Numerator of zeta $Z(C/\Bbb{F}_q, T)$ and Cayley-Hamilton theorem and Frobenius

Question

Let $C$ be a genus $g$ curve. I know numerator of congruent zeta function $Z(C/\Bbb{F}_q, T)=\sum \frac{\sharp C(\Bbb{F}_{q^n})x^n}{n}$ is product of characteristic polynomial of Frobenius, $det(1-Frob・T)$. So by Cayley-Hamilton theorem, $Frob$ is root of numerator of $Z(C/\Bbb{F}_q, T)$. But for example, let $C$ be $y^5=-x^2+x$. $Z(C/\Bbb{F}_2, T)=\frac{1+4T^4}{(1-T)(1-2T)}$, Frobenius is root of $X^4+4$, not $1+4T^4$'s. Where am I misunderstood ? Why Frobenius is root of $T→1/T$ and times $T^4$ 's polynomial ?

Answer 1

If $\phi$ is an endomorphism of a finite-dimensional vector space, there are two different definitions of characteristic polynomial that show up in the literature: $\text{det}(I - \phi x)$ and $\text{det}(xI - \phi)$.

It would be nice if there were better language around this, e.g. calling the first of these the co-characteristic polynomial, since it's "co-monic".

The fact about the Zeta function you are using is normalized for the "co-characteristic" polynomial (this seems to be standard when people are talking about zeta functions), but your appliction of Cayley-Hamilton is normalized for the characteristic polynomial. Hence the discrepancy.