# Centralizingn-Homoderivations of Semiprime Rings

Authors: M. S. Tammam El-Sayiad; A. Ageeb; A. Ghareeb
Journal: Journal of Mathematics
 (2022)

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## Abstract

We introduce the notion ofn-homoderivation on a ring ℜ and show that a semiprime ring ℜ must have a nontrivial central ideal if it admits an appropriate n-homoderivation which is centralizing on some nontrivial one-sided ideal. Under similar hypotheses, we prove commutativity in prime rings.


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## Body

## 1. Introduction

Studying the properties of rings equipped with mappings is an idea that began in the second half of the last century. Researchers have focused on two types of mappings defined on rings, namely, homomorphisms and derivations. Many authors studied the properties of prime and semiprime rings with derivations, such as commutativity of rings. It has inspired many researchers to generalize this mapping from various sides and study the commutativity of the rings and other properties, see [1–4]. At the same time, they dealt with the study of rings with homomorphisms and studied properties similar to those considered in the previous case, see [5–9].With the beginning of the new millennium, the idea of combining the two mappings into one new mapping called homoderivation arose, and several different concepts emerged from this idea.In [10], El-Sofy presented the first concept of homoderivation as a linear mapping h that verifies, for all α,β in the ring, the property hαβ=hαhβ+hαβ+αhβ. He gave examples showing the existence of such map. He showed that this map is neither a derivation nor a homomorphism. Also, he gave some results concerning the commutativity of some types of rings endowed with homoderivation and studied other properties of rings. Recently, many authors have studied and generalized the concept of homoderivation on different types of rings and near rings in several ways, and they also studiedthe properties of rings equipped with homoderivations, see [11, 12].In [13], M. Mahdi Ebrahimi and P. A. JooHesh presented the second concept, which is completely different from the first one, as a positive linear mapping h′ defined on the ℓ-ring and fulfilled the two conditions h′αβ=h′αh′β and h′αβ=h′αβ+αh′β and called that mapping homoderivation. Also, they defined antihomoderivation to be a derivation that is antihomomorphism. They generalized some results of Bell and Kappe [14].As a generalization of El-Sofy’s homoderivation in 2020, Eszter Gselmann and Gergely Kiss defined the third concept, calledε-homoderivation, as a mapping hε on a ring and satisfying, for all α,β in the ring, the condition hεαβ=hεαβ+αhεβ+εhεαhεβ, where ε belongs to the center of the ring. They characterized this mapping without additivity supposition. Obviously, if a ring has a zero center, then ε-homoderivation becomes a derivation. El-Sofy’s homoderivation can be considered as a special case of ε-homoderivation if the ring is unital. Also, they proved some results concerned with Mahdi’s homoderivation, see [15].In this paper, we introduce the new concept,n-homoderivation, as follows:Definition 1.
Letℜ be a ring. An n-homoderivation is a linear mapping hn:ℜ⟶ℜ which satisfies, for all α and β∈ℜ, hnαβ=hnαβ+αhnβ+nhnαhnβ, where n is an integer.Clearly, derivations can be considered as 0-homoderivation, the identity mapping is−1-homoderivation, the zero mapping is n-homoderivation for all integer n, and El-Sofy’s homoderivation can be considered as 1-homoderivation. So, the class of n-homoderivations is wide rather that the notion of homoderivations and derivations. Although in rings with unity the concept of n-homoderivation can be considered as a special case of ε-homoderivation, but in general, that is not true as shown in following example.Example 1.
The identity map on any nonunital ringℜ is n-homoderivation with n=−1, but it is neither homoderivation nor derivation. Also, there is no central element ε which makes the identity mapping ε-homoderivation. Therefore, the classes of n-homoderivation and ε-homoderivation are different.The following theorem shows that the notion ofn-homoderivation can be characterized in terms of homomorphism.Theorem 1.
Letℜ be a ring and assume that n is an arbitrary nonzero integer. If a linear map hn:ℜ⟶ℜ fulfills the equation(1)hnαβ=hnαβ+αhnβ+nhnαhnβ,for allα,β∈ℜ,then there exists a homomorphism ϕn:ℜ⟶ℜ such that nhnα=ϕnα−α for all α∈ℜ. Moreover, ϕnhnα=hnϕnα.Proof.
Clearly, sincehn is linear, ϕn is additive. Multiplying equation (1) with n leads to nhnαβ=nhnαβ+nαhnβ+nhnαnhnβ for all α,β∈ℜ. If we add αβ to both sides of this equation, then nhnαβ+αβ=nhnαβ+nαhnβ+nhnαnhnβ+αβ for all α,β∈ℜ. Observe, however, that nhnαβ+nαhnβ+nhnαnhnβ+αβ=nhnα+αnhnβ+β for all α,β∈ℜ and yielding exactly that the mapping ϕn:ℜ⟶ℜ defined through ϕnα=nhnα+α for all α∈ℜ is a homomorphism. Now, since ϕnhnα=nhnhnα+hnα=hnnhnα+α=hnϕnα for all α∈ℜ, this completes our proof.We give a generalization of a well-known result due to Bell and Martindale’s study [5] which stated that a semiprime ring ℜ must have a nontrivial central ideal if it admits an appropriate endomorphism or derivation which is centralizing on some nontrivial one-sided ideal. Also, they proved commutativity in prime rings under some conditions.Throughout this work,ℜ will be an associative ring with center ζℜ, and hn:ℜ⟶ℜ will be an n-homoderivation.


## 2. Results

To get our results, we will use the following facts about prime rings.Fact 1.
In a prime ringℜ, the centralizer of any nonzero one-sided ideal is equal to the center of ℜ; in particular, if ℜ has a nonzero central ideal, ℜ must be commutative.Fact 2.
LetU be a left ideal in a semiprime ring ℜ. If ϕ is an epimorphism on ℜ such that ϕu≠0 for some u∈U, then ϕℜu≠0.Proof.
Supposeϕℜu=0, then ϕℜϕu=ℜϕu=0. Therefore, ϕu=0.The following lemma plays a crucial role in our present work.Lemma 1.
Letℜ be a semiprime ring and U be a nonzero left ideal. If hn is a centralizing n-homoderivation on U, then hn is commuting on U.Proof.
For an arbitraryu∈U, we have(2)hnu,u∈ζℜ.
Therefore, we get(3)hnu2,u2∈ζℜ.
That is,(4)hnuu,u2+uhnu,u2+nhnuhnu,u2∈ζℜ.
Which gives(5)hnu,u2u+uhnu,u2+nhnuhnu,u2+nhnu,u2hnu∈ζℜ.
And again, this gives(6)2uhnu,uu+hnu,uu2+u2hnu,u+nhnuuhnu,u+nhnuhnu,uu+nuhnu,uhnu+nhnu,uuhnu∈ζℜ.
Buthnu,u∈ζℜ, hence we get(7)4u2hnu,u+2nhnuuhnu,u+2nuhnuhnu,u∈ζℜ.
Commuting (7) with hnu, we get(8)4u2hnu,u,hnu+2nhnuuhnu,u,hnu+2nuhnuhnu,u,hnu=0.
So, we have(9)4u2,hnuhnu,u+2nhnuu,hnuhnu,u+2nuhnu,hnuhnu,u=0,
and this gives(10)4uhnu,uhnu,u+4hnu,uuhnu,u+2nhnuhnu,uhnu,u+2nhnu,uhnuhnu,u=0.
After collecting, (10) can be reduced to(11)8uhnu,u2+4nhnuhnu,u2=0.
Consequently, we get(12)42u+nhnuhnu,u2=0.
But by using (12), we have(13)8hnu,u3=8hnuu−uhnuhnu,u2,=8hnuuhnu,u2−8uhnuhnu,u2,=8hnuuhnu,u2−8uhnu,u2hnu,=8hnuuhnu,u2+4nhnu,u2hnu2,=8hnuuhnu,u2+4nhnu2hnu,u2,=hnu42u+nhnuhnu,u2=0.
Therefore, we get(14)2hnu,u3=0,for allu∈U.
But2hnu,u is central and nilpotent, as well as, ℜ is semiprime, hence(15)2hnu,u=0,for allu∈U.
Moreover,(16)hnu,u2=hnu,uu+uhnu,u,=2hnu,uu=0,that is,(17)hnu,u2=0,for allu∈U.
Linearizingu in (15), we get(18)2hnu,v+hnv,u=0,for allu,v∈U.
Also, sincehnu,u∈ζℜ, then(19)hnu,v+hnv,u∈ζℜ,for allu,v∈U.
Now, using (15), (18), and (19), we get(20)hnu,uv+vu+hnv,u2=hnu,uv+uhnu,v+vhnu,u+hnu,vu+hnv,uu+uhnv,u,=2hnu,uv+uhnv,u+hnu,v+hnv,u+hnu,vu,=0+2uhnv,u+hnu,v=0,
That is,(21)hnu,uv+vu+hnv,u2=0,for allu,v∈U.
Replacingv by vu in (21), we get(22)hnu,uvu+vu2+hnvu,u2+vhnu,u2+nhnvhnu,u2=0,for allu,v∈U.
Using (17), we get(23)hnu,uv+vuu+hnv,u2u+v,u2hnu+nhnv,u2hnu=0,for allu,v∈U,
That is,(24)uv+vuhnu,u+hnu,uv+vuu+hnv,u2u+v,u2hnu+nhnv,u2hnu=0,for allu,v∈U.
Therefore, using (21), we have(25)uv+vuhnu,u+hnu,uv+vu+hnv,u2u+v,u2hnu+nhnv,u2hnu=uv+vuhnu,u+v,u2hnu+nhnv,u2hnu=0,for allu,v∈U.
Thus,(26)uv+vuhnu,u+v,u2hnu+nhnv,u2hnu=0,for allu,v∈U.
In (26), replacing uv+vu by u,v+2vu, we have(27)u,vhnu,u+2vuhnu,u+v,u2hnu+nhnv,u2hnu=0.
Therefore, using (15) in (27), we get(28)u,vhnu,u+v,u2hnu+nhnv,u2hnu=0,so,(29)u,vhnu,u+v,u2hnu+nuhnv,uhnu+nhnv,uuhnu=0.
Now, ifn is even, then n=2k for some k∈ℤ, so from (18), we have 2hnv,u=−2hnu,v, hence(30)u,vhnu,u+v,u2hnu−2kuhnu,vhnu−2khnu,vuhnu=0.
In (30), substituting hnuu instead of v, we get(31)u,hnuuhnu,u+hnu,u2uhnu−2kuhnuhnu,uhnu−2khnuhnu,uuhnu=0.
Now, by using (15), (17), (31), and hnu,u∈ζℜ, we get(32)uhnu,u2=0,for allu∈U.
But(33)hnu,u3=hnuuhnu,u2−uhnuhnu,u2,=hnuuhnu,u2−uhnu,u2hnu=0.
Hence,hnu,u=0 for all u∈U; that is, hn is commuting on U for every even integer n.
Now, ifn is an odd integer, then there exists m∈ℤ such that n=2m+1, and then (29) can be rewritten as(34)u,vhnu,u+v,u2hnu+2muhnv,uhnu+2mhnv,uuhnu+uhnv,uhnu+hnv,uuhnu=0.
So from (18), we have 2hnv,u=−2hnu,v; hence, (34) becomes(35)u,vhnu,u+v,u2hnu−2muhnu,vhnu−2mhnu,vuhnu+uhnv,uhnu+hnv,uuhnu=0.
In (35), substituting hnuu instead of v, we get(36)u,hnuuhnu,u+hnu,u2uhnu−2muhnuhnu,uhnu−2mhnuhnu,uuhnu+uhnhnuu,uhnu+hnhnuu,uuhnu=0.
Now, by using (15), (17), (36), and hnu,u∈ζℜ we get(37)uhnu,u2−uhnhnuu,uhnu−hnhnuu,uuhnu=0.
Now, (37) can be rewritten as(38)uhnu,u2−hnhnuu,u2hnu=0.
In (21), replacing v by hnuu, we get(39)hnu,uhnuu+hnuu2+hnhnuu,u2=0,for allu∈U.
That is,(40)hnu,uhnuu+uhnuhnu,u+hnu,hnuu2+hnhnuu,u2=0.
Using (17) in (40), the third term is zero, and using hnu,u∈ζℜ, we can get(41)hnuu+uhnuhnu,u=−hnhnuu,u2,for allu,v∈U.
But using (15), we get(42)hnuu+uhnuhnu,u=hnu,u+2uhnuhnu,u,=hnu,u2+2uhnuhnu,u,=hnu,u2.
So we get(43)hnuu+uhnuhnu,u=hnu,u2.
Comparing (41) and (43), we get(44)hnu,u2=−hnhnuu,u2.
Now, using (44) in (38), we get(45)uhnu,u2=−hnuhnu,u2.
Now, we have(46)hnu,u3=hnu,uhnu,u2,=hnuu−uhnuhnu,u2,=hnuuhnu,u2−uhnu,u2hnu,=hnuuhnu,u2+hnuhnu,u2hnu,=hnuuhnu,u2+hnuhnuhnu,u2,=hnuu+hnuhnu,u2,=0.
Buthnu,u is central and nilpotent, as well as, ℜ is semiprime, hence(47)hnu,u=0,for allu∈U.
That is,hn is commuting on U for every odd integer n. Actually, hn is commuting on U for every integer n.Lemma 2.
Letℜ be a ring, hn:ℜ⟶ℜ be an n-homoderivation, and U be a left ideal on ℜ. If hnu∈U for some u∈U, then(1)
hnℜu⊆U(2)
ϕnu∈U.(3)
ϕnℜu⊆U(4)
Moreover, ifhn2u∈U, then hnϕnu∈UProof.
Supposehnu=v∈U. Then, for any α∈ℜ, we have(1)
hnαu=hnαu+αhnu+nhnαhnu=hnαu+αv+nhnαv∈U.(2)
ϕnu=nhnu+u=nv+u=v′∈U. Therefore, if U is a left invariant ideal in ℜ by hn, then it is invariant by ϕn.(3)
ϕnαu=ϕnαϕnu=ϕnαv′∈U.(4)
hnϕnu=hnnhnu+u=nhnhnu+hnu=nhn2u+hnu∈U.Theorem 2.
LetU be a nonzero left ideal in a semiprime ring ℜ, then ℜ admits an n-homoderivation hn and hn centralizing on U, and the corresponding homomorphism ϕn is epimorphism. If there exists an element w∈U such that hnw,hn2w∈U, hnw≠0, and ϕnw≠0 then ℜ contains a nonzero central ideal.Proof.
By Lemma1, since hn is commuting on U, we get(48)hnu,v+hnv,u=0,for allu,v∈U.
Replacingv by vu in (48), we get v,uhnu+nhnv,uhnu=0, which may be written as(49)v,uhnu−nhnu,vhnu=0for allu,v∈U.
Substitutingvt,t∈U for v in (49) and using (49), we obtain(50)v,u−nhnu,vthnu=0.
SinceU is a left ideal, we get(51)v,u−nhnu,vℜUhnu=0,for allu,v∈Uornhnu+u,vℜUhnu=0,for allu,v∈U.
The ringℜ is semiprime; hence, we can choose a family Ω=Qλ:λ∈Λ of prime ideals of ℜ for which ∩Qλ=0. From the last relation, it follows that for each Qλ, eitheranhnu+u,v=ϕnu,v∈Qλ, for all u,v∈U orbUhnU⊆Qλ
A prime idealQλ is called a type-one prime if it satisfies a and a type-two prime otherwise; let Q1 and Q2 be, respectively, the intersections of all type-one and type-two primes. Note that Q1∩Q2=0.
We now investigate a typical type-two primeQ=Qλ. From b, we have uhnu∈Q, but uhnu=hnuu; hence, hnuu∈Q for all u∈U. Consequently, hnu+vu+v∈Q for all u,v∈U, and so, hnuv+hnvu∈Q for all u,v∈U. Putting v=uv, we get hnuuv+hnuvu∈Q, but hnuu∈Q; hence, hnuvu∈Q for all u,v∈U. Since U is a left ideal, then hnβuvβu∈Q for all β∈ℜ,u,v∈U. The fact that Q is prime yields hnℜU2⊆Q or U⊆Q. But if the latter holds true, then a holds true for Q, contradicting our definition of a type-two prime ideal; therefore, hnℜU2⊆Q. Thus, hnβuv∈Q for all β∈ℜ,u,v∈U, so we have hntuv∈Q for all t,u,v∈U. By the condition of hn, we have hntuv+thnuv+nhnthnuv∈Q, and by b, we get hntuv+nhnthnuv∈Q. But hnuvu∈Q, hence we have hntuvu∈Q for all t,u,v∈U. Therefore, hntuℜvu⊆Q for all t,u,v∈U by primness of Q,hnUU⊆Q or U2⊆Q. The latter case leads to a contradiction; therefore, hnUU⊆Q, that is, hnUℜU⊆Q. Consequently, hnU⊆Q. For any β∈ℜ and t∈U, we have hnβt∈Q, that is, hnβt+βhnt+nhnβhnt∈Q, which gives hnβt∈Q for all β∈ℜ and t∈U. Thus, ℜhnℜU⊆Q. But U⊈Q, hence ℜhnℜ⊆Q. Since Q is a typical type-two prime, then(52)hnℜU⊆Q2,ℜhnℜU⊆Q2,andℜhnℜ⊆Q2.
Sinceϕn is an epimorphism, then 0≠L=ϕnU∩U as ϕnw≠0 and 0≠W=ϕnℜw⊆L, by Fact 2, are left ideals in ℜ, where ℜw is the left ideal generated by w. Now, we claim that hnW⊆L. To prove our claim, let x=ϕnαw∈W. hnx=hnϕnαw=ϕnhnαw=ϕnu∈ϕnU by Theorem 1. On the other side, hnx=hnϕnαw=hnϕnαϕnw=hnϕnαϕnw+ϕnαhnϕnw+nhnϕnαhnϕnw∈U, by Lemma 2. Hence, hnW⊆ϕnU∩U=L.
Consider now the left idealV generated by the set hnℜL. We shall show that V is commutative and hence a two-sided central ideal. A typical element of V is a sum of elements of the form hnβu and αhnβu, where β,α∈ℜ and u∈L. Thus, we need to only show that commutators of the forms hnβ1u1,hnβ2u2=ϕnu1′,v1′, α1hnβ1u1,hnβ2u2=ϕnu2′,v2′, and α1hnβ1u1,α2hnβ2u2=ϕnu3′,v3′ are all trivial, where β1, β2,α1,α2∈ℜ, u1,u2∈L, and u1′,u2′,u3′,v1′,v2′,v3′∈U. Clearly, all three types are in Q1 by a, and they are all in Q2 by (52); hence all belong to Q1∩Q2=0.
IfV≠0, we are finished. Therefore, we assume that V=0, in which case hnℜL=0. The left ideal Lhnℜ is therefore nilpotent, so Lhnℜ=0. Thus, thnβα=0 for all t∈L and all β,α∈ℜ so that thnβα+tβhnα+nthnβhnα=0, and therefore, we get(53)Lℜhnℜ=0.
SincehnW⊂L, then hnWℜhnℜ=0. So, we get hnwℜhnw=0 for all w∈W and hence, we get hnW=0. This contradicts our initial hypothesis, so the central ideal V must in fact be nonzero.UsingFact 1, the following theorem comes from Theorem 2.Theorem 3.
Letℜ be a prime ring and U be a nonzero left ideal. Suppose ℜ admits an n-homoderivation hn which is centralizing on U and the corresponding homomorphism ϕn is epimorphism. If there exists an element w∈U such that hnw,hn2w∈U, hnw≠0, and ϕnw≠0, then ℜ is commutative.As a special case of our work whenn=0, the corresponding homomorphism ϕn will be the identity map which will be an epimorphism. Also, there exists an element u∈U such that w=u4≠0 and ϕnw≠0. Also, hnu2=hnuu+uhnu=2hnuu since hn is centralizing and hence commuting; therefore, hnu2∈U. Hence, hnw=hnu4=2hnu2u2∈U. Also, hnhnw=hn2hnu2u2=2hnhnu2u2+2hnu22∈U. So, if hnw≠0, then we get the following results.Corollary 1.
Letℜ be a semiprime ring and U be a nonzero left ideal. If ℜ admits a 0-homoderivation hn (i.e., derivation) which is centralizing on U and there exists an element u∈U, such that hnu4≠0, then ℜ contains a nonzero central ideal.Corollary 2.
Letℜ be a prime ring and U a nonzero left ideal. If ℜ admits a 0-homoderivation hn (i.e., derivation) which is centralizing on U and there exists an element u∈U, such that hnu4≠0, then ℜ is commutative.



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