# How do you determine if an elliptic curve over a finite field is cyclic?

I know the group order and the points of the elliptic curve $y^2 = x^3 + Ax + B$, but I am confused on how to determine if they from a cyclic group

The curve $y^2 = x^3 + 2x +2$ in $\Bbb F_{11}$ which has $9$ points: $(1,4),$ $(1,7),$ $(2,5),$ $(2,6),$ $(5,4),$ $(5,7),$ $(9,1),$ $(9,10)$ and the point of infinity, $(0,1)$.

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What finite field? What are the points in it? –  DonAntonio Apr 21 '13 at 23:24
Edited the question –  badosky Apr 21 '13 at 23:37
I count only $8$ points in that list. –  Cameron Buie Apr 21 '13 at 23:47
Oh, sorry; plus the point of infinity, (0,1) –  badosky Apr 21 '13 at 23:51
The point $(0,1)$ is not a point on the curve, and not the point at infinity either. The point at infinity is $[0,1,0]$ in projective coordinates. –  Álvaro Lozano-Robledo Apr 22 '13 at 18:34

Well, you're looking at an abelian group (say with operation $\oplus$) of order $9,$ correct? Then either the group will be cyclic, or it will be isomorphic to $C_3\times C_3$. Let's rewrite the non-identity elements of the group as: $$(1,4),\ominus(1,4),(2,5),\ominus(2,5),(5,4),\ominus(5,4),(9,1),\ominus(9,1)$$ (these are written in the same order as you had them, but making the inverse elements clear). Since elements have the same order as their inverses, then we need consider only $$(1,4),(2,5),(5,4),(9,1).$$ Either all four of them have order $3$, or three of the four of them have order $9$. In the latter case, we'll have a cyclic group. Regardless, you'll only need to check two of them, and if you're checking $(p,q)$, then you'll only need to check if $(p,q)\oplus(p,q)=\ominus(p,q)$. If not, then $(p,q)$ has order $9$ and we're done.

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I am totally lost. What is this symbol $\ominus$? –  curious Dec 3 '13 at 21:20
I use it to denote the $\oplus$-additive inverse. –  Cameron Buie Dec 3 '13 at 22:41

Use the following code in Magma:

F:=GF(11);

E:=EllipticCurve([0,0,0,2*F.1,2*F.1]);

P:=E![1,4];

[2*P,3*P,4*P,5*P,6*P,7*P,8*P,9*P];

The result is:

[ (2 : 5 : 1), (9 : 10 : 1), (5 : 4 : 1), (5 : 7 : 1), (9 : 1 : 1), (2 : 6 : 1), (1 : 7 : 1), (0 : 1 : 0)]

Hence $P=(1,4)$ has order $9$.

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