We know, thanks to A. Weil, that such a function is rational, and the numerator has all of its zeros on the circle $z \overline{z} = q,$ where $q$ is the order of the field. The question is: can *every* polynomial in $\mathbb{Z}[x]$ with the zeros on the circle arise in this way? Is there some obvious and/or conjectural obstruction?



Inverse problem for zeta functions of curves over finite fields
Tate and Honda show that almost all polynomials like that are the L-function of an abelian variety over the finite field. The problem with curves is much harder and it's open (for genus g>2). One necessary condition is that the expression giving the number of points on the curve ($q+1$ minus sum of zeros) has to be positive. This is not true for every polynomial with your conditions, so the answer to your first question is no. That's an obstruction but not the only one.


For genus 2, see <http://arxiv.org/abs/math/0607515> 


