Exercise 11.5

Proof. Since $q$ is odd, there is a unique character $\chi$ of order 2 over $F={𝔽}_q$, and a unique character of order 2 over $F_s={𝔽}_{q^s}$ . We first compute the number in ${𝔽}_q^{n+1}$ of solutions of the equation $f(x_0,\dots ,x_n)=0$, where $f(x_0,\dots ,x_n)=a_0{x}_0^2+\cdots +a_n{x}_n^2\in F[x_0,\dots ,x_n]$.

Indeed $J_0(𝜀,\dots ,𝜀)=q^{l-1}$, and $J_0({\chi }_0,\dots ,{\chi }_n)=0$ if some but not all of the ${\chi }_i$ are trivial (generalization of Proposition 8.5.1).

We estimate $J_0(\chi ,\dots ,\chi )$, where there are $n+1$ entries of $\chi$.

$\bullet$

If $n$ is even, then ${\chi }^{n+1}=\chi \ne 𝜀$, thus $J_0(\chi ,\dots ,\chi )=0$ (Proposition 8.5.1(c)), and so $$N(a_0{x}_0^2+\cdots +a_n{x}_n^2=0)=q^n,$$

and the number of projective points on the hypersurface is given by

$$N_1=\frac{q^n-1}{q-1}=q^{n-1}+\cdots +q+1.$$

$\bullet$

If $n$ is odd, then ${\chi }^{n+1}=𝜀$, thus $J_0(\chi ,\dots ,\chi )=\chi (-1)(q-1)J(\chi ,\dots ,\chi )$, with $n$ entries of $\chi$ (same Proposition).

By Theorem 3 of chapter 8,

$$J(\chi ,\dots ,\chi )=\frac{g{(\chi )}^n}{g(\chi )}=g{(\chi )}^{n-1}.$$

Since $g{(\chi )}^2=g(\chi )g{(\chi )}^{-1}=\chi (-1)q$ (Exercise 10.22),

= $\chi (-1)g{(\chi )}^{n+1}$ $g{(\chi )}^2$ = 1 q g ( χ ) n + 1 .

$$$$

Therefore

$$N(a_0{x}_0^2+\cdots +a_n{x}_n^2=0)=q^n+\chi {(a_0)}^{-1}\cdots \chi {(a_n)}^{-1}\frac{q-1}{q}g{(\chi )}^{n+1},$$

and

$$N_1=q^{n-1}+\cdots +q+1+\frac{1}{q}\chi {(a_0)}^{-1}\cdots \chi {(a_n)}^{-1}g{(\chi )}^{n+1}.$$

To conclude this first part,

$$\begin{array}{cc}{N}_1=q^{n-1}+\cdots +q+1\hfill & \text{if }n\text{ is even},\hfill \\ N_1=q^{n-1}+\cdots +q+1+\frac{1}{q}\chi {(a_0)}^{-1}\cdots \chi {(a_n)}^{-1}g{(\chi )}^{n+1}\hfill & \text{if }n\text{ is odd}.\hfill \end{array}$$

To compute $N_s$, we must replace $q$ by $q^s$ and $\chi$ by ${\chi }_s$, the character of order $2$ on $F_s$. Then

$$\begin{array}{cc}{N}_s=q^{s(n-1)}+\cdots +q^s+1\hfill & \text{if }n\text{ is even},\hfill \\ N_s=q^{s(n-1)}+\cdots +q^s+1+\frac{1}{q^s}{\chi }_s{(a_0)}^{-1}\cdots {\chi }_s{(a_n)}^{-1}g{({\chi }_s)}^{n+1}\hfill & \text{if }n\text{ is odd}.\hfill \end{array}$$

(These two results can also be obtained by using the equations (1) and (2) in Theorem 2 of Chapter 10.)

It remains to study ${\chi }_s$ in the odd case.

Since ${\chi }_s^2=𝜀$, for all $\alpha \in F_s$, ${\chi }_s{(\alpha )}^{-1}={\chi }_s(\alpha )$, and ${\chi }_s(\alpha )=-1\in ℂ$ if ${\alpha }^{\frac{q^s-1}{2}}=-1\in F_s,{\chi }_s(\alpha )=1$ otherwise.

If $a\in F$, $a^{\frac{q-1}{2}}=\pm 1=𝜀$. Since $q$ is odd, $1+q+\cdots +q^{s-1}\equiv s(mod2)$, thus

$$a^{\frac{q^s-1}{2}}=a^{\frac{q-1}{2}(1+q+\cdots +q^{s-1})}={𝜀}^{1+q+\cdots +q^{s-1}}={𝜀}^s,$$

so

$${\chi }_s(a)=\chi {(a)}^s(a\in F).$$

We know that $g{({\chi }_s)}^2={\chi }_s(-1)q^s$ (Ex. 10.22), thus, as $n$ is odd,

$$\begin{array}{llll}\hfill g{({\chi }_s)}^{n+1}& ={\left[g{({\chi }_s)}^2\right]}^{\frac{n+1}{2}}& \hfill & \\ \hfill & ={\chi }_s{(-1)}^{\frac{n+1}{2}}{q}^{s\frac{n+1}{2}}.& \hfill & \end{array}$$

If $q\equiv 1(mod4)$, then ${(-1)}^{\frac{q-1}{2}}=1$, so $-1$ is a square in ${𝔽}_q$. In this case, $-1$ is a square in ${𝔽}_{q^s}$, and ${\chi }_s(-1)=1$ for all $s\ge 1$. In this case, using $a_i\in F$,

$$\begin{array}{llll}\hfill N_s& =q^{s(n-1)}+\cdots +q^s+1+{\chi }_s(a_0)\cdots {\chi }_s(a_n)q^{s\frac{n-1}{2}}& \hfill & \\ \hfill & =q^{s(n-1)}+\cdots +q^s+1+{[\chi (a_0)\cdots \chi (a_n)]}^s{q}^{s\frac{n-1}{2}}& \hfill & \end{array}$$

If $q\equiv -1(mod4)$, then $\chi (-1)={(-1)}^{\frac{q-1}{2}}=-1$, and

$$\begin{array}{llll}\hfill {\chi }_s(-1)& =\chi {(-1)}^s={(-1)}^s,& \hfill & \end{array}$$

thus

$$\frac{1}{q^s}g{({\chi }_s)}^{n+1}={(-1)}^{s\frac{n+1}{2}}{q}^{s\frac{n-1}{2}}.$$

This gives for odd integers $n$, and $q\equiv -1(mod4)$,

$$\begin{array}{llll}\hfill N_s& =q^{s(n-1)}+\cdots +q^s+1+{(-1)}^{s\frac{n+1}{2}}{\chi }_s(a_0)\cdots {\chi }_s(a_n)q^{s\frac{n-1}{2}}& \hfill & \\ \hfill & =q^{s(n-1)}+\cdots +q^s+1+{[{(-1)}^{\frac{n+1}{2}}\chi (a_0)\cdots \chi (a_n)]}^s{q}^{s\frac{n-1}{2}}.& \hfill & \end{array}$$

To collect all these cases, we have proved

$$\begin{array}{cc}{N}_s=q^{s(n-1)}+\cdots +q^s+1\hfill & \text{if }n\equiv 0(2),\hfill \\ N_s=q^{s(n-1)}+\cdots +q^s+1+{[\chi (a_0)\cdots \chi (a_n)]}^s{q}^{s\frac{n-1}{2}}\hfill & \text{if }n\equiv 1(2),q\equiv +1(4),\hfill \\ N_s=q^{s(n-1)}+\cdots +q^s+1+{[{(-1)}^{\frac{n+1}{2}}\chi (a_0)\cdots \chi (a_n)]}^s{q}^{s\frac{n-1}{2}}\hfill & \text{if }n\equiv 1(2),q\equiv -1(4).\hfill \end{array}$$

If $n$ is even this gives, as in paragraph 1,

$$Z_f(u)={(1-q^{n-1}u)}^{-1}\cdots {(1-qu)}^{-1}{(1-u)}^{-1}.$$

In the case $n\equiv 1(2),q\equiv +1(4)$, we write for simplicity $𝜀=\chi (a_0)\cdots \chi (a_n)=\pm 1$. Then

$$\begin{array}{llll}\hfill \underset{s=1}{\overset{\infty }{\sum }}\frac{N_s{u}^s}{s}& =\underset{m=0}{\overset{n-1}{\sum }}\left(\underset{s=1}{\overset{\infty }{\sum }}\frac{{(q^{m}u)}^s}{s}\right)+\underset{s=1}{\overset{\infty }{\sum }}\frac{{(𝜀q^{\frac{n-1}{2}}u)}^s}{s}& \hfill & \\ \hfill & =-\underset{m=0}{\overset{n-1}{\sum }}\ln <!--FUNCTION\; APPLICATION-->(1-q^{m}u)-\ln <!--FUNCTION\; APPLICATION-->(1-𝜀q^{\frac{n-1}{2}}u).& \hfill & \end{array}$$

Therefore

$$Z_f(u)=\left[\underset{m=0}{\overset{n-1}{\prod }}{(1-q^{m}u)}^{-1}\right]{(1-\chi (a_0)\cdots \chi (a_n)q^{\frac{n-1}{2}}u)}^{-1}.$$

(Same calculation in the last case, with $𝜀={(-1)}^{\frac{n+1}{2}}\chi (a_0)\cdots \chi (a_n)$.)

We obtain

$$\begin{array}{cc}{Z}_f(u)=P(u)\hfill & \text{if }n\equiv 0(2),\hfill \\ Z_f(u)=P(u){(1-\chi (a_0)\cdots \chi (a_n)q^{\frac{n-1}{2}}u)}^{-1}\hfill & \text{if }n\equiv 1(2),q\equiv +1(4),\hfill \\ Z_f(u)=P(u){(1-{(-1)}^{\frac{n+1}{2}}\chi (a_0)\cdots \chi (a_n)q^{\frac{n-1}{2}}u)}^{-1}\hfill & \text{if }n\equiv 1(2),q\equiv -1(4),\hfill \end{array}$$

where $P(u)={(1-q^{n-1}u)}^{-1}\cdots {(1-qu)}^{-1}{(1-u)}^{-1}.$

(These results are consistent with the example $N_s=q^{2s}+q^s+1+{\chi }_s(-1)q^s$ given in paragraph 1 for the surface defined by $-y_0^2+y_1^2+y_2^2+y_3^2=0$, where $n=3$ is odd.

$$\begin{array}{llll}\hfill Z_f(u)& ={(1-q^{2}u)}^{-1}{(1-qu)}^{-1}{(1-u)}^{-1}{(1-\chi (-1)qu)}^{-1}& \hfill & \\ \hfill & =\{\begin{array}{cc}{(1-q^{2}u)}^{-1}{(1-qu)}^{-2}{(1-u)}^{-1}\hfill & \text{if }q\equiv 1(mod4),\hfill \\ {(1-q^{2}u)}^{-1}{(1-qu)}^{-1}{(1-u)}^{-1}{(1+qu)}^{-1}\hfill & \text{if }q\equiv -1(mod4).\hfill \end{array}& \hfill & \end{array}$$ □