# Commutative Algebra/Spectrum with Zariski topology

Definition 16.1:

Let $R$ be a commutative ring. The spectrum of $R$ is the set

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

i.e. the set of all prime ideals of $R$ .

On $\operatorname {Spec} R$ , we will define a topology, turning $\operatorname {Spec} R$ 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 $R$ be a ring and $S\subseteq R$ a subset of $R$ . Then define

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

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

Proposition 16.3:

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

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

Proof:

The first two items are straightforward. For the third, we use induction on $n$ . $n=1$ is clear; otherwise, the direction $\subseteq$ is clear, and the other direction follows from lemma 14.20.$\Box$ Definition 16.4: