Let us consider the curve $C$ determined by $x^2t^2 + y^2t^2 + x^2y^2 - t^4$ over $F$ ( finite field with $|F|=q$ ). Maybe it is a stupid question but how can I compute the points at infinity? The solution has to be $[0,1,0]$ and $[0,0,1]$.
Newest Update: Here is the right order $[t,x,y]$ and I I see why this two points are points at infinity. My assumption is that I set the first coordinate zero to form the line at infinity and get $x^2y^2 = 0$. I get the solutions $x=1$ and $y=0 $ and $x=0$ and $y=1$. Both can't be zero because $[0,0,0]$ is not in the projetive space!
Added some context:
We have a finite field $F$. We define $A^n(F) = { (a_1,\dots,a_n) | a_i \in F }$. On $A^{n+1}(F) -{(0,\dots,0)}$ we define an equivalence relation: $(a_0,\dots,a_n) \sim (b_0,\dots,b_n) : \Leftrightarrow \exists \alpha \in F^{*}$ with $a_i = \alpha b_i \forall i=0,\dots,n$. The equivalence classes are the points on the projective $n$-space $P^n(F)$, where $P^n(F) =(A^{n+1}(F) -{(0,\dots,0)})/ \sim$ . $H$ called hyperplane at infinity. . $P^n(F)$ is made up of two pieces, one a copy of $A^n(F)$, called the finite points, and the other a copy of $P^{n-1}(F)$, called the points at infinity.
Thank you in advance