Volume 2014, Issue 1 298632
Research Article
Open Access

A Mean Value Formula for Elliptic Curves

Rongquan Feng

Rongquan Feng

LMAM, School of Mathematical Sciences, Peking University, Beijing 100871, China pku.edu.cn

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Hongfeng Wu

Corresponding Author

Hongfeng Wu

College of Science, North China University of Technology, Beijing 100144, China ncut.edu.cn

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First published: 25 August 2014
Citations: 2
Academic Editor: Jiyou Li

Abstract

It is proved in this paper that, for any point on an elliptic curve, the mean value of x-coordinates of its n-division points is the same as its x-coordinate and that of y-coordinates of its n-division points is n times that of its y-coordinate.

1. Introduction

Let K be a field with char(K) ≠ 2,3 and let be the algebraic closure of K. Every elliptic curve E over K can be written as a classical Weierstrass equation as follows:
()
with coefficients a, bK. A point Q on E is said to be smooth (or nonsingular) if (f/x|Q, f/y|Q) ≠ (0,0), where f(x, y) = y2x3axb. The point multiplication is the operation of computing
()
for any point PE and a positive integer n. The multiplication-by-n map
()

is an isogeny of degree n2. For a point QE, any element of [n] −1(Q) is called an n-division point of Q. Assume that (char(K), n) = 1. In this paper, the following result on the mean value of the x, y-coordinates of all the n-division points of any smooth point on an elliptic curve is proved.

Theorem 1. Let E be an elliptic curve defined over K and let Q = (xQ, yQ) ∈ E be a point with . Set

()

Then

()

According to Theorem 1, let Pi = (xi, yi), i = 1,2, …, n2, be all the points such that nP = Q and let λi be the slope of the line through Pi and Q; then yQ = λi(xQxi) + yi. Therefore,
()
Thus we have
()
where , , , and are the average values of the variables λi, xi, λixi, and yi, respectively. Therefore,
()

Remark 2. The discrete logarithm problem in elliptic curve E is to find n by given P, QE with Q = nP. The above theorem gives some information on the integer n.

2. Proof of Theorem 1

To prove Theorem 1, define division polynomials [1] on an elliptic curve E : y2 = x3 + ax + b inductively as follows:
()
It can be checked easily by induction that the ψ2n’s are polynomials. Moreover, when n is odd, and when n is even. Define the polynomial
()

for n ≥ 1. Then . Since y2 = x3 + ax + b, replacing y2 by x3 + ax + b, one has . So we can denote it by ϕn(x). Note that if n and m have the same parity. Furthermore, the division polynomials ψn have the following properties.

Lemma 3. Consider

()

when n is odd and

()

when n is even.

Proof. We prove the result by induction on n. It is true for n < 5. Assume that it holds for all ψm with m < n. We give the proof only for the case for odd n ≥ 5. The case for even n can be proved similarly. Now let n = 2k + 1 be odd, where k ≥ 2. If k is even, then by induction

()

Substituting y4 by (x3 + ax + b) 2, we have

()

Therefore,

()

The case when k is odd can be proved similarly.

The following corollary follows immediately from Lemma 3.

Corollary 4. Consider

()

Proof of Theorem 1. Define ωn as

()

Then for any P = (xP, yP) ∈ E, we have ([1])

()

If nP = Q, then . Therefore, for any P ∈ Λ, the x-coordinate of P satisfies the equation . From Corollary 4, we have that

()

Since Λ = n2, every root of is the x-coordinate of some P ∈ Λ. Therefore,

()

by Vitae’s theorem.

Now we prove the mean value formula for y-coordinates. Let K be the complex number field first and let ω1 and ω2 be complex numbers which are linearly independent over . Define the lattice

()

and the Weierstrass -function by

()

For integers k ≥ 3, define the Eisenstein series Gk by

()

Set g2 = 60G4 and g3 = 140G6; then

()

Let E be the elliptic curve given by y2 = 4x3g2xg3. Then the map

()

is an isomorphism of groups and . Conversely, it is well known [1] that, for any elliptic curve E over defined by y2 = x3 + ax + b, there is a lattice L such that g2(L) = −4a, g3(L) = −4b and there is an isomorphism between groups and given by z ↦ ((z), (1/2)(z)) and 0 ↦ . Therefore, for any point , we have (x, y) = ((z), (1/2)(z)) and n(x, y) = ((nz), (1/2)(nz)) for some .

Let Q = ((zQ), (1/2)(zQ)) for a . Then for any Pi ∈ Λ, 1 ≤ in2, there exist integers j, k with 0 ≤ j, kn − 1, such that

()

Thus,

()

which comes from . Differentiate with respect to zQ, we have

()

That is,

()

Secondly, let K be a field of characteristic 0 and let E be the elliptic curve over K given by the equation y2 = x3 + ax + b. Then all of the equations describing the group law are defined over . Since is algebraically closed and has infinite transcendence degree over , can be considered as a subfield of . Therefore we can regard E as an elliptic curve defined over . Thus the result follows.

At last assume that K is a field of characteristic p. Then the elliptic curve can be viewed as one defined over some finite field , where q = pm for some integer m. Without loss of generality, let for convenience. Let be an unramified extension of the p-adic numbers of degree m, and let be an elliptic curve over K which is a lift of E. Since (n, p) = 1, the natural reduction map is an isomorphism. Now for any point QE with , we have a point such that the reduction point is Q. For any point with nPi = Q, its lifted point satisfies and whenever PiPj. Thus,

()

since K is a field of characteristic 0. Therefore the formula holds by the reduction from to E.

Remark 5.

  • (1)

    The result for x-coordinate of Theorem 1 holds also for the elliptic curve defined by the general Weierstrass equation y2 + a1xy + a3y = x3 + a2x2 + a4x + a6.

  • (2)

    The mean value formula for x-coordinates was given in the first version of this paper [2] with a slightly complicated proof. The formula for y-coordinates was conjectured by Feng and Wu based on [2] and numerical examples in a personal email communication with Moody (June 1, 2010).

  • (3)

    Recently, some mean value formulae for twisted Edwards curves [3, 4] and other alternate models of elliptic curves were given by [5, 6].

3. An Application

Let E be an elliptic curve over K given by the Weierstrass equation y2 = x3 + ax + b. Then we have a nonzero invariant differential ω = dx/y. Let ϕ ∈ End(E) be a nonzero endomorphism. Then ϕ*ω = ωϕ = cϕω for some , since the space ΩE of differential forms on E is a 1-dimensional -vector space. Since cϕ ≠ 0 and div⁡(ω) = 0, we have
()
Hence cϕ has neither zeros nor poles and . Let φ and ψ be two nonzero endomorphisms; then
()
Therefore, cφ+ψ = cφ + cψ. For any nonzero endomorphism ϕ, set ϕ(x, y) = (Rϕ(x), ySϕ(x)), where Rϕ and Sϕ are rational functions. Then
()
where is the differential of Rϕ(x). In particular, for any positive integer n, the map [n] on E is an endomorphism. Set [n](x, y) = (Rn(x), ySn(x)). From c[1] = 1 and [n] = [1]+[(n − 1)], we have
()
For any Q = (xQ, yQ) ∈ E and any
()
we have yP = yQ/Sn(xP). Therefore, Theorem 1 gives
()
Thus,
()

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This work was supported by the National Natural Science Foundation of China (no. 11101002, no. 61370187, and no. 11271129) and Beijing Natural Science Foundation (no. 1132009).

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