• 検索結果がありません。

Nate on a certain supersingular elliptic curve

N/A
N/A
Protected

Academic year: 2021

シェア "Nate on a certain supersingular elliptic curve"

Copied!
2
0
0

読み込み中.... (全文を見る)

全文

(1)

Bulletin of Faculty of Education,Nagasaki University:Natural Science No.67, 1 -- 2 (2002. 6)

Nate on a certain supersingular elliptic curve

Tadashi WASHIO and Tetsuo KODAMA'

Department of Mathematics, Faculty of Education, Nagasaki University, Nagasaki 852, Japan

(Received Mar. 15, 2(02)

Abstract

Let p be an odd prime number such that p=2 (mod 3) and denote by F a finte prime field of characteristic p. Then it is shown that an elliptic curve y2 = X(X2 +X+s) defined over F is supersinglar and so that the following equality

t C)(-k)

Sk =

a

k=O k k

holds in F where t=(p-l)/2, n=[t/2] and s=1/3EF

1. Introduction

We denote by p an odd prime number and by F a finite prime field of character- istic p. In the previous note [1], we proved that if p=5 (mod 8) then the elliptic curve P= X(X2

+

X

+

r) defined over F is supersingular and so the following equality

± (2n) (2n-k)

rk = 0

k=0 k k

holds in F where n=(p-l)/4 and r=1/8EF

In this note, we want to prove, after the manner of [1], that if p=2 (mod 3) then the following equality

± ()(-k)

Sk =

a

k=O k k

holds where t=(p-l)/2, n= [t/2] and s=1/3EF.

2. The number of rational points

Let p be an odd prime number such that p=2 (mod 3) and denote by F a finte prime field of characteristic p. Moreover we put s=1/3EF. Then it is clear that the polynomial )(2

+

X

+

s is irreducible over F and so the curve defined by Y2=

X(J<:2+ X +s) over F is elliptic.

THEOREM 1. Denote by N the number oj rational points oj elliptic curve P=j(X) over F where j(X)=X()(2+X+s). Then N=p+l holds.

PROOF. We denote by X the multiplicative quadratic characetr of F. Then N is

'Professor emeritus, Kyushu University, Fukuoka 812, Japan

(2)

2

given by

Tadahi Washio and Tetsuo Kodama

N = p+l+ I; X(f(x)).

xEF

Using our assumptions of p=2 (mod 3) and s=1/3, we can easily show that, for any x,yEF, if X=FY then f(x)=Ff(y). This means that {f(x); xEF} =F.

Therefore we get

I; X(f(x))

=

I; X(x)

=

0

xEF xEF

and so we obtain N=p+ 1.

3. Hasse invariant and binom ial coefficients

We will now show that our curves are supersingular and give the congruence rela- tions for binomial coefficients associated to these curves.

THEOREM 2. If p=2 (mod 3) and s=1/3EF=GF(p) then the elliptic curve P=

X(~+X+s) defined over F is supersigular.

PROOF. According to Theorem 1, we see that our curve has p+ 1 rational points over F. This means that the Hasse invariant of our cuve is zero and so our curve is supersingular.

Rewri ting the Hasse invariant of curve P =X(~

+

X

+

s) in terms of binomial coefficients, it is clear that Theorem 2 leads to the following result.

THEOREM 3. If p=2(mod 3) and s=1/3EF=GF(P) then

± ()C-

k) Sk = 0

k=O k k

where t=(p-l)/2 and n=[t/2], Le., in the ring Z of rational integers,

± ()C-

k) 3n- k

=

0 (mod p).

k=O k k

References

[ 1] T.Washio and T. Kodama, On a certain supersingular elliptic curve, Bull. Fac. Educ., Nagasaki Univ.:Natural Science, No. 66 (2002), 1-3.

参照

関連したドキュメント

The initial results in this direction were obtained in [Pu98] where a description of quaternion algebras over E is presented and in [GMY97] where an explicit description of

Given the topological group Π X , is it possible to determine the reduction type of the elliptic curve E over K.. Using the terminology introduced in previous talks, this is a

In Section 2 we recall some known works on the geometry of moduli spaces which include the degeneration of Riemann surfaces and hyperbolic metrics, the Ricci, perturbed Ricci and

For example, a maximal embedded collection of tori in an irreducible manifold is complete as each of the component manifolds is indecomposable (any additional surface would have to

A curve defined over a finite field is maximal or minimal according to whether the number of rational points attains the upper or the lower bound in Hasse- Weil’s

Now it makes sense to ask if the curve x(s) has a tangent at the limit point x 0 ; this is exactly the formulation of the gradient conjecture in the Riemannian case.. By the

(9) As an application of these estimates for ⇡(x), we obtain the following result con- cerning the existence of a prime number in a small interval..

Splitting homotopies : Another View of the Lyubeznik Resolution There are systematic ways to find smaller resolutions of a given resolution which are actually subresolutions.. This is