Point group operation

Euclidean plane geometry

A point group operation on an elliptic curve is derived by geometric considerations on the curve in Weierstraß normal form $y^{2}=x^{3}+ax+b$ over the real numbers.

Because it is defined by an algebraic equation in the third degree, a straight line can intersect such a curve at at most three points in the Euclidean $(x,y)$ plane.

The basic point group operation $\oplus$ consists in finding the reflection across the x-axis of the third point of intersection of the curve with a line through two given points on the curve.

Some or all of the points on the elliptic curve, together with the point group operation $\oplus$, make out an Abelian group.

An additional point at infinity is adjoined to the $(x,y)$ plane, and considered to lie on the curve, although it is not given coördinates in the real numbers. This point is the group identity. The group inverse consists in taking the reflection of a point across the x-axis.

The operation of a point with itself, called **point-doubling**, is defined similarly, but by considering the line tangent to the curve at that point rather than through two distinct points.

Algebraic derivation

Given two points that lie on a given elliptic curve,

$P_{1}=(x_{1},y_{1}),\qquad P_{2}=(x_{2},y_{2}),$

the objective is to find the "third point" and calculate the "group sum" of the first two points:

$P_{3}=(x,y),\qquad P_{1}\oplus P_{2}=(x,-y).$

Once we have defined a suitably consistent algebraic definition for the so-called **point group operation** on an elliptic curve, we are freed from geometric considerations and able to consider the same operation on the elliptic curve over other algebraic fields, such as finite fields that have no geometric analogue.

Let

$m=\left\{{\begin{array}{lr}{\dfrac {y_{2}-y_{1}}{x_{2}-x_{1}}}&\ldots \quad x_{2}\neq x_{1};\\{\dfrac {3x_{1}^{2}}{2\surd (x_{1}^{3}-x_{1})}}&\ldots \quad x_{2}=x_{1},\,y_{2}=y_{1},\,y_{2}\geq 0;\\{\dfrac {-3x_{1}^{2}}{2\surd (x_{1}^{3}-x_{1})}}&\ldots \quad x_{2}=x_{1},\,y_{2}=y_{1},\,y_{2}<0;\\\infty &\ldots \quad x_{2}=x_{1},\,y_{2}=-y_{1}\end{array}}\right.$

be the slope of the line, either through $(x_1,y_1)$ and $(x_2,y_2)$ or tangent to the curve at $(x_1,y_1)$ if $x_2=x_1$ and $y_2=y_1$.

Substitute the equation for the line into that for the elliptic curve to find the "third point:"

${\begin{aligned}y-y_{1}&=m(x-x_{1})\\y&=mx+y_{1}-mx_{1}\\y^{2}&=m^{2}x^{2}+2m(y_{1}-mx_{1})x+(y_{1}^{2}-2mx_{1}y_{1}+m^{2}x_{1}^{2})\\y^{2}&=x^{3}+ax+b\\{}&{\phantom {=}}x^{3}-m^{2}x^{2}+[a+2m(mx_{1}-y_{1})]x+(b-y_{1}^{2}+2mx_{1}y_{1}-m^{2}x_{1}^{2})=0.\end{aligned}}$

Factor out the first two points $(x-x_{1})(x-x_{2})=x^{2}-(x_{1}+x_{2})x+x_{1}x_{2}$ from $x^{3}+ax+b$ and find the remainder by long division.

To be delivered

The remainder of this page is unfinished work.

${\begin{array}{rrrrr}&x&+(x_{1}{+}x_{2})\\x^{2}-(x_{1}{+}x_{2})x+x_{1}x_{2})&x^{3}&&+ax&+b\qquad \\&x^{3}&-(x_{1}{+}x_{2})x^{2}&+x_{1}x_{2}x\\&&(x_{1}{+}x_{2})x^{2}&+(a{-}x_{1}x_{2})x&+b\qquad \\&&(x_{1}{+}x_{2})x^{2}&-(x_{1}{+}x_{2})^{2}x&+(x_{1}{+}x_{2})x_{1}x_{2}\\&&&R=&(a{+}x_{1}^{2}{+}x_{1}x_{2}{+}x_{2}^{2})x+b-x_{1}^{2}x_{2}-x_{1}x_{2}^{2}\end{array}}$

${\begin{aligned}y^{2}&=(x-x_{1})(x-x_{2})(x_{1}+x_{2}+x)+x_{1}^{2}x+x_{1}x_{2}x+x_{2}^{2}x-x_{1}^{2}x_{2}-x_{1}x_{2}^{2}+ax+b\\x^{3}&=(x-x_{1})(x-x_{2})(x_{1}+x_{2}+x)+x_{1}^{2}x+x_{1}x_{2}x+x_{2}^{2}x-x_{1}^{2}x_{2}-x_{1}x_{2}^{2}\end{aligned}}$