# Commutative Algebra/Spectrum with Zariski topology

Definition 16.1:

Let ${\displaystyle R}$ be a commutative ring. The spectrum of ${\displaystyle R}$ is the set

${\displaystyle \operatorname {Spec} (R):=\left\{p\leq R|p{\text{ prime}}\right\}}$;

i.e. the set of all prime ideals of ${\displaystyle R}$.

On ${\displaystyle \operatorname {Spec} R}$, we will define a topology, turning ${\displaystyle \operatorname {Spec} R}$ into a topological space. This topology will be called Zariski topology, although only Alexander Grothendieck gave the definition in the above generality.

## Closed sets

Definition 16.2:

Let ${\displaystyle R}$ be a ring and ${\displaystyle S\subseteq R}$ a subset of ${\displaystyle R}$. Then define

${\displaystyle V(S):=\left\{p\in \operatorname {Spec} R|S\subseteq p\right\}}$.

The sets ${\displaystyle V(S)}$, where ${\displaystyle S}$ ranges over subsets of ${\displaystyle R}$, satisfy the following equations:

Proposition 16.3:

Let ${\displaystyle R}$ be a ring, and let ${\displaystyle (S_{\alpha })_{\alpha \in A}}$ be a family of subsets of ${\displaystyle R}$.

1. ${\displaystyle V(\emptyset )=\operatorname {Spec} R}$ and ${\displaystyle V(R)=\emptyset }$
2. ${\displaystyle \bigcap _{\alpha \in A}V(S_{\alpha })=V\left(\bigcup _{\alpha \in A}S_{\alpha }\right)}$
3. If ${\displaystyle A=\{\alpha _{1},\ldots ,\alpha _{n}\}}$ is finite, then ${\displaystyle V(S_{\alpha _{1}})\cup \cdots \cup V(S_{\alpha _{n}})=V(S_{\alpha _{1}}\cap \cdots \cap S_{\alpha _{n}})}$.

Proof:

The first two items are straightforward. For the third, we use induction on ${\displaystyle n}$. ${\displaystyle n=1}$ is clear; otherwise, the direction ${\displaystyle \subseteq }$ is clear, and the other direction follows from lemma 14.20.${\displaystyle \Box }$

Definition 16.4: