 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 * EYoon Kyung Park School of Mathematics, Korea Institute for Advanced Study, Seoul, Republic of Korea Email: ykpark@math.kaist.ac.kr 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torsEK223246, 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 22313yw 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0,1,022 46,,xy0w2313 .yaxy ayxaxax a (1) Hereafter assume that K is a number field. Since the field characteristic of K is , we can study 023yxAxB 32427 (2) instead of (1.1). When the discriminant EABEEKtorsEK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 : E,122NN 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 : BTheorem 1. Let be the elliptic curve E23y1) If 0B22,ifisasquarein,4,if 4,2, otherwise.torsAEA B0A 2) If is sixth-power free and , then 436,if 1,3,if4322 3 ,orifissquarenot1,2,ifiscubicnot 1,0, otherwise.torsEBBBB  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23yxfkxgk (3) fwith ,kgk k. Then Theorem 1 yields the case when one of fk and gk is zero and max ,1deg degkkfkgk. In this paper, we deal max ,2deg degkkfkgk. 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2433 1khhh or 241331hhh. Then *This work was supported by NRF 2012-0006901. Copyright © 2013 SciRes. APM Y. K. PARK 305 2222224,20or34mod 35,suchthat31satisfying32and6652,20or34mod 35,suchthat31satisfying32and 6652,is congruentto onetorsklmmlllklEmmllk    22of theelementsof thesetandsuch that31,0, otherwise.Klkll and2issquare,and2issquare,klllmkllllm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 ppEEE if has a good reduction at the prime . pTheorem 3. (Lutz-Nagell) Let be an elliptic cur- ve (1.1) with coefficients in and pEEp be a obtained curve by reducing coefficients of modulo . And let EE,,1PyP be the discriminant of . 1) If 1a and if is in , then 0PxtorsExP and yP are integers; 2) For any 1, if is in a,PxPy,1PtorsE, then 4xP and 8yP are integers; 3) If is an odd prime such that pEp, then the restriction to of the reduction homomorphism Etors:pp is one-to-one. The same conclu- sion is valid for if pE2prE2E and ; 14) If , so that is given by 0a20E23 ,13aaayxAxB ,1EP0yP and if is in tors , then either ( and has order 2) or else  ,PxP yP0yP and divides . 2yP 3227Bx Ax 4dA 23:EyProof. See . □ Lemma 4. Let be the elliptic curve over Bp and be a point in ,PxypE,Pxy which is not a point at infinity. Then the followings are equivalent. 1) is a point of order 3 in pE4236 12; 2) 2xAx Bx,A is congruent to 0 modulo . pProof. 1)  2) Let 22xyPPP be the point . Then by the group law algorithm (), 24228xBx22222223243324222xA AxyxAxA xyxAx Byyy   and ,.Pxy3PO  means that Then 42 2228 4xAxBx Axy (5) 22232233222.4xAxAxxAxByy 23 (6) x should satisfy that yxAxBSince , 42 236 120xAxBxA in p42 236 120xAx BxAy23. 2) 1) Assume that , is not zero and yxAxB in p, . By simple calculation, such xy 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kp E5 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 2633 62yxkxkk k where is a nonzero integer. And using the natural surjection from to pp, we can get pEEp p4332394kk  by reducing the coefficients of modulo . If does not divide the discriminant then the group pE consisting of the points defined over the finite field pp with elements is 1) 559, 1mod5,6, 2mod5,3, 3mod5.kEkk 2) 7733,3mod7,6, 1,4mod7,9, 2mod7,12,6mod 7.kkEkk  Copyright © 2013 SciRes. APM Y. K. PARK 306 Table 1. Point in 55Ekmod 5. 55EO 55E55E 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, 20, 0, 6 3  3 2, 2 pp has a subgroup of E. EProof. By , 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 k0d7235mo . When , 7 becomes od7E1mk23yx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7Ey  ,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 ,2,14236 12xAxBx20A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32662xx 6, 377For , has 9 elements. But the equation giving criterion of order 3 is in and . Therefore, 94x7771 ,6,1EE. od777E3mk2x3,1,5,31xx0,3,77E33 6mod7k, 6,2, . Last, if , 77E0,1, 2,1,3,3, 4,0 ,5,1 4,0 has only one point  of order 2. It means that 77E23 2:63362Ey xkxkk231ll231lk22 469kl ll231xl20x 231,0l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kProof. 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69k36306k23kxkConversely, suppose that the equation of 32633 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 22369 1kkmmm 2ml . Since it has an integral solution, 231kll l for an integer .  and Now we show that there is no point of order 2 except 231,0l in E. Assume that 231,0lE. Then 231kll.  32222432633 62311391812 2.xkxkkxlxlx lll    22432139 18 122xlxlllQx be Let 393 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EProposition 7. Let 23 2:63362Ey xkxkk k h be the elliptic curve with in . Assume that there is no integer such that 2433 1khhh or 241331hhh. 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 231125 12123.ETXXXX kXkk  ESX be the polynomial Let 32 23112512123 .ETX XXXk Xkk   Since 21, 3kE is not in , it suffices to check whether x is a root of or not. 0SXESuppose that 0SXEx has a root k in . Then it is an integer. In other words, for an integer not the form 2433 1hh h241331hhh k or by sorting again as , we can fine an integer  such that x32 223212 51212 31212153 0.xxk xkkkxkxxx    Copyright © 2013 SciRes. APM Y. K. PARK 307From, this must be a multiple of 12 and 3253xx xx is one of 12 or 11 for a suitable inte- ger . 3,5,9m12 3mmWhen , xES3296 16mm21 13mm221hh23hh12xmk222 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 . 24 144mkm16 441m43kh31h5,12 9m 12k4112kk23hh12xm111If or then discri- minant of the quadratic equations with respect to is or 312531211m1, 4mm21m2 23m4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k3rN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, 29,32For the cases that the subgroup is nontrivial Pro- Table 2. Type of gr oup torsEk. mod 7k mod 5 0 1position 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 355l (respectively, ) if and only if 26mod 3519 (respectively, or  or 33mod 35). 231,0l is the unique point of order 2. Using duplication formula for the elliptic curve, let ,Pxy2231,0Pl be the point satisfying . By Substituting ,, 63xyk2362kk,, and for xyA 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: 2224322243221 3181861021 31818610xlxlllxlxlllFx   9C 2 6C33C40 D C B C D1 6C9C B C B C B2 C C C C C3 33CC6C C C C C C4 B C B C B5 D C B C D6 12C A C B C A where  4234322654328764221 3691812225416210854637324 972 864270605.Fxxlxl l lxlllllxlllll  To have an integral solution of 2243221 31818610xlxlll, its discrimi- nant 336 3 2ll have to be a square. Suppose that we can find an integer such that and m232mll2316xllm 2316llmml ( or ). It is easy to check that the integer satisfying the above condition exists in each case determined by . Furthermore, by sub- stituting , x31kll232mll and to the right hand side of (1.4) we get a numerical formula 322222 25432 652963266 5296652llllmlllllmlml lm 0l Since  makes the curve (1.4) singular, 266 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.torsE12k Given elliptic curve is the form in Theorem 2 and 2123 212. Therefore 2Etors . And 11,0 is the nontrivial torsion point on E. torsE is able to be applied to The method to find Copyright © 2013 SciRes. APM Y. K. PARK t © 2013 SciRes. APM 308 k7pCopyrighall 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 pp 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p16k:93674x23x7Ey with discriminant . Find 625ppE 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  B. Mazur, “Modular Curves and the Eisenstein Ideal,” Publications Mathématiques de l’Institut des Hautes Étu- des Scientifiques, No. 47, 1977, pp. 33-168.  A. Knapp, “Elliptic Curves,” Princeton University Press, Princeton, 1992.  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