COMPARABILITY, DISTRIBUTIVITY AND NON-COMMUTATIVE φ-RINGS

CHRISTIAN LOMP AND ALVERI SANT’ANA

Abstract. The purpose of this paper is to develop the theory of non-commutative φ-rings usingconcepts from distributive rings and rings with comparability. We recover the main statementsfor commutative φ-rings as obtained by Anderson and Badawi.

1. Introduction

A. Badawi considered in [4] a class of commutative rings R, called φ-rings, whose nil radicalNil(R) forms a prime ideal, comparable to any other ideal of R in order to create a frameworkto study pseudo-valuation rings. This class of rings was used by Badawi and his collaborators tostudy Dedekind, Prufer, Bezout or Krull rings with nilpotent elements (see [1, 2, 5, 6, 7, 8]). Thepurpose of this paper is to develop a similar theory for non-commutative rings using concepts fromdistributive rings and rings with comparability.

Let us first review the commutative setting from [4]. Let R be a commutative φ-ring, with T (R)its total ring of quotient, which is the localization of R at its zero divisors N(R), and considerthe canonical homomorphism φ : T (R) = RN(R) → RNil(R) induced by Nil(R) ⊆ N(R) whereφ(ab−1) = ab−1 for any a ∈ R and b ∈ R \ N(R). Restricting φ to R we have a homomorphismφ : R → RNil(R) with φ(a) = a and kernel tS(R) = {a ∈ R | ∃x 6∈ Nil(R) : ax = 0} ⊆Nil(R). From [4] it follows, that φ(R) is a φ-ring, Nil(φ(R)) = Nil(RNil(R)) = φ(Nil(R)) andT (φ(R)) = RNil(R) is a (quasi)local ring with maximal ideal Nil(φ(R)), and RNil(R)/Nil(φ(R)) =T (φ(R))/Nil(φ(R)) is the quotient field of φ(R)/Nil(φ(R)).

In this paper we use the theory of comparability and distributivity for non-commutative ringsto study non-commutative φ-rings. Note that the nil radical Nil(R) of a commutative ring Rcoincides with its prime radical, i.e., the intersection of its prime ideals. We replace Nil(R) by thegeneralized nilradical Ng(R) which is the intersection of the completely prime ideals of R. Recallthat an ideal P is completely prime in R if R/P is a domain. Non-commutative φ-rings will berings whose generalized nil radical Ng(R) is a left and right waist such that R \Ng(R) is a rightdenominator set. Hence the right localization RNg(R) exists and we denote by φ : R → RNg(R)

the canonical map.In section 2 we gather some information on non-commutative φ-rings. In particular we show in

Theorem 2.3 that φ(R) is again a φ-ring such that RNg(R) is a localization of φ(R) at zero-divisorsand that φ(R) can be understood as a pullback of a local ring and a right Ore domain (see 2.5;compare with [2, Theorem 2.6, 3.2] and [9, Theorem 2.2]). Moreover, we prove a more generalresult on comparability (2.9 which generalizes [11, 3.3]), which shows that RNg(R) is a right chainring if and only if φ(R) has right Ng(φ(R))-comparability (2.7). In section 3, we generalize [2,Theorem 2.6] and prove a characterization of φ-distributive rings showing that they coincide with

2000 Mathematics Subject Classification. 16A66, 16U20, 16N40, 13F05.Key words and phrases. Rings with comparability, distributive rings, Prufer rings, φ-rings, nil radical, completely

prime ideals.This work started in the framework of the project Interaccoes entre algebras e co-algebras between the Univer-

sidade do Porto and Universidade Federal do Rio Grande do Sul and Universidade de Sao Paulo financed throughGRICES (Portugal) and CAPES (Brazil) during the second author’s visit to the UP. The first author was partiallysupported by Centro de Matematica da Universidade do Porto (CMUP), financed by FCT (Portugal) throughthe programs POCTI (Programa Operacional Ciencia, Tecnologia, Inovacao) and POSI (Programa OperacionalSociedade da Informacao), with national and European community structural funds. The second author was sup-ported by CAPES (Projeto 135/05 - BEX 3141/05-5) and would like to thank the Departamento de MatemticaPura for its hospitality.

1

2 CHRISTIAN LOMP AND ALVERI SANT’ANA

φ-Prufer rings and that they can be characterized as those rings such that R/Ng(R) is a Pruferdomain and RNg(R) is a right chain ring (see Theorem 4.3). A similar characterization is provedin Theorem 5.4 for Bezout rings.

All rings in this paper are unital and associative. We denote the Jacobson radical of a ring Rby Jac(R). Moreover, the notations ⊂ and ⊃ will mean strict inclusions. Recall that if I is a rightideal of R and S is a multiplicative closed set then the S-saturation of I in R is defined by

tS(R/I) = {a ∈ R : ∃s ∈ S such that as ∈ I}.I is said to be S-saturated if I = tS(R/I).

2. Comparability and non-commutative φ-rings

A commutative ring R is a φ-ring if the nil radical Nil(R) is a prime ideal comparable to allideals of R. For an arbitrary ring R we will choose the generalized nil radical Ng(R) of R as asuitable substitute of Nil(R). Let H denote the class of rings with unit such that Ng(R) is acompletely prime ideal of R and a waist as a right and a left ideal of R. Recall that a waist ina lattice is an element which is comparable to any other element. An ideal is a left (resp. right)waist if it is a waist in the lattice of left (resp. right) ideals. Clearly any domain, any chain ringand any commutative φ-ring is contained in H. Since localization is an essential tool in the studyof commutative φ-ring, we define an arbitrary ring R to be a right φ-ring if R ∈ H and R \Ng(R)is a right denominator set. Recall that a multiplicatively closed subset S of a ring R is a rightdenominator set of R if the following conditions hold (i) for every a ∈ R and s ∈ S, there existt ∈ S and b ∈ R such that at = sb (right Ore condition) and (ii) for every s ∈ S and a ∈ Rsuch that sa = 0, there exists t ∈ S with at = 0 (right reversibility condition). If S ⊆ R is aright denominator set of R, then the right ring of fractions of R with respect to S exists and wedenote it by R[S−1]. The map φ : R → R[S−1] with φ(r) = r is a ring homomorphism. Note thatR[S−1] = {φ(a)φ(s)−1 : a ∈ R, s ∈ S} and

tS(R) = Ker(φ) = {a ∈ R : as = 0, for some s ∈ S}If P is a completely prime ideal of R such that S = R \ P is a right denominator set, then wewrite also RP for R[S−1]. Hence examples of right φ-rings are right Ore domains and commutativeφ-rings. We will also see that some right distributive and right Bezout domains are right φ-rings,but first we will consider rings with comparability.

2.1. A class of φ-rings comes from trivial extensions.

Example. Let A be a (not necessarily commutative) ring and M be an A-bimodule. We form thetrivial extension R = AnM whose underlying additive abelian group is A⊕M with multiplication

(a,m)(b, n) := (ab, an + mb).

Then R is a right φ-ring with Ng(R) = 0nM if and only if A is a right Ore domain and AM andMA are divisible.

In this case tR\Ng(R)(R) = 0 n Sing(MA) where Sing(MA) denotes the singular submodule ofMA.

Proof. Suppose R is a φ-ring with Ng(R) = 0nM . Since Ng(R) is completely prime, A ' R/Ng(R)is a domain. For any a, b ∈ A with b 6= 0, there exist (x, u), (y, v) ∈ R with x 6= 0 such that(a, 0)(x, u) = (b, 0)(y, v) if and only if (ax, au) = (by, bv). Hence ax = by, i.e. A is a right Oredomain. Let 0 6= a ∈ A. Then R(a, 0) ⊇ 0 nM and for all m ∈ M , there exists (b, n) ∈ R suchthat

(0,m) = (b, n)(a, 0) = (ba, na).Hence na = m for some n ∈ M , i.e., Ma = M . Similarly one shows that aM = M .

Now suppose that A is a right Ore domain and MA and AM are divisible and set I = 0nM . Let(a,m) 6∈ I, i.e., a 6= 0. We have to show that R(a,m) ⊃ I. For any x ∈ M there exists n ∈ M suchthat na = x. Hence (0, x) = (0, na) = (0, n)(a,m) ∈ R(a,m), i.e., I is a left waist. Analogously Iis a right waist and since R/I ' A is a domain, I is completely prime, i.e., I ⊇ Ng(R). On theother hand since for any completely prime ideal P of R, (I + P )/P is nilpotent since I2 = 0, we

COMPARABILITY, DISTRIBUTIVITY AND NON-COMMUTATIVE φ-RINGS 3

have I ⊆ P , i.e., I = Ng(R). To show that S = R \Ng(R) is a right denominator set we need toshow that for all (a,m) ∈ R and (b, n) ∈ S we have (a, m)S ∩ (b, n)R 6= ∅. Let (a,m), (b, n) ∈ Rwith b 6= 0. If a = 0, then there exists u ∈ M such that m = bu, because AM is divisible andhence (0,m)(1, 0) = (0,m) = (0, bu) = (b, n)(0, u) as desired. Now let a 6= 0. Since A is a rightOre domain, there exist x, y ∈ A with x 6= 0 such that ax = by. Moreover since AM is divisibleand b 6= 0, there exists u ∈ M such that bu = mx. Since a 6= 0 and AM is divisible, there existsv ∈ M such that av = ny. Thus

(a,m)(x, v) = (ax, av + mx) = (by, ny + bu) = (b, n)(y, u)

and this shows that S is a right denominator set.Hence if R is a φ-ring with Ng(R) = 0nM , then

tS(R) = {(a, m) ∈ R | ∃(b, n) ∈ S with (a,m)(b, n) = (0, 0)}.Take (a,m) ∈ tS(R), then (a,m)(b, n) = (ab, an + mb) = (0, 0) and hence ab = 0 and an + mb = 0for b 6= 0. Thus a = 0 and mb = 0, i.e., m ∈ Sing(MA). On the other hand for any m ∈ Sing(MA)there exists b 6= 0 such that mb = 0 and hence (0,m)(b, 0) = (0, 0). Thus tS(R) = 0n Sing(MA).

¤

2.2. Basic properties of φ-rings. Let R be a right φ-ring and set S = R \Ng(R). Denote byR[S−1] the localization of R at S and by φ : R → RNg(R) the canonical map φ(r) = r = r1−1.

Proposition. The following properties hold for a right φ-ring R:(1) R[S−1] is a local ring with maximal ideal equal to Jac(R[S−1]) = Ng(R)[S−1].(2) I is a completely prime ideal of R if and only if φ(I) is completely prime in φ(R).(3) Ng(φ(R)) = φ(Ng(R)) = Ng(R)[S−1].(4) Ng(φ(R)) contains all left and all right zero divisors of φ(R).

Proof. Recall that the kernel of φ is tS(R) where S = R \Ng(R).(1) This is clear since Ng(R) is a completely prime ideal of R.(2) Since φ is a ring homomorphism and any completely prime ideal contains tS(R), we have

φ(a)φ(b) = φ(ab) ∈ φ(I) if and only if ab ∈ I. Thus I is completely prime ideal of R if and only ifφ(I) is a a completely prime ideal of φ(R).

(3) Note that if K = φ(I) is a completely prime ideal of φ(R), then I is a completely primeideal of R by (2). Hence

Ng(φ(R)) =⋂{φ(P ) | P is a completely prime ideal of R} = φ(Ng(R)).

Note that

Ng(φ(R)) = {a : a ∈ Ng(R)} ⊆ Ng(R)[S−1]} = {as−1 : a ∈ Ng(R), s ∈ S}.Let as−1 ∈ Ng(R)[S−1]. Since Ng(R) is a waist as a left ideal of R and s 6∈ Ng(R), there existsr ∈ R such that a = rs. Thus as−1 = rss−1 = r and it follows Ng(φ(R)) = Ng(R)[S−1] =Jac(R[S−1]).

(4) Let φ(a) be a left zero divisor in φ(R). Then there exists 0 6= φ(b) ∈ φ(R) such thatφ(ba) = 0, i.e., ba ∈ tS(R) and there exists s ∈ S = R \ Ng(R) with bas = 0. Since φ(b) 6= 0,b 6∈ tS(R) and thus a 6∈ S, i.e., φ(a) ∈ φ(Ng(R)) = Ng(φ(R)). Analogously we can show that anyright zero divisor of φ(R) is contained in Ng(φ(R)) using the fact that S is a right denominatorset. ¤

2.3. Like in the commutative case we obtain that φ(R) is a right φ-ring and that RNg(R) is thelocalization of φ(R) at non-zero divisors.

Theorem. Let R be a right φ-ring and set S = R \Ng(R). Then φ(R) is a right φ-ring, φ(S) =φ(R) \Ng(φ(R)) consists of non-zero divisors and φ(R)[φ(S)−1] ' RNg(R).

Proof. Since φ(R) is a subring of the local ring R[S−1] we can apply [11, Lemma 2.1(i)] to obtainthat Ng(φ(R)) = Jac(R[S−1]) is a completely prime ideal of φ(R) which is a waist as a right idealand as a left ideal of φ(R). Hence, φ(R) ∈ H. Since S satisfies the right Ore condition and φ(S) =

4 CHRISTIAN LOMP AND ALVERI SANT’ANA

φ(R) \ Ng(φ(R)), also φ(S) satisfies the right Ore condition. Moreover φ(S) = φ(R) \ Ng(φ(R))consists of non-zero divisors by 2.2(4). Hence φ(S) is left and right reversible, i.e., φ(S) is a rightdenominator set, φ(R) is a right φ-ring and R[S−1] ' φ(R)φ(S). ¤2.4. Rings with comparability. Recall that a ring R is said to be a ring with right P -comparability, where P is a completely prime ideal of R contained in Jac(R), if for every elementsa, b ∈ R, we have either aR ⊆ bR, or bR ⊆ aR, or tR\P (R/aR) = tR\P (R/bR). Analogously onedefines rings R with left P -comparability and says that R is a ring with comparability if it is a ringwith right and left P -comparability for any completely prime ideal P contained in its Jacobsonradical Jac(R) (see [12]). By [13, Proposition 2.2] a ring R has right P -comparability, where P isa completely prime ideal of R contained in Jac(R) such that S = R \ P is a right Ore set if andonly if for every x, y ∈ R we have either xR ⊆ yR or y ∈ tS(R/xR).

Proposition. Any ring with comparability whose Jacobson radical contains a completely primeideal belongs to H.

Proof. Let R be a ring with comparability, then by [12, Lemma 1.3] any completely prime idealcontained in Jac(R) is a left and right waist. Thus, the set of completely prime ideals of Rcontained in Jac(R) is linearly ordered by inclusion and so Ng(R) is a completely prime ideal ofR which is a waist as a right and a left ideal. ¤2.5. We have the following result which shows that right φ-rings can be understood as pullbacks(compare with [2, Theorem 2.6, 3.2] and [9, Theorem 2.2]).

Corollary. Let R be a right φ-ring. Then φ(R) is a pullback of a local ring T with maximal idealJac(T ) and a right Ore domain D such that T/Jac(T ) is the right skew field of fraction of D.

Proof. Since φ(S) = φ(R) \Ng(φ(R)) is a right Ore set and Ng(φ(R)) is a completely prime idealof φ(R), we have that φ(R)/Ng(φ(R)) is a right Ore domain with right skew field of fractions equalto RNg(R)/Ng(R)[S−1]. Hence, φ(R) = π−1(φ(R)/Ng(φ(R))) is a pullback:

φ(R) −−−−→ φ(R)/Ng(φ(R))y

yR[S−1] π−−−−→ R[S−1]/Jac(R[S−1])

where the vertical arrows are the natural inclusions, RNg(R) is a local ring and φ(R)/Ng(φ(R)) isa right Ore domain with right skew field of fractions equal to R[S−1]/Ng(φ(R)). ¤2.6. Subrings of local rings which contain its maximal ideal. Let R be a right φ-ring. By2.2 φ(R) can be described as a subring of a local ring containing its Jacobson radical.

So let T be a local ring with maximal ideal Jac(T ) 6= 0 and suppose that R is a subring ofT such that Jac(T ) ⊆ R. By [11, Lemma 2.1], Jac(T ) is a completely prime ideal of R whichis a left and a right waist of R. More can be said about the structure of R if R is a ring withJac(T )-comparability. The next result extends [11, Lemma 2.3].

Theorem. Let T be a local ring which is not simple and let R be a subring of T such thatJac(T ) ⊆ R. Consider the following statements:

(i) R is a ring with right Jac(T )-comparability.(ii) T = RJac(T ) is a right chain ring.

(iii) R/Jac(T ) is a right Ore domain with right skew field of fractions T/Jac(T ).Then (i) ⇔ (ii) ⇒ (iii) and all conditions are equivalent if T is a right chain ring.

Proof. Let J = Jac(T ) and set S := R \ J .(i) ⇒ (ii) : Suppose (i) holds, i.e., that R is a ring with right J-comparability. By [12,

Proposition 1.4] it follows that S is a right Ore set. Since J is the set of non-units of T , any zerodivisor of R belongs to J , which shows that S is a right denominator set and that RJ exists. Also,it is easy to see that every localization of a ring with right J-comparability is a right chain ring.So, we need to show that RJ = T . Let t ∈ T and j ∈ J . Since J is a two sided ideal of T , we have

COMPARABILITY, DISTRIBUTIVITY AND NON-COMMUTATIVE φ-RINGS 5

jt, tj ∈ J . Since R is a ring with right J-comparability we have j ∈ (jt)R or jt ∈ tS(R/jR). Ifj = jtr, for some r ∈ R, then j(1− tr) = 0 and so 1− tr is not invertible in T , i.e., 1− tr ∈ J ⊆ Rso that tr ∈ R. Now we observe that r 6∈ J , because tr 6∈ J clearly. If jts = jr, for some s ∈ S andr ∈ R, then we have j(ts − r) = 0 and analogously as before, we have ts − r ∈ R and so ts ∈ R.Therefore, RJ = T as required. Thus (ii) holds.

(ii) ⇒ (i) : Suppose T = RJ is a right chain ring and let a, b ∈ R such that a 6∈ bR. Thena 6∈ bRJ and hence b ∈ aRJ . Hence b = ars−1 for some s ∈ S such that bs = ar (in R). Thusb ∈ tS(R/aR), i.e., R has J-comparability.

(ii) ⇒ (iii) : For any t ∈ T there exist a ∈ R, s ∈ S such that t = as−1. If 0 6= t + J ∈ T/J ,then (t + J)(s + J) = a + J ∈ R/J , i.e., T/J is the right skew field of fractions of R/J .

If T is a right chain ring, then (iii) ⇒ (i) by [12, Theorem 2.4]. ¤

2.7. The following Corollary is a consequence of 2.3 and 2.6.

Corollary. Let R be a right φ-ring. Then φ(R) is a ring with right Ng(φ(R))-comparability if andonly if RNg(R) is a right chain ring.

Let A = k be a field, M any vector space over k of dimension ≥ 2 and R = A nM - as inexample 2.1. Then R is a commutative local φ-ring with R[S−1] = R not being a chain ring, i.e.,R = φ(R) is not a ring with (right) Ng(φ(R))-comparability.

2.8. In [23] Tuganbaev studied (right and left) distributive rings whose zero divisors are containedin the Jacobson radical. Ferrero and Mazurek extended Tuganbaev’s main results [23, Theorems1 and 2] to Bezout rings in [11, Corollaries 3.3 and 3.4], giving also shorter proofs. Later on wewill see that every right Bezout ring containing a completely prime ideal in its Jacobson radicalis a ring with right comparability and so we can reprove [11, Corollary 3.3] by applying the leftand right versions of [12, Theorem 3.3].

Denote by Nr(R) = {x ∈ R | ∃y ∈ R \ {0} : yx = 0} the set of right zero divisors of R andlet Nl(R) be the set of left zero divisors. To generalize [11, Corollary 3.4], we need the followingLemma.

Lemma. Let R be a ring with right P -comparability and set S = R \ P .

(1) If Nl(R) ⊆ P , then S is a right denominator set.(2) If Nr(R) ⊆ P and S is right denominator set, then Nl(R) ⊆ P .

Proof. (1) Since R is a ring with right P -comparability, S is a right Ore set whose elements arenot left zero divisors since S ∩ Nl(R) = ∅. Thus for any a ∈ R and s ∈ S we have that sa = 0implies that a = 0. Hence S is a right denominator set.

(2) Take x ∈ Nl(R). Then there exists a ∈ R \ {0} such that xa = 0. If x ∈ S, then there existst ∈ S such that at = 0, because S is a right denominator set. Thus we have t ∈ Nr(R) ∩ S = ∅, acontradiction. Hence x ∈ P . ¤

2.9. Denote by N(R) = Nl(R)∪Nr(R). As a consequence we get a generalization of [11, Lemma3.1] and[11, Corollary 3.4].

Proposition. The following conditions are equivalent:

(i) R is a ring with right P -comparability for some completely prime ideal P which is a leftwaist of R and N(R) ⊆ P ⊆ Jac(R).

(ii) R is a subring of a right chain ring T such that Jac(T ) ⊆ R, R/Jac(T ) is a right Oredomain and T/Jac(T ) is the right skew field of fractions of R/Jac(T ).

Proof. (ii) ⇒ (i): This is clear.(i) ⇒ (ii): Since Nl(R) ⊆ P we have that RP exists by Lemma 2.8. Let T = RP and take

x ∈ Jac(T ) = PRP . There exist a ∈ P and t 6∈ P such that x = at−1. Since P is a left waist ofR, we have a = rt, for some r ∈ P and so x = at−1 = rtt−1 = r and consequently x ∈ P . HenceJac(T ) = P ⊆ R. Now we can apply Theorem 2.6 to get the result. ¤

6 CHRISTIAN LOMP AND ALVERI SANT’ANA

2.10. Waists. We will refer to a right multiplicative subsemigroup I of R as a right multiplicativeideal of R (i.e., for every x, y ∈ I we have xy ∈ I). A right multiplicative ideal I is called completelyprime if ab ∈ I ⇒ a ∈ I or b ∈ I. For a right multiplicative ideal I of a ring R define for any x ∈ Rthe right multiplicative ideal (I : x) := {a ∈ R : xa ∈ I}. If I is a right ideal of R, then so is (I : x).Following [10], we define the right associated prime multiplicative ideal of I as Pr(I) := ∪x6∈I(I : x)which is a completely prime right multiplicative ideal. Clearly if I is a completely prime rightmultiplicative ideal, then R\I is a multiplicatively closed set and I = Pr(I) holds if I is completelyprime. Moreover, it is known that any completely prime one-sided ideal I contained in Jac(R) istwo-sided (see [15, Lemma 2.5]).

A right multiplicative ideal I ⊂ R is called a right waist of R if it is a waist in the lattice ofright multiplicative ideals of R (see [3]). By definition a right multiplicative ideal I of R is a rightmultiplicative waist if I ⊂ aR, for every a ∈ R \ I. Moreover a multiplicative waist is containedin any maximal right ideal and therefore contained in the Jacobson radical.

The following Proposition generalizes [15, Theorem 2.6]:

Proposition. Let R be a ring and I a completely prime right multiplicative ideal of R. If I is aright waist of R, then it is a two-sided ideal.

Proof. Since I is completely prime, we have Pr(I) = I ⊆ Jac(R). By [13, Lemma 1.2] for anyx ∈ R \ I we have I = x(I : x). Thus I ⊆ xI ⊆ xJac(R) since (I : x) ⊆ Pr(I) = I ⊆ Jac(R). LetI ′ =

⋂x 6∈I xJac(R) which is clearly a right ideal. We will show that I = I ′. For any y ∈ I ′ \ I,

we have yR ⊆ I ′ ⊆ yJac(R), thus y(1− a) = 0 for some a ∈ Jac(R). Since 1− a is invertible, weget y = 0, i.e., I = I ′. Hence I is a right ideal and [15, Lemma 2.5] applies which shows that I istwo-sided. ¤

2.11. In [11, Section 3] right waists of right distributive and of right Bezout rings which aresubrings of local rings were described. We are able to improve these results to the larger class ofrings with right comparability, i.e., R has P -comparability with respect to all completely primeideals P contained in Jac(R).

Theorem. Let T be a local ring and R a subring of T containing Jac(T ). Suppose that R is aring with right comparability and let I be a proper right ideal of R. Then I is a right waist of Rif and only if Pr(I) ⊆ Q, where Q is the largest completely prime ideal of R contained in Jac(R).

Proof. Suppose that I is a right waist of R. Since R is a subring of a local ring T , it is easy to seethat Nr(R) ⊆ Jac(T ) ⊆ Q. By contradiction, we suppose that there exists a ∈ Pr(I) \Q and letb 6∈ I such that ba ∈ I. By [13, Lemma 1.2], we have b(I : b) = I ⊆ bQ and for every x ∈ (I : b)we have bx = bqx, for some qx ∈ Q. This implies that b(x− qx) = 0 so that x− qx ∈ Nr(R) ⊆ Qand it follows that (I : b) ⊆ Q, a contradiction.

Conversely, if Pr(I) ⊆ Q then we have that Pr(I) is a two-sided completely prime ideal of Rby 2.10 and so R has right Pr(I)-comparability. Hence, if a ∈ I and b 6∈ I are elements suchthat b ∈ (aR)S−1

I , where SI = R \ Pr(I), then there exists s ∈ SI such that bs ∈ aR ⊆ I andconsequently s ∈ Pr(I) ∩ SI = ∅, a contradiction. Therefore, we have a ∈ bR and the proof iscomplete now. ¤

3. Zero divisors of rings with comparability

The purpose of this section is to determine the relation between the zero divisors of a rightφ-ring R and its generalized nil radical Ng(R). Note that the right zero divisors Nr(R) form aright multiplicative ideal which is completely prime. Our purpose will be to determine, when zerodivisors form a one-sided ideal contained in the Jacobson radical.

3.1. The Proposition 2.10 has some consequences for rings with P -comparability and should becompared with [15, Theorem 2.6].

Corollary. Let I be a completely prime right multiplicative ideal of a ring R with P -comparability.Suppose that I ⊆ P . Then I is a right waist and a two-sided ideal of R.

COMPARABILITY, DISTRIBUTIVITY AND NON-COMMUTATIVE φ-RINGS 7

Proof. Set S = R \ P . For any x ∈ I and y ∈ R \ I we have xR ⊆ yR or tS(R/xR) = tS(R/yR).If y ∈ tS(R/xR) then there exist s ∈ S such that ys ∈ xR ⊆ I, hence y ∈ I as I was completelyprime and s 6∈ I - a contradiction. Thus tS(R/xR) 6= tS(R/yR) and xR ⊆ yR follows, i.e., I is aright waist of R. By Proposition 2.10, I is two-sided. ¤3.2. Corollary 3.1 implies that the set of right zero divisors Nr(R) of a ring R with P -comparabilityforms a two-sided ideal and a right waist of R if Nr(R) ⊆ P ⊆ Jac(R). In particular we have forright φ-rings using 2.7 the following

Corollary. Let R be a right φ-ring whose localization RNg(R) is a right chain ring. Then Nr(φ(R)) =Ng(φ(R)).

3.3. We finish this section with the observation that the generalized nil radical is precisely the setof right zero divisors for a prime ring with P -comparability.

Proposition. Let R be a prime ring with right P -comparability. Then Nr(R) = Ng(R) ⊆ P .

Proof. We first note that if R is a domain then nothing is to proof. Thus suppose that R is aprime ring but not a domain. Let 0 6= x ∈ R be such that the right annihilator r(x) of x isnot empty. For any a ∈ r(x) and b 6∈ r(x) one has aR ⊆ bR or tS(R/aR) = tS(R/bR), whereS = R \ P . Suppose b ∈ tS(R/aR), then there exists s ∈ S such that bs = ar for some r ∈ R.Thus we have 0 = xarP = xbsP = xbP , because sP = P , which implies xb = 0 since R is primeand P 6= 0, i.e., b ∈ r(x) - a contradiction. Hence any right annihilator r(x) is a right waist ofR. Since Nr(R) =

⋃0 6=x∈R r(x) also Nr(R) is a completely prime right multiplicative ideal which

is a right waist of R. By [15, Lemma 2.5] Nr(R) ⊆ Jac(R) and by Proposition 2.10 Nr(R) is atwo-sided ideal. Hence Ng(R) ⊆ Nr(R). Let a ∈ Nr(R). Then there exists 0 6= x ∈ R such thatxa = 0. If a 6∈ Ng(R), then Ng(R) = aNg(R) and 0 = xaNg(R) = xNg(R) and it follows thatNg(R) = 0 and R is a domain - a contradiction. ¤

4. Distributivity

A ring R is called right distributive if the lattice of right ideals of R is distributive, i.e., forevery right ideals A,B, C of R we have A∩ (B + C) = (A∩B) + (A∩C) (see [20] and [14]). R isdistributive if it is a left and right distributive ring.

4.1. It follows from [17, lemma 3.1] and [15, Lemma 3.4] that any right distributive ring R hasright P -comparability for any completely prime ideal contained in Jac(R).

Proposition. A distributive ring belongs to the class H if and only if its Jacobson radical containsa completely prime ideal. In particular any distributive ring R in H has right Ng(R)-comparability.

Proof. If R is distributive and P ⊆ Jac(R) is a completely prime ideal, then it is a left and rightwaist by [20, Proposition 2.1(ii)], Thus the set of completely prime ideals of R which is containedin the Jacobson radical of R is a linearly ordered set and so Ng(R) is a completely prime ideal ofR which is a waist as a right and a left ideal. If R ∈ H, then Ng(R) ⊆ Jac(R) since it is a rightwaist. ¤4.2. A right Prufer ring is a right distributive ring R that is right localizable, i.e., for every maximalright ideal I of R the set R \ I is a right denominator set ([22, p. 71]). Note that there are rightdistributive rings which are not right localizable (see [19] or [24]). For a commutative domainthis definition coincides with the ordinary definition of a Prufer domain (see [16, Theorem 13]).A right φ-ring R is called right φ-distributive (resp. right φ-Prufer) if φ(R) is right distributive(resp. right Prufer).

4.3. Characterizations of right φ-Prufer rings. Using results from earlier sections we havethe following Theorem. Note that (b) ⇔ (c) generalizes [2, Theorem 2.6] from the commutativeto the non-commutative setting.

Theorem. Let R be a right φ-ring and S = R \Ng(R). Consider the following statements:(a) R is right φ-distributive;

8 CHRISTIAN LOMP AND ALVERI SANT’ANA

(b) R is right φ-Prufer;(c) φ(R) is a subring of a local ring T containing Jac(T ), such that T/Jac(T ) is a distributive

right φ(R)/Jac(T )-module.(d) R/Ng(R) is a right distributive domain and R[S−1] is a right chain ring.(e) R/Ng(R) is a right distributive domain and φ(R) has Ng(φ(R))-comparability.

Then (a) ⇔ (b) ⇔ (c) ⇒ (d) ⇔ (e) hold and all statements are equivalent if S is a leftdenominator set.

Proof. (a) ⇒ (b) : Suppose that R is right φ-distributive, i.e., φ(R) is a right distributive ring.We have that φ(R) \ P is a right Ore set, for every maximal (right) ideal P of φ(R). Let P beany maximal ideal of φ(R) and take x ∈ φ(R) and s 6∈ P such that sx = 0. By 2.2(4) we havex ∈ Nl(φ(R)) ⊆ Jac(φ(R)) ⊆ P and so x = 0. Therefore, φ(R) is right localizable.

(b) ⇒ (a) : This is trivial.(b) ⇒ (d) : First we observe that R/Ng(R) ' φ(R)/Ng(φ(R)) is a right distributive domain.

Furthermore R[S−1] = φ(R)[φ(S)−1] is a right chain ring because it is a localization of a rightdistributive ring.

(a) ⇔ (c) : Suppose that R is right φ-distributive, i.e., φ(R) is a right distributive ring and thelocalization R[S−1] is a right chain ring. We can apply [11, Theorem 2.5] to obtain the desiredresult with T = R[S−1]. The converse is an immediate consequence of [11, Theorem 2.5].

(d) ⇒ (a) : If S is a left denominator set then φ(S) = φ(R)\Ng(φ(R)) is a left denominator settoo and so, in this case, we have that R[S−1]/Ng(R[S−1]) is distributive as a right φ(R)/Ng(φ(R))-module, by [20, Proposition 3.3] (resp. [11, Lemma 2.6]). Applying [11, Theorem 2.5] we obtainthat φ(R) is right distributive.

(d) ⇔ (e) : It follows from 2.7. ¤

5. Bezout rings

Griffin called a commutative ring Bezout, if any finitely generated ideal that contains a non-zero divisor is principal. In non-commutative ring theory it is rather common to define a rightR-module M over a ring R to be Bezout if every finitely generated submodule of M is cyclic;R is called right Bezout if it is Bezout as right R-module (see [11]). We will opt for the latter(stronger) definition of Bezout. By [22, p. 33] a right R-module M is Bezout if and only if forevery x, y ∈ M there exist c, d, e, f ∈ R such that x(1− ce) = yde and y(1− df) = xcf .

5.1. Bezout rings are rings with comparability. As mentioned in 4.1 any right distributivering is a ring with comparability. We will show that this is the same for right Bezout rings.

Theorem. Let R be a right Bezout ring and P a completely prime ideal contained in Jac(R).Then R has right P -comparability.

Proof. Let P be a completely prime ideal contained in the Jac(R) for a right Bezout ring R. LetS = R \ P . We will show that S is a right Ore set. Take a ∈ R and s 6∈ P . By [22, p. 33] thereexist c, d, e, f ∈ R such that a(1 − ce) = sde and s(1 − df) = acf . If c ∈ P then we have ce ∈ Pand so 1− ce 6∈ P and we are done. Hence we suppose that c 6∈ P . In this case, we observe thatif f ∈ P then 1 − df 6∈ P which implies s(1 − df) = acf ∈ P - a contradiction. Thus, f 6∈ P andso cf 6∈ P . Hence, in every possible cases there exist t ∈ S = R \ P and r ∈ R such that at = sr.Therefore, S = R \ P is a right Ore set as desired. Note that by [13, Proposition 2.2] R hasright P -comparability if and only if for every a, b ∈ R we have either aR ⊆ bR or b ∈ tS(R/aR).Thus consider a, b ∈ R such that a 6∈ bR. Again by [22, p. 33] we have c, d, e, f ∈ R such thata(1− ce) = bde and b(1− df) = acf . If c ∈ P or e ∈ P then 1− ce is invertible, since P ⊆ Jac(R)and so a ∈ bR - a contradiction. Thus c, e 6∈ P . If d ∈ P we have 1−df 6∈ P and so b ∈ tS(R/aR).Also, if d 6∈ P we obtain that de 6∈ P and b ∈ tS(R/aR) again, which completes the proof. ¤5.2. We observe that in [18, Lemma 1.6 (i)] it had been shown that every completely prime idealcontained in the Jacobson radical of a right Bezout ring is a right waist. This result can beobtained as a consequence of the fact that such a ring has right P -comparability. In [18, Lemma1.6(ii)] it had been shown that a right Bezout ring R with a completely prime ideal P contained

COMPARABILITY, DISTRIBUTIVITY AND NON-COMMUTATIVE φ-RINGS 9

in Jac(R) satisfies aR ⊆ bR or bR ⊆ aR or aP = bP for any a, b ∈ R. This kind of comparability(called weak comparability in [12]) is weaker than P -comparability (see [12, Example 2.6]).

5.3. Bezout rings in H. Analogously to 4.1 we have the following

Proposition. A right Bezout ring belongs to H if and only if its Jacobson radical contains acompletely prime ideal.

Proof. By [18, Lemma 1.6(i)], every completely prime ideal L ⊆ Jac(R) of a right Bezout ring Ris a waist as a right and a left ideal of R. Thus, the set of completely prime ideals of R whichis contained in the Jacobson radical of R is a linearly ordered set and so Ng(R) is a completelyprime ideal of R which is a waist as a right and a left ideal. ¤

5.4. Characterization of right φ-Bezout rings. A ring R is called right φ-Bezout, if φ(R) isa right Bezout ring.

Theorem. Let R be a right φ-ring and S = R \Ng(R). Consider the following statements:(a) R is right φ-Bezout;(b) φ(R) is a subring of a local ring T containing Jac(T ) such that T/Jac(T ) is right Bezout

as right φ(R)/Jac(T )-module;(c) RNg(R) is a right chain ring and R/Ng(R) is a right Bezout domain;

Then (a) ⇔ (b) ⇒ (c) holds and all statements are equivalent if S is a left denominator set.

Proof. (a) ⇒ (c) : follows by the same argument as in 4.3 because localizations of Bezout ringsare Bezout.

(a) ⇔ (b) : Suppose that R is right φ-Bezout, i.e., φ(R) is right Bezout. Then RNg(R) is a rightchain ring and we can apply [11, Theorem 2.5] to obtain the desired result with T = R[S−1]. Theconverse is immediate consequence of [11, Theorem 2.5].

(c) ⇒ (a) : Suppose S = R \ Ng(R) is a left denominator set then also φ(S) is. HenceRNg(R)/Ng(RNg(R)) is Bezout as a right φ(R)/Ng(φ(R))-module by [11, Lemma 2.6]. Now weonly need to apply [11, Theorem 2.5] to obtain that φ(R) is a right Bezout ring.

¤

5.5. As mentioned above, the notion of Bezout rings used in this paper is stronger than the notionused for commutative rings. In [1] Anderson and Badawi showed that any commutative φ-Bezoutring is Bezout in their sense and that the converse does not hold in general ([1, Example 3.6]).

We observe now that for the non-commutative notion of Bezout rings used in this paper, theimplications are reversed. Recall that we demand that any finitely generated right ideal in a rightBezout ring is principal. In particular any factor ring R/I of a right Bezout ring R is right Bezoutring, because if F/I is a finitely generated right ideal of R/I, then F = Rx1 + · · · + Rxn + I forsome xi ∈ R. Since R is right Bezout, there exists x ∈ R such that Rx = Rx1 + · · · + Rxn, i.e.,(Rx + I)/I = F/I is cyclic. Hence if R is a right φ-ring which is right Bezout, then φ(R) is rightBezout and so R is right φ-Bezout.

On the other hand let M = (Q/Z)⊕ (Q/Z) and R = ZnM be the trivial extension of Z by M ,which is a right φ-ring by 2.1. Note that Ng(R) = tS(R) = 0nM . Hence φ(R) ' R/Ng(R) ' Z isa commutative principal ideal domain, i.e., R is φ-Bezout. Although any finitely generated idealof R which contains a non-zero divisor is principal and hence Bezout in the sense of Anderson andBadawi, it is not Bezout in our sense since for m = 1/2+Z, the right ideal generated by (0, (m, 0))and (0, (0, m)) is not principal.

5.6. Zero divisors in Bezout rings. We finish this paper with some remark on zero divisorsin Bezout rings, which should be compared to [15, Theorem 2.6] for distributive rings.

Proposition. Let R be a right Bezout ring and L a completely prime right multiplicative ideal ofR contained in Jac(R). Then L is a (two-sided) completely prime ideal of R and a right waist. Inparticular if Nr(R) is contained in J(R), then Nr(R) is a completely prime ideal of R and a rightwaist.

10 CHRISTIAN LOMP AND ALVERI SANT’ANA

Proof. First we observe that we need only to prove that a multiplicative ideal L satisfying ourhypothesis is a right multiplicative waist to apply 2.10. We can repeat the arguments in [18,Lemma 1.6(i)] to obtain that L is a right multiplicative waist of R, since those arguments onlyuse the multiplicative structure of L and R. ¤

In particular Nr(φ(R)) is a completely prime two-sided ideal of φ(R) if R is right φ-Bezout.

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Departamento de Matematica Pura, Universidade do Porto, PortugalE-mail address: clomp@fc.up.pt

Instituto de Matematica, Universidade Federal do Rio Grande do Sul, BrazilE-mail address: alveri@mat.ufrgs.br