"Inverse problem" for the zeta function [duplicate]

Question

Let $C$ be a smooth, projective, geometrically irreducible curve, of genus $g$, over a finite field $\mathbb{F}_q$. By the Weil conjectures, the zeta function has the shape $$ Z_C(t)=\frac{P(t)}{(1-t)(1-qt)} $$ with $P \in 1+t\mathbb{Z}[t]$ a polynomial of degree $2g$ satisfying the functional equation and such that $|\alpha|=q^{-1/2}$ for all complex $\alpha$ such that $P(\alpha)=0$.

Now consider the "inverse problem", that is, take a polynomial $P$ with all the above properties. Does there exist a curve $C$ such that $P$ is the numerator of $Z_C$? My guess is that the answer is either no for some trivial reason that I don't see or it is not known.

Answer 1

The Honda-Tate theorem says that there exists some abelian variety $A$ over $\mathbb{F}_q$ such that $P(t)$ occurs in the zeta function of $A$, coming from the part corresponding to $H^1(A,\mathbb{Q}_\ell)$. Moreover $A$ is determined up to isogeny by $P(t)$. However $$H^1(C,\mathbb{Q}_\ell) \cong H^1(J(C),\mathbb{Q}_\ell)$$ where $J(C)$ denotes the Jacobian of $C$. A simple counting argument shows that not every abelian variety is isogenous to the Jacobian of some curve, which implies that the answer to your question is no.