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