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

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

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

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Values for $y$ :

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

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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 $mP=0$.

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