ON A CLASS OF SEMICOMMUTATIVE RINGS Abdullah Harmanci and Handan Kose 1

17 Pages·2017·0.27 MB·English

NEWZEALANDJOURNALOFMATHEMATICS
Volume47(2017),69-85
ON A CLASS OF SEMICOMMUTATIVE RINGS
Abdullah Harmanci and Handan Kose
(ReceivedJanuary1,2017)
Abstract. Let R be a ring with identity and an ideal I. In this paper, we
introduce a class of rings generalizing semicommutative rings which is called
I-semicommutative. TheringRiscalledI-semicommutativewheneverab=0
implies aRb ⊆ I for any a, b ∈ R. We investigate general properties of I-
semicommutativeringsandshowthatseveralresultsofsemicommutativerings
andJ-semicommutativeringscanbeextendedtoI-semicommutativeringsfor
thisgeneralsettings.
1. Introduction
Throughoutthispaper, aringmeansanassociativeringwithidentity. Wewrite
U(R) for the set of all units in R, T (R) stands for the ring of all n×n triangular
n
matrices over a ring R. A ring R is called semicommutative if for any a, b ∈ R,
ab = 0 implies aRb = 0, this ring is also called ZI ring in [9] and [13], while, in
[24], R is said to be central semicommutative if ab=0 implies aRb is central in R.
And in [17] a ring R is called weakly semicommutative, if for any a, b∈R, ab=0
implies arb is a nilpotent element for each r ∈ R. Another generalization is made
in [6], in which a ring R is called nil-semicommutative-II if a, b ∈ R satisfy ab ∈
Nil(R), then arb ∈ Nil(R) for any r ∈ R where Nil(R) is the set of all nilpotent
elementsofR. Asimilarconceptisnil-semicommutativityisinvestigatedin[20],in
which it is said that a ring R is nil-semicommutative-I if for all nilpotent elements
a, b of R, ab=0 implies aRb=0.
Every semicommutative ring is central semicommutative, weakly semicommu-
tative, nil-semicommutative-I and nil-semicommutative-II. These classes of rings
have been getting much attention, see namely, [6, 11, 20, 24, 25, 26]. Their rela-
tions with other classes of rings, such as Abelian rings, reduced rings, Armendariz
rings and others, have been studied in the past. Also, another generalization of
semicommutative rings is given in [33], called J-semicommutative rings. Let J(R)
denote the Jacobson radical of R. A ring R is J-semicommutative if ab=0 implies
aRb ⊆ J(R). It is well known that J(R) has some nice properties to help us to
get good results. Motivated by these considerations, a natural question arises here
when we replace J(R) by an ideal I of R. In this paper, a new kind of rings which
behavelikesemicommutativeringsisconsidered. Weinvestigatethesemicommuta-
tivity of the ring relative to an ideal I. These are called I-semicommutative rings.
That is, a ring R with an ideal I is called I-semicommutative provided ab = 0
implies aRb⊆I. In the second section of this paper, occasionally, we focus on the
2010 Mathematics Subject Classification 13C99,16D80,16U80.
Key words and phrases: Semicommutative rings, J-Semicommutative rings, I-Semicommutative
rings,Exchangerings,Idealextensions,NilSemicommutativerings.
70 ABDULLAH HARMANCI and HANDAN KOSE
general case of the ideal I fixed in R. We prove that if R is von Neumann regular,
then every I-semicommutative ring for which I is a quasiregular ideal is semicom-
mutative. If R is an abelian I-semipotent ring and I is a semiprime ideal, then R
is I-semicommutative. In the third section, the I-semicommutativity of certain ex-
tensionsofringsisstudied. ItisprovedthatRisanI-semicommutativeringifand
only if T (R) is T (I)-semicommutative for all positive integer n. Also a necessary
n n
and sufficient conditions for the ideal extensions, Dorroh extensions and Nagata
extensions to be I-semicommutative are studied. The last section is devoted to
exchange J-semicommutative rings. It is proved that if R is a J-semicommutative
exchange ring, then R/J(R) is commutative if and only if for each positive integer
n, the set GL (R) of all invertible n×n matrices is closed under transposition if
n
and only if for all a,b,c∈U(R), c+[a,b]∈U(R) if and only if for all a,b∈U(R),
[a,b]∈J(R), where U(R) is the set of invertible elements in R.
2. I-Semicommutative Rings
A ring R is defined to be nil-semicommutative-II in case for any a, b ∈ R,
ab ∈ Nil(R) implies that arb ∈ Nil(R) whenever r ∈ R (see for detail [6]). In
[33], J-semicommutative rings are studied. A ring is J-semicommutative if ab=0
implies aRb ⊆ J(R). It is well known that J(R) has some nice properties to help
us to get good results. A natural question arises here when we replace J(R) by
an ideal of R. In the preceding definitions, we replace Nil(R) and J(R) by an
ideal I and we call the ring R I-semicommutative. A ring R is called abelian
if every idempotent is central, that is, ae = ea for any e2 = e, a ∈ R. Every
local ring is J-semicommutative. Every semicommutative ring and every central
semicommutativeringisabelian[24]. However, thereareI-semicommutativerings
that are neither semicommutative nor abelian. Also both nil-semicommutative-I
rings and nil-semicommutative-II rings need not be abelian.
Example 2.1. Let Z denote the ring of integers and consider the ring R=T (Z).
2
(cid:20) Z Z (cid:21) (cid:20) a b (cid:21)
(1) Let I = . Then R is I-semicommutative. For if A = ,
0 0 0 c
(cid:20) (cid:21)
x y
B = ∈R and AB =0, we have ax=0, ay+bz =0, cz =0. Then a=0
0 z
or x=0 and c=0 or z =0. Accordingly, ARB ⊆I and R is I-semicommutative.
(cid:20) 0 Z (cid:21) (cid:20) a b (cid:21) (cid:20) x y (cid:21)
(2) Let I = . Then if A= , B = ∈R and AB =0, we
0 Z 0 c 0 z
have ax = 0, ay+bz = 0, cz = 0. Then a = 0 or x = 0 and c = 0 or z = 0. We
have ARB ⊆I and R is I-semicommutative.
(cid:20) 0 Z (cid:21)
(3)LetI = . AsimpleargumentrevealsthatRisagainI-semicommutative.
0 0
(cid:20) (cid:21) (cid:20) (cid:21)
1 1 0 1
(4) Let A= , B = ∈R. Then AB =0 and ARB (cid:54)=0. Hence
0 0 0 −1
R is not semicommutative.
Example 2.2. Let R be a commutative ring and consider the ring
     
a b c 0 b c
   
S =  0 d 0 |a,b,c,d,e∈R , ideal I =  0 0 0 |b,c∈R of
 0 0 e   0 0 0 
ON A CLASS OF SEMICOMMUTATIVE RINGS 71
   
a b c x y z
S. Let A =  0 d 0 , B =  0 t 0  with AB = 0. Then ASB =
0 0 e 0 0 u
  
0 n l
 
 0 0 0 |n,l∈R ⊆I. Hence S is I-semicommutative.
 0 0 0 
Lemma 2.3. Let R be a ring with an ideal I.
(1) If R/I is a semicommutative ring, then R is I-semicommutative.
(2) If I consists of all nilpotent elements, then R is I-semicommutative.
Proof. (1) Let a, b ∈ R with ab = 0. Then in R/I := R¯ we have a¯¯b = ¯0. By the
semicommutativity of R/I, a¯R¯¯b=¯0 and so aRb⊆I.
(2) Note first that being I nil ideal, I is contained in J(R). Let a, b ∈ R with
ab = 0. Then (bRa)2 = 0. By hypothesis on I, bRa ⊆ I and RbRa ⊆ I. Then
(aRbR)2 = a(RbRa)RbR ⊆ I. Hence (aRbR)2 consists of nilpotent elements. So
aRbR⊆I. Since R has an identity, aRb⊆I. (cid:3)
Recall that a ring R is called semiprime if R has no nonzero nilpotent ideals.
An ideal I in a ring R is called semiprime if the ring R/I is a semiprime ring.
Equivalently, the ideal I is semiprime in the ring R if and only if for any ideal J
of R and any positive integer n, Jn ⊆I implies J ⊆I if and only if for any a∈R,
aRa⊆I implies a∈I.
Proposition 2.4. Let R be a ring. If I is a semiprime ideal, then the following
are equivalent.
(1) R is I-semicommutative.
(2) For any a∈R, a2 =0 implies that a∈I.
(3) For any a, b∈R, ab=0 implies that bRa⊆I.
Proof. (1)⇒(2)Ifa2 =0,thenaRa⊆I. Hence,(cid:0)RaR(cid:1)2 ⊆I. AsI issemiprime,
it follows that RaR⊆I. Hence a∈I.
(2)⇒(3) If ab=0, then (bra)2 =0 for any r ∈R. By (2) bra∈I. Thus, bRa⊆I.
(cid:0) (cid:1)2
(3) ⇒ (1) If ab = 0, then bRa ⊆ I. Hence, RaRbR ⊆ I. As I is semiprime, we
deduce that RaRbR ⊆ I. This implies that aRb ⊆ I since R has an identity, as
required. (cid:3)
Proposition 2.5. Let R and S be rings, R×S their direct product, I an ideal of
R and L an ideal of S. Then R×S is I ×L-semicommutative if and only if R is
I-semicommutative and S is L-semicommutative.
Proof. Assume that R×S is I ×L-semicommutative and a, b ∈ R with ab = 0.
Then (a,0)(b,0) = (0,0) in R×S. By assumption (a,0)(R×S)(b,0) ⊆ I ×L.
This implies aRb⊆I. Hence R is I-semicommutative. A similar discussion reveals
that S is L-semicommutative. Conversely, suppose that R is I-semicommutative
and S is L-semicommutative. Let (r,s), (r ,s )∈R×S with (r,s)(r ,s )=(0,0).
1 1 1 1
Then rr = 0 and ss = 0. By supposition rRr ⊆ I and sSs ⊆ L. Hence
1 1 1 1
(r,s)(R×S)(r ,s )⊆I×L. (cid:3)
1 1
Corollary 2.6. Let R be an abelian ring. Then R is I-semicommutative if and
only if for any idempotent e∈R, eR and (1−e)R are both I-semicommutative.
72 ABDULLAH HARMANCI and HANDAN KOSE
Lemma 2.7. If I is a semiprime ideal of R, R is an I-semicommutative ring and
idempotents lift modulo I, then R/I is abelian.
Proof. Let y ∈R with y2 =0. Then yRy ⊆I. Hence y ∈I, since I is semiprime.
Let e2 =e∈R. For any x∈R, (ex−exe)2 =0 and (xe−exe)2 =0 which implies
ex−exe, xe−exe∈I. Since idempotents lift modulo I, hence ex−xe∈I. (cid:3)
Proposition 2.8. LetR beanI-semicommutativeringandI beanidealcontained
in J(R). If R is a von Neumann regular ring, then R is semicommutative.
Proof. Let x,y ∈ R with xy = 0. By hypothesis xry ∈ I for all r ∈ R. Since
I ⊆J(R),foralla∈R,1−a(xry)isinvertible. SinceRisavonNeumannregular,
there exists u ∈ R such that xry = (xry)u(xry). It implies xry(1−u(xry)) = 0.
By invertibility of 1−u(xry) we have xry =0 for all r ∈R. This shows that R is
semicommutative. (cid:3)
Lemma 2.9. Let R be an I-semicommutative ring. If I is an ideal contained in
J(R), then R is directly finite.
Proof. Let R be an I-semicommutative ring and a,b ∈ R with ab = 1, e = ba.
Clearly, ae = a,eb = b. Since R is I-semicommutative and e(1−e) = 0, we get
eR(1−e)⊆I. In particular, eb(1−e)∈I. Since eb=b, b(1−e)∈I. Multiplying
the latter from left by a, we have ab(1−e) ∈ I. Since ab = 1, 1−e ∈ I. Being
1−e an idempotent and I ⊆J(R), we have 1=e=ba. (cid:3)
Proposition 2.10. Let e∈R be an idempotent. If R is I-semicommutative, then
eRe is eIe-semicommutative.
Proof. Let eae, ebe ∈ eRe with (eae)(ebe) = 0. Since R is I-semicommutative,
(eae)R(ebe) ⊆ I, and so (eae)R(ebe) ⊆ eIe. That is, (eae)(eRe)(ebe) ⊆ eIe.
Therefore eRe is eIe-semicommutative. (cid:3)
Let R be a ring, n a positive integer and R denote the subring of the n×n upper
n
triangular matrix ring T (R) defined by
n
  
a a ··· a
 0 a12 ··· a12nn  
Rn = 0... 0... ·.·.·. a...  | a,aij ∈R(i<j)
  
a a ··· a
 0 a12 ··· a12nn  
Let I be an ideal of R. Then In =  0... 0... ·.·.·. a... ∈Tn(R)|a∈I is
an ideal of R .
n
Proposition 2.11. Let R be a ring and n a positive integer with 2≤n. Then the
following are equivalent.
(1) R is I-semicommutative.
(2) R is I -semicommutative.
n n
ON A CLASS OF SEMICOMMUTATIVE RINGS 73
Proof. Assume that R is I-semicommutative. Let A,B ∈R with AB =0, where
n
   
a a ··· a b b ··· b
12 1n 12 1n
 0 a ··· a2n   0 b ··· b2n 
A =  ... ... ... ... , B =  ... ... ... ... . Then ab = 0. Since R
0 0 ··· a 0 0 ··· b
is I-semicommutative, aRb ⊆ I. The diagonal entries of AR B consist of arb ∈ I
n
where r ∈R and other entries are in R. It follows that R is I -semicommutative.
n n
  
a 0 ··· 0 b 0 ··· 0
 0 a ··· 0  0 b ··· 0 
Conversely,leta,b∈Rwithab=0. Then ... ... ... ...  ... ... ... ... =
0 0 ··· a 0 0 ··· b
 
c x ··· y
 0 c ··· z 
0. For any  ... ... ... ... ∈Rn, we have
0 0 ··· c
   
a 0 ··· 0 c x ··· y b 0 ··· 0
 0 a ··· 0  0 c ··· z  0 b ··· 0 
 ... ... ... ...  ... ... ... ...  ... ... ... ... ∈In.
0 0 ··· a 0 0 ··· c 0 0 ··· b
Therefore acb∈I for all c∈R. So R is I-semicommutative. (cid:3)
Proposition 2.12. Let R be a ring and K ⊆ K ⊆ R and L ⊆ L ⊆ R be
1 1
ideals of R. If (R/K)×(R/L) is (K /K)×(L /L)-semicommutative, then R is
1 1
K ∩L -semicommutative.
1 1
Proof. Suppose that (R/K)×(R/L) is (K /K)×(L /L)-semicommutative, and
1 1
let a, b ∈ R with ab = 0. Then (a+K,0+L)(b+K,0+L) =(0+K,0+L) and
(0+K,a+L)(0+K,b+L) = (0+K,0+L) in (R/K)×(R/L). By hypothesis
(a+K,0+L)(R/K×R/L)(b+K,0+L)⊆K /K×L /Land(0+K,a+L)(R/K×
1 1
R/L)(0+K,b+L) ⊆ K /K ×L /L and so (a+K)(R/K)(b+K) ⊆ K /K and
1 1 1
(a+L)(R/L)(b+L) ⊆ L /L. Hence aRb ⊆ K and aRb ⊆ L . Thus aRb ⊆
1 1 1
K ∩L . (cid:3)
1 1
Proposition 2.13. R is an I-semicommutative ring if and only if the ring S =
{(x,y)∈R×R | x−y ∈I} is I×I-semicommutative.
Proof. S = {(x,y) ∈ R×R | x−y ∈ I} forms a subring of R×R. If R is an
I-semicommutative ring and (x,y),(a,b)∈S and (a,b)(x,y)=0, then ax=0 and
by =0 implies aRx⊆I and bRy ⊆I. Hence (a,b)S(x,y)⊆I×I.
Conversely, suppose that S is I×I-semicommutative and let a,b∈R with ab=0.
Then(a,a)(b,b)=0inS. SinceS isI×I-semicommutative,(a,a)(x,y)(b,b)liesin
I×I,forall(x,y)∈S. Inparticularfor(x,x)∈S,wehave(a,a)(x,x)(b,b)∈I×I
for all x∈R. So axb∈I. Therefore R is I-semicommutative. (cid:3)
Definition 2.14. Let R be a ring with an ideal I. R is called I-semipotent if each
leftideal(resp.,rightideal)notcontainedinI containsanon-zeroidempotent,and
R is called I-potent if, in addition, idempotents lift modulo I.
74 ABDULLAH HARMANCI and HANDAN KOSE
Proposition 2.15. Let R be an abelian I-semipotent ring. If I is a semiprime
ideal, then R is I-semicommutative.
Proof. Let ab=0. Assume that bra∈/ I for some r ∈R. Then there is a non-zero
idempotent e ∈ braR. Write e = brat. Then e = e2 = (brat)(brat) = bra(brat)t =
br(ab)rat2 =0 since e is central. This is a contradiction. So bRa⊆I. This proves
that ab=0 implies bRa⊆I. By Proposition 2.4, R is I-semicommutative. (cid:3)
3. Certain Extensions and Applications to J(R)
The goal of this section is to consider some extensions of J-semicommutative
rings. Let S and T be any rings, M an S-T-bimodule and R the formal triangular
(cid:20) (cid:21)
S M
matrix ring . Let I and I be ideals of S and T respectively. Then
0 T 1 2
(cid:20) (cid:21)
I M
I = 1 is an ideal of R.
0 I
2
Proposition 3.1. Let S,T,M,R and I be as above. Then R is I-semicommutative
if and only if S and T are I -semicommutative and I -semicommutative, respec-
1 2
tively.
(cid:20) (cid:21) (cid:20) (cid:21)
S M I M
Proof. Necessity: AssumethatR= isI = 1 -semicommutative.
0 T 0 I
2
(cid:20) (cid:21)(cid:20) (cid:21)(cid:20) (cid:21) (cid:20) (cid:21)(cid:20) (cid:21)(cid:20) (cid:21)
1 0 S M 1 0 1 0 I M 1 0
ByProposition2.10, is 1 -
0 0 0 T 0 0 0 0 0 I 0 0
2
(cid:20) (cid:21)(cid:20) (cid:21)(cid:20) (cid:21)
1 0 S M 1 0
semicommutative. Since is isomorphic to S and
0 0 0 T 0 0
(cid:20) (cid:21)(cid:20) (cid:21)(cid:20) (cid:21)
1 0 I M 1 0
1 is isomorphic to I , S is I -semicommutative. Re-
0 0 0 I 0 0 1 1
2
(cid:20) (cid:21) (cid:20) (cid:21)
1 0 0 0
placing the idempotent by , a similar discussion reveals that T
0 0 0 1
is I -semicommutative.
2
Sufficiency: SupposethatSandT areI -semicommutativeandI -semicommutative
1 2
(cid:20) (cid:21) (cid:20) (cid:21)
a x c y
respectively, and A= , B = ∈R with AB =0. Then ac=0 in
0 b 0 d
S and bd = 0 in T. By supposition, aSc ⊆ I and bTd ⊆ I . Thus ARB ⊆ I and
1 2
so R is I-semicommutative. (cid:3)
Let I be an ideal of R, for a positive integer n, and T (I) denote the ring of all
n
n×nuppertriangularmatrices[a ]witha ∈I. ThenT (I)isanidealofT (R).
ij ii n n
Theorem 3.2. Let R be a ring, I an ideal of R and n a positive integer. Then the
following are equivalent.
(1) R is I-semicommutative.
(2) T (R) is T (I)-semicommutative.
n n
Proof. (1) ⇒ (2) Let [a ], [b ] ∈ T (R) with [a ][b ] = 0. Then a b = 0
ij ij n ij ij ii ii
and so by (1), a Rb ⊆ I. Thus [a ]T (R)[b ] ⊆ T (I) and so T (R) is T (I)-
ii ii ij n ij n n n
semicommutative.
ON A CLASS OF SEMICOMMUTATIVE RINGS 75
 
a 0 0 ... 0
0 a 0 ... 0
(2) ⇒ (1) Let a, b ∈ R with ab = 0 and A = . . . . and
. . . .
. . ... . .
0 0 0 ... a
 
b 0 0 ... 0
0 b 0 ... 0
B = .. .. .. .. ∈ Tn(R). Then AB = 0. By (2),
. . ... . .
0 0 0 ... b
AT (R)B ⊆T (I). It follows that aRb⊆I. (cid:3)
n n
Corollary 3.3. Let R be a ring. Then the following are equivalent.
(1) R is I-semicommutative.
(2) R[x]/(xn) is (cid:0)I[x]+(xn)(cid:1)/(xn)-semicommutative for all n≥2.
Proof. Since
  
a a a ··· a a
R[x]/(xn)∼= 0000...1 a000...21 aa00...321 ···..···..···.. aannna0...−−−1123 aannaa...−−n212  | ai ∈R,
1
we get the result as in Theorem 3.2. (cid:3)
n s
(cid:88) (cid:88)
The ring R is called Armendariz if for any f(x)= a xi, g(x)= b xj ∈R[x],
i j
i=0 j=0
f(x)g(x)=0 implies a b =0 for all i and j (See [12]). For example every reduced
i j
ring is Armendariz.
Proposition3.4. LetRbearing. LetK beanidealoftheringR[x]ofpolynomials
over R. Let I be the ideal of R generated by the coefficients of all polynomials in K.
If R[x] is K-semicommutative, then R is I-semicommutative. The converse holds
if R is Armendariz.
Proof. Assume that R[x] is K-semicommutative. Let a,b ∈ R and ab = 0. Then
aR[x]b ⊆ K, so aRb ⊆ I. Conversely, suppose that R is an I-semicommutative
s t
Armendariz ring, f(x) = (cid:80)a xi, g(x) = (cid:80) b xj ∈ R[x] and f(x)g(x) = 0. R
i j
i=0 j=0
being Armendariz, we have a b = 0, where 0 ≤ i ≤ s, 0 ≤ j ≤ t. By hypothesis
i j
a Rb ⊆ I. Then for all i with 0 ≤ i ≤ s a xiRb ⊆ I[x] and a xiR[x]b ⊆ I[x].
i j i j i j
So f(x)R[x]b ⊆ I[x]. Similarly, f(x)R[x]g(x) ⊆ I[x]. Since all coefficients of the
j
polynomials in I[x] belong to the K, I[x]⊆K. Thus f(x)R[x]g(x)⊆K. (cid:3)
In general, J-semicommutative rings are not Armendariz.
Example3.5. LetRbethefieldofrealnumbers,R=T (R). ByTheorem3.2,Ris
3
aJ-semicommutativering, butitisnotanabelianring, soR isnotanArmendariz
ring by [11, Corollary 8].
76 ABDULLAH HARMANCI and HANDAN KOSE
Let R be a ring with an ideal I and V an R-R-bimodule which is a general ring
(possiblywithnounity)inwhich(vw)r =v(wr),(vr)w =v(rw)and(rv)w =r(vw)
holdforallv,w ∈V andr ∈I. Thentheidealextension(itisalsocalledtheDorroh
extension (see [7])) D(I;V) of I by V is defined to be the additive abelian group
I⊕V withmultiplication(r,v)(s,w)=(rs,rw+vs+vw)where(r,v),(s,w)∈I⊕V.
Since R is an ideal of itself, D(R;V) is a well defined ring and it is also called the
Dorroh extension of R by V. Note that D(I;V) is an ideal of D(R;V) (see [19]).
Proposition 3.6. Suppose that V = J(V). Then the following are equivalent for
a ring R.
(1) R is I-semicommutative.
(2) D(R;V) is D(I;V)-semicommutative.
Proof. (1) ⇒ (2) Let s = (r,v), a = (b,c) ∈ D(R;V) with sa = (0,0). Then
rb=0 and so rRb⊆I by (1). Hence sD(R;V)a⊆D(I;V).
(2) ⇒ (1) Suppose that D(R;V) is D(I;V)-semicommutative. Let a, b ∈ R with
ab = 0. Then (a,0)(b,0) = (0,0) and so (a,0)D(R;V)(b,0) ⊆ D(I;V) by (2).
Hence aRb⊆I. (cid:3)
Corollary 3.7. Suppose that V = J(V). Then the following are equivalent for a
ring R.
(1) R is J(R)-semicommutative.
(2) The ideal extension D(R;V) is J(D(R;V))-semicommutative.
Proof. Note that by the hypothesis and [19, Theorem 13], (r,v) ∈ J(D(R;V))
if and only if r ∈ J(R). Assume that R is J(R)-semicommutative. Let (r,v),
(s,u) ∈ D(R;V) with (r,v)(s,u) = 0 in D(R;V). Then rs = 0 and so rRs ⊆
J(R). Hence (r,v)D(R;V)(s,u)⊆J(D(R;V)). Conversely, suppose that D(R;V)
is J(D(R;V))-semicommutative. Let r, s ∈ R with rs = 0. Then (r,0)(s,0) =
(0,0). By supposition (r,0)D(R;V)(s,0) ⊆ J(D(R;V)). It follows that rRs ⊆
J(R). (cid:3)
Let R be a commutative ring, M be an R-module and σ be an endomorphism
of R. The abelian group R⊕M has a ring structure with multiplication
(r ,m )(r ,m )=(r r ,σ(r )m +r m ),
1 1 2 2 1 2 1 2 2 1
where r ∈ R, m ∈ M. This extension is constructed by Nagata in [21], is called
i i
the Nagata extension of R by M and σ, and is denoted by N(R,M,σ).
Theorem 3.8. Let R be a ring with an ideal I. Then, R is I-semicommutative if
and only if N(R,M,σ) is I⊕M-semicommutative.
Proof. Assume that R is I-semicommutative. Let (r ,m ),(r ,m )∈N(R,M,σ)
1 1 2 2
with (r ,m )(r ,m ) = 0; then r r = 0. By assumption, r Rr ⊆ I. Then
1 1 2 2 1 2 1 2
r rr ∈I for all r ∈R. Hence, for any (r,m)∈R⊕M, we have
1 2
(r ,m )(r,m)(r ,m )=(r rr ,x)∈I⊕M,
1 1 2 2 1 2
since x ∈ M. This implies N(R,M,σ) is an I ⊕M-semicommutative ring. The
converse is clear. (cid:3)
ON A CLASS OF SEMICOMMUTATIVE RINGS 77
Let R be a ring, and let σ be an endomorphism of R. Let T (R,σ) be the set of
2
(cid:20) (cid:21) (cid:20) (cid:21)
a b x y
all 2×2 upper triangular matrices over R. For , ∈T (R,σ), we
0 c 0 z 2
define the multiplication
(cid:20) (cid:21)(cid:20) (cid:21) (cid:20) (cid:21)
a b x y ax ay+bσ(z)
= .
0 c 0 z 0 cz
This multiplication and usual matrix addition makes T (R,σ) is a ring since σ is a
2
ring homomorphism.
Theorem 3.9. Let R be a ring with an ideal I. Then, R is I-semicommutative if
and only if T (R,σ) is I(R,σ)-semicommutative, where
2
(cid:26)(cid:20) (cid:21) (cid:27)
a b
I(R,σ)= | a, c∈I, b∈R .
0 c
(cid:20) (cid:21) (cid:20) (cid:21)
a b d e
Proof. AssumethatRisI-semicommutativeandlet , ∈T (R,σ)
0 c 0 f 2
(cid:20) (cid:21)(cid:20) (cid:21)
a b d e
with = 0. Then ad = 0, ae + bσ(f)=0 and cf = 0. By
0 c 0 f
the I-semicommutativity of R, we have aRd ⊆ I and cRf ⊆ I. This implies
T (R,σ)isI(R,σ)-semicommutative. Conversely,supposethatT (R,σ)isI(R,σ)-
2 2
(cid:20) (cid:21)(cid:20) (cid:21)
a 0 b 0
semicommutative and let a,b ∈ R with ab = 0. We have = 0.
0 0 0 0
(cid:20) (cid:21) (cid:20) (cid:21)(cid:20) (cid:21)(cid:20) (cid:21)
x y a 0 x y b 0
So for any ∈ T (R,σ), we get ∈ I(R,σ).
0 z 2 0 0 0 z 0 0
Hence we have aRb⊆I. Thus R is I-semicommutative. (cid:3)
Let R be a ring with a ring homomorphism σ : R → R and R[[x,σ]] the ring
of skew formal power series over R; that is, all formal power series in x with
coefficients from R with multiplication defined by xr = σ(r)x for all r ∈ R. In
particular, R[[x]] = R[[x,1 ]] is the ring of formal power series over R. Note that
R
J(R[[x,σ]])=J(R)+(x) (see [27, Lemma 16.10]).
Proposition 3.10. Let R be a ring, σ : R → R a ring homomorphism and I an
ideal of R. Then the following are equivalent.
(1) R is I-semicommutative.
(2) R[[x,σ]] is I+(x)-semicommutative.
Proof. (1)⇒(2)AssumethatRisI-semicommutative. Letf(x)=(cid:80)a xi,g(x)=
i
(cid:80)b xi ∈ R[[x,σ]] with f(x)g(x) = 0. Then a b = 0. By assumption a Rb ⊆ I.
i 0 0 0 0
Then f(x)R[[x,σ]]g(x)⊆a Rb +(x)⊆I+(x).
0 0
(2)⇒ (1) Let a, b ∈ R with ab = 0. Then aR[[x,σ]]b ⊆ I +(x). Hence we have
aRb⊆I. (cid:3)
Let R be a ring and a∈R. a is called a r.q.r element(r.q.r for short) in R if there
exists a(cid:48) ∈ R such that a◦a(cid:48) = 0, where a◦a(cid:48) = a+a(cid:48) −aa(cid:48). In terms of r.q.r
elements of R, J(R) = {a ∈ R | aR is r.q.r in R}, see also [18, Definition 6.6 and
Chapter 6] for details. Let A be a ring and B a subring of A and
R[A,B]={(a ,a ,··· ,a ,b,b,···):a ∈A,b∈B,n≥1,1≤i≤n}.
1 2 n i
78 ABDULLAH HARMANCI and HANDAN KOSE
ThenR[A,B]isaringunderthecomponentwiseadditionandmultiplication. Since
the ring R[A,B] need not have an identity, we use the preceding characteriza-
tion of the Jacobson radical in the next lemma. Next we first show the equality
J(R[A,B]) = R[J(A),J(A) ∩ J(B)] and then we prove necessary and sufficient
conditions for R[A,B] to be J-semicommutative.
Lemma 3.11. Let A be a ring and B a subring of A. Then J(R[A,B]) =
R[J(A),J(A)∩J(B)].
Proof. Let X = (a ,a ,··· ,a ,b,b,···) ∈ J(R[A,B]). Then for any
1 2 n
Y =(a(cid:48),a(cid:48),··· ,a(cid:48) ,b(cid:48),b(cid:48),···) ∈ R[A,B], XY is r.q.r for all Y ∈ R[A,B]. So
1 2 m
for such X and Y, there exists Z = (x ,x ,··· ,x ,y,y,···) ∈ R[A,B] such that
1 2 t
(XY)◦Z =0. We prove that the components a ,a ,··· ,a of X are in J(A) and
1 2 n
b∈J(A)∩J(B). Forthisitisenoughtoshowa ∈J(A). Foranya(cid:48) ∈A, setY =
1 1
(a(cid:48),0,··· ,0,0,,···) ∈ R[A,B]. Then there exists Z = (x ,x ,··· ,x ,y,y,···) ∈
1 1 2 n
R[A,B] such that (XY)◦Z =0. This implies (a a(cid:48))◦x =0, that is, a ∈J(A).
1 1 1 1
In the same way, we may show that the other components a , ..., a of X belong
2 n
to J(A). As for the other components b, let b(cid:48) ∈ B be an arbitrary element. Set
Y(cid:48) = (0,0,··· ,0,b(cid:48),b(cid:48),···). Then XY(cid:48) is r.q.r. Hence bb(cid:48) is r.q.r for each b(cid:48) ∈ B.
Thus b∈J(B). To prove b∈J(A), let a(cid:48) ∈A be an arbitrary element in A and
n+1
set Y(cid:48)(cid:48) = (a(cid:48),a(cid:48),··· ,a(cid:48) ,b(cid:48),···) ∈ R[A,B]. Then XY(cid:48)(cid:48) is r.q.r. Hence ba(cid:48) is
1 2 n+1 n+1
r.q.r in A for all a(cid:48) ∈A. So b∈J(A). Thus b∈J(A)∩J(B). This implies that
n+1
J(R[A,B])⊆R[J(A),J(A)∩J(B)].
For the converse inclusion, let X = (a ,a ,··· ,a ,b,b,···) ∈ R[J(A),J(A)∩
1 2 n
J(B)]. Then a ∈ J(A) for each 1 ≤ i ≤ n, and b ∈ J(A) ∩ J(B). Let Y =
i
(x ,x ,··· ,x ,y,y,···) be an arbitrary element in R[A,B]. We prove that there
1 2 m
exists Z = (z ,z ,··· ,z ,z,z,···) such that (XY)◦Z = 0. We divide the proof
1 2 t
into cases.
Case I: n=m. Since a ∈J(A) and b∈J(A) and b∈J(B), a x are r.q.r in A for
i i i
1≤i≤nandbyisr.q.rinbothAandB. Sothereexistz ∈Aandz ∈B suchthat
i
(a x )◦z =0 where 1≤i≤n and (by)◦z =0. Set Z =(z ,z ,··· ,z ,z,z,···).
i i i 1 2 n
Since the ring operations in R[A,B] is componentwise, (XY)◦Z =0.
Case II: n<m. a x are r.q.r for 1≤i≤n and bx is r.q.r in A for n+1≤i≤m
i i i
and by is r.q.r in B. There exist z ∈A, z ∈A and z ∈B such that (a x )◦z =0
i j i i i
for1≤i≤nand(bx )◦z =0forn+1≤j ≤m. SetZ =(z ,z ,··· ,z ,z,z,···).
j j 1 2 m
Hence (XY)◦Z =0 holds.
Case III: n > m. a x is r.q.r for 1 ≤ i ≤ m and a y is r.q.r for m+1 ≤ i ≤ n
i i i
in A and by is r.q.r in B. There exist z ∈ A and z ∈ B such that (a x )◦z = 0
i i i i
for 1 ≤ i ≤ m and (a y) ◦ z = 0 for m + 1 ≤ i ≤ n and (by) ◦ z = 0. Let
i i
Z = (z ,z ,··· ,z ,z,z,···). By using componentwise operations in R[A,B], we
1 2 n
have (XY)◦Z =0. So R[J(A),J(A)∩J(B)]⊆J(R[A,B]) holds. This completes
the proof. (cid:3)
Proposition 3.12. Let A be a ring with B a subring of A. Then the following are
equivalent.
(1) A and B are J-semicommutative.
(2) R[A,B] is J-semicommutative.