# How to Find the Generators of an Elliptic Curve

Could someone explain how can I find the generator points of an elliptic curve?

For example the generators of the EC: $$y^2= x^3+x+6, Z_7$$.

• Is the answer of Is every point on an elliptic curve of a prime order group a generator? satisfying you? Jan 22, 2019 at 12:22
• What do you mean by the generator points of an elliptic curve? An elliptic curve group is not necessarily cyclic. Jan 22, 2019 at 13:43
• You are right Daniel, but the problem is if the elliptic curve is circular, how can we find the generators. Jan 23, 2019 at 16:35

First, find points of the curve : (When working on $$\mathbb{Z}_n$$, with $$n$$ prime, I used to write numbers in $$[-(n-1)/2 ; (n-1)/2 ]$$ rather than in $$[0,n]$$, so $$x^3+x+6 = x^3+x-1 \mod 7$$)

Values for $$x$$ :

$$x \qquad \quad \qquad | -3 \quad -2 \quad -1 \qquad 0 \qquad 1 \qquad 2 \qquad 3$$ $$------------------------$$ $$x^3 + x -1 \quad | -3 \qquad 3 \ \quad -3 \quad -1 \ \ \ \quad 1 \qquad 2 \qquad 1$$

$$~$$

Values for $$y$$ :

$$y \qquad \quad \qquad | -3 \quad -2 \quad -1 \qquad 0 \qquad 1 \qquad 2 \qquad 3$$ $$------------------------$$ $$y^2 \qquad \qquad \ \ |\quad 2 \ \quad -3 \qquad 1 \ \qquad 0 \qquad 1 \ \quad -3 \ \ \quad 2$$

$$~$$

Then we can see that we have $$y^2=x^3+x-1$$ for the following couples : $$\{(-3,-2);(-3,2),(-1,-2);(-1,2);(1,-1);(1,1);(2,-3);(2,3);(3,-1);(3,1)\}$$.

So all these points belong to the curve. To have all the points belonging to the curve just add the point at infinity $$\mathcal O$$. So you have $$11$$ points on the curve.

Then, the comment of @kelalaka tell you that all those points (except $$\mathcal O$$) are a generator because $$11$$ is prime.

In a more general way, when the order of an elleptic curve is $$m$$, a point $$P$$ is a generator if, and only if for all divisors $$d$$ of $$m$$ , $$dP \neq 0$$ and $$mP=0$$.