Number of solutions of $x_{0}x_{1} - x_{2}x_{3}$ over $\mathbb{F}_{p^{s}}$ I'm trying to solve the following exercise:
Compute the zeta function of $x_{0}x_{1} - x_{2}x_{3}$ over $\mathbb{F}_{p}$. Well, for this, I need to find $N_{s}$, the number of solutions in the field $\mathbb{F}_{p^{s}}$. What I did:
First, looking at the points at infinity ($x_{0} = 0$), we have that
$x_{2}x_{3} = 0$ and it can only happen if $x_{2}$ or $x_{3}$ equals $0$.
Then we have that the solutions are:
$[0,x_{1},0,x_{3}]$ and $[0,x_{1},x_{2},0]$ In each case we have $p^{2s}$ solutions ($p^{s}$ for $x_{1}$ and $p^{s}$ for $x_{2}$). Now we have to remove the intersection, which is $[0,x_{1},0,x_{3}] = [0,x_{1},x_{2},0]$ if and only if $x_{2} = 0$ and $x_{3} = 0$. So in this case we have $2p^{2s} - p^{s}$ solutions.
Now the finite points ($x_{0} \neq 0$):
Dividing all by $x_{0}$ we have that
$x_{1} = \dfrac{x_{2}x_{3}}{x_{0}}$, and then the solutions are
$[x_{0},\dfrac{x_{2}x_{3}}{x_{0}},x_{2},x_{3}]$ giving us $(p^{s}-1)p^{s}p^{s} = p^{3s} - p^{2s}$ solutions. Putting all together we have that 
$N_{s} = p^{3s} + p^{2s} - p^{s}$ solutions. 
But this is wrong, in the book it says that $N_{s} = 3p^{2s} - p^{s} - 1$ and in another source in the internet I found that $N_{s} = (p^{s} + 1)^{2}$. Which on is the right one? And what is wrong with my resolution? 
 A: I computed the number $N_s$ and got the same number as you i.e. $N_s = p^{3s} + p^{2s} - p^s$.  I use that there is a bijection between the product of projective spaces $\mathbb{P}^1 \times \mathbb{P}^1$ and the projective solutions of $x_0x_1 - x_2x_3 = 0$ in $\mathbb{P}^3$ with homogeneous coordinates $[x_0 : x_1 : x_2 : x_3]$. Such map is called the Segre map $s :\mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^3$: 
$$s([u:v],[u':v']) = [u u' : v v' : u v' :v u']$$ 
So the number of projective solutions is $(\# \mathbb{P}^1)^2 = (p^s + 1)^2$. By the way, notice that this is the number you got in Internet. So perhaps in the webplace you see it the problem was to compute the projective solutions. If you want the non projective solutions you multiply each projective by a non zero number of $\mathbb{F}_{p^s}$ to get $$(p^s + 1)^2.(p^s - 1)$$ non projective solutions. But you need to add the most trivial solution i.e. $x_0=x_1=x_2=x_3=0$ so the total number of solutions is: $$N_s = (p^s + 1)^2.(p^s - 1) + 1 =  p^{3s} + p^{2s} - p^s $$
You can check that the number $3p^{2s} - p^s - 1$ is wrong for $p=2$ and $s=1$ by hand. 
