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I asked this question on Mathematics stack exchange but didn't get a response, so I ask here too.

Let $\chi$ be the non-trivial quadratic character of $\mathbb{F}_q$, and let $f(x)$ be a square-free polynomial over $\mathbb{F}_q[x]$. Then by the Weil bound, we have the generic estimate $|\sum_{x\in\mathbb{F}_q}\chi(f(x))|\le \deg(f)q^{1/2}$. Say we now deal with a function $f(x,y)\in \mathbb{F}_q[x,y]$, so that we look for a bound on $|\sum_{x,y\in\mathbb{F}_q}\chi(f(x,y))|$. Do we have any similar generic bound, perhaps using the full power of the Weil conjectures? This is basically the number of points on the surface $z^2=f(x,y)$, but I don't know how to use the Weil conjectures to bound the number of points on this surface.

One could approach this by fixing $x$ and showing that the resulting polynomial in $y$ is generically squarefree (or not a square). This gives a bound of order $q^{3/2}$ if works. I wonder if there's some tighter bound, probably with some extra demands on the polynomial $f$. On M.SE., Jyrki Lahtonen suggested that it has something to do with cohomology of the surface. If someone has any good reference to the connection, this would be very helpful too.

Thanks!

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As you note, as long as your polynomial is non-constant and square-free when restricting a variable, from Weil's bound, you can get an upper bound of $deg(f) |F_q|^{3/2}$. For a typical bivariate polynomial one should expect square root cancelation, that is your sum should be bounded by $c|F_q|$ where $c$ depends on the degree of $f(x)$. However one needs to check that certain geometric degeneracies do not occur, and this quickly leads to questions involving concepts from algebraic geometry. See my question Why are Deligne-type exponential sum estimates so hard to use, and Will Sawin's insightful answer. There Will gives an example of quadratic character composed with a bivariate polynomial that fails to have square root cancelation for a highly-nontrivial geometric reason.

As I reference in that question, in the bivariate case there are essentially necessary and sufficient conditions due to Hooley, however it requires geometric concepts to state. See:

C Hooley, On exponential sums and certain of their applications, in: Number Theory Days, 1980 (Exeter 1980), London Math Soc. Lecture Note Ser. 56, Cambridge Univ. Press, Cambridge, 1982, 92–122.

There are also a number "off the shelf" necessary conditions that can be used in many circumstances for specific polynomials. See the chapter on exponential sums in Analytic Number Theory by Iwaniec and Kowalski for some examples.

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    $\begingroup$ Your example of $x^2 + 2xy + y^2$ actually shows that $f$ must still be assumed squarefree: because $x^2 + 2xy + y^2 = (x+y)^2$, the character value $\chi(x^2+2xy+y^2)$ is always $1$ (except for the $p$ cases of $x+y=0$), so the sum grows as $|F|^2$. For an example of $|F|^{3/2}$ we can take $f(x,y) = g(x)$ where $g$ is a cubic polynomial whose character sum grows as $|F|^{1/2}$, or more generally $g(h(x,y))$ where $h$ is a polynomial of a form such as $x+P(y)$ that guarantees that each equation $h(x,y) = c$ has $|F|$ solutions. $\endgroup$
    – Noam D. Elkies
    Commented Jun 15 at 15:54
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    $\begingroup$ I was using $e(\cdot)$ as an additive character, in which case the estimate is correct, but the statement is less responsive the question. I'll remove the statement to avoid confusion. $\endgroup$
    – Mark Lewko
    Commented Jun 15 at 15:59
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    $\begingroup$ Ah, I was wondering why you changed $\chi(\cdot)$ to $e(\cdot)$. $\endgroup$
    – Noam D. Elkies
    Commented Jun 15 at 17:17
  • $\begingroup$ Thanks for the references! In the reference to Iwaniec and Kowalski, perhaps you mean the chapter about sums over finite fields (chapter 11)? $\endgroup$
    – Madarb
    Commented Jun 16 at 18:53
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    $\begingroup$ Madarb: Yes Chapter 11. I had in mind Theorem 11.43 there, but I see now that is more concerned with additive and not multiplicative characters. $\endgroup$
    – Mark Lewko
    Commented Jun 16 at 23:21

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