TITLE: Two definitions of Zariski Topology
QUESTION [10 upvotes]: $\DeclareMathOperator{\Spec}{Spec}$
Given a ring $R$ the closed sets of the Zariski Topology on the Spectrum of $R$, $\Spec(R)$, is:
$$
V(I)= \{P\in \Spec(R) \mid I \subset P\} \, .
$$
Given an affine space $A^n$ the closed sets of the Zariski Topology of $A^n$ is: 
$$
V(I)= \{x \in A^n \mid f(x)=0, \forall f \in I \}
$$
What is a concrete relationship between the two definitions if any? I.e., is there some $R$ so that we have a one to one correspondence between closed sets?

REPLY [2 votes]: To say a bit more about Dustan and Tsemo's answers, there are some things you can see.$\DeclareMathOperator{\Spec}{Spec}$

You can take the family of the quotients of the form $R/I$ where $I$ is a radical ideal of $R$ and $R$ is a polynomial algebra over an algebraically closed field $k$. It'll be a category which morphisms will be the $k-$algebra homomorphisms.
For each of these algebras $R/I$ and by Nullstellensatz you can associate to them an algebraic set $V=V(I)$, embedded in some affine space $\mathbb{A}^n(k)$. You'll get $I=I(V)$ and then denote $R/I$ as $k[V]$ and call it its coordinate ring. All the algebraic sets will also form a category, and morphisms will be the restrictions of polynomial functions. And this way you can define a "functor" between these two categories where each coordinate ring is mapped to its corresponding algebraic set and each $k-$algebra homomorphism is mapped to some morphism of algebraic sets. In less words: $k[V]\mapsto V$ determines a functor.
We know that there is an homeomorphism between $V$ and $\text{maxSpec}(k[V])$, but there is a bit more structure here. In general, we have that the prime spectrum association $R\mapsto\text{Spec}R$ determines a functor from $\textbf{Ring}$ to $\textbf{Top}$ (To each ring homomorphism $f:R\to S$ you can assign a continuous functions $f^{-1}:\Spec(S)\to\Spec(R)$ given that the inverse image of a prime ideal is also prime). In general we can't do that with the maximal spectrum (The inverse image of a maximal ideal is not always maximal). However, if you restrict the rings to coordinate rings and the homomorphisms to $k-$algebra homomorphisms, we get that the inverse image of a maximal ideal is a maximal ideal and then the maximal spectrum $k[V]\mapsto \text{maxSpec} (k[V])$ determines a functor (We won't need the fact that $k$ is algebraically closed here, but it makes the proof easier and we will need it in 4 anyway).
And last, the homeomorphism from Nullstellensatz will determine a natural isomorphism between the two functors, $k[V]\mapsto V$ and $k[V]\mapsto\text{maxSpec}(k[V])$.

(Note, all the functors here are contravariant)