Recently I've heard about the Riemann hypothesis for one-variable exponential sums, which states as

For a polynomial $f\in\mathbb{F}_{p^k}[x]$ of degree $d$ and a character $\chi$ of $(\mathbb{F}_{p^k},+)$, provided $(d,p)=1$, we have$$\left|\sum_{x\in\mathbb{F}_{p^k}}\chi(f(x))\right|\le(d-1)\sqrt{p^k}$$

I also know the $L$-function associated to $f$ is$$L(f,T)=\exp\left(\sum_{n\ge1}S_n(f,\chi)\frac{T^n}{n}\right)$$ where $S_n(f,\chi)=\sum_{x\in\mathbb{F}_{p^{kn}}}\chi(\operatorname{tr}_{\mathbb{F}_{p^{kn}}/\mathbb{F}_{p^{k}}}(f(x)))$.

My question is, what is the relation of the Riemann hypothesis for one-variable exponential sums, and Riemann hypothesis for the associated $L$-function? I guess they are equivalent forms, but I can't prove it.

My thought: This is similar to the relation in the elliptic curve version. The Hasse theorem

For an elliptic curve $E$ over $\mathbb{F}_p$, $|\#E(\mathbb{F}_p)-p-1|\le 2\sqrt p$.

is equivalent to the Riemann hypothesis for elliptic curve:

The zeros of $\zeta(E,s)$ has real part $\frac1{2}$.

This equivalence is due to computing the zeta function $\zeta(E,s)$, involving Riemann-Roch theorem. But as for the exponential sum case, I have no idea how to compute the $L$-function associated to $f\in\mathbb{F}_{p^k}[x]$.

Any help will be appreciated.

  • 7
    $\begingroup$ You need to change $T$ to $q^{-s}$ to relate to the real part of the zeros. $L(f,T)$ is a factor of the zeta function of the Artin-Schreier curve $y^q-y=f(x)$. See, e.g., Weil "Basic Number Theory". $\endgroup$ Aug 12, 2020 at 5:47
  • $\begingroup$ @FelipeVoloch Thanks, I think I get the idea. $\endgroup$ Aug 12, 2020 at 13:37

1 Answer 1


One can show by a nontrivial but elementary argument that $L(f,T)$ is a polynomial in $T$ of degree $d-1$.

The Riemann hypothesis in this case says all zeroes of $L(f,T)$ have absolute value $p^{-k/2}$.

It follows from the Riemann hypothesis that we can write $L(f,T) = \prod_{i=1}^{d-1} (1 - \alpha_i T)$ where $|\alpha_i|= p^{k/2}$. The bound $S_n(f,x) \leq (d-1) \sqrt{p^{kn}}$ follows from this by taking logs.

Conversely, if we have the bound $S_n(f,x) \leq (d-1) \sqrt{p^{kn}}$ for all $n$, we can check that the roots have absolute value $\geq p^{-k/2}$ using the radius of convergence for the power series, and then check that they have absolute value $p^{-k/2}$ by using the functional equation. Unlike in the elliptic curve case, we need all $n$ here instead of just one.


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