This is a problem from my coding theory book which I am trying to wrap my head around.
Consider the curve $f_3F(q)$ given by $x^3+y^3+z^3=0$
A) Find the three projective points (x:y:z) of $P^2(F_2)$ on $f_3(F_2)$
B) Find the nine projective points (x:y:z) of $P^2(F_4)$ on $f_3(F_4)$
Attempt:
A) I think the answer is (1:0:1);(1:1:0);(0:1:1).
B) I am getting a little stuck here since there are 16 points which hold with the curve. I know that we have to concern ourselves with equivalences here but when I look at equivalences I end up with only 6 points not 9.
Work part B) Since $1^3 = 1$ $2^3 = 0$ and $3^3 = 3$ over $F_4$
(1:0:3)~(2:0:2)~(3:0:1) ; (1:2:3)~(2:0:2)~(3:2:1) ; (1:3:0)~(2:2:0)~(3:1:0) ; (1:3:2)~(2:2:0)~(3:1:2) ; (0:1:3)~(0:2:2)~(0:3:1) ; (2:1:3)~(0:2:2)~(2:3:1) ; (3:1:0)~(2:2:0)~(1:3:0)
So what am I missing/doing wrong here?
The key ideas are already in the comments, so just wrapping this up.
Indeed for the finite field $\Bbb{F}_4=\{0,1,A,B\}$, where $B=A+1=A^2$ we have that any non-zero element cubed is equal to $1$. Thus the equation $x^3+y^3+z^3=0$ forces exactly one of the projective coordinates to be zero.
If, say, $z=0, x\neq0\neq y$, then we can scale $x$ to be $=1$, and then have three different choices, $1,A,B$, for $y$ each giving a solution and a distinct point of the projective plane. The same thing happens when $y=0$ and when $x=0$, and we can scale the first non-zero homogeneous coordinate to $=1$.
This gives us the following nine points: $$ \begin{array}{ccc} (1:1:0),&(1:A:0),&(1:B:0),\\ (1:0:1),&(1:0:A),&(1:0:B),\\ (0:1:1),&(0:1:A),&(0:1:B). \end{array} $$