Exercise 11.8

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

If $f$ is a nonhomogeneous polynomial, we can consider the zeta function of the projective closure of the hypersurface defined by $f$ (see Chapter 10). One way to calculate this is to count the number of points on $H_f(F_q)$ and then add to it the number of points at infinity. For example, consider $y^2 = x^3$ over $F_{p^s}$. Show that there is one point at infinity. The origin $(0,0)$ is clearly on this curve. If $x 
e 0$, write $(y/x)^2 = x$ and show that there are $p^s$ more points on this curve. Altogether we have $p^s$ points and the zeta function over $F_p$ is $(1-pu)^{-1}$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Consider the polynomial $f(x,y) = y^2 - x^3$ and $g(x,y) = y^2 - x$, and
\begin{align*}
\Gamma =H_f(F_q) &= \{(x,y) \in F_p^2 \mid y^2 = x^3\},\
\Gamma_1 = H_g(F_q) &= \{(x,y)\in F_q^2 \mid y^2 = x\}.
\end{align*}
Then 
$$
\varphi
\left\{
\begin{array}{ccl}
\Gamma \setminus \{(0,0)\}  & 	o &\Gamma_1\setminus \{(0,0)\} \
(x,y) & \mapsto & \left(x,\frac{y}{x} ight)
\end{array}
ight.
$$
is defined, since $\left(\frac{y}{x} ight) ^2 = x$ for $(x,y) \in \Gamma \setminus \{(0,0)\}$, thus $ \left(x,\frac{y}{x} ight) \in \Gamma_1$.
Moreover
$$
\psi
\left\{
\begin{array}{ccl}
 \Gamma_1\setminus \{(0,0)\} & 	o & \Gamma \setminus \{(0,0)\} \
(x,y) & \mapsto & \left(x,xy ight)
\end{array}
ight.
$$
is correctly defined, since for each $(x,y) \in \Gamma_1\setminus \{(0,0)\}, y^2 = x$, then $x 
e 0$, thus $(xy)^2 =x^3$, and $(x,xy) \in \Gamma$, where $(x,xy) 
e (0,0)$.

Moreover $\psi$ satisfies $\psi \circ \varphi = \mathrm{id}, \varphi \circ \psi = \mathrm{id}$:
\begin{align*}
(\psi \circ \varphi)(x,y) &= \psi\left(x,\frac{y}{x} ight) =\left (x, x \frac{y}{x}ight) = (x,y) \qquad ((x,y) \in \Gamma \setminus \{(0,0)\}),\
(\varphi \circ \psi)(x,y) &= \varphi(x,xy) = \left( x, \frac{xy}{x} ight) = (x,y) \qquad ((x,y) \in \Gamma_1 \setminus \{(0,0)\}).
\end{align*}
So $\varphi$ is a bijection. This shows that $| \Gamma \setminus \{(0,0)\} | =  |\Gamma_1\setminus \{(0,0)\}|$, where $(0,0) \in \Gamma$ and $(0,0) \in \Gamma_1$, thus
$$|\Gamma_1| = |\Gamma|.$$ 
To count the points on $\Gamma_1$, we consider
$$\lambda
\left\{
\begin{array}{ccl}
F_q & 	o & \Gamma_1\
y & \mapsto & (y^2,y).
\end{array}
ight.
$$
Then $\lambda$ is bijective, with inverse $\mu : (x,y) \mapsto y$. This show that $$|\Gamma| = |\Gamma_1| = q = p^s.$$

Therefore the zeta function of the affine curve $y^2 = x^3$ over $F_p$ is 
$$Z_f(u) = (1-pu)^{-1}.$$

But the projective closure $H_{\overline{f}}(F_q)$ of this curve has $p^s + 1$ points, with only one point at infinity, since $ty^2 = x^3$ has only one point $[t,x,y]$ satisfying $t= 0$, the point $[0,0,1]$.

The zeta function of the curve with homogeneous equation $\overline{f}(t,x,y) = ty^2 - x^3$ over $F_p$ is 
$$Z_{\overline{f}}(u) = (1-u)^{-1} (1-pu)^{-1}.$$
\end{proof}

Exercise 11.8

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Exercise 11.8

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