Vol. 19 (2018), No. 2, pp. 969–981 DOI: 10.18514/MMN.2018.2585
A GENERALIZATION OF NIL-CLEAN RINGS
ALEKSANDRA KOSTI ´C, ZORAN Z. PETROVI ´C, ZORAN S. PUCANOVI ´C, AND MAJA ROSLAVCEV
Received 06 April, 2018
Abstract. The conditions that allow an element of an associative, unital, not necessarily commut- ative ringR, to be represented as a sum of (commuting) idempotents and one nilpotent element are analyzed. Some applications to group rings are also presented.
2010Mathematics Subject Classification: 16U99; 16S34 Keywords: idempotents, nilpotent element, group ring
1. INTRODUCTION
An elementain an associative unital ringRis called clean if it can be represented as a sumaDeCu, wheree is an idempotent element anduis a unit. This notion was introduced by Nicholson in [8]. If one can find such elementseandusuch that aDeCuandeuDue, the elementais called strongly clean. The ringRitself is called (strongly) clean if every element inRis (strongly) clean.
Many families of clean rings were investigated in previous decades. In recent years, a particular attention has been paid to the nil-clean rings and its relatives. A nil-clean ring (see [5]) is a ring in which every element is nil-clean, which means that every element can be written as a sum of an idempotent element and a nilpotent one.
Analogously, we have a notion of strongly nil-clean elements (and rings). For some of the results concerning this class of rings and some of the related classes of rings, the reader may wish to consult also [1,3,7,10].
A class of strongly2-nil-clean rings was introduced in [4]. Namely, an element ain a ring Ris called strongly2-nil-clean if it can be represented in the form aD eCf Cn, wheree and f are idempotents, n is a nilpotent element and they all commute with each other.
In this paper we analyze elements of a ring which can be written as a sum of finitely many idempotents and one nilpotent element which are pairwise commutative. If the number of idempotents which appear in this sum iss, we call these elements strongly
First, second and fourth author are partially supported by Ministry of Education, Science and Envir- onmental Protection of Republic of Serbia Project #174032.
c 2018 Miskolc University Press
s-nil-clean. It turns out that if every element in a ring is strongly s-nil-clean for somes, this ring has finite characteristic and every element in this ring is strongly .p 1/-nil-clean, wherepis the largest prime dividing the characteristic of this ring (see Theorem1). These rings are naturally called strongly.p 1/-nil-clean rings and they are all strongly clean (see Corollary 3and the discussion preceding it). There are many examples of strongly .p 1/-nil-clean rings. Theorem 2 shows that in every commutative ringR of finite characteristick, elements which are stronglys- nil-clean for somes, form a subring which is.p 1/-nil clean (wherepis the largest prime dividingk) and if this ring contains idempotents or nilpotents not belonging toZk (which is necessarily contained inR) we have a non-trivial example of such a ring. Proposition5provides examples of finite commutative local rings which are .p 1/-nil-clean.
The plan of the paper is as follows. In Section 2, we analyze sums of idempotents and one nilpotent element and derive our main criteria for strongly.p 1/-nil-clean elements in a ring. Section 3 deals with some structure theorems. It is shown that in analyzing strongly.p 1/-nil-clean rings, we may reduce this analysis to the case when p is a nilpotent element in a ring under investigation. We also show in this section that strongly.p 1/-nil-clean rings are strongly-regular and, consequently, strongly clean.
Section 4 deals with the investigation of group ringsRGwhereRis a.p 1/-nil- clean commutative ring andG is a group. For example, we show that, for the ring RG to be strongly.p 1/-nil-clean, when the characteristic ofRis of the formps, for a prime integerp, it is necessary that the order of any element of a groupG is of the form dpk, for somed j.p 1/ andk0 (see Lemma3). In the case of a commutative groupG this condition is also sufficient (see Theorem4).
We emphasize that we work in associative, unital rings which need not be com- mutative. When a ring is commutative, we drop the adjective “strongly” since it is unnecessary. We use the same symbolkto denote the integerkand to denote the ring elementk1R. It will always be clear what we mean. Finally, we denote the field with pelements byZp, the Jacobson radical byJ.R/and the set of nilpotent elements by N.R/.
2. BASIC RESULTS
A ring in which every element is a sum of certain number of idempotents and one nilpotent element, that commute with each other, is a generalization of strongly nil- clean rings and strongly 2-nil-clean rings. In view of this, we introduce the following definition.
Definition 1. An element aof a ring R iss-nil-clean if it can be written in the following form:
aDe1C CesCn; (2.1)
where elementse1; : : : ; esare idempotents andnis nilpotent. If an elementacan be written in the form (2.1) so that elements in this sum are pairwise commutative, we say that this element is stronglys-nil-clean. If every element in R is (strongly)s-nil clean, we say thatRis a (strongly)s-nil-clean ring.
Of course, ifais (strongly)s-nil-clean ands < t,ais alsot-nil-clean — we simply addt szeroes to the presentation ofaas a (strongly)s-nil-clean element.
Remark1. It is clear that, if fWR!S is a ring homomorphism and an element a2Ris (strongly)s-nil-clean, thenf .a/2Sis (strongly)s-nil-clean. Similarly, an elementaD.a1; : : : ; al/2R1 Rl is (strongly)s-nil-clean iffai is (strongly) s-nil-clean for alli. So, homomorphic images and finite direct products of (strongly) s-nil-clean rings are itself (strongly)s-nil-clean. Also, a subring of a stronglys-nil- clean ring has the same property, as we shall see later. However, this does not hold for s-nil-clean rings. Namely, if we have a ring which iss-nil-clean, but it is not strongly s-nil-clean, it is enough to take an element awhich is not strongly s-nil-clean and look at the subring generated by this element. This subring is commutative, so it cannot bes-nil-clean – if it were, this element would also be stronglys-nil-clean.
We begin our analysis with a useful result concerning sum of several idempotents and one nilpotent element.
Proposition 1. LetR be a ring and suppose that elementa2Ris stronglys-nil clean. Thena.a 1/ .a s/is nilpotent.
Proof. LetaDe1Ce2C CesCn, wheree1; e2; : : : ; esare idempotents andn is nilpotent that commute with each other. Observe that
1D..1 e1/Ce1/..1 e2/Ce2/ ..1 es/Ces/:
After multiplication, we get a sum of products of the form ei1 eik.1 ej1/ .1 ejs k/;
where1ks,i1< < ik,j1< < js kandfi1; : : : ; ik; j1; : : : ; js kg D f1; : : : ; sg. Next, we get that
.k a/ei1 eik.1 ej1/ .1 ejs k/
D..1 ei1/C C.1 eik/ ej1 ejs k n/ei1 eik.1 ej1/ .1 ejs k/ D nei1 eik.1 ej1/ .1 ejs k/ :
This follows from the fact that .1 e/eDe e2D0, for an idempotent e. Since nis nilpotent, so is this product. Thus, when 1 is multiplied by a.a 1/ .a s/, we get a sum of nilpotent elements that commute with each other. Therefore,a.a
1/ .a s/is nilpotent.
The following corollary is simple, but important.
Corollary 1. a) If a ringRis such that 1is stronglys-nil-clean for somes1, then this ring has finite characteristic.
b) Ifchar.R/Dk, then 1is strongly.p 1/-nil-clean, wherepis the largest prime dividingk.
Proof. a) From the previous proposition we conclude that . 1/. 2/ . .sC 1//D. 1/sC1.sC1/Šis nilpotent, so..sC1/Š/mD0for somem1and the char- acteristic of the ringRis not 0.
b) It is enough to show that 1is.p 1/-nil-clean in the ringZk, which is contained inR. IfkDp1˛1 pl˛l is the prime factorization ofk, wherep1< < pl Dp, then Zk ŠZp˛1
1 Zp˛l
l . Since 17!. 1; : : : ; 1/under this isomorphism, this reduces the proof to show that 1 is .p 1/-nil-clean in Zp˛i
i . This is clear since 1Dpi˛i 1D1C C1
„ ƒ‚ …
pi 1
Cpi.p˛ii 1 1/, andpi.pi˛i 1 1/is nilpotent in Zp˛i
i .
Example 1. The ring Zk is (strongly) .p 1/-nil-clean, where p is the largest prime integer dividing k. Namely, using the notation from the previous corollary, this reduces to show that Zpi˛i is .p 1/-nil-clean. Since any element in Zp˛i i
can be written in the form 1C C1
„ ƒ‚ …
s
Cpit, for some s2 f0; : : : ; pi 1g and t 2 f0; : : : ; p˛ii 1 1g, and sincepi is nilpotent inZp˛i
i , we are done.
Lemma 1. IfkDchar.R/Dp1˛1 pl˛l is the prime factorization of the charac- teristic of the ring R, then RŠR1 Rl, whereRi DR=p˛iiR. In particular, char.Ri/Dp˛ii.
Proof. This follows easily from the Chinese remainder theorem taking into ac-
count the fact that elementspi˛i are central.
Since the products of the forma.a 1/ .a s/are important for our investiga- tion, we introduce the symbol.a/k WDa.a 1/ .a .k 1//(falling factorial, as is known in combinatorics), wherekis a positive integer.
We have the following corollary.
Corollary 2. LetRbe a ring. Suppose that elementa2Ris stronglys-nil clean andk.< s/is nilpotent. Then.a/k is a nilpotent element.
Proof. Clearly
.a/sDa.a 1/ .a .s 1//Dat0.a 1/t1 .a .k 1//tk 1Ckq.a/;
for some non-negative integers ti, such that Pk 1
iD0ti Ds and polynomial q.X /2 ZŒX . Taking into account thatkis nilpotent, the result follows.
In order to see what the fact that.a/sis nilpotent implies, we start with a simple lemma.
Lemma 2. Letpbe a prime integer andr; m1arbitrary positive integers. Then the element x in the ring ZprŒX =hXm.X 1/m .X .p 1//mi is (strongly) .p 1/-nil-clean, wherexis the class ofX in this quotient ring.
Proof. Ideals hX ii and hX ji are coprime inZprŒX , for all 0i < j <
p, since.X i / .X j /Dj i and j i is invertible in the ringZprŒX . So, h.X i /mi;h.X j /mi are coprime as well. By applying the Chinese remainder theorem we obtain the isomorphism
ZprŒX =hXm .X .p 1//mi ŠZprŒX =hXmi ZprŒX =h.X .p 1//mi; such thatx7!.x; : : : ; x/. Thus, it is enough to show thatxhas the desired presenta- tion in all factors and sincexD1C C1
„ ƒ‚ …
i
C.x i /in the factorZprŒX =h.X i /mi,
this is true.
Proposition 2. LetR be a ring. Suppose that the elementp is nilpotent, where pis a prime integer, and leta2Rbe such that.a/p is nilpotent. Thenais strongly .p 1/-nil-clean.
Proof. Consider the homomorphismfWZŒX !R, given byf .X /Da. An im- mediate consequence of the fact that.a.a 1/ .a .p 1///mD0and thatprD0 inR, for somem; r 1, is the existence of an induced homomorphism
fNWZprŒX =hXm.X 1/m .X .p 1//mi !R; such thatx7!a:
Sincexis strongly.p 1/-nil-clean, so is its imagea.
Proposition 3. LetRbe a ring of characteristick.> 0/. Ifpis the largest prime dividingkanda2Ris such that.a/sis nilpotent for somes1, thenais strongly .p 1/-nil-clean.
Proof. Under isomorphism R ŠR1 Rl, where Ri DR=pi˛iR and kD p1˛1 pl˛l, implied by Lemma 1, the element a goes to .a1; : : : ; al/. Also, a is strongly.p 1/-nil-clean iffai is such for alli. However, from the fact that.a/sis nilpotent, it follows that.ai/s2Ri is nilpotent for alli. From this, it easily follows that.ai/pi is also nilpotent for alli. Namely, ifs < pi, this follows since.ai/pi D .ai/s.ai s/pi sand ifs > pi, it follows from Corollary2. Sincepi is nilpotent in Ri, Proposition2gives thatai 2Ri is strongly.pi 1/-nil-clean for alli. From the fact thatpi p for alli, it follows that these elementsai are all.p 1/-nil-clean
and so isa.
Theorem 1. LetRbe a ring such that every elementa2Ris stronglys-nil-clean for some s. Then R has finite characteristic and R is strongly .p 1/-nil-clean, wherepis the largest prime dividingchar.R/.
Proof. Corollary1shows thatR has finite characteristic and from Proposition 1 and Proposition3it follows that every element is strongly.p 1/-nil-clean, wherep
is the largest prime dividing this characteristic.
Theorem 1 shows that one needs only to investigate strongly .p 1/-nil-clean rings, wherep is a prime integer. For example, the class of all strongly 3-nil-clean rings is the same as the class of strongly 2-nil-clean rings, and the class of strongly 9-nil-clean-rings is the same as the class of strongly 6-nil-clean rings. Namely, if a ring is, say, strongly9-nil-clean, then.a/10is nilpotent for alla2R. So, this is true foraD10. Consequently, 10ŠD.10/10 is nilpotent, so char.R/j.10Š/m, for some m1. We conclude that the largest prime dividing char.R/is at most 7 (it may be smaller), so our ring is strongly6-nil-clean.
Proposition 4. A subring of a strongly.p 1/-nil-clean ring is also strongly.p 1/-nil-clean.
Proof. LetS be a subring of a strongly .p 1/-nil-clean ring R and leta2S. Sincea2R,ais strongly.p 1/-nil-clean, and according to Proposition1element .a/p is nilpotent. Ris of finite characteristic, sayk, which means that char.S /Dk, withpbeing the largest prime dividingk. When we apply Proposition3, we get that
ais strongly.p 1/-nil-clean inS.
Theorem 2. Let R be a commutative ring of finite characteristic k and p the largest prime dividingk. LetSD fa2RWaiss-nil-clean for somesg. ThenSis the largest subring ofRwhich is.p 1/-nil-clean.
Proof. Since kD0inR, we have: 1D1C C1
„ ƒ‚ …
k 1
, so 1 is.k 1/-nil-clean and 12S. Also, ifa; b2S, thenaDe1C CesCn,bDf1C CftCn0and we getabDP
i;jeifjCN, whereNis nilpotent and alleifj are idempotents (Ris a commutative ring). So,ab2S. Similarly,aCbDe1C CesCf1C CftC nCn02S. Finally, sincea bDaC. 1/bwe conclude thata b2S as well. So, S is a ring in which every element iss-nil-clean for somes. From Theorem1, we conclude thatSis actually.p 1/-nil-clean ring. It is clear thatS is the largest such
subring.
Remark2. From the Theorem2it is clear that, in order to show that a commutative ring of finite characteristickis.p 1/-nil-clean, it is enough to check only its ring generators overZk. For example, the ringZprŒX =hXm.X 1/m .X .p 1//mi, appearing in Lemma2, is.p 1/-nil-clean, since it is generated byxand this element is.p 1/-nil-clean.
The following proposition provides us with a lot of examples of commutative.p 1/-nil-clean rings.
Proposition 5. LetRbe a finite commutative local ring,M its maximal ideal and R=M ŠZp. ThenRis.p 1/-nil-clean.
Proof. We know that every element inM is nilpotent. Ifx2R, thenxCM D sCM, for somes2 f0; : : : ; p 1g. So,xD1C C1
„ ƒ‚ …
s
Cm, wheremis nilpotent.
Remark3. In the case of non-commutative rings, the set of all elements which are stronglys-nil-clean for somesdo not necessarily form a subring. Such examples will be given in Section 3 and Section 4. However, a simple application of Zorn’s lemma shows that there exist maximal subrings which are strongly.p 1/-nil-clean.
The following proposition gives another characterization of strongly.p 1/-nil- clean elements.
Proposition 6. Let R be a ring of finite characteristic k, p the largest prime dividingkanda2R. The following conditions are equivalent.
(1) .a/pis nilpotent.
(2) ais strongly.p 1/-nil-clean.
(3) aDbCn, whereb2Ris such that.b/p D0,nis nilpotent andbnDnb.
Proof. .1/ H) .2/. This is contained in Proposition3.
.2/ H) .3/. LetaDe1C Cep 1Cnbe a.p 1/-nil-clean decomposition ofa.
TakebWDe1C Cep 1. The proof of Proposition1shows that.b/pD0.
.3/ H) .1/. Assume thataDbCn. So, we have
.a/p D.bCn/..b 1/Cn/ ..b pC1/Cn/:
Therefore, since n andb commute, .a/p D.b/pCnq.n; b/Dnq.n; b/, for some polynomialq.X; Y /2ZŒX; Y . Sincenis nilpotent, so is.a/p.
3. STRUCTURE THEOREMS
The purpose of this section is to discuss the structure of (strongly).p 1/-nil-clean rings, for prime numberp.
The following proposition sums up the discussion from the previous section.
Proposition 7. Suppose thatchar.R/DkDp1˛1 p˛ll, wherep1< < plDp.
ThenRis strongly.p 1/-nil-clean if and only ifRi is strongly.pi 1/-nil-clean, whereRiDR=pi˛iRand1il.
This shows that in investigation of strongly.p 1/-nil-clean rings, for p prime, we can reduce our analysis to the case whenp is nilpotent (equivalently, when the characteristic of the ring is a power of a prime).
Let us recall that a ringRis called strongly-regular if for every elementa2R there existsn1andx2Rsuch thatanDanC1x.
Theorem 3. Every strongly.p 1/-nil-clean ring is strongly-regular.
Proof. It is enough to consider the case when p is a nilpotent element. Then .p 1/Šis invertible. Leta2R. Since.a/p is nilpotent, we have that..a/p/sD0 for somes. But
0D..a/p/sD.a.a 1/ .a .p 1///sDas..p 1/Š/sCasC1y;
for somey2R. Since.p 1/Šis invertible, we get thatasDasC1x, for somex2R
and we are done.
It is well known that strongly-regular ring is strongly clean (see [2, Proposition 2.6], [9, Theorem 1], [5, Corollary 2.4]). Also, Jacobson radical of a strongly - regular ring is nil and commutative strongly-regular rings have Krull dimension 0 (see [2]). So, we have the following corollary.
Corollary 3. Every strongly.p 1/-nil-clean ring is strongly clean.
The following proposition is rather useful.
Proposition 8. LetR be a ring,a2Rand letpbe a nilpotent element, wherep is prime. Thenap ais nilpotent if and only if.a/p is nilpotent.
Proof. It is well-known that Xp X D.X /p in ZpŒX . So, ap a .a/p D pr.a/, for some polynomialr.X /2ZŒX . From this fact, the proof follows imme-
diately.
For future reference, we formulate the following corollary which directly follows from Proposition6and Proposition8.
Corollary 4. LetRbe a ring. Ifpis nilpotent, then Ris a strongly.p 1/-nil- clean ring if and only ifap ais nilpotent for everya2R.
Let us proceed with some of the special properties of.p 1/-nil-clean rings.
Proposition 9. LetR be a ring and let I be any nil ideal of R. Then R is.p 1/- nil-clean if and only ifR=I is.p 1/-nil-clean.
Proof. .H)/As observed before, this is trivial sinceR=I is a homomorphic im- age ofR.
.(H/ Suppose thatR=I is.p 1/-nil-clean. Let x2R. ThenxCI is .p 1/- nil-clean. Thus,xCI D.x1CI /C.x2CI /C C.xp 1CI /C.yCI /, where xiCI are idempotents, 1i p 1, and yCI is nilpotent. It is well known that idempotents lift modulo nil ideals (see, e.g. [6, Theorem 21.28]) so there are idempotentsei such thatxiCI DeiCI. So,x e1 e2 ep 1 y2I, i.e., xDe1Ce2C Cep 1CyCn, for somen2I. ElementyCnis nilpotent. Indeed, sinceyCI is nilpotent,yk 2I for somek2N. Every element different fromyk in the sum one gets in the expansion of.yCn/k, is inI and we can conclude that .yCn/k2I, soyCnis nilpotent. ThereforeRis.p 1/-nil-clean.
An analogous result holds for the strongly.p 1/-nil-clean rings.
Proposition 10. LetRbe a ring and let I be any nil ideal of R. ThenRis strongly .p 1/-nil-clean if and only ifR=I is strongly.p 1/-nil-clean.
Proof. .H)/Again, this is trivial sinceR=I is a homomorphic image ofR.
.(H/Leta2R. SinceR=I is strongly.p 1/-nil-clean, one has..aCI /p/k DI, for somek2N. Consequently,..a/p/k 2I:AsI is nil ideal,..a/p/k 2N.R/:So, .a/p is nilpotent. Since .p/p is also nilpotent, the characteristic k ofR is finite.
The characteristiclofR=I has the property thatp is the largest prime dividing this characteristic, but this also holds fork. Namely,l 2I and thereforel is nilpotent in R. So we have thatkjlsfor somes, and alsoljk. It follows that the sets of primes dividingkandlare the same. Now the result follows from Proposition6.
The following corollary follows directly from the fact thatJ.R/is nil for a strongly .p 1/-nil-clean ring and Proposition10.
Corollary 5. A ringRis strongly.p 1/-nil-clean if and only ifJ.R/is nil and R=J.R/is strongly.p 1/-nil-clean.
4. GROUP RINGS
Let us recall the notion of a group ring. LetGbe a group, written multiplicatively, and letRbe a commutative ring. The group ring ofG overR, denoted byRG, is a freeR-module with generating setG, i.e.:
RGDM
g2G
Rg:
So, elements ofRGare formal finite sums of the formP
irigi, withri 2R,gi 2G, while the multiplication is implied by multiplication inG. The identity of this ring is 1Re, where1R is the identity inRandeis the neutral element ofG. We denote the identity simply by 1.
Our main interest here is focused on strongly.p 1/-nil-clean group ringsRG. It is obvious that ifRGis strongly.p 1/-nil-clean, so isR. Since we assume that the coefficient ringRis commutative, we refrain from using adjective “strongly” when referring toR, we use it only forRGwhen appropriate. We begin by discussing rings R, such that char.R/is a power of a prime.
Lemma 3. LetRbe a.p 1/-nil-clean ring such thatchar.R/Dps, for a prime pand somes1and letGbe a group. For the list of conditions:
(1) RGis strongly.p 1/-nil-clean;
(2) For eachg2G, the elementgp 1 1is nilpotent;
(3) For eachg2G, there existsk0anddj.p 1/such that!.g/Ddpk, the following holds:.1/ H) .2/and.2/ ” .3/. Here,!.g/denotes the order of ginG.
Proof. .1/ H) .2/. Letg2G. Since RG is strongly.p 1/-nil-clean, from Corollary 4 it follows that gp g is nilpotent. Since g is invertible, we get that gp 1 1is nilpotent.
.3/ H) .2/. Letg2G. Then!.g/Ddpk, wherek0anddj.p 1/. Letp 1D ds. Since!.gd/Dpkand gcd.s; pk/D1, we have that!.gp 1/D!..gd/s/Dpk. Therefore,
.gp 1 1/pk D.gp 1/pk 1
„ ƒ‚ …
0
Cpu; for someu2RG:
Sincepis nilpotent, the elementgp 1 1is nilpotent as well.
.2/ H) .3/. Letg2G. The order ofgcannot be infinite — in that case, it would not be possible forgp 1 1to be nilpotent. Namely, the elementg.p 1/sin the sum .gp 1 1/sDg.p 1/sC C. 1/scannot be cancelled out.
So, let us suppose that !.g/Dtpk, for some k0 and t such that p −t and gcd.t; p 1/Dd¤t. Also, lettDdt1,p 1Dd´1andhDgp 1. Since!.gd/D t1pk and gcd.´1; t1pk/D1, it follows that
!.h/D!.gp 1/D!..gd/´1/Dt1pk:
Sinceh 1is nilpotent,hpk 1is nilpotent as well. Leth1Dhpk. Then!.h1/D t1. Let us focus on the polynomialf .X /D.X 1/t1 .Xt1 1/, which is clearly divisible byX 1:
f .X /D.X 1/..X 1/t1 1 .Xt1 1C CXC1//D.X 1/f1: This follows from the fact thatt1¤1(which also implies thath1 1¤0). We can see that
f .X /D.X 1/. t1C.X 1/q.X //;
for some polynomialq2ZŒX , sincef1.1/D t1. We can conclude that f .h1/D.h1 1/t1 .ht11 1
„ ƒ‚ …
0
/D.h1 1/. t1C.h1 1/q.h1//:
We know thath1 1is nilpotent. Asp−t1andpis nilpotent, elementt1is invertible inR. So
.h1 1/t1Du.h1 1/;
for an invertibleu2RG. So
.h1 1/..h1 1/t1 1 u/D0;
and since.h1 1/t1 1 uis invertible, we have thath1 1D0. That is a contradic- tion.
It is easy to check that the proof of2 ” 3is valid, although shorter, even for
pD2.
In the previous lemma, G was an arbitrary group. If we add commutativity, we actually get equivalence.1/ ” .2/.
Theorem 4. LetRbe a.p 1/-nil-clean ring such thatchar.R/Dps, for a prime p and somes1and letG be an Abelian group. ThenRGis.p 1/-nil-clean iff gp 1 1is nilpotent for everyg2G.
Proof. We only need to prove the “if” part. It follows directly from the Remark following Theorem2since elements of the groupG form a generating set forRG
overR.
Example2. The previous theorem does not hold for non-commutative groups. Let us takeRDZ5andGDD4, the dihedral group of order 8 generated by elementss andr such thats2D1Dr4,s rDr3s. In this group,g4D1for all g2G, so the condition thatg4 1is nilpotent is trivially satisfied. However, direct computation shows that
.sCs r/5D2sC2rC3s r2C3r3 ..sCs r/5/8D3C2r2
.3C2r2/2D3C2r2;
so.sCs r/5is not nilpotent and the group ringZ5D4is not strongly4-nil-clean.
Let us concentrate now on the general case.
Proposition 11. LetR be a.p 1/-nil-clean ring andchar.R/Dp1˛1 pl˛l, so thatl > 1,p1< < plDp.
(1) IfGis an elementary Abelian 2-group, thenRGis strongly.p 1/-nil-clean.
(2) IfGis an elementary Abelian group in which every element has orderqand qjgcd.p1 1; : : : ; pl 1/, thenRGis strongly.p 1/-nil-clean.
Proof. .1/As we know,RŠR1 Rl, wherepi is nilpotent inRi, and con- sequentlyRGŠR1G RlG. So, all the ringsRiare strongly.p 1/-nil-clean, and sincepi is nilpotent inRi,Ri is actually strongly.pi 1/-nil-clean. We will use Theorem4. Since2j.pi 1/for alli 2 andg2D1for allg2G, we have thatgpi 1 1D0for allg2G. Ifp1> 2, the same holds forp1. Ifp1D2, then .g 1/2Dg2 2gC1D2.1 g/. Since 2 is nilpotent inR1G,g 1is also nilpo- tent inR1G. In this case also, fromG being Abelian, we can conclude thatRiG is .pi 1/-nil-clean. Therefore,RGis.p 1/-nil-clean.
.2/Similarly, fromgqD1, andqjgcd.p1 1; : : : ; pl 1/, we conclude thatgpi 1
1D0for allg2G, and proceed as in.1/.
Theorem 5. Letchar.R/Dp˛11 pl˛l, wherel > 1,p1< < plDpand letG be a group. Suppose thatRGis strongly.p 1/-nil-clean ring.
(1) For allg2Gthe following holds:!.g/jgcd.p2 1; : : : ; pl 1/and!.g/D d1ps1, such thatd1j.p1 1/ands0.
(2) If there existsh2Gsuch that!.h/p1, thenp1j.pi 1/, for2i l.
(3) If for allg2G,!.g/ < p1, then!.g/jgcd.p1 1; : : : ; pl 1/.
(4) Ifp1D2, then every element inG has order2sfor somes0.
(5) Ifp1D2andpi3 .mod 4/for at least onei2, thenGis an elementary Abelian 2-group.
Proof. .1/ As before, RG ŠR1G RlG, where pi is nilpotent in RiG. Lemma3shows thatgpi 1 1is nilpotent inRiG. It follows that!.g/Ddipiki for someki 0anddi j.pi 1/. So,
d1p1k1Dd2p2k2D Ddlplkl:
Since d1< p1, it is clear thatd1p1k1 cannot be divisible by any prime greater than p1. Soki D0fori2. Therefore,
!.g/Dd1p1k1Dd2D Ddl; (4.1) for all g2G, where di j.pi 1/. Hence,!.g/j.p2 1/; : : : ; !.g/j.pl 1/, and we are done.
.2/Ifh2G is such that!.h/p1, we have!.h/Dd1p1k1Dd2D Ddl, where k11. Sincedi j.pi 1/, it follows thatp1j.pi 1/.
.3/Under this assumption, we get thatk1D0in4.1, hence for allg2G
!.g/Dd1Dd2D Ddl: We conclude that!.g/jgcd.p1 1; : : : ; pl 1/.
.4/The fourth assertion follows easily. Namely, in this casep1D2, so!.g/Dd12k1, whered1j.2 1/. So,!.g/D2k1.
.5/It is enough to show that there are no elements of order 4 inG. If it were, then for an elementg2G, we would have equalities
4Dd2D Ddl;
wheredij.pi 1/, for2il. This would imply that4j.pi 1/, that ispi 1 .mod 4/, for2i l, which is a contradiction. Hence, we can conclude thatG is
an elementary Abelian 2-group.
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Authors’ addresses
Aleksandra Kosti´c
University of Belgrade, Faculty of Mathematics, Studentski trg 16, Belgrade, Serbia E-mail address:alex@matf.bg.ac.rs
Zoran Z. Petrovi´c
University of Belgrade, Faculty of Mathematics, Studentski trg 16, Belgrade, Serbia E-mail address:zoranp@matf.bg.ac.rs
Zoran S. Pucanovi´c
University of Belgrade, Faculty of Civil Engineering, Bulevar kralja Aleksandra 73, Belgrade, Serbia E-mail address:pucanovic@grf.bg.ac.rs
Maja Roslavcev
University of Belgrade, Faculty of Mathematics, Studentski trg 16, Belgrade, Serbia E-mail address:roslavcev@matf.bg.ac.rs
