Exercise 11.13

Proof. Here $p\equiv 1(\mathit{mod}4)$, thus $p^s\equiv 1(\mathit{mod}4)$. We consider here the two fields $F={𝔽}_p$ and $F_s={𝔽}_{p^s}$, where $|F|=p$ and $|F_s|=p^s$.

Let ${\rho }^{\prime }=\rho \circ N_{F_s∕F}$, and ${\chi }^{\prime }=\chi \circ N_{F_s∕F}$. The results of ¤3 show that the map $\xi \mapsto {\xi }^{\prime }=\xi \circ N_{F_s∕F}$ induces a group isomorphism between the group cyclic $C_n$ of characters on $F$ whose order divides $n$ on the group cyclic $C_n^{\prime }$ of characters on $F_s$ whose order divides $n$ (see Exercise 16). Thus the order of ${\rho }^{\prime }$ is $2$ and the order of ${\chi }^{\prime }$ is $4$, and ${\chi {}^{\prime }}^2={\rho }^{\prime }$.

Replacing $\chi ,\rho$ by ${\chi }^{\prime },{\rho }^{\prime }$, and $p$ by $p^s$, we obtain by the same reasoning that the number of projective point of $C_1$ in ${\bar{H}}_f(F_s)$ is

To compute ${\chi }^{\prime }(-4)$ and ${\chi }^{\prime }(D)$ we use the property (c) of ¤3. Since $-4$ and $D$ are in $F$,

Therefore

It remains to compute $J({\chi }^{\prime },{\chi {}^{\prime }}^2)$. Since ${\chi }^{\prime }\ne 𝜀,{\chi {}^{\prime }}^2\ne 𝜀,{\chi {}^{\prime }}^3\ne 𝜀$,

The Hasse-Davenport relation gives $g({\chi {}^{\prime }}^k)=-{(-g({\chi }^k))}^s$, thus

where $\pi =-J(\chi ,{\chi }^2)\in \mathbb{Z}[i]$. To conclude,

Then Exercise 2 gives

Since $|\pi {\text{|}}^2=|J(\chi ,{\chi }^2){\text{|}}^2=p$ (corollary of Theorem 1, chapter 8), expanding the numerator, we obtain

Note: Since $Z_f(u)=\exp <!--\; FUNCTION\; APPLICATION\; -->(N_{1}u+\cdots )=1+N_{1}u+\cdots$, and

the comparison of the coefficient of $u$ in the two power series gives

This gives anew $a=-\mathrm{tr}\left(\overline{\chi (D)}\pi \right)$. □