Elliptic curve over a finite ring generated by 1 and an idempotent element $e$ with coefficients in the finite field $\mathbb{F}_{3^d}$

A Boulbot,A Mouhib,Abdelhakim Chillali

doi:10.5269/bspm.43654

Abstract

An elliptic curve over a ring $\mathcal{R}$ is a curve in the projective plane $\mathbb{P}^{2}(\mathcal{R})$ given by a specific equation of the form $f(X, Y, Z)=0$ named the Weierstrass equation, where $f(X, Y, Z)=Y^2Z+a_1XYZ+a_3YZ^2-X^3-a_2X^2Z-a_4XZ^2-a_6Z^3$ with coefficients $a_1, a_2, a_3, a_4, a_6$ in $\mathcal{R}$ and with an invertible discriminant in the ring $\mathcal{R}.$ %(see \cite[Chapter III, Section 1]{sil1}). In this paper, we consider an elliptic curve over a finite ring of characteristic 3 given by the Weierstrass equation: $Y^2Z=X^3+aX^2Z+bZ^3$ where $a$ and $b$ are in the quotient ring $\mathcal{R}:=\mathbb{F}_{3^d}[X]/(X^2-X),$ where $d$ is a positive integer and $\mathbb{F}_{3^d}[X]$ is the polynomial ring with coefficients in the finite field $\mathbb{F}_{3^d}$ and such that $-a^3b$ is invertible in $\mathcal{R}$.

Highlights

In the forth section, we classify the points of the elliptic curve Ea,b(F3d [e]) on the fact that the third projective coordinate of an element in Ea,b(F3d [e]) is invertible or not
The ring F3d [e], e2 = e where d is a positive integer can be constructed as an extension of the finite field F3d by using the quotient ring of the polynomial ring F3d [X] by the polynomial X2 − X
An element X = x0 + x1e is invertible in the ring F3d [e] if and only if x0 ≡ 0 mod 3 and x0 + x1 ≡ 0 mod 3

Summary

Introduction

In the forth section, we classify the points of the elliptic curve Ea,b(F3d [e]) on the fact that the third projective coordinate of an element in Ea,b(F3d [e]) is invertible or not. Using the Corollary 2.6, we show that the set of non invertible elements in the ring F3d[e] is the union of the two distinct ideals e and 1 − e , e ∪ 1 − e is not an ideal, which prove the lemma.

Results

Conclusion