Suppose is a ring(not necessarily unital nor abelian) and denote all (proper) prime ideals of by . For each subset of , define
Proof. In fact, (a), .
(b) This can be seen straightforwardly by definition without using any property of an ideal.
(c).
If , then or , thus since is an ideal, and hence .
Suppose ,i.e., . If and , then there exists some and , thus since the complement of a prime ideal is multiplication closed, a contradiction. Hence . ◻
We call this topology Zariski topology of , and call with Zariski topology the spectrum of .
We can verify the following properties directly: (For the empty set of ,define .)
Proof. Suppose for some , then , and thus . Therefore, is the smallest closed set in that contains . ◻
For each semiprime ideal of , Recall that an ideal of a ring is called semiprime if is the intersection of some prime ideals of .
If is a unital abelian ring, is an ideal of , then where .