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Let $f_1, \dots, f_n$ be a finite set of polynomials in the polynomial ring $Z[x_1, \dots, x_m]$. At a prime $p$, let $N_p$ be the number of solutions $x=(x_1, \dots, x_m)\in (\mathbb{Z}/p\mathbb{Z})^m$ to the system $$f_1(x)=f_2(x)=f_3(x)=\dots=f_n(x)=0$$ (with coordinates $x_1, \dots, x_m\in \mathbb{Z}/p\mathbb{Z}$).

As a function of $p$, are there any conditions (on the set of polynomials $f_1, \dots, f_n$) under which $N_p$ is a polynomial in $p$ for large enough values (or perhaps all values) of $p$?

The system that I'd like this to be true for (all primes $p$) is $$f_1=(a_3^2-a_3)-a_2(c_1^2-c_1)-a_1(b_1^2-b_1)=0$$ $$f_2=(a_4^2-a_4)-a_2(c_2^2-c_2)-a_1(b_2^2-b_2)=0$$ $$f_3=a_3a_4-a_2c_1c_2-a_1b_1b_2=0 $$

in the $8$ variable polynomial ring $\mathbb{Z}[a_1,a_2, a_3,a_4, b_1, b_2, c_1, c_2]$.

Computations done for $p\leq 19$ lead us to expect $N_p=p^5 + 12 p^4 - 20 p^3 + 30 p^2 - 10 p$.

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    $\begingroup$ This is the kind of thing that zeta functions are for, isn't it? $\endgroup$
    – David Loeffler
    Commented Nov 11, 2022 at 17:14
  • $\begingroup$ @DavidLoeffler I suppose that the zeta function one usually considered is that attahced to a variety over a fixed finite field, and it measures the variation of N_q where q runs over all powers of a fixed prime p, so maybe if one were to ask a similar question but count number of solutions in $\mathbb{F}_{p^r}$ and let r go to infinity, perhaps the zeta function would play a role. I guess the zeta function of interest in this context when p varies would be associated to the scheme over $\mathbb{Z}$, however, I'm not well aware of what is known about their Euler product factorization. $\endgroup$
    – Anwesh Ray
    Commented Nov 11, 2022 at 18:27
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    $\begingroup$ The Hasse-Weil zeta function might be the one David Loeffler was alluding to, although, from experience, if you have a conjectural polynomial formula it suggests there is a nice parameterization for the solutions that can be found by hand. It seems you have 8 variables (not 7) which suggests the count should grow like $p^{8-3}$ and not like $p^4$, so I am confused. $\endgroup$
    – Ofir Gorodetsky
    Commented Nov 11, 2022 at 18:37
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    $\begingroup$ mathoverflow.net/questions/167934/… $\endgroup$
    – Felipe Voloch
    Commented Nov 11, 2022 at 21:39
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    $\begingroup$ If the variety has no odd cohomology and the even cohomology groups are all spanned by algebraic cycles in characteristic zero, you will have polynomial count for varying $p$ as well. $\endgroup$
    – Felipe Voloch
    Commented Nov 12, 2022 at 8:40

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