# On Nil-Symmetric Rings

Authors: Uday Shankar Chakraborty; Krishnendu Das
Journal: Journal of Mathematics
 (2014)

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## Abstract

The concept of nil-symmetric rings has been introduced as a generalization of symmetric rings and a particular case of nil-semicommutative rings. A ringR is called right (left) nil-symmetric if, for a,b,c∈R, where a,b are nilpotent elements, abc=0(cab=0) implies acb=0. A ring is called nil-symmetric if it is both right and left nil-symmetric. It has been shown that the polynomial ring over a nil-symmetric ring may not be a right or a left nil-symmetric ring. Further, it is also proved that if R is right (left) nil-symmetric, then the polynomial ring R[x] is a nil-Armendariz ring.


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## Body

## 1. Introduction

Throughout this paper, all rings are associative with unity. Given a ringR, nil(R) and R[x] denote the set of all nilpotent elements of R and the polynomial ring over R, respectively. A ring R is called reduced if it has no nonzero nilpotent elements; R is said to be Abelian if all idempotents of R are central; R is symmetric [1] if abc=0 implies acb=0 for all a,b,c∈R. An equivalent condition for a ring to be symmetric is that whenever product of any number of elements of the ring is zero, any permutation of the factors still gives the product zero [2]. R is reversible [3] if ab=0 implies ba=0 for all a,b∈R; R is called semicommutative [4] if ab=0 implies aRb=0 for all a,b∈R. In [5], Rege-Chhawchharia introduced the concept of an Armendariz ring. A ring R is called Armendariz if whenever polynomials f(x)=a0+a1x+⋯+anxn, g(x)=b0+b1x+⋯+bmxm∈R[x] satisfy f(x)g(x)=0, then aibj=0 for each i,j. Liu-Zhao [6] and Antoine [7] further generalize the concept of an Armendariz ring by defining a weak-Armendariz and a nil-Armendariz ring, respectively. A ring R is called weak-Armendariz if whenever polynomials f(x)=a0+a1x+⋯+anxn,  g(x)=b0+b1x+⋯+bmxm∈R[x] satisfy f(x)g(x)=0, then aibj∈nil(R) for each i,j. A ring R is called nil-Armendariz if whenever f(x)=a0+a1x+⋯+anxn,  g(x)=b0+b1x+⋯+bmxm∈R[x] satisfy f(x)g(x)∈nil(R)[x], then aibj∈nil(R) for each i,j. Mohammadi et al. [8] initiated the notion of a nil-semicommutative ring as a generalization of a semicommutative ring. A ring R is nil-semicommutative if ab=0 implies aRb=0 for all a,b∈nil(R). In their paper it is shown that, in a nil-semicommutative ring R, nil(R) forms an ideal of R. Getting motivated by their paper we introduce the concept of a right (left) nil-symmetric ring which is a generalization of symmetric rings and a particular case of nil-semicommutative rings. Thus all the results valid for nil-semicommutative rings are valid for right (left) nil-symmetric rings also. We also prove that if a ring R is right (left) nil-symmetric and Armendariz, then R[x] is right (left) nil-symmetric. In the context, there are also several other generalizations of symmetric rings (see [9, 10]).


## 2. Right (Left) Nil-Symmetric Rings

For a ringR, Mn(R) and Tn(R) denote the n×n full matrix ring and the upper triangular matrix ring over R, respectively. We observe that if R is a ring, then
(1)nil(Tn(R))=(nil(R)RR⋯R0nil(R)R⋯R00nil(R)⋯R⋮⋮⋮⋱⋮000⋯nil(R)).Definition 1.
A ringR is said to beright (left) nil-symmetric if whenever, for every a,b∈nil(R) and for every c∈R, abc=0(cab=0), then acb=0. A ring R is nil-symmetric if it is both right and left nil-symmetric.Example 2.
letk be a field, and let R be the path algebra of the quiver
(2)1⟵x2↺y,
over k, modulo the relation y2=0. Let e1 and e2 be the paths of length 0 at vertices 1 and 2, respectively. Composing arrows from left to right, xy is a nonzero path, while yx is not.
Then any nilpotent element is a linear combination ofx, y, and xy.
Let(ax+by+cxy) and (dx+ey+fxy) be two such elements and let (ge1+he2+ix+jy+lxy) be an arbitrary element. We have
(3)(ax+by+cxy)(dx+ey+fxy)(ge1+he2+ix+jy+lxy)=(aeh)xy,(ax+by+cxy)(ge1+he2+ix+jy+lxy)(dx+ey+fxy)=(aeh)xy.
Thus R is a right nil-symmetric ring. However, we have that e2xy=0, while xe2y=xy≠0. Hence, R is not a left nil-symmetric ring.
Similarly by considering the opposite ring ofR, one can have a left nil-symmetric ring which is not right nil-symmetric.
Clearly every symmetric ring is nil-symmetric but the converse is not true by Example3 and that every subring of a right (left) nil-symmetric ring is right (left) nil-symmetric.Example 3.
For a reduced ringR, T2(R) is a nil-symmetric ring which is not symmetric. This can be verified as follows.
Let(4)(0a00),(0b00)∈nil(T2(R));let(cd0e)∈T2(R).
Then
(5)(0a00)(0b00)(cd0e)=0.
Also
(6)(0a00)(cd0e)(0b00)=0.
Thus T2(R) is a right nil-symmetric ring. Similarly it can be shown that T2(R) is a left nil-symmetric ring. But
(7)(1101)(1000)(0100)≠0
whereas
(8)(1101)(0100)(1000)=0.
Thus T2(R) is not symmetric.From the above example we observe that a nil-symmetric ring need not be Abelian, as(1100) is an idempotent in T2(R), but
(9)(1100)(1101)≠(1101)(1100).Remark 4.
An Abelian ring also need not be either a right nil-symmetric or a left nil-symmetric ring as shown by the following example.Example 5.
We consider the ring in [11, Example 2.2]
(10)R={(abcd):a,b,c,d∈Z,a-d≡b≡c≡0(mod⁡2)}.R is an Abelian ring as (0000) and (1001) are the only idempotents. Again we have
(11)(0020)∈nil(R),(0020)(0020)(2222)=0=(2222)(0020)(0020),
but
(12)(0020)(2222)(0020)≠0.
Hence, R is neither right nil-symmetric nor left nil-symmetric.Proposition 6.
LetR be a reduced ring. Then
(13)S={(abc0ad00a):a,b,c,d∈R}
is a nil-symmetric ring.Proof.
Let(14)(0b1c100d1000),(0b2c200d2000)∈nil(S),(a3b3c30a3d300a3)∈S
be such that
(15)(0b1c100d1000)(0b2c200d2000)(a3b3c30a3d300a3)=0.
This implies
(16)(00b1d2a3000000)=0,thatis,b1d2a3=0.
Since R is reduced, b1a3d2=0. Thus
(17)(0b1c100d1000)(a3b3c30a3d300a3)(0b2c200d2000)=(00b1a3d2000000)=0.
Hence, S is a right nil-symmetric ring. Similarly it can be shown that S is a left nil-symmetric ring.LetS be a reduced ring and we define a new ring as follows:
(18)Rn={(aa12a13⋯a1n0aa23⋯a2n00a⋯a3n⋮⋮⋮⋱⋮000⋯a):a,aij∈S},
where n≥2. Based on Proposition 6, one may think that Rn may also be nil-symmetric for n≥4, but the following example nullifies that possibility.Example 7.
LetR be a reduced ring and let
(19)R4={(aa12a13a140aa23a2400aa34000a):a,aij∈R}.
Now
(20)(01-10000000000000)(0000000100010000)(0000001000000000)=0,(0000001000000000)(01-10000000000000)(0000000100010000)=0
but
(21)(01-10000000000000)(0000001000000000)(0000000100010000)≠0.
Thus R4 is neither a right nil-symmetric ring nor a left nil-symmetric ring.
For a ringR, let
(22)V(R)={(aa12a13a140aa23a2400a0000a):a,aij∈R}.
Then V(R) forms a subring of R4.Example 8.
For every reduced ringR, V(R) is nil-symmetric.
Let(23)(0a12a13a1400a23a2400000000),(0b12b13b1400b23b2400000000)∈nil(V(R))
and let
(24)(cc12c13c140cc23c2400c0000c)∈V(R)
be such that
(25)(0a12a13a1400a23a2400000000)(0b12b13b1400b23b2400000000)×(cc12c13c140cc23c2400c0000c)=0.
This gives
(26)(00a12b23ca12b24c000000000000)=0.
Thus a12b23c=0,  a12b24c=0. Since R is reduced, we have a12cb23=0,  a12cb24=0. Therefore,
(27)(00a12cb23a12cb24000000000000)=(0a12a13a1400a23a2400000000)×(cc12c13c140cc23c2400c0000c)×(0b12b13b1400b23b2400000000)=0.
Hence, V(R) is a right nil-symmetric ring. Similarly, it can be shown that V(R) is a left nil-symmetric ring.We also observe that every right (left) nil-symmetric ring is nil-semicommutative.Proposition 9.
Every right (left) nil-symmetric ring is nil-semicommutative.Proof.
LetR be a right nil-symmetric ring and a,b∈nil(R) such that ab=0. Let c∈R be arbitrary; then abc=0. By right nil-symmetric property of R, acb=0. Thus aRb=0. Hence, R is nil-semicommutative. Proceeding similarly one can show that every left nil-symmetric ring is nil-semicommutative.Remark 10.
The converse is however not true, as shown by the following example.Example 11.
For every reduced ringR, T3(R) is a nil-semicommutative ring which is neither a right nil-symmetric ring nor a left nil-symmetric ring. This can be verified as follows.
We have(28)(010000000),(000001000)∈nil(T3(R)),(010000000)(000001000)(000010000)=0=(000010000)(010000000)(000001000),
but
(29)(010000000)(000010000)(000001000)≠0.
Thus T3(R) is neither a right nil-symmetric ring nor a left nil-symmetric ring. But T3(R) is nil-semicommutative by [8, Example 2.2].Remark 12.
Semicommutativity and nil-symmetry do not follow each other. In Example3, T2(R) is a nil-symmetric ring but not Abelian (and so not semicommutative [12]). The following example [13, Example 2.8] shows that a semicommutative ring need not be a right or left nil-symmetric ring.Example 13.
LetQ8={1,x-1,xi,x-i,xj,x-j,xk,x-k} be the quaternion group and let Z2 be the ring of integers modulo 2. Consider the group ring R=Z2Q8. By [14, Corollary 2.3], R is reversible and so semicommutative. Let a=1+xj, b=1+xi, c=1+xi+xj+xk. Then a,b∈nil(R) and c∈R such that abc=cab=0, but acb≠0. Hence, R is neither a right nil-symmetric ring nor a left nil-symmetric ring.Proposition 14.
For a reduced ringR and for n≥2, (30)Vn(R)={(a1a2a3a4⋯an0a1a2a3⋯an-100a1a2⋯an-2⋮⋮⋮⋮⋱⋮0000⋯a20000⋯a1):a1,…,an∈R}isanil-symmetricring.Proof.
LetR be a reduced ring. Then by [9, Theorem 2.3], R[x]/(xn) is a symmetric ring and hence a nil-symmetric ring, where (xn) is the ideal generated by xn for any positive integer n. Also by [15], R[x]/(xn)≅Vn(R) for n≥2. Hence, for n≥2,Vn(R) is nil-symmetric.Since the class of nil-symmetric rings is contained in the class of nil-semicommutative rings, the results which are valid for nil-semicommutative rings are also valid for nil-symmetric rings. Mohammadi et al. [8, Example 2.8] have shown that T5(R) is not a nil-semicommutative ring, where R is a reduced ring. Thus T5(R) is not nil-symmetric. Now we give an example of a weak-Armendariz ring which is not nil-symmetric.Example 15.
LetR be a reduced ring and let
(31)R4={(aa12a13a140aa23a2400aa34000a):a,aij∈R}.
By [6, Example 2.4], R4 is weak-Armendariz. By Example 7, R4 is neither a right nor a left nil-symmetric ring.Proposition 16.
Finite product of right (left) nil-symmetric rings is right (left) nil-symmetric.Proof.
It comes from the fact thatnil(∏i=1nRi)=∏i=1nnil(Ri) [8, Proposition 2.13]. Let (a1,a2,…,an),(b1,b2,…,bn)∈nil(∏i=1nRi) and (c1,c2,…,cn)∈∏i=1nRi such that (a1,a2,…,an)(b1,b2,…,bn)(c1,c2,…,cn)=0. Thus, for each i=1,2,…,n, aibici=0. Since Ri is right nil-symmetric, aicibi=0 for each i=1,2,…,n. So, we get (a1,a2,…,an)(c1,c2,…,cn)(b1,b2,…,bn)=0. The result can be similarly proved for left nil-symmetric rings.Proposition 17.
LetR be a ring and let Δ be a multiplicatively closed subset of R consisting of central nonzero-divisors. Then R is right (left) nil-symmetric if and only if Δ-1R is right (left) nil-symmetric.Proof.
It suffices to prove the necessary condition because subrings of right (left) nil-symmetric rings are also right (left) nil-symmetric. Letαβγ=0 with α=u-1a, β=v-1b∈nil(Δ-1R), and γ=w-1c∈Δ-1R; then u,v,w∈Δ, a,b∈nil(R), and c∈R. Since Δ is contained in the center of R, we have 0=αβγ=u-1av-1bw-1c=(uvw)-1abc and so abc=0. It follows that acb=0, since R is right nil-symmetric. Thus αγβ=(uvw)-1abc=0. Hence, Δ-1R is right nil-symmetric. Similarly, Δ-1R can be shown to be left nil-symmetric if R itself is a left nil-symmetric ring.Corollary 18.
For a ringR, R[x] is a right (left) nil-symmetric ring if and only if R[x;x-1] is a right (left) nil-symmetric ring.Proof.
It directly follows from Proposition17. If Δ={1,x,x2,…}, then Δ is clearly a multiplicatively closed subset of R[x] and R[x;x-1]=Δ-1R[x].Proposition 19.
LetR be a ring. Then eR and (1-e)R are right (left) nil-symmetric for some central idempotent e of R if and only if R is right (left) nil-symmetric.Proof.
It suffices to prove the necessary condition because subrings of right (left) nil-symmetric rings are also right (left) nil-symmetric. LeteR and (1-e)R be right (left) nil-symmetric rings for some central idempotent e of R. Since, R≅eR⊕(1-e)R, R is right (left) nil-symmetric by Proposition 16.Since the class of right (left) nil-symmetric rings is closed under subrings, therefore, for any right (left) nil-symmetric ringR and for any e2=e∈R, eRe is a right (left) nil-symmetric ring. The converse is, however, not true, in general as shown by the following example.Example 20.
LetS be any reduced ring. Then by Example 11, R=T3(S) is neither a right nil-symmetric nor a left nil-symmetric ring.
But for(32)e2=e=(100000000)∈R,eRe={(a00000000):a∈S}
is a reduced ring and so a nil-symmetric ring.For any nonempty subsetsA,B,C of a ring R, ABC denotes the set of all finite sums of the elements of the type abc, where a∈A, b∈B, c∈C.Proposition 21.
A ringR is right (left) nil-symmetric if and only if ABC=0 implies ACB=0 (CAB=0 implies ACB=0) for any two nonempty subsets A,B of nil(R) and any subset C of R.Proof.
LetR be a right nil-symmetric ring and let A,B be nonempty subsets of nil(R); let C be a nonempty subset of R such that ABC=0. Then abc=0 for all a∈A, b∈B, c∈C. Right nil-symmetric property of R gives acb=0 for all a∈A, b∈B, c∈C. Thus ACB=0. Similar proof can be given for left nil-symmetric rings. The converse is straightforward.The following result shows that, for a semiprime ring, the properties of reduced, symmetric, reversible, semicommutative, nil-semicommutative, and nil-symmetric rings coincide. Note that a ringR is said to be semiprime if, for a∈R, aRa=0 implies that a=0.Proposition 22.
For a semiprime ringR, the following statements are equivalent.(1)
Ris reduced.(2)
R is symmetric.(3)
R is reversible.(4)
R is semicommutative.(5)
R is nil-semicommutative.(6)
R is right (left) nil-symmetric.Proof.
(1)–(4) are equivalent by [16, Lemma 2.7]. (1)⇔(5) by [8, Proposition 2.18]. (2)⇒(6) is clear. (6)⇒ (1): let a2=0 for a∈R. Then a2c=0 for any c∈R, and so aca=0, since R is right nil-symmetric. Thus a=0 by semiprimeness of R and, therefore, R is reduced.Given a ringR and a bimodule  MRR, the trivial extension of R by M is the ring T(R,M)=R⊕M with the usual addition and the following multiplication:
(33)(r1,m1)(r2,m2)=(r1r2,r1m2+m1r2).
This is isomorphic to the ring of all matrices:
(34)(rm0r),
where r∈R and m∈M and the usual matrix operations are used.Proposition 23.
For a reduced ringR, T(R,R) is a nil-symmetric ring.Proof.
LetR be a reduced ring. Since T(R,R) is a subring of S in Proposition 6 and the class of right(left) nil-symmetric rings is closed under subrings, thus T(R,R) is a nil-symmetric ring.Considering the above proposition one may conjecture that if a ringR is nil-symmetric, then T(R,R) is nil-symmetric. However, the following example eliminates the possibility.Example 24.
LetH be the Hamilton quaternions over the real number field and let
(35)R={(abc0ad00a):a,b,c,d∈H}.
Then by Proposition 6, R is a nil-symmetric ring. Let S be the trivial extension of R by itself. Then S is not a right nil-symmetric ring. Note that
(36)((0i0000000)(000000000)(000000000)(0i0000000)),((00000j000)(-i000-i000-i)(000000000)(00000j000))∈nil(S),((0i0000000)(000000000)(000000000)(0i0000000))×((00000j000)(-i000-i000-i)(000000000)(00000j000))×((000001000)(k000k000k)(000000000)(000001000))=0.
However we have
(37)((0i0000000)(000000000)(000000000)(0i0000000))×((000001000)(k000k000k)(000000000)(000001000))×((00000j000)(-i000-i000-i)(000000000)(00000j000))=((000000000)(002000000)(000000000)(000000000))≠0.
Thus S=T(R,R) is not a right nil-symmetric ring.Example 25.
LetR be a ring and let I be an ideal of R such that R/I is nil-symmetric. Then R may not be nil-symmetric. This can be verified as follows. Let S be any reduced ring. Then by Example 11, R=T3(S) is not nil-symmetric but nil-semicommutative. Thus
(38)I=nil(R)={(0bc00d000):b,c,d∈S}
is an ideal of R and R/I is reduced, so nil-symmetric.Homomorphic image of a right (left) nil-symmetric ring need not be a right (left) nil-symmetric ring. This is discussed after Example26.


## 3. Polynomial Extension of Nil-Symmetric Rings

Anderson-Camillo [17] proved that a ring R is Armendariz if and only if R[x] is Armendariz; Huh et al. [12] have shown that polynomial rings over semicommutative rings need not be semicommutative; Kim-Lee [16] showed that polynomial rings over reversible rings need not be reversible. Recently Mohammadi et al. [8] have given an example of a nil-semicommutative ring R for which R[x] is not nil-semicommutative. Based on the above findings, it is natural to check whether the polynomial ring over a nil-symmetric ring is nil-symmetric. However, the answer is given in the negative through the following example.Example 26.
LetZ2 be the field of integers modulo 2 and let A=Z2[a0,a1,a2,b0,b1,b2,c] be the free algebra of polynomials with zero constant terms in noncommuting indeterminates a0,  a1,  a2,  b0,  b1,  b2, and c over Z2. Consider an ideal of the ring Z2+A, say I, generated by the following elements: a0b0, a0b1+a1b0, a0b2+a1b1+a2b0, a1b2+a2b1, a2b2, a0rb0, a2rb2, b0a0, b0a1+b1a0, b0a2+b1a1+b2a0, b1a2+b2a1, b0ra0, b2ra2, (a0+a1+a2)r(b0+b1+b2), (b0+b1+b2)r(a0+a1+a2), and r1r2r3r4, where r,r1,r2,r3,r4∈A. Now R=(Z2+A)/I is symmetric by [9, Example 3.1] and so a nil-symmetric ring. By [8, Example 3.6], we have a0+a1x+a2x2, b0+b1x+b2x2∈nil(R[x]). Now (a0+a1x+a2x2)(b0+b1x+b2x2)c, c(a0+a1x+a2x2)(b0+b1x+b2x2)∈I[x], but (a0+a1x+a2x2)c(b0+b1x+b2x2)∉I[x] because a0cb1+a1cb0∉I. Hence R[x] is neither a right nil-symmetric ring nor a left nil-symmetric ring.Remark 27.
The above example also helps in showing that homomorphic image of a right (left) nil-symmetric ring need not be a right (left) nil-symmetric ring. This is verified as follows.Example 28.
In Example26, (Z2+A)[x] is a domain [16] and so a nil-symmetric ring. But the quotient ring (Z2+A)[x]/I[x]≅R[x] is neither a right nil-symmetric ring nor a left nil-symmetric ring.Now we study some conditions under which the answer may be given positively. Since every right (left) nil-symmetric ring is nil-semicommutative by Proposition9, therefore, by [8, Theorem 3.3] for each right (left) nil-symmetric ring R, nil(R[x])=nil(R)[x]. The converse is, however, not true, in general. Now we give an example of a ring R which satisfies nil(R[x])=nil(R)[x], but R is neither a right nil-symmetric ring nor a left nil-symmetric ring.Example 29.
We use the ring in [7, Example 4.8]. Let K be a field, n≥2 and R=K〈a,b|bn=0〉. Then nil(R) is not an ideal of R. Thus R is neither a right nil-symmetric nor a left nil-symmetric ring by Proposition 9 and [8, Theorem 2.5]. But R is a nil-Armendariz ring and hence by [7, Corollary 5.2], nil(R[x])=nil(R)[x].Proposition 30.
IfR is a right (left) nil-symmetric and Armendariz ring, then the polynomial ring R[x] is right (left) nil-symmetric.Proof.
LetR be a right nil-symmetric and Armendariz ring and let f(x)=∑i=0maixi,g(x)=∑j=0nbjxj∈nil(R[x]) and h(x)=∑k=0pakxk∈R[x] such that f(x)g(x)h(x)=0. Since R is right nil-symmetric, nil(R[x])=nil(R)[x] by Proposition 9 and [8, Theorem 3.3]. Thus ai,bj∈nil(R) for i=0,1,2,…,m; j=0,1,2,…,n. Since R is Armendariz, therefore, aibjck=0 by [17, Proposition 1]. Thus by right nil-symmetric property of R, aickbj=0. Therefore, f(x)h(x)g(x)=0. Hence, R[x] is a right nil-symmetric ring. Similarly it can be shown that R[x] is a left nil-symmetric ring if R is a left nil-symmetric and Armendariz ring.Proposition 31.
IfR is a right (left) nil-symmetric ring, then R[x] is nil-Armendariz.Proof.
LetR be a right (left) nil-symmetric ring. Thus by Proposition 9, R is nil-semicommutative. By [8, Corollary 2.9], R is a nil-Armendariz ring. Again by [8, Theorem 3.3], nil(R[x])=nil(R)[x]. Thus by [7, Theorem 5.3], R[x] is nil-Armendariz.



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*Source: 483784-2014-10-16.xml*