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: