Journal of Pure and Applied Algebra 218 (2014) 919–924

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Journal of Pure and Applied Algebra

www.elsevier.com/locate/jpaa

On classical rings of quotients of duo rings

Michał Ziembowski

Faculty of Mathematics and Information Science, Warsaw University of Technology, 00-662 Warsaw, Poland

a r t i c l e i n f o a b s t r a c t

Article history:Received 26 April 2013Received in revised form 17 September2013Available online 28 October 2013Communicated by S. Iyengar

MSC:Primary: 16S85; 16U20; secondary: 16S99;16U80

Keywords:Duo ringsClassical right ring of quotients

In this work we will construct a duo ring R such that the classical right ring of quotientsQ r

cl(R) of R is neither right nor left duo.© 2013 Elsevier B.V. All rights reserved.

1. Introduction and preliminaries

Throughout this paper all rings are associative with identity.Recall that a ring R is right duo (respectively left duo) if any right (resp. left) ideal of R is a two-sided ideal. If R is left

and right duo, then we say that R is a duo ring. In [4] A.J. Diesl et al. posed the following: If a ring R is duo, is Q rcl(R) duo?

In this paper we want to show that there exists a duo ring R such that its classical right ring of quotients Q rcl(R) is left duo

and not right duo. Building on this construction, we will construct a duo ring with classical right ring of quotients which isneither right nor left duo. In our opinion, the existence of the examples is very surprising. At this point we would also liketo mention that in [2] H.H. Brungs constructed a duo ring R such that the localization at a prime ideal P of R is not a duoring.

Let R be a ring and S a multiplicative set in R (i.e. S · S ⊆ S , 1 ∈ S , and 0 /∈ S). Then a ring R S−1 is called a right ring ofquotients of R with respect to S if there exists a ring homomorphism ϕ : R → R S−1 such that

(a) For any s ∈ S , ϕ(s) is a unit of R S−1.(b) Every element of R S−1 has the form ϕ(a)ϕ(s)−1 for some a ∈ R and s ∈ S .(c) kerϕ = {r ∈ R: rs = 0 for some s ∈ S}.

It is well known (e.g. see [5, Theorem 10.6]) that the ring R has a right ring of quotients with respect to S if and only ifthe following conditions are satisfied:

(1) For any a ∈ R and s ∈ S , aS ∩ sR �= ∅.(2) For a ∈ R , if ta = 0 for some t ∈ S , then as = 0 for some s ∈ S .

E-mail address: m.ziembowski@mini.pw.edu.pl.

0022-4049/$ – see front matter © 2013 Elsevier B.V. All rights reserved.http://dx.doi.org/10.1016/j.jpaa.2013.10.009

920 M. Ziembowski / Journal of Pure and Applied Algebra 218 (2014) 919–924

A multiplicative set S satisfying the above conditions (1) and (2) is called a right denominator set. If the set S consists of allregular elements of R (i.e. all elements a ∈ R such that a is neither a left zero-divisor nor a right zero-divisor of R), thenthe right ring of quotients R S−1 is called the classical right ring of quotients of R and is denoted by Q r

cl(R).In the same way we can consider left sided version of above and get a left ring of quotients S−1 R of R with respect to

a left denominator set S and the classical left ring of quotients of R which is denoted by Q lcl(R).

Proposition 1.1. Let R be a right (resp. left) duo ring and P an ideal of R such that S = R \ P is a right (resp. left) denominator setin R. Then R S−1 (resp. S−1 R) is right (resp. left) duo if and only if for any a ∈ R we have Sa ⊆ aS or as = 0 (resp. aS ⊆ Sa or sa = 0)for some s ∈ S.

Proof. Obviously it is enough to prove right sided version of the result.Assume that R S−1 is right duo, and suppose that for an element a ∈ R , Sa � aS . Then for some t ∈ S we have ta /∈ aS .

Since R is right duo we deduce that ta = ap for some p ∈ P . Hence in R S−1 we have ϕ(a) = ϕ(t)−1ϕ(ap), where ϕ : R →R S−1 is a ring homomorphism satisfying conditions (a)–(c) of the definition of R S−1, recalled at the beginning of thissection. Since R S−1 is right duo, it follows that ϕ(a) = ϕ(ap)α for some α ∈ R S−1. Since P is an ideal of R , R S−1 is a localring and the Jacobson radical of R S−1 is equal to J (R S−1) = {ϕ(q)ϕ(z)−1: q ∈ P , z ∈ S}. Hence ϕ(p)α ∈ J (R S−1), and thusϕ(a) = ϕ(a)ϕ(p)α ∈ ϕ(a) J (R S−1), which implies ϕ(a) = 0. Therefore, for some s ∈ S we have as = 0.

Now suppose that for any a ∈ R we have Sa ⊆ aS or as = 0 for some s ∈ S . To see that R S−1 is right duo it is enoughto show that for any σ ,β ∈ R S−1, σβ ∈ βR S−1. It is well known that for fixed σ and β there exist a,b ∈ R and s ∈ S suchthat σ = ϕ(a)ϕ(s)−1 and β = ϕ(b)ϕ(s)−1. Since R is right duo, as = sa′ and bs = sb′ for some a′,b′ ∈ R . Thus we have

ϕ(sb′) = ϕ(bs) = ϕ(b)ϕ(s)−1ϕ(s)2 ∈ βR S−1. (1.1)

Moreover, by our assumption there exists some t ∈ S such that sb = bt so

ϕ(b) = ϕ(s)2ϕ(b′)ϕ(ts)−1. (1.2)

Finally, using (1.1), (1.2), the fact that R is right duo, and asb′ = sb′a′ for some a′ ∈ R , we get

σβ = ϕ(a)ϕ(s)−1ϕ(b)ϕ(s)−1 = ϕ(a)ϕ(s)ϕ(b′)ϕ(ts)−1ϕ(s)−1

= ϕ(asb′)ϕ(ts)−1ϕ(s)−1 = ϕ

(sb′)ϕ(

a′)ϕ(ts)−1ϕ(s)−1 ∈ βR S−1.

Thus R S−1 is right duo. �Now, we would like to recall the construction of the generalized power series [7] which we will use to construct our

main example.Let (T , ·) be a monoid and � an order relation on T . We say that (T , ·,�) is an ordered monoid if for any s1, s2, t ∈ T ,

s1 � s2 implies s1t � s2t and ts1 � ts2. Furthermore, if s1 < s2 implies s1t < s2t and ts1 < ts2, then (T , ·,�) is said to be astrictly ordered monoid. If for any s1, s2 ∈ T , either s1 � s2 or s2 � s1, then we say that (S, ·,�) is totally ordered.

Given a ring R and a strictly ordered monoid (T ,�), consider the set A of all maps f : T → R whose support supp( f ) ={s ∈ T | f (s) �= 0} is artinian (i.e., it does not contain any infinite strictly decreasing chains of elements) and narrow (i.e.,it does not contain infinite subsets of pairwise order-incomparable elements). It is easy to see that if (T ,�) is a totallyordered monoid, then the subsets which are both artinian and narrow are precisely the well-ordered subsets of (T ,�).

If f , g ∈ A and s ∈ T , it turns out that the set

Xs( f , g) = {(x, y) ∈ supp( f ) × supp(g): s = xy

}is finite. Thus one can define the product f g: T → R of f , g ∈ A as follows:

( f g)(s) =∑

(x,y)∈Xs( f ,g)

f (x)g(y) for any s ∈ T

(by convention, a sum over the empty set is 0). With pointwise addition and multiplication as defined above, A becomes aring, called the ring of generalized power series with coefficients in R and exponents in T , and denoted by R[[T ]].

We will use the symbol 1 to denote the identity elements of the monoid T , the ring R and the ring R[[T ]]. To each r ∈ Rand s ∈ S , we associate elements cr,es ∈ R[[T ]] defined by

cr(x) ={

r if x = 1,

0 if x ∈ T \ {1}, es(x) ={

1 if x = s,0 if x ∈ T \ {s}.

It is clear that r → cr is a ring embedding of R into R[[T ]] and s → es is a monoid embedding of T into the multiplicativemonoid of the ring R[[T ]].

M. Ziembowski / Journal of Pure and Applied Algebra 218 (2014) 919–924 921

2. The construction of an example of a duo ring with classical right ring of quotients that is left duo and is not right duo

The aim of this section is construction of a duo ring such that its classical right ring of quotients is left duo and notright duo. Therefore, we want to agree with the reader that all statements and notation we are making remain in force forthe remainder of the section.

Let G be the free abelian group with operation written as multiplication and generated by the set {xi: i ∈ Z} and let ψ

be an endomorphism of G defined by

ψ(xi) = xi+1 for any i ∈ Z.

Definition 2.1. Given an element g ∈ G , we can write g = xk1�1

xk2�2

· · · xkn�n

where �1 < �2 < · · · < �n and ki ∈ Z − {0}. We callthis the canonical representation for g . We call kn the final exponent and we call x�n the final component. (The canonicalrepresentation of g = 1 is somewhat special, being an empty product of such terms, and we write g = 1.)

For any g1, g2 ∈ G we write g1 ≺ g2 if g1 �= g2 and g−11 g2 has a (strictly) positive final exponent.

It is easy to see that (G,�) is a totally ordered group and for any g1, g2 ∈ G , g1 ≺ g2 implies ψ(g1) ≺ ψ(g2).To construct the desired ring R , we first consider the set T of all pairs (m, g) ∈ Z× G such that either m > 0, or m = 0

and g � 1. We define a multiplication and order relation on T in the following way. For (m1, g1), (m2, g2) ∈ T we define

(m1, g1)(m2, g2) = (m1 + m2,ψ

m2(g1)g2),

and

(m1, g1) � (m2, g2) ⇔ either m1 < m2 or m1 = m2 and g1 � g2.

It is easy to verify that (T ,�) is a positively strictly totally ordered monoid with (0,1) as a unity (an ordered monoid(T , ·,�) is positively ordered if s � 1 for any s ∈ T ).

A ring (resp. monoid) R is said to be a right chain ring (resp. monoid) if the right ideals of R are totally ordered by setinclusion [1,3], i.e., if aR ⊆ bR or bR ⊆ aR for any a,b ∈ R . Left chain rings (resp. monoids) are defined similarly. If R is leftand right chain, then we say that R is a chain ring (resp. monoid).

Claim 2.2. The monoid T is right chain.

Proof. Since the order � on T is total, and (m1, g1) � (m2, g2) implies

(m2, g2) = (m1, g1)(m2 − m1,ψ

m2−m1(

g−11

)g2

) ∈ (m1, g1)T ,

it follows that T is a right chain monoid. �Claim 2.3. The monoid T is left chain.

Proof. To show that T is left chain, for elements (m1, g1), (m2, g2) ∈ T we will consider two cases.Case 1. m1 < m2. It is enough to show that there exists y ∈ G (notice that since m2 − m1 > 0 we do not have any

restriction on y) such that (m2, g2) = (m2 − m1, y)(m1, g1) = (m2,ψm1(y)g1). Thus we want to find y such that ψm1 (y) =

g−11 g2. But by the construction of G and the construction of ψ obviously such a y exists.

Case 2. m1 = m2 = m. In this situation we can assume that g1 ≺ g2 and then for the canonical representation xk1�1

xk2�2

· · · xkn�n

of g−11 g2 the final exponent kn is positive. Notice that, what is very crucial, also xk1

�1−mxk2�2−m · · · xkn

�n−m has a positive canonical

exponent since still 0 < kn . Hence (0, xk1�1−mxk2

�2−m · · · xkn�n−m) ∈ T . Now it is easy to check that

(m, g2) = (0, xk1

�1−mxk2�2−m · · · xkn

�n−m

)(m, g1)

and it follows that T is a left chain monoid. �Let D be a division ring. We will consider the generalized power series ring R = D[[T ]].

Claim 2.4. The generalized power series ring R = D[[T ]] is a duo chain domain.

Proof. Since (T ,�) is a strictly, positively and totally ordered chain monoid, the claim follows from [6, Corollary 24] andits right-handed version. �

Since (T ,�) is a positively strictly totally ordered monoid, for any f ∈ R \ {0} the set supp( f ) is well-ordered andthus there exists a unique minimal element π( f ) of supp( f ). It is obvious that for any f ,h ∈ R \ {0}, if f + h �= 0 andπ( f ) � π(h), then π( f + h)� π(h). Furthermore, since D is a division ring, for any f ,h ∈ R with f h �= 0 we have

922 M. Ziembowski / Journal of Pure and Applied Algebra 218 (2014) 919–924

π( f h) = π( f )π(h), (2.1)

and since (T ,�) is positively ordered, it follows that π( f h)� π( f ) and π( f h)� π(h). Now it is clear that the set

I = {0} ∪ {f ∈ R \ {0}: π( f ) >

(1, xi

1x j2x3

)for any i, j ∈ Z

}is a proper ideal of R . In what follows the “bars” refer to modulo I , that is R stands for the factor ring R/I , and f = f + Ifor any f ∈ R . Since R is a duo chain ring, so is R . Furthermore, since in any chain ring the set of right zero-divisors aswell as the set of left zero-divisors are ideals (see [1, Lemma 2.3(i)]), the set P of those elements of R that are right or leftzero-divisors is an ideal of R , and S = R \ P is a right and left denominator set in R . Hence we can consider a right and leftring of quotients R S−1 and S−1 R . Note that since S coincides with the set of regular elements of R , R S−1 and S−1 R arethe classical right and left rings of quotients, respectively, and we have

R S−1 = Q rcl(R) = Q l

cl(R) = S−1 R.

The ring Q rcl(R) will just be denoted by Q .

Claim 2.5. For an element f ∈ R we have f ∈ S if and only if π( f ) � (0, xk1) for some non-negative integer k.

Proof. Assume that for f ∈ R the element f is neither right nor left zero-divisor in R . To get a contradiction, supposethat π( f ) > (0, xk

1) for any non-negative k. Set π( f ) = (m, g). If m � 1 then π( f e(1,1)) = π( f )π(e(1,1)) = (m, g)(1,1) =(m + 1,ψ(g)) � (1, xi

1x j2x3) for any i and j. Thus f e(1,1) = f e(1,1) = 0 in R which gives a contradiction since e(1,1) is

non-zero in R and f is not zero-divisor. Hence it must be that m = 0. So if we consider the canonical representationxk1�1

xk2�2

· · · xkn�n

of g we have kn > 0. As (0, g) > (0, xk1) for any k, we have always 1 ≺ x−k

1 · xk1�1

xk2�2

· · · xkn�n

which implies that

�n � 2. Indeed, if �n = 1, then for k = kn +1 we have x−k1 ·xk1

�1xk2�2

· · · xkn�n

= x−11 xk1

�1xk2�2

· · · xkn−1�n−1

≺ 1 since �1 < · · · < �n−1 < �n = 1,a contradiction. Similarly we can show that �n < 1 is impossible.

Using the fact that kn > 0 and �n � 2 we get

π( f c(1,x3)) = π( f )π(c(1,x3)) = (0, g)(1, x3) = (1,ψ

(xk1�1

xk2�2

· · · xkn�n

)x3

)= (

1, xk1�1+1xk2

�2+1 · · · xkn�n+1x3

)>

(1, xi

1x j2x3

)for any i and j. Thus f c(1,x3) = f c(1,x3) = 0, and it follows that f is a left zero-divisor since e(1,x3) �= 0 in R , a contradiction.Thus π( f ) � (0, xk

1) for some k.Now, assume that for an element f ∈ R , π( f ) � (0, xk

1) for some non-negative k. Notice that if we are able to show thate(0,xk

1)is not zero-divisor in R then the fact that f is not zero-divisor follows. Indeed, assume that e

(0,xk1)

is not zero-divisor

in R . Then for any h ∈ R such that h �= 0 we have π( f h) = π( f )π(h) � π(e(0,xk

1))π(h) = π(e

(0,xk1)

h) � (1, xi1x j

2x3) for some

i and j, and π(hf ) = π(h)π( f ) � π(h)π(e(0,xk

1)) = π(he

(0,xk1)

) � (1, xi1x j

2x3) for some i′ and j′ . Thus f is not zero-divisor

in R .Suppose that for some h ∈ R , e

(0,xk1)

h = 0 and h �= 0. Then for π(h) = (m, g) we have (0, xk1)(m, g) = (m,ψm(xk

1)g) >

(1, xi1x j

2x3) for any i and j. So, it is easy to deduce that m = 1 (if m > 1 then h = 0) and we have (1, xk2 g) > (1, xi

1x j2x3) for

any i, j. Thus, setting g = xk1�1

xk2�2

· · · xkn�n

with the final component x�n and the final exponent kn we get

1 ≺ x−i1 xk− j

2 x−13 · xk1

�1xk2�2

· · · xkn�n

(2.2)

for any i and j (notice that k − j can take any value). We deduce that �n = 3. Indeed, if �n > 3 and kn > 0 then forany i and j, (1, xi

1x j2x3) < (1, g) = (1, xk1

�1xk2�2

· · · xkn�n

) so it follows that h = 0, a contradiction. If �n > 3 and kn < 0 then

x−i1 xk− j

2 x−13 · xk1

�1xk2�2

· · · xkn�n

≺ 1 which contradicts (2.2). Finally, if �n < 3 then x−i1 xk− j

2 x−13 · xk1

�1xk2�2

· · · xkn�n

≺ 1, a contradiction.Now we want to show that kn � 2. To see that, we consider the situation with kn � 1. Then using (2.2)

1 ≺ x−i1 xk− j

2 x−13 · xk1

�1xk2�2

· · · xkn�n

= x−i1 xk− j

2 xkn−13 · xk1

�1xk2�2

· · · xkn−1�n−1

and �n − 1 < 3, kn − 1 � 0. If kn − 1 = 0 then x−i1 xk− j

2 · xk1�1

xk2�2

· · · xkn−1�n−1

≺ 1 for small enough k − j and we get a contradiction.Obviously the situation kn − 1 < 0 also gives a contradiction. So we have justified that kn � 2.

Using �n = 3 and kn � 2, for any i and j we have

x−i1 x− j

2 x−13 · xk1

�1xk2�2

· · · xkn�n

= x−i1 x− j

2 · xk1�1

xk2�2

· · · xkn−1�n−1

xkn−13 � 1,

since kn −1 � 1, which implies that π(h) = (1, g) > (1, xi1x j

2x3). Thus h = 0, a contradiction. Thus we have proved that e(0,xk

1)

is not left zero-divisor.

M. Ziembowski / Journal of Pure and Applied Algebra 218 (2014) 919–924 923

By similar consideration we can see that e(0,xk

1)is not a right zero-divisor. Thus our proof is completed. �

Claim 2.6. The ring Q is not right duo.

Proof. By Proposition 1.1, to show that Q rcl(R) is not right duo it suffices to prove that e(0,x1) ∈ S and e(0,x1)e(1,1) /∈ e(1,1) S .

The first fact follows from Claim 2.5. Now we will show that e(0,x1)e(1,1) /∈ e(1,1) S . For a contradiction, suppose thate(0,x1)e(1,1) = e(1,1) f for some f ∈ R such that f ∈ S . Then, since e(0,x1)e(1,1) = e(1,x2) , it follows that e(1,x2) − e(1,1) f ∈ I .This and the definition of I imply that if π(e(1,x2)) �= π(e(1,1) f ), then e(1,x2) ∈ I , a contradiction. Hence (1, x2) = π(e(1,x2)) =π(e(1,1) f ) = (1,1)π( f ), and thus π( f ) = (0, x2). But since (0, xk

1) ≺ (0, x2) for any k, by Claim 2.5 we have f /∈ S , a contra-diction. Thus our proof that the ring Q r

cl(R) is not right duo is finished. �Claim 2.7. The ring Q is left duo.

Proof. Firstly, we will show that for any (m, g), (0, p) ∈ T such that (m, g)� (1, xi1x j

2x3) for some i and j, and (0, p)� (0, xk1)

for some k, there exists (0,q) ∈ T such that (0,q) � (0, xk′1 ) for some k′ and

(m, g)(0, p) = (0,q)(m, g).

Obviously we have m � 1 and therefore we consider the following two cases:Case 1. m = 0. Then we have (0, g)(0, p) = (0, p)(0, g) and this case is clear.Case 2. m = 1. If p = 1 then (m, g)(0,1) = (0,1)(m, g) and we are done. So assume that p �= 1 and consider the canonical

representation xk1�1

xk2�2

· · · xkn�n

of p. As (0, p) ∈ T we have kn > 0 for the canonical exponent of p. Since p � xk1 for some k we

have �n � 1. Consider q = ψ−1(p). Since 0 < kn we have 1 ≺ q and (0,q) ∈ T . As ψ(q) = p, using �1 − 1 < · · · < �n − 1 < 1we get the information that (0,q) < (0, x1). Now we check that (0,q)(1, g) = (1,ψ(q)g) = (1, pg) and taking k′ = 1 we endthis part of the proof.

To show that Q is left duo, using Proposition 1.1 it is enough to show that for any a ∈ R , aS ⊆ Sa.Consider non-zero a ∈ R and f ∈ S . Then π(a) = (m, g) � (1, xi

1x j2x3) for some i and j, and by Claim 2.5 π( f ) � (0, xk

1)

for some k. Since R is left duo there exists an element f ′ ∈ R such that af = f ′a. Moreover, π(a)π( f ) = π(af ) = π( f ′a) =π( f ′)π(a). By the first part of our proof we know that there exists (0,q) ∈ T such that (0,q) � (0, xk′

1 ) for some k′ , andπ( f ′)π(a) = π(a)π( f ) = (0,q)π(a). Since T is strictly totally ordered it is easy to see that π( f ′) = (0,q) and it follows thatf ′ ∈ S . Thus we have proved that aS ⊆ Sa and it follows that Q is left duo. �Remark 2.8. The ring Q being left duo is semicommutative, so also NI and 2-primal.

3. There exists a duo ring such that its classical right ring of quotients is neither right duo nor left duo

Example 3.1. By R we denote the ring which was the main object of our interest in the previous section. Recall that by Swe denote the set of all regular elements of R .

Using the construction of the opposite ring Rop of R we can consider the ring B = R × Rop which is obviously duo. Notethat the regular elements are S × S . It is easy to see that for every s ∈ S and p ∈ R \ S , we have s − p ∈ S . Thus the setB \ S × S is not an ideal of B , and we can not use Proposition 1.1 in this place. Fortunately some ideas from the proof ofthis proposition will work here.

Consider the following elements a = e(1,1), p = e(0,x2) ∈ R \ S and s = e(0,x1), q = e(0,x0) ∈ S . Then in B we have(s, s)(a,a) = (a,a)(p,q) and (a,a)(s, s) = (q, p)(a,a). Suppose that the classical right ring of quotients Q r

cl(B) of B is rightduo. Using equation (s, s)(a,a) = (a,a)(p,q), for the ring homomorphism ϕ : B → B(S × S)−1(= Q r

cl(B)) satisfying conditions(a)–(c) of the definition of a right ring of quotients, recalled at the beginning of Section 1 we get

ϕ((a,a)

) = ϕ((s, s)

)−1ϕ

((a,a)

)ϕ

((p,q)

).

As Q rcl(B) is right duo, for some w, v ∈ R and s1, s2 ∈ S

ϕ((a,a)

) = ϕ((a,a)

)ϕ

((p,q)

)ϕ

((w, v)

)ϕ

((s1, s2)

)−1

and finally

ϕ((a,a)

)ϕ

((s1, s2)

) = ϕ((a,a)

)ϕ

((p,q)

)ϕ

((w, v)

).

By above (a,a)(s1, s2) − (a,a)(p,q)(w, v) ∈ ker ϕ so

(a,a)(s1, s2)(s3, s4) − (a,a)(p,q)(w, v)(s3, s4) = 0

for some s3, s4 ∈ S . Thus looking at the first coordinate we get a(s1s3 − pws3) = 0 which is impossible since p ∈ R \ S ands1s3 − pws3 ∈ S . Thus the ring Q r (B) is not right duo.

cl

924 M. Ziembowski / Journal of Pure and Applied Algebra 218 (2014) 919–924

Using equation (a,a)(s, s) = (q, p)(a,a), similar arguments to those above and focusing on the second coordinate we canshow that the classical left ring of quotients Q l

cl(B)(= Q rcl(B)) is not left duo.

Summarizing, we have constructed a duo ring B such that its classical right ring of quotients Q rcl(B) is neither right nor

left duo.

Acknowledgements

The author is highly grateful to Pace P. Nielsen for correcting many errors and valuable suggestions. The author wouldalso like to express his deep gratitude to the referee for comments and suggestions which greatly improved the clarityof this paper. This research was supported by the Foundation for Polish Science – Homing Plus Programme, Homing PlusBIS/2011-3/2.

References

[1] C. Bessenrodt, H.H. Brungs, G. Törner, Right Chain Rings. Part 1, Schriftenreihe des Fachbereichs Math., vol. 181, Duisburg Univ., 1990.[2] H.H. Brungs, Three questions on duo rings, Pac. J. Math. 58 (1975) 345–349.[3] M. Ferrero, R. Mazurek, A. Sant’Ana, On right chain semigroups, J. Algebra 292 (2005) 574–584.[4] A.J. Diesl, C.Y. Hong, N.K. Kim, P.P. Nielsen, Properties which do not pass to classical rings of quotients, J. Algebra 379 (2013) 208–222.[5] T.Y. Lam, Lectures on Modules and Rings, Grad. Texts Math., vol. 189, Springer-Verlag, New York, 1999.[6] R. Mazurek, M. Ziembowski, On Bezout and distributive generalized power series rings, J. Algebra 306 (2006) 387–401.[7] P. Ribenboim, Semisimple rings and von Neumann regular rings of generalized power series, J. Algebra 198 (1997) 327–338.