Exercise 11.10

Proof.

Here we compute the zeta function of the projective closure ${\bar{H}}_f(F_q)$, with equation $f(x,y,t)=y^{2}t=x^3+A{t}^3$. If $t=0$, then $x=0$, thus there is only one point $[0,1,0]$ at infinity (over $F_q$ or over $F_{q^s}$).

We assume that the characteristic is not 2. Then $q$ is odd, and so $q\equiv 1(\mathit{mod}6)$. Therefore, there are characters of order $2$ and $3$ on $F_q$. Write $\rho$ the unique character of order $2$, and write $\chi$ a character of order $3$. As $\chi$ is a character of order 3, the characters whose order divides 3 are $𝜀,\chi ,{\chi }^2$.

We compute first $N_1$. We write $N(y^2=x^3+A)$ for the number of points of the affine cubic over $F_q$, and $N_1$ for the number of points of the projective cubic, so that $N_1=N(y^2=x^3+A)+1$. We recall the results obtained in Ex. 8.15.

The map $x\mapsto -x$ is a bijection between the set of roots of $x^3=b$ and the set of roots of ${(-x)}^3=b$, so $N(x^3=b)=N({(-x)}^3=b)=N(x^3=-b)$.

Using Prop. 8.1.5, we obtain, since $A\ne 0$,

We know (generalization of Theorem 1, Chapter 8) that $J(\chi ,𝜀)=J({\chi }^2,𝜀)=J(𝜀,\rho )=0,$ and $J(𝜀,𝜀)=q$, so

As ${\chi }^2(A)={\chi }^{-1}(A)=\overline{\chi (A)}$, and as $\overline{\rho (A)}=\rho (A)$, then $J({\chi }^2,\rho )=J(\bar{\chi },\bar{\rho })=\overline{J(\chi ,\rho )}$, and

therefore

Since the orders of $\chi ,\rho$, and $\mathit{\chi \rho }$ are $3,2$ and $6$, $\chi \ne 𝜀,\rho \ne 𝜀,\mathit{\chi \rho }\ne 𝜀$, thus Theorem 1 of Chapter 6 gives

Write ${\chi }^{\prime }=\chi \circ N_{F_{q^s}∕F_q},{\rho }^{\prime }=\rho \circ N_{F_{q^s}∕F_q}$. Then ${\chi }^{\prime },{\rho }^{\prime }$ are characters on $F_{q^s}$, and the orders of ${\chi }^{\prime },{\rho }^{\prime }$ are $3$ and $2$ (by properties (a), (b) of ¤3). The same reasoning in $F_{q^s}$ gives

Since $A\in F_q$, the property (c) of ¤3 gives ${\chi }^{\prime }(A)=\chi {(A)}^s,{\rho }^{\prime }(A)=\rho {(A)}^s$. Using the Hasse-Davenport Relation, and ${(\mathit{\chi \rho })}^{\prime }={\chi }^{\prime }{\rho }^{\prime }$, we obtain

This gives $N_s$ in the appropriate form:

Using the converse to Proposition 11.1.1 given in Exercise 2, we obtain

Note that $\pi \bar{\pi }=|\pi {\text{|}}^2=q$ (by Exercise 10.22). Expanding the numerator, this gives

where $a=\pi +\bar{\pi }$.

For all $t\in F_q^{\text{∗}}$, ${\chi }^3(t)=1$, thus $\chi (t)\in \{1,\omega ,{\omega }^2\}\subset \mathbb{Z}[\omega ]$, and $\rho (t)=\pm 1$, therefore $\pi =\rho (A)\chi (A){\sum <!--\; FUNCTION\; APPLICATION\; -->}_{t\in F_q}\chi (t)\rho (t)\in \mathbb{Z}[\omega ]$. Writing $\pi =u+\mathit{v\omega },u,v\in \mathbb{Z}$, we obtain $a=\pi +\bar{\pi }=2u-v\in \mathbb{Z}$.

Moreover,

To conclude,

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