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I want to know how to apply Deligne's theorem to estimate exponential sums. Let $k$ be the finite field of order $q=p^k$ and $f(x,y)\in k[x,y]$ a polynomial. Deligne's theorems imply that $$ \sum_{(x,y) \in k^2} e^{\frac{2\pi i f(x,y)}{q}} \le C_f q, $$ under certain conditions, where $C_f$ depends on $f$. What are these 'certain conditions' in the simplest terms? For example if I chose a random polynomial such as $f(x,y)= x^4y^3+2x^2y^2+3x$, and $p>10$ say, then how can I check the conditions of Deligne's theorem?

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    $\begingroup$ Katz and Adolphson-Sperber have produced a huge number of "ready-to-use" versions of Deligne's general machine. Look up their papers, or start with the nice survey chapter in Iwaniec and Kowalski's book. $\endgroup$ Apr 20, 2011 at 14:36
  • $\begingroup$ Perhaps the following paper is helpful. Katz-Fouvry: A general stratification theorem for exponential sums, and applications, J. reine angew. Math. 540 (2001), 115-166. Note also that often you can get weaker but still nontrivial upper bounds with more elementary tools. Deligne's bound is kind of optimal when it works. $\endgroup$
    – GH from MO
    Apr 20, 2011 at 16:42

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Theorem 8.4 of Deligne's paper (Weil I) gives what you want with $C_f=(d-1)^2, d = \deg f$, provided the homogeneous part of degree $d$ of $f$ defines a "smooth variety" in $\mathbb{P}^1$, which (in our case of $n=2$) simply means that the homogeneous part of degree $d$ is a product of $d$ distinct linear factors. Unfortunately, in your example, the term of degree $7$ is $x^4y^3$, is not a product of distinct linear factors.

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  • $\begingroup$ Thank you... since it is not smooth at x=0 and y=0, which define sets of dimension q, are there ways to ignore these sets and still get the desired estimate of Deligne? $\endgroup$
    – Jussmar
    Apr 20, 2011 at 15:18
  • $\begingroup$ Is q meant to be 0? The non-smooth locus in this situation will always have dimension zero, so I don't think they can be ignored. It's possible that there might be some other trick. $\endgroup$ Apr 20, 2011 at 16:17

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