Zariski topology

Let $R$ be a commutative ring , and consider its prime spectrum \operatorname{Spec}(R) .

For an ideal $I\subseteq R$, define

The Zariski topology on $\operatorname{Spec}(R)$ is the topology for which the sets $V(I)$ are precisely the closed subsets, i.e. a subset $Z\subseteq \operatorname{Spec}(R)$ is closed if and only if $Z=V(I)$ for some ideal $I$.

Equivalently, the sets

form a basis of open sets. These basic opens interact tightly with localization : the correspondence of prime ideals under localization yields a natural homeomorphism between $D(f)$ and the spectrum of the localized ring $R_f$ (compare the prime correspondence for localization and the fact that localization inverts a multiplicative set ).

A useful geometric intuition is: $V(I)$ consists of those primes where all elements of $I$ “vanish,” while $D(f)$ consists of those primes where $f$ “does not vanish.”

Examples

1. A field.
   If $k$ is a field , then $\operatorname{Spec}(k)=\{(0)\}$. The only closed sets are $\emptyset$ and $\{(0)\}$, so the Zariski topology is the trivial one-point topology.

2. The integers.
   In $\operatorname{Spec}(\mathbb{Z})$, the closed set $V((n))$ consists of all prime ideals containing $n$, i.e. the points $(p)$ where $p$ divides $n$. Thus the basic open $D(n)$ is the set of primes not dividing $n$, together with the generic point $(0)$.

3. A principal ideal domain: $\operatorname{Spec}(k[x])$.
   Let $k$ be a field and $R=k[x]$. If $f\neq 0$, then $V((f))$ is a finite set: it consists of the prime ideals generated by the irreducible factors of $f$. Consequently, the complements $D(f)$ are cofinite open subsets (plus the generic point behavior), illustrating why Zariski opens are typically “large.”