[Math] cheap proof of power savings for exponential sums over finite fields

exponential-sumsnt.number-theoryweil-conjectures

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.)

Best Answer

For your main question, see L. J. Mordell, On a sum analogous to a Gauss's sum, Quart. J. Math. Oxford Ser. 3 (1932), 161–167. It is a similar trick as in your 2.

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.

Edit: For an explicit statement see L. Carlitz and S. Uchiyama Duke Math. J. 24, (1957), 37-41.

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