Can I calculate congruent zeta function of given hyperelliptic curve by hand?

Question

How can I calculate the numerator of congruent zeta function of given hyperelliptic curve ?

For example, let $C:y^2=(x^2+1)(x^4-8x^3+2x^2+8x+1)$. numerator of congruent zeta function mod$23$ of this is known to be $1+46X^2+529X^4$.

My text reads ''Counting up rations points of the reduction of $C$ at $23$, we obtain the coefficients of numerator of congruent zeta function'', but I don't come up with good idea to calculate it by hand.

If I could specify Frobenius of $J(C)$, we obtain its character polynomial, and numerator of congruent zeta, but this takes much efforts.

Answer 1

Well, I am such an outsider to this field that I hesitate to reveal how fragmentary my knowledge is. So fragmentary, in fact, that I can only tell you the sketchiest facts, without any explanation at all.

For a nonsingular curve of genus $g$ over $\Bbb F_q$, the Zeta function has the form $$ Z(u)=\frac{1+c_1u+c_2u^2+\cdots +q^{g-2}c_2u^{2g-2} +q^{g-1}c_1u^{2g-1} +q^gu^{2g}}{(1-u)(1-qu)}\,, $$ so that for $0\le n\le g$, the coefficient of $u^{2g-n}$ is $q^{g-n}$ times the coefficient of $u^n$. In particular, for your genus-two curve over $\Bbb F_{23}$, the Zeta function must look like $$ Z(u)=\frac{1+c_1u+c_2u^2+23c_1u^3+23^2u^4}{(1-u)(1-23u)}\,. $$ Now, to attend to your purposes, the logarithmic derivative of $Z$, that is $Z'/Z$, will expand as a series $N_1+N_2u+N_3u^2+\cdots$, where $N_{m-1}$ is the number of rational points of your curve over $\Bbb F_{q^m}$. Thus all you need to know for your curve over $\Bbb F_{23}$ are $N_0$ and $N_1$.