Could the Weil zeroes of curves be evenly distributed?

Question

If $X$ is a smooth, geometrically connected, projective curve of genus $g$ over $\mathbb{F}_q$, then the zeta function of $X$ is of the form $P(s)/(1 - s)(1 - qs)$, where $P(s)$ is a polynomial of degree $2g$. Denote the $P(s)$ of $X$ by $P_X$.

By the Riemann hypothesis for curves, the zeroes of $P_X$ all have complex absolute value $1/\sqrt q$.

Question: Is it possible to find a family of curves $\{X_n\}_{n=1}^{\infty}$ with $g(X_n) \rightarrow \infty$ ($g(X_n)$ is the genus of $X_n$) such that the zeros of $P_{X_n}$ tends to being evenly distributed, i.e. if $Y_n$ denotes the set of phase angle differences (normalized into $[0,2\pi)$) of consecutive zeros of $P_{X_n}$, then $\max{Y_n}-\min{Y_n} = o(1/g(X_n))$?

Answer 1

If $q$ is a prime for which $2$ is a primitive root, then I claim that the Frobenius eigenvalues of the curve $y^2-y = x^q$ over $\mathbb{F}_2$ have spacing exactly $\tfrac{2 \pi}{q-1}$. This, if Artin's conjecture holds, such examples exist over $\mathbb{F}_2$.

Let $C$ be the affine curve $y^2-y = x^q$ over $\mathbb{F}_2$. We note that $C$ is smooth, by taking the partial derivative with respect to $y$; that $C$ has one point at $\infty$ and that $C$ has genus $\tfrac{q-1}{2}$. So the number of points on $C$ over $\mathbb{F}_{2^k}$ is $$2^k - \sum_{i=1}^{q-1} \lambda_i^k$$ where $\lambda_i$ are the eignvalues of Frobenius. To show that the $\lambda_i$ are spaced as claimed, we must show that $\#C(\mathbb{F}_{2^k}) = 2^k$ for $1 \leq k < q-1$.

The hypothesis that $2$ is a primitive root modulo $q$ means that $x \mapsto x^q$ is bijective on $\mathbb{F}_{2^k}$ for $k$ not divisible by $q-1$. So, for each $y \in \mathbb{F}_{2^k}$, there is exactly one solution to $y^2-y = x^q$ in $\mathbb{F}_{2^k}$. $\square$

The same argument works if $p$ is an odd prime which is a primitive root for infinitely many $q$ and we consider the hyperelliptic curves $y^2 = x^q-1$ over $\mathbb{F}_p$.

Answer 2

For $q$ odd, and any natural number $n$, let $\alpha$ be an element of $\mathbb F_{q^n}$ of order $q^n-1$ and let $ \sum_{i=0}^n a_i x^i$ be the minimal polynomial of $\alpha$.

Then I think the hyperelliptic curve $$ y^2 = \sum_{i=0}^n a_i x^{q^i}$$ has $q^n-1$ zeroes, all evenly spaced.

It certainly has $q^n-1$ zeroes since it has genus $\frac{q^n-1}{2}$ by Riemann-Hurwitz. To check they are evenly spaced, we must check it has exactly $q^d+1$ rational points over $\mathbb F_{q^d}$ whenever $d$ is not a multiple of $q^n-1$.

To do this, note that, over any field $\mathbb F_{q^d}$ where $ \sum_{i=0}^n a_i x^{q^i}$ lacks nonzero roots, it is an additive homomorphism with trivial kernel, hence bijective, so $y^2= \sum_{i=0}^n a_i x^{q^i}$ has exactly $q^d$ solutions, plus the one at $\infty$. So it suffices to show $\sum_{i=0}^n a_i x^{q^i}$ has no nonzero roots over these fields.

If $x$ is a nonzero root in $\mathbb F_{q^d}$ then the Frobenius map $F \colon x \to x^q$ is a $\mathbb F_q$-linear endomorphism of the vector space of roots in $\mathbb F_q^d$, and we have $\sum_{i=0}^n a_i F^i =0$ as an endomorphism of the vector space of roots, so the eigenvalues of Frobenius are conjugates of $\alpha$ and have order $q^{n}-1$, so Frobenius has order a multiple of $q^{n}-1$, and thus $d$ is divisible by $q^n-1$, as desired.