Exercise 11.4

Proof. Here $F={𝔽}_p$, and $F_s={𝔽}_{p^s}$.

To calculate $N_s$, we calculate the number of points at infinity (such that $x_0=0$), and the numbers of affine points of the curve ${\bar{H}}_f({𝔽}_{p^s})$ associate to

$\text{∙}$

To estimate the number of points at infinity, we calculate first the cardinality of the set $$U=\{({\alpha }_0,{\alpha }_1,{\alpha }_2,{\alpha }_3)\in F_s^4\mid {\alpha }_0{\alpha }_1-{\alpha }_2{\alpha }_3=0,{\alpha }_0=0\}.$$

Then ${\alpha }_1$ takes an arbitrary value $a\in F_s$. Write

$$U_a=\{({\alpha }_0,{\alpha }_1,{\alpha }_2,{\alpha }_3)\in U\mid {\alpha }_1=a\}.$$

Then $U_a=\{({\alpha }_0,{\alpha }_1,{\alpha }_2,{\alpha }_3)\in F_s^4\mid {\alpha }_0=0,{\alpha }_1=a,{\alpha }_2{\alpha }_3=0\}$ , thus $U_a=A\cup B$, where

$$\begin{array}{llll}\hfill A& =\{({\alpha }_0,{\alpha }_1,{\alpha }_2,{\alpha }_3)\in U_a\mid {\alpha }_2=0\},& \hfill & \\ \hfill B& =\{({\alpha }_0,{\alpha }_1,{\alpha }_2,{\alpha }_3)\in U_a\mid {\alpha }_3=0\}.& \hfill & \end{array}$$

Since ${\alpha }_0,{\alpha }_1,{\alpha }_2$ are fixed in $A$, the map $A\to F_s$ defined by $({\alpha }_0,{\alpha }_1,{\alpha }_2,{\alpha }_3)\mapsto {\alpha }_3$ is a bijection, therefore $|A|=p^s$, and similarly $|B|=p^s$. But $A\cap B=\{(0,0,0,0)\}$, thus

$$|U_a|=|A|+|B|-|A\cap B|=2p^s-1.$$

Since $U$ is the disjoint union of the $U_a$, thus

$$|U|=\underset{a\in F_s}{\sum }|U_a|=\underset{a\in F_s}{\sum }(2p^s-1)=2p^{2s}-p^s.$$

Therefore the number of projective points $[{\alpha }_0,{\alpha }_1,{\alpha }_2,{\alpha }_3]\in P^3(F_s)$ at infinity (such that ${\alpha }_0=0$) is

$$N_{\infty }=\frac{|U|-1}{p^s-1}=\frac{2p^{2s}-p^s-1}{p^s-1}=2p^s+1.$$

$\text{∙}$

Now we calculate the number of points of the affine surface $H_f({𝔽}_s)$ associate to the equation $y_1=y_2{y}_3$ (where $y_i=x_i∕x_0$).

The maps

$$u\{\begin{array}{ccc}\hfill F_s^2\hfill & \hfill \to \hfill & H_f(F_s)\hfill \\ \hfill (\beta ,\gamma )\hfill & \hfill \mapsto \hfill & (\mathit{\beta \gamma },\beta ,\gamma )\hfill \end{array}v\{\begin{array}{ccc}\hfill H_f(F_s)\hfill & \hfill \to \hfill & F_s^2\hfill \\ \hfill (\alpha ,\beta ,\gamma )\hfill & \hfill \mapsto \hfill & (\beta ,\gamma )\hfill \end{array}$$

satisfy $u\circ v=\mathrm{id},v\circ u=\mathrm{id}$, so $u$ is a bijection. With more informal words, the arbitrary choice of $\beta ,\gamma \in F_s$ gives the affine point $(\alpha ,\beta ,\gamma )$, where $\alpha =\mathit{\beta \gamma }$.

This gives $|H_f(F_s)|=p^{2s}$.

Therefore

We obtain in $ℂ[[u]]$

This gives

Note: The result for $N_s$ is verified with the naive and very slow following code in Sage:

def N(p,s):
    Fs = GF(p^s)
    counter = 0
    for x in Fs:
        for y in Fs:
            for z in Fs:
                for t in Fs:
                    if x*y == z*t:
                        counter += 1
    return (counter - 1)//(p^s - 1)

p, s = 5, 3
print N(p,s), p^(2*s) + 2*p^s +1

There is a misprint in the “Selected Hints for the Exercises” in Ireland-Rosen p.371. □