JMATH Journal of Mathematics 2314-4785 2314-4629 Hindawi Publishing Corporation 10.1155/2014/483784 483784 Research Article On Nil-Symmetric Rings http://orcid.org/0000-0002-7719-9834 Chakraborty Uday Shankar 1 http://orcid.org/0000-0002-5694-900X Das Krishnendu 2 Guo Li 1 Department of Mathematics Albert Einstein School of Physical Sciences Assam University Silchar, Assam 788011 India aus.ac.in 2 Department of Mathematics Netaji Subhas Mahavidyalaya Udaipur, Tripura 799120 India nsmahavidyalaya.org.in 2014 16102014 2014 04 05 2014 17 09 2014 16 10 2014 2014 Copyright © 2014 Uday Shankar Chakraborty and Krishnendu Das. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The concept of nil-symmetric rings has been introduced as a generalization of symmetric rings and a particular case of nil-semicommutative rings. A ring R is called right (left) nil-symmetric if, for a,b,cR, 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.

1. Introduction

Throughout this paper, all rings are associative with unity. Given a ring R, 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  if abc=0 implies acb=0 for all a,b,cR. 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 . R is reversible  if ab=0 implies ba=0 for all a,bR; R is called semicommutative  if ab=0 implies aRb=0 for all a,bR. In , 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++bmxmR[x] satisfy f(x)g(x)=0, then aibj=0 for each i,j. Liu-Zhao  and Antoine  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++bmxmR[x] satisfy f(x)g(x)=0, then aibjnil(R) for each i,j. A ring R is called nil-Armendariz if whenever f(x)=a0+a1x++anxn,  g(x)=b0+b1x++bmxmR[x] satisfy f(x)g(x)nil(R)[x], then aibjnil(R) for each i,j. Mohammadi et al.  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,bnil(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 ring R, 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)RRR0nil(R)RR00nil(R)R000nil(R)).

Definition 1.

A ring R is said to be right (left) nil-symmetric if whenever, for every a,bnil(R) and for every cR, abc=0(cab=0), then acb=0. A ring R is nil-symmetric if it is both right and left nil-symmetric.

Example 2.

let k be a field, and let R be the path algebra of the quiver (2)1x2y, 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 of x, 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=xy0. Hence, R is not a left nil-symmetric ring.

Similarly by considering the opposite ring of R, 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 Example 3 and that every subring of a right (left) nil-symmetric ring is right (left) nil-symmetric.

Example 3.

For a reduced ring R, 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,dZ,a-dbc0(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.

Let R be a reduced ring. Then (13)S={(abc0ad00a):a,b,c,dR} 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,that  is,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.

Let S be a reduced ring and we define a new ring as follows: (18)Rn={(aa12a13a1n0aa23a2n00aa3n000a):a,aijS}, where n2. Based on Proposition 6, one may think that Rn may also be nil-symmetric for n4, but the following example nullifies that possibility.

Example 7.

Let R be a reduced ring and let (19)R4={(aa12a13a140aa23a2400aa34000a):a,aijR}. 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 ring R, let (22)V(R)={(aa12a13a140aa23a2400a0000a):a,aijR}. Then V(R) forms a subring of R4.

Example 8.

For every reduced ring R, 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.

Let R be a right nil-symmetric ring and a,bnil(R) such that ab=0. Let cR 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 ring R, 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 Example 3, T2(R) is a nil-symmetric ring but not Abelian (and so not semicommutative ). The following example [13, Example 2.8] shows that a semicommutative ring need not be a right or left nil-symmetric ring.

Example 13.

Let Q8={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,bnil(R) and cR such that abc=cab=0, but acb0. Hence, R is neither a right nil-symmetric ring nor a left nil-symmetric ring.

Proposition 14.

For a reduced ring R and for n2, (30)Vn(R)={(a1a2a3a4an0a1a2a3an-100a1a2an-20000a20000a1):a1,,anR}isanil-symmetricring.

Proof.

Let R 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 , R[x]/(xn)Vn(R) for n2. Hence, for n2,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.

Let R be a reduced ring and let (31)R4={(aa12a13a140aa23a2400aa34000a):a,aijR}.

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 that nil(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.

Let R 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. Let αβγ=0 with α=u-1a, β=v-1bnil(Δ-1R), and γ=w-1cΔ-1R; then u,v,wΔ, a,bnil(R), and cR. 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 ring R, 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 Proposition 17. 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.

Let R 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. Let eR and (1-e)R be right (left) nil-symmetric rings for some central idempotent e of R. Since, ReR(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 ring R and for any e2=eR, eRe is a right (left) nil-symmetric ring. The converse is, however, not true, in general as shown by the following example.

Example 20.

Let S 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):aS} is a reduced ring and so a nil-symmetric ring.

For any nonempty subsets A,B,C of a ring R, ABC denotes the set of all finite sums of the elements of the type abc, where aA, bB, cC.

Proposition 21.

A ring R 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.

Let R 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 aA, bB, cC. Right nil-symmetric property of R gives acb=0 for all aA, bB, cC. 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 ring R is said to be semiprime if, for aR, aRa=0 implies that a=0.

Proposition 22.

For a semiprime ring R, the following statements are equivalent.

R  is reduced.

R is symmetric.

R is reversible.

R is semicommutative.

R is nil-semicommutative.

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 aR. Then a2c=0 for any cR, and so aca=0, since R is right nil-symmetric. Thus a=0 by semiprimeness of R and, therefore, R is reduced.

Given a ring R and a bimodule  MRR, the trivial extension of R by M is the ring T(R,M)=RM 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 rR and mM and the usual matrix operations are used.

Proposition 23.

For a reduced ring R, T(R,R) is a nil-symmetric ring.

Proof.

Let R 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 ring R is nil-symmetric, then T(R,R) is nil-symmetric. However, the following example eliminates the possibility.

Example 24.

Let H be the Hamilton quaternions over the real number field and let (35)R={(abc0ad00a):a,b,c,dH}. 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.

Let R 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,dS} 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 Example 26.

3. Polynomial Extension of Nil-Symmetric Rings

Anderson-Camillo  proved that a ring R is Armendariz if and only if R[x] is Armendariz; Huh et al.  have shown that polynomial rings over semicommutative rings need not be semicommutative; Kim-Lee  showed that polynomial rings over reversible rings need not be reversible. Recently Mohammadi et al.  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.

Let Z2 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,r4A. 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+b2x2nil(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+a1cb0I. 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 Example 26, (Z2+A)[x] is a domain  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 Proposition 9, 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, n2 and R=Ka,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.

If R is a right (left) nil-symmetric and Armendariz ring, then the polynomial ring R[x] is right (left) nil-symmetric.

Proof.

Let R be a right nil-symmetric and Armendariz ring and let f(x)=i=0maixi,g(x)=j=0nbjxjnil(R[x]) and h(x)=k=0pakxkR[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,bjnil(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.

If R is a right (left) nil-symmetric ring, then R[x] is nil-Armendariz.

Proof.

Let R 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.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors are deeply indebted to Dr. Pierre-Guy Plamondon, Laboratory of Mathematics, University of Paris, France, for providing Example 2 and Professor Mangesh B. Rege, Department of Mathematics, NEHU, Shillong, India, for his valuable suggestions.

Lambek J. On the representation of modules by sheaves of factor modules Canadian Mathematical Bulletin 1971 14 359 368 MR0313324 Anderson D. D. Camillo V. Semigroups and rings whose zero products commute Communications in Algebra 1999 27 6 2847 2852 10.1080/00927879908826596 MR1687281 ZBL0929.16032 2-s2.0-0033247247 Cohn P. M. Reversible rings Bulletin of the London Mathematical Society 1999 31 6 641 648 10.1112/S0024609399006116 MR1711020 2-s2.0-0012989052 Motais de Narbonne L. Anneaux semi-commutatifs et unis riels anneaux dont les id aux principaux sont idempotents Proceedings of the 106th National Congress of Learned Societies (Perpignan, 1981) 1982 Paris, France Bib Necklace 71 73 Rege M. B. Chhawchharia S. Armendariz rings Proceedings of the Japan Academy 1997 73 1 14 17 10.3792/pjaa.73.14 MR1442245 2-s2.0-21744441266 Liu Z. Zhao R. On weak Armendariz rings Communications in Algebra 2006 34 7 2607 2616 10.1080/00927870600651398 MR2240395 2-s2.0-33745624288 Antoine R. Nilpotent elements and Armendariz rings Journal of Algebra 2008 319 8 3128 3140 10.1016/j.jalgebra.2008.01.019 MR2408310 ZBL1157.16007 2-s2.0-39749107315 Mohammadi R. Moussavi A. Zahiri M. On nil-semicommutative rings International Electronic Journal of Algebra 2012 11 20 37 MR2876884 ZBL1253.16024 Huh C. Kim H. K. Kim N. K. Lee Y. Basic examples and extensions of symmetric rings Journal of Pure and Applied Algebra 2005 202 1–3 154 167 10.1016/j.jpaa.2005.01.009 MR2163406 2-s2.0-23944515451 Ouyang L. Chen H. On weak symmetric rings Communications in Algebra 2010 38 2 697 713 10.1080/00927870902828702 MR2598907 2-s2.0-77950892607 Agayev N. Güngöroğlu G. Harmanci A. Halıcıoğlu S. Central Armendariz rings Bulletin of the Malaysian Mathematical Sciences Society 2011 34 1 137 145 MR2783785 Huh C. Lee Y. Smoktunowicz A. Armendariz rings and semicommutative rings Communications in Algebra 2002 30 2 751 761 10.1081/AGB-120013179 MR1883022 ZBL1023.16005 2-s2.0-0036002743 Kafkas G. Ungor B. Halicioglu S. Harmanci A. Generalized symmetric rings Algebra and Discrete Mathematics 2011 12 2 72 84 MR2952903 ZBL1259.16042 Courter R. C. Finite-dimensional right duo algebras are duo Proceedings of the American Mathematical Society 1982 84 2 157 161 10.2307/2043655 MR637159 ZBL0495.16013 Lee T.-K. Zhou Y. Armendariz and reduced rings Communications in Algebra 2004 32 6 2287 2299 10.1081/AGB-120037221 MR2100471 ZBL1068.16037 2-s2.0-22744439140 Kim N. K. Lee Y. Extensions of reversible rings Journal of Pure and Applied Algebra 2003 185 1–3 207 223 10.1016/S0022-4049(03)00109-9 MR2006427 2-s2.0-0141649250 Anderson D. D. Camillo V. Armendariz rings and Gaussian rings Communications in Algebra 1998 26 7 2265 2272 10.1080/00927879808826274 MR1626606 2-s2.0-22044453683