# Riemann hypothesis for zeta function of definable sets over finite fields

Hi,

Consider the zeta function of a definable set over a finite field. More precisely, let $\varphi$ be a formula in the language of fields and let $X$ be a definable subset of $F^n$ given by $X=\{(a_1,\dots,a_n) \in F^n| F\models \varphi(a_1,\dots,a_n) \}$ where $F$ is a finite field of characteristic $p$, then $$Z_X(t)= \exp \sum_{i=1}^ \infty \frac{N_i(X)}{i} t^i,$$ where $N_i(X)= card \{ (a_1,\dots,a_n) \in F_i^n| F_i \models \varphi[a_1,\dots,a_n]\}$ with $F_i$ is the unique extension of $F$ of degree $i$.

My question is:

Does the Riemann hypothesis for $Z_X(t)$ hold, i.e. if we make the change of variables $t=q^{-s}$ does it follow that all zeroes of $Z_X(q^{-s})$ lie on the line $Re(s)=1/2$?

Thank you

• It seems to me that $Z_X(t)$ is not well-defined. A single set $X\subseteq F^n$ could have several equivalent definitions over $F$ that become inequivalent over extension fields. So "the formula defining $X$" is ambiguous. – Andreas Blass Jul 29 '12 at 22:19

The zeta function of the set $\{x|x \neq 0\}$ over $\mathbb F_p$ is:
$e^{\sum_{i=1}^\infty \frac{p^i-1}{i} t^i} = \frac{1-t}{1-pt}=\frac{1-p^{-s}}{1-p^{1-s}}$
It has a zero where $s=0$.
More generally, every affine algebraic curve minus a point has a zeta function that is zero at $s=0$.