In my algebraic geometry course I found the following problem:
Let $E$ an elliptic curve over the field $\mathbb{F}_5$ given by the equation: $$y^2z=x^3+xz^2-z^3$$ Find the group asociated to $E$.
Doing some computations and with quadratic residues I found that $E$ has the following points:
$$[0:1:0],[0:2:1],[0:3:1],[1:1:1],[1:4:1],[2:2:1],[2:3:1],[3:2:1],[3:3:1]$$
Thus the group asociated to $E$ has $9$ elements, so it must be either $C_9$ or $C_3\times C_3 $. I thought that maybe we can compute the order of the elements to choose which of the groups is the solution, but it looks like a very long computational problem. My question is whether there is a more elegant solution avoiding all these computations or not.
Thanks for your help :)
The computation is not so complicated.
We write $X = \frac x z, Y = \frac y z$ for the affine coordinates.
By doubling formula, we have $X(2P) = \frac{X^4 - 2X^2 + 8X + 1}{4(X^3 + X - 1)}$.
Now let $A$ be the point $(0, 2)$. Plug the value $X = 0$ into the doubling formula, we see that $X(2A) = 1$. This shows that the group $E(\Bbb F_5)$ cannot be isomorphic to $C_3 \times C_3$: otherwise we would have $2A = -A = (0, 3)$.
Therefore we have $E(\Bbb F_5) \simeq C_9$.