# Polynomials over Z evaluated with finite field arguments

A) Given a non-constant polynomial $q\in\mathbb{Z}[\alpha_1,\alpha_2,\ldots,\alpha_n],$ if we pick random $\omega_i\in\mathbb{F}$ (a finite field) uniformly and independently across $1\leq i\leq n,$ then, we know that $q(\omega_1,\omega_2,\ldots,\omega_n)\neq 0$ with high probability (i.e. the probability goes to 1 as $|\mathbb{F}|\rightarrow\infty$).

B) Given another polynomial $r\in\mathbb{Z}[\alpha_1,\alpha_2,\ldots,\alpha_n],$ I am interested in determining if there exists a field $\mathbb{F}$ and a choice of $\omega_i\in\mathbb{F}$ which simultaneously satisfy $q(\omega_1,\omega_2,\ldots,\omega_n)\neq 0$ and $r(\omega_1,\omega_2,\ldots,\omega_n)= 0.$ Is there a theorem that gives necessary or sufficient conditions for this to happen? Is it true that if it happens over some field, then it happens over all sufficiently large finite fields?

Is it true that if there is a point which satisfies $r=0$ and $q\neq 0,$ then "most" of the points satisfying $r=0$ also satisfy $q\neq 0,$ in similar spirit to the result A which is the case of $r$ being the zero polynomial?

I am interested only in solutions over finite fields and not over their algebraic closures.

Thanks a lot.

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The last remark is a little jarring: any point over the algebraic closure is defined over some finite field. –  Charles Matthews Oct 12 '10 at 10:50
Sorry about that. I had no idea how the algebraic closure of a finite field looks like. From your comment, I see that if one can find a choice of $\omega_i$ over the algebraic closure of a prime finite field, then one can find such $\omega_i$ over some finite extension of the prime field. Thanks. –  Hedonist Oct 12 '10 at 19:39
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## 2 Answers

This should work out for you, with a bit more geometry and perhaps a little trickery. You are asking for points over finite fields avoiding a given hypersurface, in affine space. It is conceptually a little easier to think of points in a projective space avoiding both a hypersurface and the hyperplane at infinity, because the typical estimates refer immediately to projective varieties. Your (A) works because the points on the hypersurface have main term which is the size of the finite field (often written q but your notation clashes) to the power of its dimension, which is n - 1. The error term is smaller, naturally, in its power, with implied constant that comes from Betti numbers that are fixed by the choice of q (your polynomial). The points on the hyperplane at infinity are easy to count.

For your (B) there will have to be some conditions at least as strong as r not dividing q. One trick that can be useful is to introduce a further variable y for which one requires yq(x) - 1 = 0: then that's a second equation to add to r(x) = 0 (rather than the inequation on q). The general theory applied to the simultaneous equations should then produce adequate numbers of solutions of the kind you want, over all large finite fields. Some care is needed to discuss the possibility that the simultaneous equations produce a singular variety over Z. When q and r are in "general position" this should go through easily. The old Lang-Weil paper on counting points might have enough to deal with it.

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Not without additional hypotheses. Charles has already mentioned that you need to assume that r does not divide q, otherwise whenever r=0 you also have q=0. But even that is not enough. If r factors in an extension field without factoring over the ground field, then r=0 will have few (or no) solutions in the ground field and it is easy to construct q vanishing at all those few solutions but not being divisible by r. Note that if you start with your polynomials over the integers and r is as as above, it will remain so for infinitely reductions mod p but will factor over the field of p elements for another infinite set of p. For the latter set you will get points mod p with r=0,q not zero as soon as p is large.

A good set of hypotheses to get what what you want is to assume r absolutely irreducible and q not divisible by r. Then, as Charles also pointed out, Lang-Weil will give what you want.

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Thanks Charles and Felipe for your help. I am currently going through Felipe's lecture notes which have been very useful too. –  Hedonist Oct 12 '10 at 19:44
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