Generalizations of r-ideals of commutative rings
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Generalizations of r -ideals of commutative rings Emel Aslankarayigit Ugurlu
Received: 29 February 2020 / Accepted: date
Abstract
In this study, we present the generalization of the concept of r -ideals in commutative rings with nonzero identity. Let R be a commutativering with 0 = 1 and L ( R ) be the lattice of all ideals of R. Suppose that φ : L ( R ) → L ( R ) ∪ {∅} is a function. A proper ideal I of R is called a φ − r -ideal of R if whenever ab ∈ I and Ann ( a ) = (0) imply that b ∈ I for each a, b ∈ R. In addition to giving many properties of φ − r -ideal, we also examinethe concept of φ − r -ideal in trivial ring extension and use them to characterizetotal quotient rings. Keywords r -ideal · φ − prime ideal · φ − r -ideal Mathematics Subject Classification (2010) · In this article, we focus only on commutative rings with nonzero identity andnonzero unital modules. Let R always denote such a ring and M denote suchan R -module. L ( R ) denotes the lattice of all ideals of R. Also the radical of I is defined as √ I := { r ∈ R | r k ∈ I for some k ∈ N } .A proper ideal I of a commutative ring R is prime if whenever a , a ∈ R with a a ∈ I , then a ∈ I or a ∈ I, [4]. In 2003, the authors said that ifwhenever a , a ∈ R with 0 R = a a ∈ I , then a ∈ I or a ∈ I, a proper ideal I of a commutative ring R is weakly prime [2]. In [5], Bhatwadekar and Sharmadefined a proper ideal I of an integral domain R as almost prime (resp. n -almost prime) if for a , a ∈ R with a a ∈ I − I , (resp. a a ∈ I − I n , n ≥ a ∈ I or a ∈ I . Later, Anderson and Batanieh, in [1], introduced a E. Aslankarayigit UgurluDepartment of Mathematics, Marmara University, Istanbul, Turkey.E-mail: [email protected] Emel Aslankarayigit Ugurlu concept which covers all the previous definitions in a commutative ring R asfollowing: let φ : L ( R ) → L ( R ) ∪{∅} be a function, where L ( R ) denotes the setof all ideals of R . A proper ideal I of a commutative ring R is called φ - prime if for a , a ∈ R with a a ∈ I − φ ( I ), then a ∈ I or a ∈ I. They defined themap φ α : L ( R ) → L ( R ) ∪ {∅} as follows: φ ∅ ( I ) = ∅ prime ideal φ ( I ) = 0 R weakly prime ideal φ ( I ) = I almost prime ideal φ n ( I ) = I n n -almost prime ideal ( n ≥ φ ω ( I ) = ∩ ∞ n =1 I n ω -prime ideal φ ( I ) = I any ideal.The principal ideal generated by a ∈ R is denoted by < a > . For a subset S of R and an ideal I of R, we define ( I : R S ) := { r ∈ R : rS ⊆ I } . In particular,we use
Ann ( S ) instead of (0 : R S ) . Also, for any a ∈ R and any ideal I of R, we use ( I : a ) and Ann ( a ) to denote ( I : R { a } ) and Ann ( { a } ) , respectively. Anelement a ∈ R is called a regular (resp., zerodivisor) element if Ann ( a ) = (0)(resp., Ann ( a ) = (0)) . The set of all regular (resp., zerodivisor) elements of R isdenoted by r ( R ) (resp., zd ( R )) . Let S be a multiplicatively closed of R. S − R denotes the quotient ring of R at S. Particularly, ( r ( R )) − R = T ( R ) is thetotal quotient ring of R. A ring R is said to be a total quotient ring if R = T ( R ) , equaivalently, every element a ∈ R is either zerodivisor or unit. Anideal I of R is called a regular ideal if it contains at least a regular element.For a subset S of R, a ∈ S is called a von Neumann regular elemet if thereis b ∈ S such that a = a b. If all elements of R are von Neumann regularelement, R is called a von Neumann regular ring. Similarly, if all elements in asubset S (resp., an ideal I ) of R are von Neumann regular element, S (resp., I ) is called a von Neumann regular subset (resp., ideal). An ideal I is saidto be pure ideal if for every a ∈ I there is b ∈ I such that a = ab. A ring R is said to satisfy strongly annihilator condition (briefly, s. a. c.) if for everyfinitely generated (briefly, f. g.) ideal I of R there is an element b ∈ I such that Ann ( I ) = Ann ( b ) . For more information about the above notions, we refer to[6], [7].In 2015, R. Mohamadian [8] present the notion of r -ideals in commutativerings with nonzero identity as follows: an ideal I is a commtative ring withidentity R is called r -ideal (resp., pr -ideal), if whenever ab ∈ I and a is regularelement imply that b ∈ I (resp., b n ∈ I, for some natural number n ) , for each a, b ∈ R. In this paper, our aim is to introduce the generalization of the concepts of r -ideal, pr -ideal, pure ideal and von Neumann regular ideal in commutative rings eneralizations of r -ideals of commutative rings 3 with nonzero identity. For this, firstly with Definition 1 we give the definitionof φ − r -ideal. Similarly, in Definition 3 we give the definitions of φ − pr -ideal, φ − pure ideal and φ − von Neumann regular ideal in R. Then we investigatethe basic properties of φ − r -ideal of R, see Proposition 1 and Remark 1 . InTheorem 2, we give a method for constructing φ − r -ideal in a commutativering with non-zero identity. Also, if φ ( I ) is an r -ideal of R, we prove that I isa φ − r -ideal of R ⇔ I is r -ideal of R, see Theorem 3. Let I be a proper idealof R with p φ ( I ) = φ ( √ I ) . If I is a φ − r -ideal of R , then √ I is a φ − r -idealof R. With Theorem 4, under the condition p φ ( I ) = φ ( √ I ) , we obtain if I is a φ − r -ideal of R , then √ I is a φ − r -ideal of R. In Theorem 5, we showthat if φ preserves the order, then S k ∈ ∆ I k is a φ − r -ideal of R where { I k } k ∈ ∆ is a collection of φ − r -ideal of R. Moreover, in Theorem 8, we examine theconcept of φ − r -ideal in R (+) M, that is, the trivial ring extension, where M isan R -module. Also, we examine the notion of φ − r -ideal in S − R, where S isa multiplicatively subset of R , see Theorem 9. Finally, we characterize totalquotient rings in terms of φ − r -ideals. φ − r -ideals of Commutative RingsDefinition 1 Let R be a commutative ring with nonzeo identity and I be aproper ideal of R . Let φ : L ( R ) → L ( R ) ∪{∅} be a function. If ab ∈ I − φ ( I ) and Ann ( a ) = (0) imply that b ∈ I for each a, b ∈ R, then I is called a φ − r -ideal. Definition 2
If we define the map φ α : L ( R ) → L ( R ) ∪ {∅} as the followings, φ ∅ ( I ) = ∅ φ ∅ − r -ideal ( r -ideal) φ ( I ) = 0 R φ − r -ideal (weakly r -ideal) φ ( I ) = I φ − r - ideal (any ideal) φ ( I ) = I φ − r -ideal (almost r -ideal) φ n ( I ) = I n φ n − r -ideal ( n -almost r -ideal) ( n ≥ φ ω ( I ) = ∩ ∞ n =1 I n φ ω − r -ideal ( ω − r -ideal)Observe that 2-almost r -ideals are exactly almost r -ideals. Definition 3
Let R be a commutative ring with nonzero identity and I be aproper ideal of R . Let φ : L ( R ) → L ( R ) ∪ {∅} be a function.1. If ab ∈ I − φ ( I ) and Ann ( a ) = (0) imply that b n ∈ I for some naturalnumber n and a, b ∈ R, then I is called a φ − pr -ideal.2. If for every a ∈ I − φ ( I ) there is b ∈ I such that a = ab, then I is called a φ − pure ideal. Emel Aslankarayigit Ugurlu
3. If for every a ∈ I − φ ( I ) there is b ∈ I such that a = a b, then I is calleda φ − von Neumann regular ideal.Moreover, we define the concepts in Definition 2 for φ − pr -ideal, φ − pureideal and φ − von Neuman regular ideal. Remark 1
The elementary properties satisfy:1. An ideal is weakly r -ideal if and only if it is an r -ideal.2. Every φ − r -ideal is a φ − pr -ideal.3. Every φ − pure ideal is a φ − r -ideal.4. Every φ − von Neumann regular ideal is a φ − r -ideal.Throughout this paper φ : L ( R ) → L ( R ) ∪{∅} is a function. Since I − φ ( I ) = I − ( I ∩ φ ( I )), for any ideal I of R , without loss of generality, assume that φ ( I ) ⊆ I. Moreover, let ψ , ψ : L ( R ) → L ( R ) ∪ {∅} be two functions, if ψ ( I ) ⊆ ψ ( I ) for each I ∈ L ( R ) , we denote ψ ≤ ψ . Thus clearly, we havethe following order: φ ∅ ≤ φ ≤ φ ω ≤ · · · ≤ φ n +1 ≤ φ n ≤ · · · ≤ φ ≤ φ . Proposition 1
Let R be a ring and I be a proper ideal R. Let ψ , ψ : L ( R ) → L ( R ) ∪ {∅} be two functions with ψ ≤ ψ .
1. If I is a ψ - r -ideal of R, then I is a ψ - r -ideal of R. I is a r -ideal ⇔ I is a weakly r -ideal ⇒ I is an ω - r -ideal ⇒ I is an ( n + 1) -almost r -ideal ⇒ I is an n - r -ideal ( n ≥ ⇒ I is an almost r -ideal.3. I is an ω - r -ideal if and only if I is an n -almost r -ideal for each n ≥ . I is an idempotent ideal of R ⇒ I is an φ n − r -ideal of R for every n ≥ . Proof (1): It is evident.(2): Follows from (1).(3): Every ω - r -ideal is an n -almost r -ideal ideal for each n ≥ φ ω ≤ φ n . Now, let I be an n -almost r -ideal for each n ≥ . Choose two elements a, b ∈ R such that ab ∈ I − ∩ ∞ n =1 I n and a is regular. Then we have ab ∈ I − I n for some n ≥ . Since I is an n -almost r -ideal of R and a is regular, weconclude b ∈ I. Therefore, I is an ω - r -ideal.(4): Since I n = I for n ≥ , it is clear. Theorem 1
Let I be a proper ideal R.
1. Let φ ( I ) ⊆ r ( R ) . If I is a φ − r -ideal of R, then I/φ ( I ) is an r -ideal of R/φ ( I ) .
2. Let φ ( I ) be an r -ideal of R . If I/φ ( I ) is an r -ideal of R/φ ( I ) , then I is a φ − r -ideal of R Proof (1) : Suppose that I is a φ − r -ideal of R. Let a + φ ( I ) , b + φ ( I ) ∈ R/φ ( I )such that ( a + φ ( I ))( b + φ ( I )) ∈ I/φ ( I ) and a + φ ( I ) be a regular element of R/φ ( I ). Therefore ab ∈ I − φ ( I ) . Since a + φ ( I ) is regular element, a is a eneralizations of r -ideals of commutative rings 5 regular element of R . Indeed, if a is non-regular, there exists 0 = x ∈ R suchthat xa = 0 . Then we have ( a + φ ( I ))( x + φ ( I )) = 0 + φ ( I ) . Since a + φ ( I ) isregular, x + φ ( I ) = 0 + φ ( I ) , i.e., x ∈ φ ( I ) ⊆ r ( R ) . This gives us
Ann ( x ) = (0) , a contradiction with 0 = a ∈ Ann ( x ). Thus as I is since φ − r -ideal of R, weobtain b ∈ I. This implies that b + φ ( I ) ∈ I/φ ( I ) . It is done.(2) : Let
I/φ ( I ) be an r -ideal of R/φ ( I ) . Choose a, b ∈ R such that ab ∈ I − φ ( I ) which a is regular. Then ( a + φ ( I ))( b + φ ( I )) ∈ I/φ ( I ) . Also, since a isregular, a + φ ( I ) is regular. Indeed, if a + φ ( I ) is non-regular, there exists0 = x + φ ( I ) ∈ R/φ ( I ) such that ( a + φ ( I ))( x + φ ( I )) = 0 + φ ( I ) . Then ax ∈ φ ( I ) . Since φ ( I ) is r -ideal and a is regular, we get x ∈ φ ( I ) , a contradiction.Therefore, since a + φ ( I ) is regular and I/φ ( I ) is an r -ideal, we have b + φ ( I ) ∈ I/φ ( I ) , so b ∈ I. Theorem 2
Let I be a proper ideal R. Then the followings are equivalent:1. I is a φ − r -ideal of R.
2. For every regular element x of R , ( I : R x ) = I ∪ ( φ ( I ) : R x ) .
3. For every regular element x of R , ( I : R x ) = I or ( I : R x ) = ( φ ( I ) : R x ) . Proof (1) ⇒ (2) : It is clear that I ⊆ ( I : R x ) and ( φ ( I ) : R x ) ⊆ ( I : R x ) . So, the first containment is obtained. For the other, choose y ∈ ( I : R x ) . Then xy ∈ I. If xy ∈ φ ( I ) , y ∈ ( φ ( I ) : R x ) . If xy / ∈ φ ( I ) , as I is φ − r -ideal and x isregular, y ∈ I. Consequently, y ∈ I ∪ ( φ ( I ) : R x ) . (2) ⇒ (3) : It is obvious.(3) ⇒ (1) : Choose y, z ∈ R such that yz ∈ I − φ ( I ) which y is a regularelement. By (3), we conclude that ( I : R y ) = I or ( I : R y ) = ( φ ( I ) : R y ) . Let ( I : R y ) = I . Then as yz ∈ I, we get z ∈ ( I : R y ) = I, as needed. Let( I : R y ) = ( φ ( I ) : R y ) . Since z ∈ ( I : R y ) , we say yz ∈ φ ( I ) . This gives us acontradiction.
Proposition 2
Let I be a proper ideal R. If I is a φ − r -ideal of R, then I − φ ( I ) ⊆ zd ( R ) . Proof
Suppose that I − φ ( I ) * zd ( R ) . Then there exists a regular elementwhich x ∈ I − φ ( I ) . Consider x as x · R . Then as I is φ − r -ideal, we get1 R ∈ I. This contradicts with the choice of I . As conclusion, I − φ ( I ) ⊆ zd ( R ) . Theorem 3
Let φ ( I ) be an r -ideal of R. Then I is a φ − r -ideal of R ⇔ I is r -ideal of R. Proof ⇒ : Assume that a, b ∈ R such that ab ∈ I and a is a regular element.If ab / ∈ φ ( I ) , since I is φ − r -ideal, we have b ∈ I, as desired. If ab ∈ φ ( I ) , as φ ( I ) is r -ideal, we conclude that b ∈ φ ( I ) ⊆ I. Thus it is done. ⇐ : Obvious. Theorem 4
Let I be a proper ideal of R with p φ ( I ) = φ ( √ I ) . If I is a φ − r -ideal of R , then √ I is a φ − r -ideal of R. Emel Aslankarayigit Ugurlu
Proof
Choose a, b ∈ R such that ab ∈ √ I − φ ( √ I ) and a is regular. As ab ∈ √ I, there exists a natural number k which ( ab ) k = a k b k ∈ I. On the other hand, ab / ∈ φ ( √ I ) = p φ ( I ) , one can say a k b k / ∈ φ ( I ) . Also, since a is regular, it isobvious that a k is regular. As I is φ − r -ideal, we get b k ∈ I, i.e., b ∈ √ I. Theorem 5
Let { I k } k ∈ ∆ be a collection of ascending chain of φ − r -ideals of R. If φ preserves the order, then S k ∈ ∆ I k is a φ − r -ideal of R. Proof
Choose x, y ∈ R such that xy ∈ S k ∈ ∆ I k − φ ( S k ∈ ∆ I k ) which x is a regularelement. Then for some i ∈ ∆, xy ∈ I i . It is clear that xy / ∈ φ ( I i ) , since φ preserves the order. Thus, as I i is a φ − r -ideal of R, we see y ∈ I i ⊆ S k ∈ ∆ I k . Proposition 3
Let I be a proper ideal of R and φ ( I ) be an r -ideal of R. If I is a φ − r -ideal of R, then I ⊆ zd ( R ) . Proof
Suppose that I * zd ( R ) . There is an element x ∈ I but x / ∈ zd ( R ) . So x is a regular element. Consider x · R ∈ I. If x / ∈ φ ( I ) , then 1 R ∈ I, acontradiction. If x ∈ φ ( I ) , we obtain 1 R ∈ φ ( I ) ⊆ I, since φ ( I ) is an r -ideal of R. Proposition 4
Let φ ( I ) be an r -ideal of R and I be a prime ideal of R. Then I is φ − r -ideal ⇔ I ⊆ zd ( R ) . Proof
Suppose I is a prime ideal and φ ( I ) is an r -ideal of R . ⇒ : It is clear by Proposition 3. ⇐ : Choose a, b ∈ R such that ab ∈ I − φ ( I ) and a is a regular elemet . Since I is prime, either a ∈ I or b ∈ I. The first option contradicts with a ∈ r ( R ) , since I ⊆ zd ( R ) . Consequently, b ∈ I, as desired. Proposition 5
Let I be a proper ideal of R and x ∈ R − I. Let ( φ ( I ) : R x ) ⊆ φ (( I : R x )) . If I is a φ − r -ideal of R , then ( I : R x ) is φ − r -ideal of R .Proof Set y, z ∈ R such that yz ∈ ( I : R x ) − φ (( I : R x )) and y is regular. Thenwe get yzx ∈ I and yzx / ∈ φ ( I ) by ( φ ( I ) : R x ) ⊆ φ (( I : R x )) . Since y is regularand I is a φ − r -ideal, we get zx ∈ I, i.e., z ∈ ( I : R x ) , as required. Definition 4
Let I be a proper ideal of R . If for every ideal X and Y of R such that XY ⊆ I, XY * φ ( I ) and Ann ( X ) = (0) implies Y ⊆ I, then I iscalled strongly φ − r -ideal of R. Proposition 6
Every strongly φ − r -ideal is a φ − r -ideal.Proof Suppose that I is a strongly φ − r -ideal. Choose x, y ∈ R such that xy ∈ I − φ ( I ) and x is regular. Then it is clear that < x >< y > ⊆ I and < x >< y > * φ ( I ) . Now let us observe that
Ann ( < x > ) = (0) . Suppose thatfor 0 = u ∈ R, u < x > = (0) . This means that for all r ∈ R, urx = 0 . Since x is regular, we obtain ur = 0 for all r ∈ R. This implies u = 0 , a contradiction.Thus as I is strongly φ − r -ideal, < y > ⊆ I , so y ∈ I. eneralizations of r -ideals of commutative rings 7 Theorem 6
Let R satisfy the s. a. c. and I be a proper ideal. Let φ ( I ) be an r -ideal of R. I is a φ − r -ideal if and only if for every f. g. ideal X of R andevery ideal Y of R such that XY ⊆ I, XY * φ ( I ) and Ann ( X ) = (0) implies Y ⊆ I. Proof ( ⇒ ): Assume that I is a φ − r -ideal and φ ( I ) is a r -ideal. Let X be af. g. ideal of R and Y be an ideal of R such that XY ⊆ I, XY * φ ( I ) and Ann ( X ) = (0) . Suppose Y * I. Then there is an element 0 = y ∈ Y and y / ∈ I. On the other hand, since R satisfies the s. a. c. and X is f.g., thereis z ∈ X with Ann ( X ) = Ann ( z ) . This means that z is a regular element, as Ann ( X ) = (0) . Consider zy ∈ I. If zy / ∈ φ ( I ) , since I is a φ − r -ideal, we obtain y ∈ I, a contradiction. If zy ∈ φ ( I ) , as φ ( I ) is a r -ideal, again we conclude thesame contradiction. Thus it must be Y ⊆ I. ( ⇐ ): It is straightforward.In [8] by the help of Proposition 2.22, R. Mohammadian proved that if J/I is an r -ideal of R/I, then J is also an r -ideal of R, where I is an r -ideal of R contained in ideal J. Now, let us examine the proposition for the concept of φ − r -ideal.Let I be an ideal of R. Define φ I : L ( R/I ) → L ( R/I ) ∪ {∅} by φ I ( J/I ) =( φ ( J ) + I ) /I for every ideal I ⊆ J and φ I ( J/I ) = ∅ if φ ( J ) = ∅ . Notice that φ I ( J/I ) ⊆ J/I.
Theorem 7
Let I ⊆ r ( R ) be an ideal of R contained in ideal J. If J is a φ − r -ideal of R , then J/I is a φ I − r -ideal of R/I.
Proof
Let a + I, b + I ∈ R/I such that ( a + I )( b + I ) ∈ J/I − φ I ( J/I ) and a + I is a regular element. Then we see ab ∈ J − φ ( J ) . Also, since a + I isa regular element, a is a regular element. Indeed, if a is non-regular, there is0 = x ∈ R such that xa = 0 . Then we have ( a + I )( x + I ) = 0 + I. Since a + I is regular, x + I = 0 + I, i.e., x ∈ I ⊆ r ( R ) . This gives us a contradiction as0 = a ∈ Ann ( x ). Thus, as J is a φ − r -ideal, b ∈ J, so b + I ∈ J/I.
Proposition 7
Let I be a r -ideal of R contained in ideal J. If J/I is an r -idealof R/I, then J is a φ − r -ideal of R. Proof
Choose x, y ∈ R such that xy ∈ J − φ ( J ) which x is a regular element.Then we have 2 cases:Case 1: Let xy ∈ I. Then as I is a r -ideal, we get y ∈ I ⊆ J, as desired.Case 2: Let xy ∈ J − I. This implies that xy + I = ( x + I )( y + I ) ∈ J/I . Also, as x is a regular element, x + I is regular. Indeed, if x + I is non-regular, there is an element 0 = a + I ∈ R/I such that ( x + I )( a + I ) = 0 + I. This means that xa ∈ I. But as x regular and I is r -ideal, we see a ∈ I, acontradiction. Therefore, since J/I is an r -ideal of R/I and x + I is regular,we see y + I ∈ J/I, so y ∈ J. Emel Aslankarayigit Ugurlu
Let R be a commutative ring with nonzero identity and M be an R -module.Then the idealization, R (+) M = { ( a, m ) : a ∈ R, m ∈ M } is a commuta-tive ring with componentwise addition and the multiplication ( a, m )( b, n ) =( ab, an + bm ) for each a, b ∈ R and m, n ∈ M. In addition, if J is an ideal of R and N is a submodule of M, then J (+) N is an ideal of R (+) M if and onlyif JM ⊆ N [3]. Also note that z ( R (+) M ) = { ( r, m ) : r ∈ z ( R ) ∪ z ( M ) } , where z ( M ) = { r ∈ R : rm = 0 for some 0 = m } [3]. Theorem 8
Let M be an R -module such that z ( R ) = z ( M ) . Let ψ : L ( R ) → L ( R ) ∪ {∅} and ψ : L ( R (+) M ) → L ( R (+) M ) ∪ {∅} be two function such that ψ ( I (+) M ) = ψ ( I )(+) M for a proper ideal I of R. If I (+) M is a ψ − r -idealof R (+) M, then I is a ψ − r -ideal of R .Proof Let a, b ∈ R such that ab ∈ I − ψ ( I ) and a is regular. Since ψ ( I (+) M ) = ψ ( I )(+) M, we have ( a, b, ∈ I (+) M − ψ ( I (+) M ) . Also, since z ( R ) = z ( M ) , it is clear that ( a,
0) is a regular element of R (+) M. This means that( b, ∈ I (+) M, so b ∈ I, as needed.Let S be a multiplicatively closed subset of R . Then S − R := { rs : r ∈ R, s ∈ S } is called the localization of R at the multiplicatively closed set S and it is denoted by R S . Let φ : L ( R ) → L ( R ) ∪ {∅} be a function. Define φ S : L ( R S ) → L ( R S ) ∪ {∅} by φ S ( J ) = S − φ ( J ∩ R ) for every ideal J of R S and φ S ( J ) = ∅ if φ ( J ∩ R ) = ∅ . Notice that φ S ( J ) ⊆ J. Theorem 9
Let φ : L ( R ) → L ( R ) ∪ {∅} be a function and S be a multiplica-tively closed subset of R such that S ⊆ r ( R ) . Suppose that S ∩ I = ∅ for anideal I of R. If I is a φ − r -ideal and φ ( I ) S ⊆ φ S ( I S ) , then the followings arehold:1. I S is a φ S − r -ideal of R S .
2. If I S = φ ( I ) S , then I S ∩ R ⊆ zd ( R ) . Proof (1): Choose a/s, b/t ∈ R S such that as bt ∈ I S − φ S ( I S ) and as is regular.Then there is s ∈ S such that s ab ∈ I. Also, it is clear that since as is regular, a is regular. On the other hand, as bt / ∈ φ S ( I S ) implies that s ab / ∈ φ S ( I S ) ∩ R for all s ∈ S. By φ ( I ) S ⊆ φ S ( I S ), we have s ab / ∈ φ ( I ) for all s ∈ S. Thus wecan say as b ∈ I − φ ( I ) . Then as I is φ − r -ideal, s b ∈ I. Hence bt = s bs t ∈ I S , as required.(2): Set a ∈ I S ∩ R. Then there is s ∈ S such that as ∈ I. Also, s / ∈ I by S ∩ I = ∅ . If as / ∈ φ ( I ) , then we obtain a is non-regular since s / ∈ I. If as ∈ φ ( I ) , then a ∈ φ ( I ) S ∩ R. Thus I S ∩ R ⊆ zd ( R ) ∪ ( φ ( I ) S ∩ R ) . Then byour assumption I S = φ ( I ) S , we obtain I S ∩ R ⊆ zd ( R ) . Let φ i : L ( R i ) → L ( R i ) ∪ {∅} be a function for each i = 1 , , . . . , m and R = R × R × · · · × R m , where R , R , . . . , R m are commutative rings . Then φ × : L ( R ) → L ( R ) ∪ {∅} , defined by φ × ( I × I × · · · × I m ) = φ ( I ) × φ ( I ) ×· · ·× φ m ( I m ) , becomes a function. Particularly, if each φ i is the function φ n , then φ × − r -ideal is denoted by φ × n − r -ideal. eneralizations of r -ideals of commutative rings 9 Theorem 10
Let m ≥ and n ≥ be two integers. Suppose that R = R × R × · · · × R m , where R , R , . . . , R m are commutative rings. The followingstatements are equivalent:(i) R , R , . . . , R m are total quotient rings.(ii) Every proper ideal of R is an φ × n − r -ideal.Proof ( i ) ⇒ ( ii ) : Suppose that R , R , . . . , R m are total quotient rings so thatevery element of R is either zerodivisor or unit. Thus every ideal ideal of R isa trivially φ × n − r -ideal.( ii ) ⇒ ( i ) : Assume that every ideal of R is an φ × n − r -ideal. Let a ∈ R be a regular element. Now, we will show that a is unit. Suppose to thecontrary. Put I = ( a ) × R × · · · × R n . Then I is proper so is a φ × n − r -ideal by assumption. Now, put x = ( a, , , . . . ,