# Riemann hypothesis for exponential sum

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.

• 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". Aug 12, 2020 at 5:47
• @FelipeVoloch Thanks, I think I get the idea. Aug 12, 2020 at 13:37

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.