On the Torsion Subgroups of Certain Elliptic Curves over *

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 certain elliptic curves depending on their coefficients. E   tors E 


Introduction
A is fourth-power free and , then A cubic curve over the field in Weierstrass form is given by projectively Hereafter assume that K is a number field.Since the field characteristic of K is , we can study 0 4 27 (2) instead of (1.1).When the discriminant 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 isomorphic to one of the following 15 types [1]: 2 This paper is focused on knowing how the coefficients A and of (1.2) determine tors .For the earlier work, we see the cases is zero in [2]: B Theorem 1.Let be the elliptic curve E tors is sixth-power free and , then It is too hard to determine the group  tors with- out any relation between the coefficients.Hence we consider the elliptic curve as follows: . Then Theorem 1 yields the case when one of

   
f k and g k is zero and In this paper, we deal be the elliptic curve with in .Suppose that is an integer such that  and there is no of the elements of the set and such that 3 1 , 0, otherwise.

The Proof of Theorem 2
We which is not a point at infinity.Then the followings are equivalent.
1) is a point of order 3 in x y P P P   be the point .Then by the group law algorithm ( x should satisfy that . By simple calculation, such x y P satisfy (5) and (6) and if is the point   . We are done.□  , Here we choose two rational primes and calculate the groups   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 .
. Table 1 is the proof of (1).Both cases can be calculated as using simple calculation.For 2), since and  , can not be congruent to and 5 mo .When , 7 becomes . Since it is an abelian group with 6 elements, 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 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 For , has 9 elements.But the equation giving criterion of order 3 is in and .Therefore, 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 Conversely, suppose that the equation of 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 . Since it has an integral solution, and  Now we show that there is no point of order 2 except . It gives us the value  or 1 .Hence and is singular.
be the elliptic curve with in  .Assume that there is no integer such that   Proof.As we mentioned in the proof of the previous lemma, the point , it suffices to check whether x is a root of or not.
Then it is an integer.In other words, for an integer not the form or by sorting again as , we can fine an integer  such that 12 5 12 12 3 12 12 1 5 3 0.
From, this must be a multiple of 12 and x is one of 12 or 11 for a suitable integer . 3,5,9 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 .
or then discriminant of the quadratic equations with respect to is or

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 , i.e., and .In fact, tors is a subgroup of both 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.

 
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 .

C n   
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

 
have to be a square.Suppose that we can find an integer such that and m It is easy to check that the integer satisfying the above condition exists in each case determined by .Furthermore, by substituting to the right hand side of (1.4) we get a numerical formula 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.
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