Which polynomials in $\mathbb{F}_p[x,y,z]$ evaluate to 0 at all points of $\mathbb{F}_p^3$? (Spin off from Roots modulo a prime - I thought it might be useful to post this proof in case somebody else was curious about it in the future...)
Let $p$ be prime.  Then the set of polynomials in $\mathbb{F}_p[x,y,z]$ which vanish at all points of $\mathbb{F}_p^3$ is easily seen to be an ideal, and to be equal to $\bigcap_{a,b,c\in \mathbb{F}_p} \langle x-a, y-b, z-c \rangle$.  Can you give a set of generators of this ideal?
 A: It is easily seen that $x^p - x, y^p - y, z^p - z$ are all in this ideal.  We claim that, in fact, the ideal is generated by these elements.
Let $\mathrm{Fun}(\mathbb{F}_p^3, \mathbb{F}_p)$ denote the ring of functions $\mathbb{F}_p^3 \to \mathbb{F}_p$ (with the ring operations being pointwise addition, multiplication, negation; the element 1 is the constant function with value one; and the element 0 is the constant function with value zero). Then we have an induced ring homomorphism $$\phi : \mathbb{F}_p[x,y,z] / \langle x^p - x, y^p - y, z^p - z \rangle \to \mathrm{Fun}(\mathbb{F}_p^3, \mathbb{F}_p).$$
Now, the left hand side has basis $\{ x^a y^b z^c : 0 \le a, b, c < p \}$ as an $\mathbb{F}_p$-module.  (This is because $\{ x^p - x, y^p - y, z^p - z \}$ is a Gröbner basis of the ideal $\langle x^p - x, y^p - y, z^p - z \rangle$ of $\mathbb{F}_p[x,y,z]$, since the leading monomials are pairwise relatively prime.)  Therefore, the left hand side has cardinality $p^{p^3}$, which is also equal to the cardinality of the right hand side.
We claim that in fact, $\phi$ is a ring isomorphism.  To show this, by the cardinality comparison, it suffices to show $\phi$ is surjective.  But suppose for each $a,b,c \in \mathbb{F}_p$, we define
$$Q_{a,b,c} := \left( \prod_{a' \ne a} \frac{x-a'}{a-a'} \right) \left( \prod_{b' \ne b} \frac{y-b'}{b-b'} \right) \left( \prod_{c'\ne c} \frac{z-c'}{c-c'} \right) \in \mathbb{F}_p[x,y,z].$$
Then $\phi(Q_{a,b,c})$ is the function which evaluates to 1 at $(a,b,c)$ and 0 elsewhere.  Thus, for $f \in \mathrm{Fun}(\mathbb{F}_p^3, \mathbb{F}_p)$, we have:
$$ f = \phi \left( \sum_{a,b,c \in \mathbb{F}_p} f(a,b,c) Q_{a,b,c} \right). $$
This concludes the proof that $\phi$ is surjective and therefore a ring isomorphism.  Now, from the fact that $\phi$ is injective, we conclude that the kernel of the ring homomorphism $\mathbb{F}_p[x,y,z] \to \mathrm{Fun}(\mathbb{F}_p^3, \mathbb{F}_p)$ is equal to $\langle x^p - x, y^p - y, z^p - z \rangle$.
(Obviously, this argument easily generalizes to other numbers of variables than 3, and also to other finite fields $\mathbb{F}_q$ with $q$ a prime power.)
