Suppose I have $R$ homogeneous polynomials $F_1, ..., F_R$ with integer coefficients. Let $V$ be the affine variety defined by these polynomials over $\mathbb{C}$. I was wondering if some bound that looks like the following was known? $$ \#\{ \mathbf{x} \in (\mathbb{Z}/q \mathbb{Z})^n: F_{i}(\mathbf{x}) \equiv 0 (\text{mod }q) \text{ for each } 1 \leq i \leq R \} \leq C q^{\dim V + \delta} $$ for some $\delta > 0$. Here $C>0$ is independent of $q$. Of course if $\delta = n  dim V$ then it's trivial so I was hoping for something smaller. I would greatly appreciate any comments on this. Thank you very much.

$\begingroup$ Thank you very much for all the helpful solutions and comments. I greatly appreciate it. $\endgroup$ – Johnny T. Mar 3 '18 at 14:10

2$\begingroup$ So to be sure about what you are asking and wanting, $\delta$ is a real number, not necessarily an integer? And it is important for you to have $C$ completely independent of $q$, isn't it? or is it okay if $C$ depends only on the number of prime factors of $q$ (in particular, is independent on $q$ when $q$ is prime or a prime power)/ $\endgroup$ – Joël Mar 3 '18 at 15:55

$\begingroup$ @Joël Yes, I was thinking $\delta$ to be a real number. And I was hoping $C$ to be completely independent of $q$'s and $p$'s and the number of prime factors of $q$ etc. At least that's what I was hoping.. $\endgroup$ – Johnny T. Mar 4 '18 at 12:27

$\begingroup$ But since $C^{\omega(q)} \ll q^{\varepsilon}$ I wasn't too concerned about the dependency on the number of prime factors. $\endgroup$ – Johnny T. Mar 4 '18 at 12:33
First note that you can't do better than the trivial bound $q^n$ in general; for example if $q$ divides all the coefficients of the $F_i$.
However this silly problem only occurs for finitely many primes (by the Chinese remainder theorem we can reduce to the case of prime powers). So let $p$ be a prime such that $V_{\mathbb{F}_p}$ has dimension $d:=\dim V$ (here I denote by $V_{\mathbb{F}_p}$ the reduction of $V$ modulo $p$). By Noether normalisation there exists a finite map $$f: V_{\mathbb{F}_p} \to \mathbb{A}^d_{\mathbb{F}_p}.$$ Let $C$ be the degree of $f$. Then we have $$\#V(\mathbb{F}_p) \leq C\#\mathbb{A}(\mathbb{F}_p) = Cp^{d}.$$ This is much easier and weaker than the LangWeil estimates, which are much deeper and require the Riemann hypothesis for curves.
To treat prime powers, there are some subtleties. One needs to understand the cardinalities of the fibres of the map $$\#V(\mathbb{Z}/p^k\mathbb{Z}) \to \#V(\mathbb{F}_p) \quad (*).$$ If $V_{\mathbb{F}_p}$ is smooth, then a version of Hensel's lemma states that the fibres of the map $(*)$ all have cardinality exactly $p^{d(k1)}$. This gives the upper bound $$\#V(\mathbb{Z}/p^k\mathbb{Z}) \leq Cp^{kd}.$$ By "spreading out arguments" (essentially applying Noether Normalistion to the variety over $\mathbb{Q}$ then spreading out the map), one finds an absolute constant $C$ such that $$ \#\{ \mathbf{x} \in (\mathbb{Z}/q \mathbb{Z})^n: F_{i}(\mathbf{x}) \equiv 0 (\text{mod }q) \text{ for each } 1 \leq i \leq R \} \leq C^{\omega(q)} q^{\dim V} $$ holds for those $q$ which satisfy the above conditions on the dimension of $V \bmod q$ and smooth reduction. Here $\omega(q)$ denotes the number of prime divisors of $q$.

$\begingroup$ So the answer to the initial question is yes for any $\delta>0$. Good job! $\endgroup$ – Joël Mar 5 '18 at 2:09

$\begingroup$ Indeed, providing though that your original variety is smooth over $\mathbb{Q}$! The nonsmooth case is also interesting; I don't know whether one can always take any $\delta > 0$ in this case. I would guess that you need some conditions on the singularities in general. For example the equation $x^2 = 0$ has many solutions modulo $p^k$ (though this example is nonreduced, so I don't know whether you would count this as a variety). $\endgroup$ – Daniel Loughran Mar 5 '18 at 9:55

$\begingroup$ @DanielLoughran I think in the nonsmooth case you cannot always take $\delta > 0$. Consider the variety defined by $x_1^3 f(x_3, \cdots, x_n)  x_2^3 g(x_3, \cdots, x_n)$, where $f,g$ are arbitrary polynomials. Then for any $k$ the number of points mod $p^{3k}$ is at least $p^{3k(n2/3)}$, while the dimension is $n1$. Thus one cannot take $\delta$ to be less than $k$. $\endgroup$ – Stanley Yao Xiao May 29 '18 at 18:12
The magic words are LangWeil (actually, the upper bound is easy, it is the lower bound that occupies most of Lang and Weil's paper
Lang, Serge; Weil, André, Number of points of varieties in finite fields, Am. J. Math. 76, 819827 (1954). ZBL0058.27202.
The result is for prime moduli, for composite moduli you can use chinese remaindering. For more on this, see Terry Tao's blog post.

3$\begingroup$ This works fine modulo a prime but modulo prime powers a different argument is required. There are some subtleties here are not every $\mathbb{F}_p$ solution lifts to a unique $\mathbb{Z}/p^r\mathbb{Z}$ in general. The links you give do not address this. $\endgroup$ – Daniel Loughran Mar 2 '18 at 20:12

1$\begingroup$ @DanielLoughran That was my thinking (Hensel's lemma). This paper: arxiv.org/pdf/1502.07004 implies that there are some pathological case, but I could not think of an example offhand. $\endgroup$ – Igor Rivin Mar 2 '18 at 20:24

2$\begingroup$ Also, knowing the case of all prime powers (say with $\delta=0$, to be optimistic) would not exactly gives the desired result through the Chinese remainder theorem. For if you know the result asked for all prime power $q$ with the same constant $C$, then you now the result for a general $q$ with the constant $C^{\omega (q)}$, with $\omega(q)$ the number of prime factors of $q$. $\endgroup$ – Joël Mar 3 '18 at 4:00

$\begingroup$ @IgorRivin If I apply LangWeil (for the prime modulus case) then the implied constant on the inequality would depend on the number of irreducible components, degree and the dimension. Are these things controlled uniformly as one considers points modulo $p$ for all primes $p$? (I am guessing they are, but I would appreciate any reference for this.) Thank you very much. $\endgroup$ – Johnny T. Mar 3 '18 at 14:07

1$\begingroup$ @JohnnyT. Yes, they are, they are all controlled by the degrees of the equations  I think Terry's blog post talks about that. $\endgroup$ – Igor Rivin Mar 3 '18 at 16:09
The case of $q$ a prime being answered by Igor, let us assume that $q=p^k$ is a power of a prime $p$. In this case there is the following result of Serre, which gives a very partial answer:
Let $d_p$ be the dimension of the $p$adic analytic space $V$ defined in $\mathbb Z_p^R$ by the equations $F_1=\dots=F_r=0$. Then $$ \#\{ \mathbf{x} \in (\mathbb{Z}/q \mathbb{Z})^n: F_{i}(\mathbf{x}) \equiv 0 (\text{mod }q) \text{ for each } 1 \leq i \leq R \} \leq C_p q^{d_p} $$ where $C_p$ is some constant depending on $p$ but not on $k$.
The result is Serre's Théorème 8 in his IHES paper on Chebotarev.
Comparing to what you ask, there are two differences: the exponent is $d_p$ instead of $\dim_{\mathbb C} V$ (but they are the same for almost all $p$, aren't they?). And the constant $C$ depends on $p$, though it is the same for all powers $q$ of a given $p$.

1$\begingroup$ As mentioned in my comment under Igor's answer, even knowing that $C_p$ is independent on $p$ would not immediately give the result for all integers $q$. Looking at Serre's proof, it was not immediately clear to me that you can take the same $C_p$ for all $p$. Of course, having this fr almost all $p$ would be enough, so you can throw away bad $p$'s and assume your $p$adic space $V$ is smooth, in which case Serre's proof is much simpler (in the nonsmooth case, it uses Hironaka's resolution of singularities). But even in the smooth case, it is not clear to me how to control $C_p$. $\endgroup$ – Joël Mar 3 '18 at 4:05

2$\begingroup$ I just realized that everything I said was already said in Daniel's deleted answer (only visible to highrep users I suppose). Why was this fine answer deleted? $\endgroup$ – Joël Mar 3 '18 at 4:11

$\begingroup$ I don't think my proof completely works for prime powers, so I deleted my answer with a view to possibly fixing it later when I have more time. $\endgroup$ – Daniel Loughran Mar 3 '18 at 7:32