generalized semicommutative rings

Bull. Korean Math. Soc. 45 (2008), No. 2, pp. 285–297

GENERALIZED SEMICOMMUTATIVE RINGSAND THEIR EXTENSIONS

Muhittin Baser, Abdullah Harmanci, and Tai Keun Kwak

Reprinted from theBulletin of the Korean Mathematical Society

Vol. 45, No. 2, May 2008

c©2008 The Korean Mathematical Society

Bull. Korean Math. Soc. 45 (2008), No. 2, pp. 285–297

GENERALIZED SEMICOMMUTATIVE RINGSAND THEIR EXTENSIONS

Muhittin Baser, Abdullah Harmanci, and Tai Keun Kwak

Abstract. For an endomorphism α of a ring R, the endomorphism α iscalled semicommutative if ab = 0 implies aRα(b) = 0 for a ∈ R. A ring Ris called α-semicommutative if there exists a semicommutative endomor-phism α of R. In this paper, various results of semicommutative ringsare extended to α-semicommutative rings. In addition, we introduce thenotion of an α-skew power series Armendariz ring which is an extensionof Armendariz property in a ring R by considering the polynomials in theskew power series ring R[[x; α]]. We show that a number of interestingproperties of a ring R transfer to its the skew power series ring R[[x; α]]and vice-versa such as the Baer property and the p.p.-property, whenR is α-skew power series Armendariz. Several known results relating toα-rigid rings can be obtained as corollaries of our results.

1. Introduction

Throughout this paper R denotes an associative ring with identity and αdenotes a nonzero non identity endomorphism of a given ring, unless specifiedotherwise.

Recall that a ring is reduced if it has no nonzero nilpotent elements. Lambek[13] called a ring R symmetric provided abc = 0 implies acb = 0 for a, b, c ∈R, Habeb [5] called a ring R zero commutative if R satisfies the condition:ab = 0 implies ba = 0 for a, b ∈ R, while Cohn [4] used the term reversiblefor what is called zero commutative. A generalization of a reversible ring is asemicommutative ring. A ring R is semicommutative if ab = 0 implies aRb = 0for a, b ∈ R. Historically, some of the earliest results known to us aboutsemicommutative rings (although not so called at the time) was due to Shin[15]. He proved that (i) R is semicommutative if and only if rR(a) is an idealof R where rR(a) = {b ∈ R | ab = 0} [15, Lemma 1.2]; (ii) every reduced ring issymmetric [15, Lemma 1.1] (but the converse does not hold [1, Example II.5]);and (iii) any symmetric ring is semicommutative but the converse does not

Received May 16, 2007.2000 Mathematics Subject Classification. 16U80, 16W20, 16W60.Key words and phrases. semicommutative rings, rigid rings, skew power series rings, ex-

tended Armendariz rings, Baer rings, p.p.-rings.

c©2008 The Korean Mathematical Society

285

286 MUHITTIN BASER, ABDULLAH HARMANCI, AND TAI KEUN KWAK

hold ([15, Proposition 1.4 and Example 5.4(a)]. Semicommutative rings werealso studied under the name zero insertive by Habeb [5].

Another generalization of a reduced ring is an Armendariz ring. Rege andChhawchharia [14] called a ring R Armendariz if whenever any polynomialsf(x) = a0 + a1x + · · · + amxm, g(x) = b0 + b1x + · · · + bnxn ∈ R[x] satisfyf(x)g(x) = 0, then aibj = 0 for each i and j. The Armendariz property of aring was extended to one of the skew polynomial ring in [7]. For an endomor-phism α of a ring R, a skew polynomial ring (also called an Ore extension ofendomorphism type) R[x; α] of R is the ring obtained by giving the polynomialring over R with the new multiplication xr = α(r)x for all r ∈ R, while R[[x; α]]is called a skew power series ring. A ring R is called α-skew Armendariz [7,Definition] if for p = a0 + a1x + · · · + amxm and q = b0 + b1x + · · · + bnxn

in R[x; α], pq = 0 implies aiαi(bj) = 0 for all 0 ≤ i ≤ m and 0 ≤ j ≤ n.

Recall that an endomorphism α of a ring R is called rigid [12] if aα(a) = 0implies a = 0 for a ∈ R. If R is an α-rigid ring then for p =

∑∞i=0 aix

i andq =

∑∞j=0 bjx

j in R[[x; α]], pq = 0 if and only if aibj = 0 for all 0 ≤ i, 0 ≤ j

[6, Proposition 17]; and R is an α-rigid ring if and only if a skew power seriesring R[[x; α]] of R is reduced and α is monomorphism [6, Corollary 18].

Motivated by the above, we introduce the notion of an α-semicommutativering with the endomorphism α (see Definition 2.1 in Section 2), as both a gen-eralization of α-rigid rings and an extension of semicommutative rings, andstudy characterizations of α-semicommutative rings and their related proper-ties. And then for some condition with respect to α, say α-sps Armendarizproperty, in a skew power series ring R[[x;α]] of R which is an extension of theArmendariz property of a ring R, the relationship between R and R[[x;α]] isstudied, and the existence of strong connections among such rings and theirvarious properties are also investigated. Moreover, we show that a number ofinteresting properties of a ring R satisfying α-sps Armendariz property trans-fer to its the skew power series ring R[[x; α]] and vice-versa such as the Baerproperty and the p.p.-property. Several known results relating to α-rigid ringscan be obtained as corollaries of our results.

2. α-semicommutative rings and related rings

Our focus in this section is to introduce the concept of an α-semicommutativering and study its properties. Observe that the notion of α-semicommutativerings not only generalizes that of α-rigid rings, but also extends that of semi-commutative rings. We also investigate connections to other related conditions.Examples to illustrate the concepts and results are included. We start with thefollowing definition.

Definition 2.1. An endomorphism α of a ring R is called semicommuta-tive if whenever ab = 0 for a, b ∈ R, aRα(b) = 0. A ring R is called α-semicommutative if there exists a semicommutative endomorphism α of R.

GENERALIZED SEMICOMMUTATIVE RINGS AND THEIR EXTENSIONS 287

It is clear that a ring R is semicommutative if R is IR-semicommutative,where IR is the identity endomorphism of R. It is easy to see that every subringS with α(S) ⊆ S of an α-semicommutative ring is also α-semicommutative.

Remark 2.2. Let R be an α-semicommutative ring with ab = 0 for a, b ∈ R.Then aRα(b) = 0 and, in particular, aα(b) = 0. Since R is α-semicommutative,we get aRα2(b) = 0. So, by induction hypothesis, we obtain aRαk(b) = 0 andaαk(b) = 0 for any positive integer k.

Notice that in general the reverse implication in the above definition does nothold by the following example which also shows that there exists an endomor-phism α of a semicommutative ring R such that R is not α-semicommutative.

Example 2.3. Let Z2 be the ring of integers modulo 2 and consider a ringR = Z2 ⊕ Z2 with the usual addition and multiplication. Then R is semicom-mutative, since R is commutative reduced. Now, let α : R → R be defined byα((a, b)) = (b, a). Then α is an automorphism of R. For a = (1, 0) = b ∈ R,aRα(b) = 0 but ab = (1, 0) 6= 0. Moreover, R is not α-semicommutative: Infact, for (1, 0), (0, 1) ∈ R, (1, 0)(0, 1) = (0, 0) but (0, 0) 6= (1, 0)(1, 1)α((0, 1)) ∈(1, 0)Rα((0, 1)).

Theorem 2.4. A ring R is α-rigid if and only if R is a reduced α-semicommut-ative ring and α is a monomorphism.

Proof. Let R be an α-rigid ring. Then R is reduced and α is a monomorphismby [6, p.218]. Assume that ab = 0 for a, b ∈ R. Let r be an arbitrary elementof R. Then ba = 0 and arα(b)α(arα(b)) = arα(ba)α(r)α2(b) = 0. Since R isα-rigid, arα(b) = 0 and so aRα(b) = 0. Thus R is α-semicommutative.

Conversely, assume that aα(a) = 0 for a ∈ R. Since R is reduced and α-semicommutative, α(a)a = 0 and so α(a)Rα(a) = 0. Hence α(a2) = 0 andso a = 0, since α is a monomorphism of a reduced ring R. Therefore R isα-rigid. ¤

The following example shows that the conditions “R is a reduced ring” and“α is a monomorphism” in Theorem 2.4 cannot be dropped respectively.

Example 2.5. (1) Let Z be the ring of integers. Consider a ring R ={( a b

0 a ) | a, b ∈ Z} . Let α : R → R be an endomorphism defined by α (( a b0 a )) =(

a −b0 a

). Note that α is an automorphism. Clearly, R is not reduced and

hence R is not α-rigid. Let AB = O for A = ( a b0 a ) , B = ( c d

0 c ) ∈ R. Thenac = 0 and ad + bc = 0. For an arbitrary

(h k0 h

) ∈ R, ( a b0 a )

(h k0 h

)α (( c d

0 c )) =(ahc −ahd+akc+bhc0 ahc

). Since ac = 0, a = 0 or c = 0. If a = 0 then bc = 0. So

ARα(B) = O. If c = 0 then ad = 0. Again ARα(B) = O. Thus R is anα-semicommutative ring.

(2) Let F be a field and R = F [x] the polynomial ring over F . Defineα : R → R by α(f(x)) = f(0) where f(x) ∈ R. Then R is a commutativedomain (and so reduced) and α is not a monomorphism. If f(x)g(x) = 0 for


f(x), g(x) ∈ R then f(x) = 0 or g(x) = 0, and so f(x) = 0 or α(g(x)) = 0.Hence f(x)R α(g(x)) = 0, and thus R is α-semicommutative. Note that R isnot α-rigid, since xα(x) = 0 for 0 6= x ∈ R.

Observe that if R is a domain then R is both semicommutative and α-semi-commutative for any endomorphism α of R. Example 2.5(1) also shows thatthere exists an α-semicommutative ring R which is not a domain.

Proposition 2.6. Let R be an α-semicommutative ring. Then(1) α(1) = 1 where 1 is the identity of R if and only if α(e) = e for any

e2 = e ∈ R.(2) If α(1) = 1, then R is abelian (i.e., all its idempotents are central).

Proof. (1) Suppose that α(1) = 1. If e2 = e ∈ R, then we get e(1 − e) = 0and (1 − e)e = 0. Then eRα(1 − e) = 0 and (1 − e)Rα(e) = 0 because Ris α-semicommutative. Hence eα(1 − e) = 0 implies e(1 − α(e)) = 0 and soeα(e) = e. From (1 − e)α(e) = 0, we get α(e) = eα(e) and so α(e) = e, sinceeα(e) = e. The converse is obvious.

(2) Assume that α(1) = 1. Let e be an arbitrary idempotent in R. By thesame method as in (1), we get eRα(1 − e) = 0 and (1 − e)Rα(e) = 0, and soeR(1− e) = 0 and (1− e)Re = 0 by (1). Hence er(1− e) = 0 and (1− e)re = 0for all r ∈ R. So er = ere = re. Therefore R is an abelian ring. ¤

The concepts of α-semicommutative rings and abelian rings are independenton each other by Example 2.3 and Example 2.7 which also shows that thecondition “α(1) = 1” in Proposition 2.6(2) is not superfluous.

Example 2.7. Let Z be the ring of integers. Consider a ring R = {( a b0 c ) | a, b, c

∈ Z} . Let α : R → R be an endomorphism defined by α (( a b0 c )) = ( a 0

0 0 ) . ForA = ( a b

0 c ) , B =(

a′ b′0 c′

) ∈ R, if AB = O then we obtain aa′ = 0, and so a = 0or a′ = 0. This implies ARα(B) = O, and thus R is an α-semicommutativering. Note that α(1) 6= 1 and R is not abelian.

Corollary 2.8. Semicommutative rings are abelian.

Given a ring R and an (R,R)-bimodule M , the trivial extension of R byM is the ring T (R, M) = R ⊕ M with the usual addition and the followingmultiplication: (r1,m1)(r2, m2) = (r1r2, r1m2 + m1r2). This is isomorphic tothe ring of all matrices ( r m

0 r ), where r ∈ R and m ∈ M and the usual matrixoperations are used.

For an endomorphism α of a ring R and the trivial extension T (R, R) of R,α : T (R, R) → T (R, R) defined by α (( a b

0 a )) =(

α(a) α(b)0 α(a)

)is an endomorphism

of T (R,R). Since T (R, 0) is isomorphic to R, we can identify the restriction ofα by T (R, 0) to α.

Notice that the trivial extension of a semicommutative ring is not semicom-mutative by [8, Example 11]. Now, we may ask whether the trivial extension


T (R,R) is α-semicommutative if R is α-semicommutative. But the followingexample erases the possibility, in general.

Example 2.9. Consider an α-semicommutative ring R = {( a b0 a ) | a, b ∈ Z}

with an endomorphism α defined by α (( a b0 a )) =

(a −b0 a

)in Example 2.5(1).

For

A =(

( 0 10 0 )

(−1 10 −1

)( 0 0

0 0 ) ( 0 10 0 )

), B =

(( 0 1

0 0 ) ( 1 10 1 )

( 0 00 0 ) ( 0 1

0 0 )

)∈ T (R, R),

we have AB = O. However, for C =(

( 1 00 1 ) ( 0 0

0 0 )( 0 0

0 0 ) ( 1 00 1 )

)∈ T (R, R), we obtain

O 6=(

( 0 00 0 ) ( 0 2

0 0 )( 0 0

0 0 ) ( 0 00 0 )

)= ACα(B) ∈ AT (R, R)α(B),

and therefore T (R, R) is not α-semicommutative.

However we have the following:

Proposition 2.10. Let R be a reduced ring. If R is an α-semicommutativering, then T (R, R) is an α-semicommutative ring.

Proof. We freely use the condition that R is reduced α-semicommutative andthe fact that reduced rings are semicommutative. Note that R is a reducedring if and only if for any a, b ∈ R, ab2 = 0 implies ab = 0. Let AB =O for A = ( a b

0 a ) , B = ( c d0 c ) ∈ T (R, R). Then ac = 0 and ad + bc = 0.

So 0 = ad + bc = (ad + bc)c = bc2 implies bc = 0, and so ad = 0. ThenaRα(c) = 0, bRα(c) = 0 and aRα(d) = 0. Thus for any C =

(h k0 h

) ∈ T (R,R),

ACα(B) =(

ahα(c) ahα(d)+akα(c)+bhα(c)0 ahα(c)

)= O. Hence AT (R,R)α(B) = O, and

thus T (R, R) is α-semicommutative. ¤

Corollary 2.11. If R is an α-rigid ring, then T (R, R) is an α-semicommuta-tive ring.

Proof. It follows from Theorem 2.4 and Proposition 2.10. ¤

The trivial extension T (R,R) of a ring R is extended to

S3(R) =

a b c0 a d0 0 a

∣∣∣∣ a, b, c, d ∈ R

and an endomorphism α of a ring R is also extended to the endomorphism αof S3(R) defined by α((aij)) = (α(aij)). There exists a reduced ring R suchthat S3(R) is not α-semicommutative by the following example.

Example 2.12. We consider the commutative reduced ring R = Z2 ⊕ Z2 andthe automorphism α of R defined by α((a, b)) = (b, a), in Example 2.3. ThenS3(R) is not α-semicommutative. For, let


A =

(1, 0) (0, 0) (0, 0)(0, 0) (1, 0) (0, 0)(0, 0) (0, 0) (1, 0)

and B =

(0, 1) (0, 0) (0, 0)(0, 0) (0, 1) (0, 0)(0, 0) (0, 0) (0, 1)

∈ S3(R).

Then AB = O, but AAα(B) = A 6= O. Thus AS3(R)α(B) 6= O, and thereforeS3(R) is not α-semicommutative.

However, we obtain that S3(R) is α-semicommutative for a reduced α-semicommutative ring R by the similar method to the proof of Proposition2.10 as follows:

Proposition 2.13. Let R be a reduced ring. If R is an α-semicommutativering, then

S3(R) =

a b c0 a d0 0 a

∣∣∣∣ a, b, c, d ∈ R

is an α-semicommutative ring.

Proof. Let AB = O for A =(

a b c0 a d0 0 a

), B =

(a′ b′ c′0 a′ d′0 0 a′

)∈ S3(R). Then we have

the following equations:

(1) aa′ = 0,

(2) ab′ + ba′ = 0,

(3) ac′ + bd′ + ca′ = 0,

(4) ad′ + da′ = 0.

From Eq.(1), we get aRα(a′) = 0. In Eq.(2), 0 = (ab′ + ba′)a′ = ba′2, and soba′ = 0 and ab′ = 0. Similarly, from Eq.(4), we have da′ = 0 and ad′ = 0. Also,in Eq.(3), 0 = (ac′+ bd′+ ca′)a′ = ca′2 implies ca′ = 0 and ac′+ bd′ = 0. Then0 = a(ac′ + bd′) = a2c′, and so ac′ = 0 and bd′ = 0. Hence, these yield thataRα(a′) = 0, aRα(b′) = 0, bRα(a′) = 0, aRα(c′) = 0, aRα(d′) = 0, bRα(d′) =0, cRα(a′) = 0, and dRα(a′) = 0. Thus AS3(R)α(B) = O, and therefore S3(R)is an α-semicommutative ring. ¤

Corollary 2.14 ([11, Proposition 1.2]). Let R be a reduced ring. Then S3(R)is a semicommutative ring.

For an α-rigid ring R and n ≥ 2, let

Sn(R) =

a a12 a13 · · · a1n

0 a a23 · · · a2n

0 0 a · · · a3n

......

.... . .

...0 0 0 · · · a

∣∣∣∣ a, aij ∈ R

.


From Proposition 2.13, we may suspect that Sn(R) may be α-semicommutativefor n ≥ 4. But the possibility is eliminated by the next example.

Example 2.15. Let R be an α-rigid ring and

S4(R) =

a a12 a13 a14

0 a a23 a24

0 0 a a34

0 0 0 a

∣∣∣∣ a, aij ∈ R

.

Note that if R is an α-rigid ring, then α(e) = e for e2 = e ∈ R by [6, Proposition5]. In particular α(1) = 1. For

A =(

0 1 −1 00 0 0 00 0 0 00 0 0 0

), B =

(0 0 0 00 0 0 10 0 0 10 0 0 0

)∈ S4(R),

we obtain AB = O. But we have

O 6=(

0 0 0 10 0 0 00 0 0 00 0 0 0

)= ACα(B) ∈ S4(R) for C =

(0 0 0 00 0 1 00 0 0 00 0 0 0

)∈ S4(R).

Thus AS4(R)α(B) 6= O and so S4(R) is not α-semicommutative. Similarly, itcan be proved that Sn(R) is not α-semicommutative for n ≥ 5.

Observe that let Ri be a ring and αi an endomorphism of Ri for each i ∈ Γ.Then, for the product Πi∈ΓRi of Ri and the endomorphism α : Πi∈ΓRi →Πi∈ΓRi defined by α((ai)) = (αi(ai)), Πi∈ΓRi is α-semicommutative if andonly if each Ri is αi-semicommutative.

3. Related topics

Following [14], a ring R is called Armendariz if whenever two polynomialsf(x) =

∑mi=0 aix

i, g(x) =∑n

j=0 bjxj ∈ R[x] satisfy f(x)g(x) = 0 we have

aibj = 0 for every i and j. We extend the Armendariz property of a ring R tothe skew power series ring R[[x; α]] of R. In [6, Proposition 17 and Corollary18], if R is an α-rigid ring, then for p =

∑∞i=0 aix

i and q =∑∞

j=0 bjxj in

R[[x; α]], pq = 0 if and only if aibj = 0 for all 0 ≤ i, 0 ≤ j; and the skew powerseries ring R[[x; α]] is reduced.

Hence, we define the following:

Definition 3.1. Let α be an endomorphism of a ring R. A ring R is calleda skew power series Armendariz ring with the endomorphism α (simply, an α-sps Armendariz ring) if whenever pq = 0 for p =

∑∞i=0 aix

i, q =∑∞

j=0 bjxj ∈

R[[x; α]], then aibj = 0 for all i and j.

It can be easily checked that if R is an α-rigid ring then R is α-sps Armen-dariz by [6, Proposition 17], and that every subring S with α(S) ⊆ S of anα-sps Armendariz ring is also α-sps Armendariz. We remark that in generalthe reverse implication in the above definition does not hold by the followingexample.


Example 3.2. Let R = Z2⊕Z2 where Z2 is the ring of integers modulo 2, andα : R → R be an endomorphism defined by α((a, b)) = (b, a) as in Example 2.3.Then for p = (1, 0)x and q = (0, 1) in R[[x; α]], we have (1, 0)(0, 1) = (0, 0) ∈ R,but pq 6= 0. Moreover, R is not α-sps Armendariz: In fact, for p = (1, 0) +(1, 0)x and q = (0, 1) + (1, 0)x in R[[x; α]], pq = 0 but (1, 0)(1, 0) 6= (0, 0) ∈ R.

Theorem 3.3. Let R be a ring. Then we have the following.(1) R is an α-rigid ring if and only if R is a reduced α-sps Armendariz ring.(2) If R[[x;α]] is a semicommutative ring, then R is an α-semicommutative

ring.(3) Let R be an α-sps Armendariz ring. Then

(i) if R is α-semicommutative, then R[[x;α]] is semicommutative;(ii) if ab = 0 for a, b ∈ R, then αn(a)b = 0 for any positive integer n;(iii) if aαm(b) = 0 for a, b ∈ R and some positive integer m, then ab = 0.

Proof. (1) It is enough to show that R is α-rigid when R is a reduced α-spsArmendariz ring. Assume aα(a) = 0 for a ∈ R. Then for p = ax and q = a inR[[x; α]], pq = axa = aα(a)x = 0. Since R is α-sps Armendariz, a2 = 0. Thusa = 0 because R is reduced. Therefore R is an α-rigid ring.

(2) Assume that R[[x; α]] is a semicommutative ring. Let ab = 0 for a, b ∈ R.Then aR[[x;α]]b = 0. Thus arxb = 0 for any r ∈ R. Hence arα(b)x = 0 andso aRα(b) = 0. Therefore R is α-semicommutative.

(3) Let R be an α-sps Armendariz ring. (i) Assume that R is α-semicommu-tative. First we show that R is semicommutative. Let ab = 0, then aRα(b) = 0since R is α-semicommutative. Let f = arx and g = b ∈ R[[x; α]] for any r ∈ R.Then fg = arxb = arα(b)x = 0 since aRα(b) = 0, and so arb = 0 since R is α-sps Armendariz. Therefore aRb = 0, and thus R is semicommutative. Now, letpq = 0 for p =

∑∞i=0 aix

i, q =∑∞

j=0 bjxj ∈ R[[x;α]]. Then aibj = 0 for all i, j,

since R is α-sps Armendariz. Hence aiRα(bj) = 0 and so aiRαk(bj) = 0 for alli, j and positive integer k because R is α-semicommutative. This implies thatfor ckxk ∈ R[[x; α]], p(ckxk)q = (

∑∞i=0 aix

i)ckxk(∑∞

j=0 bjxj) = a0ckαk(b0)xk

+ (a0ckαk(b1) + a1 α(ck)αk+1(b0))xk+1 + · · · = 0, since aiRαk(bj) = 0 for alli, j and positive integer k. By this fact and R is semicommutative, we getphq = 0 for all h ∈ R[[x; α]]. Therefore pR[[x;α]]q = 0, and so R[[x; α]] is asemicommutative ring. (ii) Suppose that ab = 0 for a, b ∈ R. It is enoughto show that α(a)b = 0. Let p = α(a)x and q = bx in R[[x;α]]. Then pq =α(a)α(b)x2 = α(ab)x2 = 0. Since R is α-sps Armendariz, α(a)b = 0. (iii)Suppose that aαm(b) = 0 for some positive integer m. Let p = axm and q = bxin R[[x;α]]. Then pq = aαm(b)xm+1 = 0, and thus ab = 0 since R is α-spsArmendariz. ¤

It can be easily checked that if R is an α-sps Armendariz ring, then α isalways a monomorphism and aα(b) = 0 for a, b ∈ R implies α(a)b = 0 byTheorem 3.3(3)(ii) and (iii). Moreover, we have the following.


Corollary 3.4. If R is an α-sps Armendariz ring, then α(1) = 1, where 1 isthe identity of R. In this case, α(e) = e for any e2 = e ∈ R.

Proof. (1− α(1))α(1) = 0 implies α(1− α(1))α(1) = 0 by Theorem 3.3(3)(ii).Hence we have (α(1)− α(α(1)))α(1) = 0, and so α(1) = α(α(1)). Since α is amonomorphism, α(1) = 1.

Now, let e2 = e ∈ R. Then e(1− e) = 0 implies α(e)(1− e) = 0 by Theorem3.3(3)(ii). Hence α(e) = α(e)e. Similarly, (1 − e)e = 0 implies e = α(e)e.Consequently, α(e) = e. ¤

Lemma 3.5. Let R be an α-sps Armendariz ring. Then the set of all idem-potents in R[[x; α]] coincides with the set of all idempotents of R and R[[x; α]]is abelian.

Proof. Let e2 = e ∈ R[[x;α]], where e =∑∞

i=0 eixi. Since e(1−e) = 0 = (1−e)e,

we have (e0 + e1x + · · · + enxn + · · · )((1 − e0) − e1x − · · · − enxn − · · · ) = 0and ((1− e0)− e1x− · · · − enxn − · · · )(e0 + e1x + · · ·+ enxn + · · · ) = 0. SinceR is an α-sps Armendariz ring, e0(1− e0) = 0, e0ei = 0 and (1− e0)ei = 0 for1 ≤ i. Thus ei = 0 for 1 ≤ i, and so e = e0 = e2

0.Now, we claim that R is abelian. Note that α(e) = e for any e2 = e ∈ R

by Corollary 3.4. We adapt the method in the proof of [7, Proposition 20].For idempotents e and e′ ∈ R, ee′R

⋂(1 − e′)(1 − e)α(R) = 0. Suppose that

0 6= ee′(−t) = (1− e′)(1− e)α(s) ∈ ee′R⋂

(1− e′)(1− e)α(R) for some s, t ∈ R.Then ((1− e′)x + e)(e′tx + (1− e)s) = (1− e′)α(e′t)x2 + (ee′t + (1− e′)α((1−e)s)x+e(1−e)s = (1−e′)e′α(t)x2 +(ee′t+(1−e′)(1−e)α(s))x+e(1−e)s = 0,since α(e′) = e′ and α(1 − e) = 1 − e. But ee′t 6= 0; which is a contradictionsince R is α-sps Armendariz. Furthermore, suppose that e′e = 0. Then ee′ =(1− e′)(1− e)(−e′) = (1− e′)(1− e)(−α(e′)) ∈ ee′R

⋂(1− e′)(1− e)α(R) = 0.

Thus, for any idempotent e ∈ R and any r ∈ R, e′′ = e + er(1 − e) is anidempotent in R with (1− e)e′′ = 0. Hence e′′(1− e) = 0 and so er(1− e) = 0,i.e., er = ere. Similarly, e′′′ = (1 − e) + (1 − e)re is an idempotent in R withee′′′ = 0. Thus e′′′e = 0 and so (1− e)re = 0, i.e., re = ere. Hence, er = re forany r ∈ R and so R is abelian. Therefore R[[x;α]] is abelian. ¤

Now we turn our attention to the relationship between the Baerness andp.p.-property of a ring R and these of the skew power series ring R[[x; α]] incase R is α-sps Armendariz.

In [10], Kaplansky introduced the concept of Baer rings as rings in which theright (left) annihilator of every nonempty subset is generated by an idempotent.According to Clark [3], a ring R is called quasi-Baer if the right annihilatorof each right ideal of R is generated (as a right ideal) by an idempotent. It iswell-known that these two concepts are left-right symmetric.

A ring R is called a right (resp., left) p.p.-ring if the right (resp., left) annihi-lator of an element of R is generated by an idempotent. R is called a p.p.-ringif it is both a right and left p.p.-ring.


On the other hand, Birkenmeier, Kim and Park [2] called R a right (resp.,left) principally quasi-Baer (or simply right (resp., left) p.q.-Baer) ring if theright (resp., left) annihilator of a principal right (resp., left) ideal of R is gen-erated by an idempotent. R is called a p.q.-Baer ring if it is both right andleft p.q.-Baer. The class of p.q.-Baer rings has been extensively investigatedby them [2]. This class includes all biregular rings, all (quasi-) Baer rings andall abelian p.p.-rings. Various extensions of Baer, quasi-Baer, p.q.-Baer andp.p.-rings have been studied by many authors [2, 6, 7, 8].

For a nonempty subset A of a ring R, we write rR(A) = {c ∈ R | dc = 0 forany d ∈ A} which is called the right annihilator of X in R.

Theorem 3.6. Let R be an α-sps Armendariz ring.(1) R is a Baer (resp., quasi-Baer) ring if and only if R[[x; α]] is a Baer

(resp., quasi-Baer) ring.(2) If R[[x; α]] is a right p.p.-ring, then R is a right p.p.-ring.

Proof. (1) Assume that R is Baer. Let A be a nonempty subset of R[[x; α]] andA∗ be the set of all coefficients of elements of A. Then A∗ is a nonempty subsetof R and so rR(A∗) = eR for some idempotent e ∈ R. Since e ∈ rR[[x;α]](A)by Corollary 3.4, we get eR[[x;α]] ⊆ rR[[x;α]](A). Now, we let 0 6= q = b0 +b1x + · · · + btx

t + · · · ∈ rR[[x;α]](A). Then Aq = 0 and hence pq = 0 for anyp ∈ A. Since R is α-sps Armendariz, b0, b1, . . . , bt, . . . ∈ rR(A∗) = eR. Hencethere exists c0, c1, . . . , ct, . . . ∈ R such that q = ec0 + ec1x + · · ·+ ectx

t + · · · =e(c0 +c1x+ . . .+ctx

t + · · · ) ∈ eR[[x; α]]. Consequently eR[[x; α]] = rR[[x;α]](A),and therefore R[[x; α]] is Baer.

Conversely, assume that R[[x; α]] is Baer. Let B be a nonempty subset ofR. Then rR[[x;α]](B) = eR[[x;α]] for some idempotent e ∈ R by Lemma 3.5.Thus rR(B) = rR[[x;α]](B) ∩R = eR[[x; α]] ∩R = eR, and therefore R is Baer.

The proof for the case of the quasi-Baer property follows in a similar fashion:In fact, for any right ideal A of R[[x;α]], take A∗ as the right ideal generatedby all coefficients of elements of A.

(2) Assume that R[[x;α]] is a right p.p.-ring. Let a ∈ R, then there existsan idempotent e ∈ R such that rR[[x;α]](a) = eR[[x; α]] by Lemma 3.5. HencerR(a) = eR, and therefore R is a right p.p.-ring. ¤

As a consequence we obtain:

Corollary 3.7 ([6, Theorem 21]). Let R be an α-rigid ring. Then R is a Baerring if and only if R[[x; α]] is a Baer ring.

There exists an α-sps Armendariz and right p.q.-Baer ring R such thatR[[x; α]] is not right p.q.-Baer by the next example.

Example 3.8 ([6, p.225]). For a field F , let

R =

{(an) ∈

∞∏n=1

Fn | an is eventually constant

},


which is the subring of∏∞

n=1 Fn, where Fn = F for n = 1, 2, 3 . . . . ThenR is right p.q.-Baer and IR-rigid (and so IR-sps Armendariz), where IR is theidentity endomorphism of R. But R[[x; IR]] is not right p.q.-Baer. Furthermore,R[[x; IR]] is neither right p.p. nor left p.p..

Finally, we have the following result which can be compared with Lemma 3.5.

Proposition 3.9. If R is an α-semicommutative ring with α(1) = 1, then theset of all idempotents in R[[x;α]] coincides with the set of all idempotents of Rand R[[x;α]] is abelian.

Proof. Note that R is an abelian ring with α(e) = e for any e2 = e ∈ R byProposition 2.6. Let p2 = p ∈ R[[x; α]], where p = e0 + e1x + e2x

2 + · · · . Sinceα(e) = e for any e2 = e ∈ R, p2 = p implies the following system of equations:

e20 = e0 ;

e0e1 + e1α(e0) = e1 ;e0e2 + e1α(e1) + e2α

2(e0) = e2 ;...

e0ek + e1α(ek−1) + . . . + ek−1αk−1(e1) + ekαk(e0) = ek ;

...From e2

0 = e0, we see that e0 is an idempotent of R, so e0 is central. Then weget the following:

(5) e0e1 + e1e0 = e1

(6) e0e2 + e1α(e1) + e2e0 = e2

...

(7) e0ek + e1α(ek−1) + . . . + ek−1αk−1(e1) + eke0 = ek

...From Eq.(5)×(1− e0), we obtain e1(1− e0) = 0, and so e1 = e1e0 = 0. HenceEq.(6) becomes 2e0e2 = e2. Similarly, 2e0e2(1 − e0) = e2(1 − e0) impliese2 = 0. Continuing this procedure yields ei = 0 for i ≥ 1. Consequentlyp = e0 = e2

0 ∈ R and also R[[x; α]] is abelian by Proposition 2.6. ¤

Corollary 3.10. If a ring R has one of the following such that R[[x; α]] is aright p.p.-ring;

(1) R is an α-sps-Armendariz ring,(2) R is an α-semicommutative ring with α(1) = 1.

Then R[[x; α]] is semicommutative.

Proof. If R has one of the conditions, then R is abelian by Lemma 3.5 andProposition 3.9, respectively. Note that abelian right p.p.-rings are semicom-mutative by [2, Proposition 1.14]. ¤


Observe that the following example shows that the converse of Theorem3.3(2) does not hold; and the condition “R is α-sps Armendariz” in Theorem3.3(3) and the condition “α(1) = 1” in Proposition 3.9 cannot be dropped,respectively.

Example 3.11. Consider the α-semicommutative ring R = {( a b0 c ) | a, b, c ∈ Z}

in Example 2.7, where α (( a b0 c )) = ( a 0

0 0 ) . Then α(1) 6= 1, and so R is not α-sps Armendariz by Corollary 3.4. Moreover, R[[x; α]] is not semicommutative,either: In fact, For p =

(1 −10 0

)+ ( 1 1

0 1 )x and q = ( 0 10 1 ) in R[[x;α]] and ( 1 1

0 1 ) ∈R[[x; α]], pq = 0 but p ( 1 1

0 1 ) q 6= 0 and so pR[[x;α]]q 6= 0. Therefore R[[x; α]]is not semicommutative. Note that let f = ( 1 0

0 1 ) + ( 0 00 1 )x ∈ R[[x;α]], then

f2 = f ∈ R[[x; α]], but f /∈ R. Finally, for A =(

1 −10 0

), B = ( 0 1

0 1 ), andC = ( 1 0

0 1 ) ∈ R, we get AB = O but αn(A)B 6= O for any positive integer nand BC 6= O even if Bα(C) = O.

References

[1] D. D. Anderson and V. Camillo, Semigroups and rings whose zero products commute,Comm. Algebra 27 (1999), no. 6, 2847–2852.

[2] G. F. Birkenmeier, J. Y. Kim, and J. K. Park, Principally quasi-Baer rings, Comm.Algebra 29 (2001), no. 2, 639–660.

[3] W. E. Clark, Twisted matrix units semigroup algebras, Duck Math. J. 34 (1967), 417–423.

[4] P. M. Cohn, Reversible rings, Bull. London Math. Soc. 31 (1999), no. 6, 641–648.[5] J. M. Habeb, A note on zero commutative and duo rings, Math. J. Okayama Univ. 32

(1990), 73–76.[6] C. Y. Hong, N. K. Kim, and T. K. Kwak, Ore extensions of Baer and p.p.-rings, J.

Pure Appl. Algebra 151 (2000), no. 3, 215–226.[7] , On skew Armendariz rings, Comm. Algebra 31 (2003), no. 1, 103–122.[8] , Extensions of generalized reduced rings, Algebra Colloq. 12 (2005), no. 2, 229–

240.[9] C. Y. Hong, T. K. Kwak, and S. T. Rizvi, Rigid ideals and radicals of Ore extensions,

Algebra Colloq. 12 (2005), no. 3, 399–412.[10] I. Kaplansky, Rings of Operators, W. A. Benjamin, Inc., New York-Amsterdam, 1968.[11] N. K. Kim and Y. Lee, Extensions of reversible rings, J. Pure Appl. Algebra 185 (2003),

no. 1-3, 207–223.[12] J. Krempa, Some examples of reduced rings, Algebra Colloq. 3 (1996), no. 4, 289–300.[13] J. Lambek, On the representation of modules by sheaves of factor modules, Canad.

Math. Bull. 14 (1971), 359–368.[14] M. B. Rege and S. Chhawchharia, Armendariz rings, Proc. Japan Acad. Ser. A Math.

Sci. 73 (1997), no. 1, 14–17.[15] G. Y. Shin, Prime ideals and sheaf representation of a pseudo symmetric ring, Trans.

Amer. Math. Soc. 184 (1973), 43–60.

Muhittin BaserDepartment of MathematicsKocatepe UniversityAfyonkarahisar 03200, TurkeyE-mail address: [email protected]


Abdullah HarmanciDepartment of MathematicsHacettepe UniversityAnkara, TurkeyE-mail address: [email protected]

Tai Keun KwakDepartment of MathematicsDaejin UniversityPocheon 487–711, KoreaE-mail address: [email protected]

generalized semicommutative rings

Documents