# 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.

• The last remark is a little jarring: any point over the algebraic closure is defined over some finite field. Oct 12, 2010 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. Oct 12, 2010 at 19:39