# 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, $(0,1)$.

• 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.

• 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
• I don't have much background on group/field theory so I am finding it difficult to interpret the statements like "either the group will be cyclic, or it will be isomorphic". Could you please provide any link/tutorial to help me? – Gaurav Nov 16 '14 at 6:12
• @Gaurav: Suppose $G$ is an abelian group of order $9.$ Since the order of an element of $G$ must divide the order of $G,$ then every non-identity element must have order $3$ or $9.$ If an element has order $9,$ then the group is cyclic. Suppose all non-identity elements have order $3.$ Take any non-identity element $x\in G,$ and take any element $y\in G$ such that $y$ is not in the cyclic subgroup generated by $G.$ One can then show that $G$ is generated by $x$ and $y$, which is a bit easier since $G$ is abelian. After that, we can construct an isomorphism to $C_3\times C_3.$ – Cameron Buie Nov 16 '14 at 17:12
• Let me know how much of that you followed, and any particular questions you might have regarding terminology I used or claims I made. If you're looking for a reference, try googling "structure theorem for abelian groups" for more info. It's a powerful result, but isn't really necessary here. – Cameron Buie Nov 16 '14 at 17:14

A shorter solution depending on a bit more machinery is the following.

As others have said, an abelian group of order $9$ is either cyclic or isomorphic to $C_3\times C_3$.

It is known (see e.g. Silverman) that the $3$-torsion of this curve (over an algebraic closure of $\Bbb{F}_{11}$ is isomorphic to $C_3\times C_3$. Furthermore, the Weil pairing on that $3$-torsion is non-degenerate. So if all of the $3$-torsion points have coordinates in a field $K$, then the multiplicative group $K^*$ contains a primitive third root of unity.

But $\Bbb{F}_{11}^*$ is cyclic of order ten, so there are no elements of order three there. The conclusion is that $K$ must be a proper extension of $\Bbb{F}_{11}$.

This leaves the cyclic group as the only alternative.

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 $9$.