## Zariski Topology

Suppose $R$ is a ring(not necessarily unital nor abelian) and denote all (proper) prime ideals of $R$ by $\operatorname{Prim}\left( R \right)$. For each subset $S$ of $R$, define $\vee \left( S \right): = \left\{ P \in \operatorname{Prim}\left( R \right) \middle| S \subset P \right\}.$

Theorem 1. $\left\{ \operatorname{Prim}\left( R \right) \backslash \vee \left( S \right) \middle| S \subset R \right\}$ forms a topology of $R$.

Proof. In fact, (a)$\vee \left\{ 0 \right\} = \operatorname{Prim}\left( R \right)$, $\vee \left( R \right) = \varnothing$.

(b)$\cap_{\lambda} \vee \left( S_{\lambda} \right) = \vee \left( {\cup_{\lambda}S_{\lambda}} \right),S_{\lambda} \subset R.$ This can be seen straightforwardly by definition without using any property of an ideal.

(c)$\vee \left( S_{1} \right) \cup \vee \left( S_{2} \right) = \vee \left( {S_{1}S_{2}} \right),S_{1},S_{2} \subset R$.

If $P \in \vee \left( S_{1} \right) \cup \vee \left( S_{2} \right)$, then $S_{1} \subset P$ or $S_{2} \subset P$, thus $S_{1}S_{2} \subset P$ since $P$ is an ideal, and hence $\vee \left( S_{1} \right) \cup \vee \left( S_{2} \right) \subset \vee \left( {S_{1}S_{2}} \right)$.

Suppose $P \in \vee \left( {S_{1}S_{2}} \right)$,i.e., $S_{1}S_{2} \subset P$. If $S_{1} \nsubseteq P$ and $S_{2} \nsubseteq P$, then there exists some $s_{1} \in S_{1} \backslash P$ and $s_{2} \in S_{2} \backslash P$, thus $s_{1}s_{2} \notin P$ since the complement of a prime ideal is multiplication closed, a contradiction. Hence $\vee \left( {S_{1}S_{2}} \right) \subset \vee \left( S_{1} \right) \cup \vee \left( S_{2} \right)$. □

We call this topology Zariski topology of $\operatorname{Prim}\left( R \right)$, and call $\operatorname{Prim}\left( R \right)$ with Zariski topology the spectrum of $R$.

We can verify the following properties directly:

$\begin{matrix} & {\cap \vee S \supset S,S \subset R;} & & \\ & {\vee \cap \mathcal{S} \supset \mathcal{S},\mathcal{S} \subset \operatorname{Prim}\left( R \right);} & & \\ & {\vee S_{1} \supset \vee S_{2},S_{1} \subset S_{2};} & & \\ & {\cap \mathcal{S}_{1} \supset \cap \mathcal{S}_{2},\mathcal{S}_{1} \subset \mathcal{S}_{2}.} & & \\ \end{matrix}$

(For the empty set $\varnothing$ of $\operatorname{Prim}\left( R \right)$,define $\cap \varnothing = R$.)

Property 2. For each subset $\mathcal{S}$ of $\operatorname{Prim}\left( A \right)$, $\vee \cap \mathcal{S}$ is the closure of $\mathcal{S}$.

Proof. Suppose $\mathcal{S} \subset \vee S$ for some $S \subset R$, then $\cap \mathcal{S} \supset \cap \vee S$, and thus $\vee \cap \mathcal{S} \subset \vee \cap \vee S = \vee S$(here is the reason for last equation). Therefore, $\vee \cap \mathcal{S}$ is the smallest closed set in $\operatorname{Prim}\left( R \right)$ that contains $\mathcal{S}$. □

For each semiprime ideal $P$ of $R$,

$\cap \vee P = P.$

Recall that an ideal $P$ of a ring $R$ is called semiprime if $P$ is the intersection of some prime ideals of $R$.

If $R$ is a unital abelian ring, $I$ is an ideal of $R$, then

$\cap \vee I = \sqrt{I},$

where $\sqrt{R} = \left\{ r \in R \middle| r^{n} \in I\text{ for some positive interger }n \right\}$.