How can I show that the Zariski topology is the discrete topology for any finite field?

Question

In other words, that every set is closed.

I know that the closed sets in the Zariski topology are the zero sets of sets of polynomials, so they could be points, curves (polynomials, hyperbolas, etc...) and surfaces in higher dimensions. But I'm not sure how to tie any of that to proof of the above statement.

Thank you for your insight.

Answer 1

Let $x_1,...,x_n$ in the finite field $F$, $V((x-x_1)...(x-x_n))$ is $\{x_1,...,x_n\}$ so any (finite) subset of $F$ is closed. This implies that every subset is open since its complementary is a finite subset thus closed.