The Prime Spectra of Regular Rings

صخلملا  ةمظتنملا تاقلحلل يلولأا فيطلا انـسرد، ثحبلا اذه يف . مك ميهافملا ضعب انـسرد ا ك ةيجولوبتلا ا ) لباق ريغ ، ايلك طبارتم ريغ ، صارتم ، صارتم هبـش لازتخلال ( لوصحلل ةمظتنملا ةقلحلل يلولأا فيطلا نع ةديدجلا جئاتنلا ضعب ىلع .   ABSTRACT In this work, we study the prime spectrum of regular rings. Also we study some topological concepts as quasi-compact, compact, totally disconnected, and irreducible topological space in order to prove some new results on the prime spectrum of regular ring.


Introduction:
In this work, all rings are assumed to be commutative with identity.Let R be a ring.The set of all prime ideals of R is called the prime spectrum of R and is denoted by Spec(R).The prime spectrum of rings has been studied by many mathematician like Heinzer and Roitman [6], and Lemmermeyer [10].The topology on Spec(R) is called the Zariski topology.
For every ideal I of R we denote by V(I) the subset of Spec(R) consisting of all prime ideals that contain I.
For every ideal I of R we write U(I) = Spec(R)/V(I) for the set of prime ideals not containing I. The collection of subset U(I) of Spec(R) for all ideals I in R form a topological space with open sets U(I).The closed sets of the topology are the sets V(I) for all ideals I in R. A ring R is called a regular ring if for every a in R there exists b in R such that a = aba.Regular rings are studied by many authors in recent years, see for example, Fisher [5], and Ming [11], [12], [13] .

Open Sets in Spectrum of Regular Rings and Homeomorphism:
In this section, we give some basic properties of open sets in the spectrum of regular ring .We prove that the spectrum topology of regular ring is homeomorphic to the spectrum topology of S.

Proposition 2.1:[10]
Suppose that R is a ring , I and J are ideals of R , then: i.
Let R be a regular ring.Then R.of Ifor every ideal The open sets are {U(I): I is an ideal of R}, where U(I) is the complement of V(I).

Proposition 2.4:
Let R be a regular ring .Then U(I) = U(J) if and only if I = J Proof: .By Proposition (2.1) we have The Prime Spectra … 97 Therefore I = J .
Since R is regular then by Proposition (2.
Now, Ra is an ideal generated by a, implies that ,where (a) is an ideal generated by a .Now, if P  V(a) then aP .Thus (a)P since (a) is the smallest ideal containing a. Therefore P  V((a)).If P  V((a)) ,then a(a)P.
Note that any Boolean ring is a commutative regular ring.Now, we prove the following result.

Theorem 2.9:
If M is a maximal ideal in R, then M ∩ S is a maximal ideal in S .

Proof:
Let M ∩ S is not a maximal ideal in S, then there exists an ideal J in S such that M∩S  J S, so there exists an element a in J and aM∩ S , therefore Thus M∩S is a maximal ideal in S .

◼
Now, we give the major result of this section .

Theorem 2.10:
Let R be a regular ring.Then the spectrum topology of R is homeomorphic to the spectrum topology of S which is a subset of all idempotent elements of R. Proof: , for a prime ideal P in ) (R Spec .Since R is a regular ring and P is a prime ideal in R, so P is a maximal ideal.By Theorem (2.9) P∩S is a maximal ideal in S. Thus P∩S is a prime ideal in S, that is ) (S Spec (P)   .
To show that  is a homeomorphism we must prove that: For every prime ideal P of R the image under  of

Irreducible Spectrum of Regular Rings:
This section is devoted to exhibit several properties of irreducible spectrum of regular ring.
We begin this section by the following definition: Definition 3.1: [4] Let I be an ideal of the ring R. Then I is said to be irreducible if it is not a finite intersection of ideals of R properly containing I; otherwise, I is termed reducible.

Proposition 3.2:[8]
Let I be a proper ideal of R. Then the ideal I is irreducible if and . Now, we give the following Proposition which is due to in Lemmermeyer [10].

Proposition 3.3:
Let R be a ring.Then 1) For any pair of ideals I, J in R we have ) If R is a regular ring, then X = Spec(R) is irreducible if and only if R is an integral domain.

Proof:
Let R be an integral domain .To show that Spec(R) is irreducible we must prove that for each pair of elements a,b  R we have The ) (R Spec is irreducible if and only if the nil radical N(R) is a prime ideal.

Lemma 4.6:[7]
Let R be a ring, N is a nil ideal in R, and . Then there exists an idempotent e in R such that a e = .Moreover, e is unique.
Bourbaki [3] , has shown that for the prime Spectrum of a ring R to be connected , it is necessary and sufficient that R contains no idempotents other than 0 and 1 .Now, we are going to give a different proof of that theorem .

Theorem 4.7:
Let R be a ring.Then the following conditions are equivalent: a) X = Spec(R) is disconnected; b) R  R1  R2 for rings R1, R2 , neither of which is the zero ring; c) R contains an idempotent other than 0 and 1.

Proof: a  b
Let V(I), V(J) be a non-empty closed sets such that then there is no prime ideal contains both I and J. Therefore R = I+J that is I and J are co-prime, and so there exists an element a I with 1-a  J.By Proposition (3.3) we have , that is every prime ideal contains and so  If R is regular, then every prime ideal is maximal.
Before closing this section ,we present the following result.

Theorem 4.9:
If R is regular ,then every point in Spec(R) is closed.

Proof:
Since R is a regular ring, then by Proposition (4.8) we have every prime ideal is maximal .That is every point in Spec(R) is maximal.
Recall Lemma (3.7) we get that every point in Spec(R) is closed.◼ prove the following theorem which gives a connection between open sets and radical ideal.Theorem 2.5: Let R be a ring and a, b  R .Then U(a) = U(b) if and only if


and hence (U(a)) = U((a)).This implies that the image of any open subset of ) (R Spec under  is an open subset of ) (S Spec .So f is open.◼ then e is a unit, but then e =1, also if = 1 and e = 0. Contradiction, therefore V(e), V(1-e) are non-empty closed sets.◼ Proposition 4.8:[5] 5) we get and only if I  J Let V(I) be irreducible, then by Proposition (3.2) we have I is irreducible.Since R is regular then by Proposition (2.2)