Let $\mathbb{F}_p$ denote the field with $p$ elements. I will write $\mathbf{x}$ for $(x_1,\dots,x_n)$. We're interested in the structure of the ideal $I=\{f\in\mathbb{F}_p[x_1,\dots,x_n]\mid \text{ev}_f(\mathbf{x})=0\,,\forall \mathbf{x}\in\mathbb{F}_p^n\}$, i.e. the ideal of polynomials that vanish at every point. I'm looking for a minimal system of generators for $I$. The question is originally for $n=2$ and $p=2$ but I don't think this matters and I want to solve the general case.
It seems pretty natural to me to consider the polynomials $0$ and the Frobenius automorphisms minus $x$, that is to say $P(x) = x^p - x$. It is clear that this $P$ vanishes at every point on $\mathbb{F}_p$ from Fermat's little theorem. I thus think a suitable minimal system of generators would be $I=(x_1^p - x, x_2^p - x_2,\dots,x_n^p - x_n)$, however I don't see how to prove that. It's clear to me that those generators are in $I$, but how do I show any polynomial $P\in I$ is generated by those polynomials? Is it actually true that $0$ and those Frobenius-like polynomials are the only one that vanish everywhere?
Let $f(x_1, \ldots, x_n)\in \mathbb{F}_p[x_1, \ldots, x_n] = \mathbb{F}_p[ x_2, \ldots, x_n][x_1]$. Divide f to $x_1^p - x_1$ and get $$f = (x_1^p - x_1) Q(x_1, \ldots, x_n) + R(x_1, \ldots, x_n)$$ where $R(x_1, \ldots, x_n) = \sum_{k=0}^{p-1} c_k(x_2, \ldots, x_n)\cdot x_1^k$
Fix $(a_2, \ldots, a_n) \in \mathbb{F}^{n-1}$. For every $a\in \mathbb{F}$ we have $f(a, a_2, \ldots, a_n) = 0$. This implies $$\sum_{k=0}^{p-1} c_k(a_2, \ldots, a_n) a^k = 0$$ for all $a \in \mathbb{F}_p$. This implies all the values $c_k(a_2, \ldots, a_n) \in \mathbb{F}$ are $0$. Now we can use induction.
In similar way we can show the formula for the ideal of a product: $$I(X\times Y) = \langle I(X), I(Y)\rangle$$
( consider a complementary basis to $I(X)$ )