Calculating zeta functions over a field

Question

I am learning about zeta functions and have been trying the following example: Calculate the zata function of $x_0x_1-x_2x_3$ over $\mathbb{F}_p$. Does there exist an easy formula for calculating $N_s$, number of zeros of $f\in P^n(\mathbb{F}_s)$. I can then use the definition of zeta function of the hypersurface given by a exponent.

Answer 1

I assume that you are interested in the variety defined in the projective space $P^3\Bbb{F}_s$. Presumably $s=p^n$ for some $n>0$.

This one is relatively straightforward because it is linear in all the variables. As we are interested in the number of points in a projective space we need to first find the number of solutions $A_n$ of $x_0x_1=x_2x_3$ in $\Bbb{F}_s^4\setminus\{(0,0,0,0)\}$ and then afterward divide by $p^n-1$ to correctly tally the solutions in the projective spaces. For example:

• If $x_0\neq 0$ ($p^n-1$ choices) then equation $x_0x_1-x_2x_3=0$ is equivalent to $x_1=x_2x_3/x_0=x_2x_3$, so we can select $x_2$ and $x_3$ any which way we want ($p^{2n}$ choices), and the value of $x_1$ is uniquely determined. Altogether there are $p^{2n}(p^n-1)$ solutions of this type.
• If $x_0=0$ then the value of $x_1$ is irrelevant ($p^n$ choices). We are left with the equation $x_2x_3=0$ that holds iff at least one of the two remaining variables $=0$ ($2p^n-1$ choices). The unwanted solutions $x_0=x_1=x_2=x_3=0$ is included in this tally, so we get a contribution $p^n(2p^n-1)-1=2p^{2n}-p^n-1=(p^n-1)(2p^n+1)$ to $A_n$ from solutions of this type.

Therefore $ A_n=(p^{2n}+2p^n+1)(p^n-1). $ As always $$ N_s=\frac{A_n}{p^n-1}=p^{2n}+2p^n+1=s^2+2s+1=(s+1)^2. $$

Can you find the zeta function now?

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As an extra check we observe that $N_s$ is the square of the number of points over the projective line. This fits together with the fact that the projective variety you described is also familiar as the image of the Segre embedding $\Bbb{P}^1\times\Bbb{P}^1\to \Bbb{P}^3$.