I'm just looking for some direction here (researching an open problem from: survey, namely equivalence of polynomials upto change of basis).
Define the set of zero-valued polynomials in $k[X_1, \cdots, X_n]$ to be all polynomials $f$ such that $f(k^n) = 0$ or $f(x) = 0$ for all $x \in k^n$. So for instance if $k = \Bbb{F}_2$ then $X^2 + X$ is one such polynomial. The set of all such polynomials clearly forms an ideal of $k[X_1, \cdots, X_n]$ which must be finitely generated by Hilbert's basis theorem (every field being Noetherian vacuously).
The end goal is compare two polynomials $f, g$ by value (not by formal standard form equivalence). So $f \equiv g \iff f - g \in I$ the ideal in the first paragraph, or alternatively $\overline{f} = \overline{g} \pmod I$, either of which are amenable to Grobner basis techniques.
Question: How do I compute $I$ for any field $k$ assuming I have a representation for $k$'s arithmetic on a computer? Can we say that if $k = \Bbb{Q}$, then $I = (0)$?
For an infinite field the kernel is trivial. Proof : because univariate polynomials have finitely many roots, if $f\ne 0\in k[Y,t],Y=(y_1,\ldots,y_{n-1})$ and its coefficient of $Y^s t^m$ is non-zero, then there is some $a_n\in k$ such that the coefficient of $Y^s $ of $f(Y,a_n)$ is non-zero, ie. it is not the zero polynomial, repeating $n$ times we obtain some $a\in k^n$ such that $f(a_1,\ldots,a_n)\ne 0$.
For a finite field the kernel of the evaluation map $$e:\Bbb{F}_q[X_1,\ldots,X_n] \to (\text{the set of functions }(\Bbb{F}_q)^n\to \Bbb{F}_q)$$ is the ideal $$(X_1^q-X_1,\ldots,X_n^q-X_n)\subset \Bbb{F}_q[X_1,\ldots,X_n]$$
Proof : because $\prod_{m=1}^n ((X_m-a_m)^{q-1}-1)$ is non-zero only at $(a_1,\ldots,a_n)$, we get that $e$ is surjective, thus as a $\Bbb{F}_q$-vector space $\Bbb{F}_q[X_1,\ldots,X_n]/\ker(e)$ is of dimension $q^n$.
On the other hand it is clear that $(X_1^q-X_1,\ldots,X_n^q-X_n)$ is in $\ker(e)$ and as a $\Bbb{F}_q$-vector space $$\Bbb{F}_q[X_1,\ldots,X_n]/(X_1^q-X_1,\ldots,X_n^q-X_n)$$ is of dimension $q^n$ (since in the quotient, only powers $<q$ of the $X_j$ appear).