infinity. Then the followings are
equivalent.
1) is a point of order 3 in
p
E
42
36 12
;
2) 2
x
Ax Bx

,
A is congruent to 0 modulo
.
pProof. 1)  2) Let 22
x
y
PPP
be the point
. Then by the group law algorithm ([2]), 2


42
28xBx
2
22
2
2
2
2
3
2
4
3
32
42
22
xA A
xy
xA
xA x
y
x
Ax B
yyy





 


 



and


,.Pxy


3PO

means that Then
42 22
28 4
x
AxBx Axy


(5)

2
2
232
2
3
3222.
4
xA
x
AxxAxBy
y



 



23
(6)
x
should satisfy that
y
xAxBSince ,

42 2
36 120xAxBxA in

p

42 2 36 120xAx BxA
y23
.
2) 1) Assume that ,
is not zero and
y
xAxB in
p


,
. By simple
calculation, such
x
y P satisfy (5) and (6) and if is
the point


y2PP then
x

5,7
. We are done. □

,
Here we choose two rational primes and calcu-
late the groups


E

Ek
p E
5 and 7. For the integer
unmentioned in our main theorem, we can take another
prime and apply it as same manner.
Lemma 5. Let be the rational prime and be
the elliptic curve defined as



23 2
633 62yxkxkk

 
k

where is a nonzero integer. And using the natural
surjection from to pp

 , we can get
p
E
Ep p

433
2394kk 
by
reducing the coefficients of modulo . If does
not divide the discriminant then
the group
p
E consisting of the points defined over the
finite field
p
p

with elements is


1)

55
9, 1mod5,
6, 2mod5,
3, 3mod5.
k
Ek
k
 












2)


77
33,3mod7,
6, 1,4mod7,
9, 2mod7,
12,6mod 7.
k
k
Ek
k
 






 




Y. K. PARK 306


Table 1. Point in 55
E
k

mod 5

.



55
EO

55 E

55
E generators in
1 




,2, 2,3,20,0, 2,1, 1 9

 
0,0, 2,1, 1,2, 2
2



2 ,1,0 ,3,1

3, 1

2, 2
0, 0, 6
3 


3 2, 2


p
p has a subgroup of


E. E Proof. By [3], every
3
7p

94pkk
. Table 1 is the proof of (1).
Both cases can be calculated as using simple calcu-
lation. For 2), since and

, can
not be congruent to and
k
0


d7
23
5mo . When
, 7 becomes

od7

E1mk23
y
xx. By sub-
stituting all elements of to
7

x
in 7, we can find
that
E




,

3,2, 0,6,2



77 . Since it is an
abelian group with 6 elements,
1,E
77 6

E

. Like
this, if ,
77 has 6 elements.
Hence it is isomorphic to

7
 
5,0,6, 1,
4mk
4,
od

1,

E
6.
In the case has a
torsion subgroup

d7

2mok23 :2x x
7
Ey

  
,6,3 ,0,3 ,

over 7. To find the point of order 3 in the elliptic curve
as the form ((2) in Section 1), we have to get the root of
the equation in given field
and it is the
1,3 ,2,1
42
36 12xAxBx


2
0A
x
-coordinate of the order 3 point by
Lemma 4. In this case, the equation is
in 7. Hence there is no
point of order 3 except and

31xx

32
662xx 

6, 3

77
For , has 9 elements. But the
equation giving criterion of order 3 is
in and
. Therefore,
9

4x7 
 
77
1 ,6,1E
E.

od7

77
E3mk

2x

3,1,5,
31xx

0,3,

77
E33 

6mod7k
 
, 6,2, 



.
Last, if ,

77
E

0,1, 2,1,3,3, 4,0 ,5,1 

4,0 has only one point

of order 2. It means that
77
E


23 2
:63362Ey xkxkk

2
31ll

2
31lk
22 46
9kl ll2 31xl
20
x 

2
31,0l
12 .
To get 1), we use the same process as 2), I omit it. □
Propositions 6 and 7 give the necessary and sufficient
condition to have order 2 and 3 points.
Proposition 6. Let be the elliptic
curve with in . There is a point of order 2 if and
only if is an integer of the form . More-
over, the point of order 2 is unique.
k
k
Proof. Assume that is an integer of the form
. Through easy calculation, satisfies
. Then is a root of
and is
the point of order 2 in
k

l
69 k

3
63
0
6k

2
3kxk
Conversely, suppose that the equation of




32
633 620xkxkk

 
k
has a solution in .
To have solution of the equation with respect to ,
x
should be congruent to 2 modulo 3. By substituting
x
, the equation becomes 31m

to




22
369 1kkmmm 
2
ml . Since it has an integral
solution,

2
31kll l for an integer .

and

Now we show that there is no point of order 2 except


2
31,0l in


E. Assume that



2
31,0lE.
Then


2
31kll.





 

32
222432
633 62
311391812 2.
xkxkk
xlxlx lll
 
 

 
22432
139 18 122xlxlll


Qx be Let



3
93 11ll with discriminant . If the solution of







Qx exists, then 311 0ll


0l. It gives us the value

or 1. Hence and is singular. □ 0kE
Proposition 7. Let




23 2
:63362Ey xkxkk

 
k h
be the elliptic
curve with in . Assume that there is no integer
such that


2
433 1khhh or





2
41331hhh


. Then

E has no point of
order 3.
Proof. As we mentioned in the proof of the previous
lemma, the point


,Pxy is of order 3 if and only if
x
is the root of


 
 
32 2
31125 12123.
E
TX
XXX kXkk  


E
SX be the polynomial Let






32 2
31
12512123 .
E
TX X
XXk Xkk

  
Since

2
1, 3k

E

is not in , it suffices to
check whether
x
is a root of or not.

0SX
E
Suppose that


0SX
E
x

has a root 
k
in . Then
it is an integer. In other words, for an integer not the
form


2
433 1hh h




2
41331hhh 
k or by sorting again as , we can fine an integer  such that
x




32 2
232
12 51212 3
1212153 0.
xxk xkk
kxkxxx
 
 


 

Y. K. PARK 307
From, this must be a multiple of 12
and
32
53xx x


x
is one of 12 or 11 for a suitable inte-
ger .
3,5,9m
12 3m
m
When , x
E
S

32
96 16mm

2
1 13mm

2
21hh
23hh
12xm
k


22
2 5m
becomes
. Because it
has integral solutions as a quadratic equation with respect
to , its discriminant is a square.
That means that for an integer .
Through this we get or
.

2
4 144mkm
16 4
41m
43kh


31h
5,12 9m 
12
k
41
12kk
2
3hh
12xm

1
11
If or then discri-
minant of the quadratic equations with respect to is
or

3125

31211m


1, 4mm
 
2
1m

2 23m
4

6
respectively. Neither case has a
perfect square discriminant and admit any integral root. □
Proof of Theorem 2. Use the Lemma 5 and Theorem 3
3), we can determine which finite abelian group has a
subgroup of


E for the case


1mod35
k

E
k

1mod7
, i.e.,
and. In fact, tors is a
subgroup of both

od51mk
9 and 6. It yields that it is
3 or trivial. Since our group has no point of order 3,
it is trivial.
Note that tors is a subgroup of order , if it is
a subgroup of order with

, then. So it
is resolved as trivial group in many cases.

EN

3, 1N
k
3rN
To observe easily, we can refer Table 2: In this table,
takes the value modulo 5 at the horizontal line and
modulo 7 at the vertical line respectively. The groups
n in the brackets at top line and at the very left
line are result from Lemma 5 .
Cn
Each entry implies that the type of group: “A”, “B” or
“C” implies one of subgroups of 4, 2 or trivial,
respectively. The same alphabet does not mean the same
group. And “D” means that both curves


55
and
77
are singular. In this table since “C” is trivial, it
remains that a few cases
or 34 .
E

E

mod 35

4,7,12,15, 20,k22, 25, 27, 29,32
For the cases that the subgroup is nontrivial Pro-
Table 2. Type of gr oup

tors
E

k
.
mod 7k

mod 5

0 1
position 6 makes us know which curve has the point of
order 2 or not. Hence, it is sufficient to check the value
having order 4 points. k

Assume that

20,34mod 35k
l

31kll
and there exists an
integer such that . In fact



20mod 35k

34mod 35
5l (respectively, ) if and only if


26mod 3519 (respectively, or

or



33mod 35).

2
31,0l is the unique point of order 2.
Using duplication formula for the elliptic curve, let


,Pxy




2
231,0Pl be the point satisfying . By
Substituting


,, 63xyk




2
362kk
,,
and for
x
yA B

and in (in the formulas for x2 and y2 in the
proof of Lemma 4), we get two equations affirming the
existence of point of order 4:





2
22432
22432
21 31818610
21 31818610
xlxlll
xlxlllFx

  
 


9
C 2

6
C3


3
C4
0 D C B C D
1

6
C

9
C

B C B C B
2 C C C C C
3

33
CC

6
C

C C C C C
4 B C B C B
5 D C B C D
6

12
C A C B C A


where

 

4234322
65432
87642
21 36918122
25416210854637
324 972 864270605.


F
xxlxl l lx
lllllx
lllll
 


To have an integral solution of


22432
21 31818610xlxlll


, its discrimi-
nant


3
36 3 2ll

have to be a square. Suppose that we
can find an integer such that and m

232mll
2
316
x
llm
 2
316llm
ml
( or ). It is easy to check
that the integer satisfying the above condition exists
in each case determined by . Furthermore, by sub-
stituting



,
x
31kll



232mll and to the
right hand side of (1.4) we get a numerical formula







32
22
22 2
5432 652
963266 52
96652
llllm
lllllm
lml lm


 

0l Since

makes the curve (1.4) singular,


2
66 52llm


23
:75506Ey xx
is a square of a suitable integer if and
only if there exists a point of order 4.
So we are done. □
3. Conclusions
By the help of Theorem 2, we explicitly calculate the
torsion part of Modell-Weil group.
Example 8. Let be the elli-
ptic curve. Then


2.
tors
E
12k Given elliptic curve is the form in Theorem 2
and


2
123 212

. Therefore

2E
tors .
And


11,0 is the nontrivial torsion point on


E.


tors
E is able to be applied to The method to find

Y. K. PARK
t © 2013 SciRes. APM
308
k
7p
Copyrigh
all elliptic curve without a condition for by choosing
another prime .
For example, in Theorem 2, there is a condition


359 4kk

for . This is one for nonsingular curve.
For the case that
k


35 9kk4

, choose the another
prime such that 7p


94kpk

. Calculate

p
p and eliminate the order 3 point and check the
condition for having order 2 point. Since
E

21p
p, the smaller gives simpler nece-
ssary condition. For example, if then the ellip-
tic curve is
Ep
16k
:93674x

23
x
7
Ey
with discriminant . Find
6
25
p
p
E

with
and, 11 17p

11 11 and
15E

17 1718E

.
Using Lemma 4, we observe that


E
 
23
:Eyxfkx gk 
k
has no point of
order 3. So it is a trivial group.

 
and

for


maxdeg,deg2fkgk . We can
use the criterion for the quadratic equation to find a point
of order 2 or 3. Of course, it is indispensable to consider
some exceptional cases in the similar way to Proposition
7.
REFERENCES
[1] B. Mazur, “Modular Curves and the Eisenstein Ideal,”
Publications Mathématiques de l’Institut des Hautes Étu-
des Scientifiques, No. 47, 1977, pp. 33-168.
[2] A. Knapp, “Elliptic Curves,” Princeton University Press,
Princeton, 1992.
[3] D. Kim, J. K. Koo and Y. K. Park, “On the Elliptic
Curves Modulo p,” Journal of Number Theory, Vol. 128,
No. 4, 2008, pp. 945-953. doi:10.1016/j.jnt.2007.04.015
Remark 9. Generalize our elliptic curve