I'm an analyst who needs to use Deligne's Theorem 8.4 in **1**, but I feel lost in the maze of definitions, and I don't trust my geometric intuition here.



> 
> Theorem 8.4: Let $Q$ be a polynomial in $n$ variables and of degree $d$ over $\mathbb{F}\_q$, $Q\_d$ the homogeneous part of degree $d$ of $Q$, and $\psi:\mathbb{F}\_q\to\mathbb{C}^\*$ a non-trivial additive character over $\mathbb{F}\_q$. Suppose that:
> 
> 
> * $d$ is comprime to $p$;
> * The hypersurface $H\_0$ in $\mathbb{P}^{n-1}\_{\mathbb{F}\_q}$ defined by $Q\_d$ is smooth.
> 
> 
> Then
> $$\Big|\sum\_{x\_1,\ldots,x\_n\in\mathbb{F}\_q}\psi(Q(x\_1,\ldots,x\_n))\Big|\le (d-1)^nq^{n/2}$$
> 
> 
> 


I'm only interested in the case $q = p$ a prime. I need clarification about the statement.



> 
> How are smooth hypersurfaces defined?
> 
> 
> 


I guess smoothness here is equivalent to $\nabla Q\_d(x) \neq 0$ for every $x\in \overline{\mathbb{F}}\_q\setminus\{0\}$ (I'm using $x\cdot\nabla Q\_d = dQ\_d$). Am I right?


Some books I've glanced (p. ej. Shafarevich or Hartshorne) define smoothness over closed fields, and in [these notes](https://www.jmilne.org/math/CourseNotes/LEC.pdf) (p. 91) it says that we should extend the field, I think. All sources define "smoothness" in two ways, one using [regular rings](https://en.wikipedia.org/wiki/Regular_scheme) and other using the [gradient](https://en.wikipedia.org/wiki/Smooth_scheme). These definitions, which should be equivalent in my case, seem to work over any field, so do I really need to extend to $\overline{\mathbb{F}\_q}$? It would be nice to find references about all this which I could understand.


I had the idea that to define smoothness we had to take $H\_0 = \{Q\_d = 0\}\subset \mathbb{P}^{n-1}(\overline{\mathbb{F}}\_q)$, define the ideal $I\_H = \langle P\rangle$ of polynomials vanishing there, where $P$ is a polynomial, and then to verify that $\nabla P$ doesn't vanish. It seems wrong. If I take $Q = X\_1^d$, $d\ge 2$, it seems that Deligne's theorem doesn't hold, even though $\{X\_1 = 0\}$ looks smooth.



> 
> If $Q\_d\in\mathbb{Z}[X\_1,\ldots,X\_n]$ and $\nabla Q\_d(x)\neq 0$ for every $x\in\mathbb{C}^n\setminus\{0\}$, does Theorem 8.4 hold for all but finitely many primes?
> 
> 
> 


From what I've gathered until now, arithmetic-geometers find it trivial (should I too?), probably they'd point to proposition A.9.1.6 in **2**, but again, I don't know if all these definitions of smoothness coincide. I'd be nice (for me at least) to have a proof without using schemes, but [it is probably painful](https://mathoverflow.net/a/59129/90189).


I stated my question over the complex numbers because it is closed, but can I weaken the assumptions? On the other hand, is the converse true?


I'd appreciate any reference or help.


References
----------


**1** Deligne, Pierre. La conjecture de Weil. I. (French) Inst. Hautes Études Sci. Publ. Math. No. 43 (1974), 273--307.


**2** Hindry, Marc; Silverman, Joseph H. Diophantine geometry. An introduction. Graduate Texts in Mathematics, 201. Springer-Verlag, New York, 2000.



Deligne's theorem on exponential sums
Yes, smoothness is equivalent to the gradient being nonzero for every $x \in \overline{\mathbb F}\_q^n \setminus \{0\}$.


I would define smoothness of the hypersurface defined by $Q\_d$ as the condition that for each such $x$, either the gradient or the value of $Q\_d$ at $x$ is nonzero, but as you note, by homogeneity and the fact that $d$ is prime to $p$, if the value is nonzero then the gradient is nonzero.


Checking over $\overline{\mathbb F}\_q$ and not $\mathbb F\_q$ is really necessary.


For example, fix a basis for the field extension $\mathbb F\_{q^n}$, giving a bijection between $\mathbb F\_{q^n}$ and $\mathbb F\_q$. The norm map that sends each element of $\mathbb F\_{q^n}$ to the determinant over $\mathbb F\_q$ of its multiplication action on $\mathbb F\_{q^n}$ can be expressed as a polynomial of degree $n$ in the $n$ coordinates. Set $d=n$ and $Q\_= Q\_d=$ this polynomial. Then because the norm is a surjective homomorphism of multiplicative groups, there are $\frac{q^n-1}{q-1}$ elements of norm $a$ for each $a \in \mathbb F\_q^\times $, so
$$\sum\_{x \in \mathbb F\_q^n } \psi(Q(x))= \sum\_{x\in \mathbb F\_{q^n}} \psi (\operatorname{Norm}(x))= 1 + \frac{q^n-1}{q-1} \sum\_{x \in \mathbb F\_q^\times} \psi(x) = 1 - \frac{q^n-1}{q-1} $$ which does not satisfy Deligne's bound for $n>2$ and $q$ large.


However, $Q\_d$ never vanishes at any $x \in \mathbb F\_q^n$, and thus neither does its gradient. So checking over the algebraic closure is really necessary!


You can avoid working over the algebraic closure only by using definitions that involve checking point-by-point. For example, smoothness is equivalent to the claim that the ideal generated by $Q\_d$, and $\frac{ \partial Q\_d}{\partial x\_i}$ for all $i$ contains $x\_1^N, \dots, x\_n^N$ for some $N$.




---


You are right that we do not want to pass to the vanishing set of $Q\_d$ and then to the ideal - we just want to look at $Q\_d$ itself and whether it satisfies the gradient condition. The vanishing *scheme* of $Q\_d$ is sensitive to taking powers in this sense, which is why this is referred to as a smoothness condition on the hypersurface rather than a smoothness condition on $Q\_d$. If you like, this is equivalent to smoothness of the vanishing set plus the fact that $Q\_d$ isn't divisible by the square of a nontrivial polynomial.




---


It is indeed true that if the gradient of $Q\_d$ is nowhere vanishing over $\mathbb C^n \setminus \{0\}$, then it is nowhere vanishing on $\overline{\mathbb F}\_p^n \setminus \{0\}$ for all but finitely many $p$. The same is true for any algebraically closed field of characteristic zero.


The proof actually doesn't require scheme there. Here are two approach:


(1) By the Nullstellensatz, this implies that the ideal generated $\frac{ \partial Q\_d}{\partial x\_1}, \dots, \frac{ \partial Q\_d}{ \partial x\_n}$ contains a power of the ideal $(x\_1,\dots, x\_n)$ and in particular contains $x\_1^N, \dots, x\_n^N$ for some $N$. So some $\mathbb C[x\_1,\dots,x\_n]$-linear combination of those derivatives is equal to $x\_1^N$, some linear combination is $x\_2^N$, etc.


We can choose a linear map $\mathbb C \to \mathbb Q$ whose composition with $\mathbb Q \to \mathbb C$ is the identity. (In fact we only need to do this for the finite-dimensional subset of $\mathbb C$ generated by the coefficients of the linear combination.) Plugging the coefficients of all the polynomials in $\mathbb C[x\_1,\dots,x\_n]$ into this map, we see that some $\mathbb Q[x\_1,\dots,x\_n]$-linear combination of those derivatives is equal to $x\_1^N$, and so on.


Clearing denominators, we see that there is a natural number $M$ such that some $\mathbb Z[x\_1,\dots,x\_n]$-linear combination of those derivatives is equal to $M x\_1^N$, $\mathbb Z[x\_1,\dots,x\_n]$-linear combination of those derivatives is equal to $M x\_2^N$, and so on.


It follows for all $p$ not dividing $M$ that, if $x\_1,\dots, x\_n \in \overline{\mathbb F}\_p$ are not all zero, then one of these derivatives is nonvanishing, as desired.


(2) There exists a universal polynomial, the discriminant, in the coefficients of a homogeneous polynomial $f$ of degree $d$, which vanishes if and only if the derivatives of $f$ all vanish at some nonzero point over an algebraically closed field of characteristic not dividing $d$. This can be obtained as the [Macauley resultant](https://en.wikipedia.org/wiki/Resultant#Macaulay%27s_resultant) of the partial derivatives of $f$.


If the nonvanishing condition holds over $\mathbb C$, then the discriminant is nonzero, so it is nonzero mod $p$ for all but finitely many $p$, so the nonvanishing condition holds for all but finitely many $p$.


