Exercise 11.9

Proof. The curve $\Gamma$ defined by the equation $y^2=x^3+x^2$ has a singularity at the origine, as in the previous exercise. The same method applies here: if we use $z=y∕x$, then $z^2=x+1$.

Watch out! Here there are two points $(x,z)\in {\Gamma }_1$ such that $x=0$, the points $(0,1)$ and $(0,-1)$ (here we assume that $p\ne 2$). The curve ${\Gamma }_1$ defined by the equation $z^2=x+1$ is such that

is bijective, thus $|\Gamma |=|{\Gamma }_1|-1$. Since each point of ${\Gamma }_1$ is determined by its coordinate $z$, $|{\Gamma }_1|=q=p^s$, and $|\Gamma |=p^s-1$.

Therefore the zeta function of the affine curve $y^2=x^3+x^2$ over $F_p$ is

There is only one point $p$ at infinity, given by $y^{2}t=x^3+x^{2}t,t=0$, i.e. $p=[0,0,1]$. Thus $N_s=p^s$, and the zeta function of the projective closure of $\Gamma$ is

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