# Is there a cheap proof of power savings for exponential sums over finite fields?

Let $p$ be a large prime, and let $f(x) = P(x)/Q(x)$ be a non-constant rational function over ${\Bbb F}_p$ of bounded degree. From the Weil conjectures for curves, we have a bound of the form

$$|\sum_{x \in {\Bbb F}_p}^* e_p( f(x) )| \ll p^{1/2}$$

where the asterisk denotes that the summation excludes those $x$ for which $Q(x)$ vanishes, and $e_p(x) := e^{2\pi i x/p}$ is the standard additive character. (As far as I can tell, this particular bound first appeared explicitly in this paper of Perelmuter; an elementary proof based on Stepanov's method, though still using the rationality of the zeta function, also appears in this paper of Cochrane and Pinner.)

My question is whether there is a "cheap" argument, avoiding the Weil conjectures (i.e. not using either the Riemann hypothesis or rationality of the zeta function) that gives a weaker power savings bound

$$|\sum_{x \in {\Bbb F}_p}^* e_p( f(x) )| \ll p^{1-c}$$

over the trivial bound for some $c>0$ (which is now allowed to depend on the degree of the polynomials $P,Q$)? I know of two special cases in which this is possible:

1. If $f$ is a polynomial, then one can use the Weyl differencing method to obtain a bound of this form (with $c$ decaying exponentially in the degree of $f$).
2. In the case of Kloosterman sums $f(x) = ax + \frac{b}{x}$, Kloosterman obtained a power savings of $c=1/4$ by computing (or more precisely, upper bounding) the fourth moment $$\sum_{a,b \in {\Bbb F}_p} |\sum_{x \in {\Bbb F}_p}^* e_p( ax + \frac{b}{x} )|^4$$ and exploiting the dilation symmetry $$\sum_{x \in {\Bbb F}_p}^* e_p( ax + \frac{b}{x} ) = \sum_{x \in {\Bbb F}_p}^* e_p( cax + \frac{b/c}{x} )$$ for any $c \in {\Bbb F}_p^\times$. (EDIT: as pointed out by Felipe, Mordell extended this argument to the case when $f$ is a linear combination of monomials $x^n$. For Mordell-type sums there are also recent papers of Bourgain and co-authors using the sum-product phenomenon to get power savings estimates in certain high-degree cases, but these techniques rely again on multiplicative structure and do not seem to be available in general.)

However, in the positive genus case it does not appear that Weyl differencing can be used to reduce the complexity of the exponential sum even if combined with changes of variable (although I do not have a rigorous proof of this assertion), while I have been unable to extract a power saving from the Kloosterman argument in the absence of a symmetry such as dilation symmetry (although the argument does give the very weak upper bound of $(1-c)p$ for some $c>0$ in general). The only other arguments I know of go through the rationality of the zeta function, and in particular on linking the above exponential sum to the companion sums

$$\sum_{x \in {\Bbb F}_{p^n}}^* e_p( \operatorname{Tr}(f(x)) )$$

for large $n$, the key point being that one can now lose powers of $p$ in estimates on these sums as they can be recovered using the tensor power trick. However it does require a little bit of computation to attain this rationality (e.g. the Riemann-Roch theorem, or a bit of messy linear algebra, as done in Cochrane and Pinner), and I would be interested to know if there was a more direct proof of a power saving (or even a qualitative improvement $o(p)$ over the trivial bound of $p$) that did not require the introduction of the companion sums.

(My motivation here is to explore the minimal background needed to establish Zhang's recent theorem on bounded gaps between primes; currently, the argument requires the Weil conjectures for curves, but a cheap power savings for general rational function phases would eliminate the dependence on these conjectures.)

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Cochrane and Zheng's survey article math.ksu.edu/~cochrane/research/ill00p.pdf seems relevant. In particular see section 6. I think all of the results stated there require n>1, but the references may be worth looking at. –  Mark Lewko Jul 30 '13 at 21:31
It looks like Cochrane and Zheng are not quite studying the companion sums, but rather the sums over the cyclic groups ${\bf Z}/p^n {\bf Z}$ for $n>1$, where Hensel's lemma becomes available. Unfortunately this doesn't seem to say much about the ${\bf Z}/p{\bf Z}$ case... –  Terry Tao Jul 30 '13 at 21:41
Maybe it is better to replace "these conjectures" with "these theorems" at the end of the final parenthetical. :) –  user36938 Jul 31 '13 at 1:32
Cochrane and Pinner aren't assuming rationality of the $\zeta$ function, they are proving it. Lemma 2.1 of their paper proves polynomiality of the $L$-function in the case they need. This argument is more concrete and requires less background than rationality of the $\zeta$ function for a standard curve. –  David Speyer Jul 31 '13 at 2:11
What is going on secretly is the following: We are dealing with the $\zeta$ function of $y^p-y = P(x)/Q(x)$, which is an abelian extension of $\mathbb{F}_p(x)$. By class field theory, we can write the $L$-function as $\sum_{f \in \mathbb{F}_p[x]/\mathbb{F}_p^{\times}} \psi(f) T^{\deg f}$ where $\psi$ depends only on $f$ modulo a conductor $\mathfrak{m}$. In this specific case, one can do this without class field theory. Once $\deg f$ is large enough, $f$ takes on every value modulo $\mathfrak{m}$ equally often and the higher degree terms drop out. See mathoverflow.net/questions/14627 . –  David Speyer Jul 31 '13 at 2:21

I am sure that Hasse (in the 30's) stated the fact that RH for the curve $y^p-y = f(x)$ implies the bound with $p^{1/2}$. Weil didn't feel he needed to bother to state that as a consequence of his result.
Thanks for the references! But it appears that Mordell's paper only handles the polynomial case (where again there is a dilation symmetry after splitting into monomials), or a bit more generally the case when the denominator $Q(x)$ is monomial (so that $f$ splits into monomials $x^n$ for integer $n$). I don't see how to handle, for instance, $\sum_{x \in {\Bbb F}_p}^* e_p( \frac{1}{x(x+1)} + x)$ by these methods. (Incidentally, the Carlitz-Uchiyama paper also only explicitly mentions the polynomial case, although in this case the extension to the rational setting is routine.) –  Terry Tao Jul 30 '13 at 20:24
@joro I meant $O(p^{1/2})$. The constant is not $1$. –  Felipe Voloch Aug 4 '13 at 13:33