Generalized symmetric rings

In this paper, we introduce a class of rings which is a generalization of symmetric rings. Let R be a ring with identity. A ring R is called central symmetric if for any a, b,c∈R, abc=0 implies bac belongs to the center of R. Since every symmetric ring is central symmetric, we study sufficient c...

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Datum:	2011
Hauptverfasser:	Kafkas, G., Ungor, B., Halicioglu, S., Harmanci, A.
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Sprache:	English
Veröffentlicht:	Інститут прикладної математики і механіки НАН України 2011
Schriftenreihe:	Algebra and Discrete Mathematics
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spelling	irk-123456789-1547592019-06-16T01:32:27Z Generalized symmetric rings Kafkas, G. Ungor, B. Halicioglu, S. Harmanci, A. In this paper, we introduce a class of rings which is a generalization of symmetric rings. Let R be a ring with identity. A ring R is called central symmetric if for any a, b,c∈R, abc=0 implies bac belongs to the center of R. Since every symmetric ring is central symmetric, we study sufficient conditions for central symmetric rings to be symmetric. We prove that some results of symmetric rings can be extended to central symmetric rings for this general settings. We show that every central reduced ring is central symmetric, every central symmetric ring is central reversible, central semmicommutative, 2-primal, abelian and so directly finite. It is proven that the polynomial ring R[x] is central symmetric if and only if the Laurent polynomial ring R[x,x−1] is central symmetric. Among others, it is shown that for a right principally projective ring R, R is central symmetric if and only if R[x]/(xn) is central Armendariz, where n≥2 is a natural number and (xn) is the ideal generated by xn 2011 Article Generalized symmetric rings / G. Kafkas, B. Ungor, S. Halicioglu, A. Harmanci // Algebra and Discrete Mathematics. — 2011. — Vol. 12, № 2. — С. 72–84. — Бібліогр.: 21 назв. — англ. 1726-3255 2010 Mathematics Subject Classification:13C99, 16D80, 16U80 http://dspace.nbuv.gov.ua/handle/123456789/154759 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України
institution	Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection	DSpace DC
language	English
description	In this paper, we introduce a class of rings which is a generalization of symmetric rings. Let R be a ring with identity. A ring R is called central symmetric if for any a, b,c∈R, abc=0 implies bac belongs to the center of R. Since every symmetric ring is central symmetric, we study sufficient conditions for central symmetric rings to be symmetric. We prove that some results of symmetric rings can be extended to central symmetric rings for this general settings. We show that every central reduced ring is central symmetric, every central symmetric ring is central reversible, central semmicommutative, 2-primal, abelian and so directly finite. It is proven that the polynomial ring R[x] is central symmetric if and only if the Laurent polynomial ring R[x,x−1] is central symmetric. Among others, it is shown that for a right principally projective ring R, R is central symmetric if and only if R[x]/(xn) is central Armendariz, where n≥2 is a natural number and (xn) is the ideal generated by xn
format	Article
author	Kafkas, G. Ungor, B. Halicioglu, S. Harmanci, A.
spellingShingle	Kafkas, G. Ungor, B. Halicioglu, S. Harmanci, A. Generalized symmetric rings Algebra and Discrete Mathematics
author_facet	Kafkas, G. Ungor, B. Halicioglu, S. Harmanci, A.
author_sort	Kafkas, G.
title	Generalized symmetric rings
title_short	Generalized symmetric rings
title_full	Generalized symmetric rings
title_fullStr	Generalized symmetric rings
title_full_unstemmed	Generalized symmetric rings
title_sort	generalized symmetric rings
publisher	Інститут прикладної математики і механіки НАН України
publishDate	2011
url	http://dspace.nbuv.gov.ua/handle/123456789/154759
citation_txt	Generalized symmetric rings / G. Kafkas, B. Ungor, S. Halicioglu, A. Harmanci // Algebra and Discrete Mathematics. — 2011. — Vol. 12, № 2. — С. 72–84. — Бібліогр.: 21 назв. — англ.
series	Algebra and Discrete Mathematics
work_keys_str_mv	AT kafkasg generalizedsymmetricrings AT ungorb generalizedsymmetricrings AT halicioglus generalizedsymmetricrings AT harmancia generalizedsymmetricrings
first_indexed	2025-07-14T06:51:44Z
last_indexed	2025-07-14T06:51:44Z
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fulltext	Algebra and Discrete Mathematics RESEARCH ARTICLE Volume 12 (2011). Number 2. pp. 72 – 84 c© Journal “Algebra and Discrete Mathematics” Generalized symmetric rings G. Kafkas, B. Ungor, S. Halicioglu, A. Harmanci Communicated by V. Mazorchuk Abstract. In this paper, we introduce a class of rings which is a generalization of symmetric rings. Let R be a ring with identity. A ring R is called central symmetric if for any a, b, c ∈ R, abc = 0 implies bac belongs to the center of R. Since every symmetric ring is central symmetric, we study sufficient conditions for central symmetric rings to be symmetric. We prove that some results of symmetric rings can be extended to central symmetric rings for this general settings. We show that every central reduced ring is central symmetric, every central symmetric ring is central reversible, central semmicommutative, 2-primal, abelian and so directly finite. It is proven that the polynomial ring R[x] is central symmetric if and only if the Laurent polynomial ring R[x, x−1] is central symmetric. Among others, it is shown that for a right principally projective ring R, R is central symmetric if and only if R[x]/(xn) is central Armendariz, where n ≥ 2 is a natural number and (xn) is the ideal generated by xn. 1. Introduction Throughout this paper all rings are associative with identity unless otherwise stated. A ring is reduced if it has no nonzero nilpotent elements, similarly a ring R is called central reduced [2] if every nilpotent element of R is central. A weaker condition than “reduced” is defined by Lambek in [16]. A ring R is symmetric if for any a, b, c ∈ R, abc = 0 implies acb = 0 if and only if abc = 0 implies bac = 0. An equivalent condition 2010 Mathematics Subject Classification: 13C99, 16D80, 16U80. Key words and phrases: symmetric rings, central reduced rings, central symmet- ric rings, central reversible rings, central semicommutative rings, central Armendariz rings, 2-primal rings. G. Kafkas, B. Ungor, S. Halicioglu, A. Harmanci 73 on a ring is that whenever a product of any number of elements is zero, any permutation of the factors still yields product zero. A ring R is right (left) principally quasi-Baer [8] if the right (left) annihilator of a principal right (left) ideal of R is generated by an idempotent. Finally, a ring R is called right (left) principally projective if the right (left) annihilator of an element of R is generated by an idempotent [7]. Throughout this paper, Z denotes the ring of integers. We write R[x] and R[x, x−1] for the polynomial ring and the Laurent polynomial ring, respectively. 2. Central symmetric rings In this section we introduce a class of rings, called central symmetric rings, which is a generalization of symmetric rings. We investigate which properties of symmetric rings hold for the central case. Clearly, symmet- ric rings are central symmetric and central symmetric rings are central reversible. We supply an example to show that all central symmetric rings need not be symmetric. We provide another example to show that all central reversible rings may not be central symmetric. Therefore the class of central symmetric rings lies strictly between classes of symmetric rings and central reversible rings. We obtain sufficient conditions for central symmetric rings to be symmetric. It is shown that the class of central symmetric rings is closed under finite direct sums. We have an example which shows that the homomorphic image of a central symmetric ring is not central symmetric. Then we determine under what conditions a homomorphic image of a ring is central symmetric. We now give our main definition. Definition 2.1. A ring R is called central symmetric if for any a, b, c ∈ R, abc = 0 implies bac belongs to the center of R. All commutative rings, reduced rings, central reduced rings and symmet- ric rings are central symmetric. One may suspect that central symmetric rings are symmetric. But the following example erases the possibility. Example 2.2. Let x, y and z be indeterminates and consider the set R = {a0 + a1x+ a2y + a3z \| a0, a1, a2, a3 ∈ Z} with componentwise addition and defining multiplication (a0 + a1x+ a2y + a3z)(b0 + b1x+ b2y + b3z) = a0b0 + (a0b1 + a1b0)x+ (a0b2 + a2b0)y + (a0b3 + a3b0 + a1b2)z. Then R is a ring with identity. From this multiplication all products are zero except that xy = z and that 1 acts as an identity (see [21, 74 Generalized symmetric rings Example 5.1]). Since x2 = y2 = 0, xz = xxy = 0 = xyx = zx and zy = xyy = 0 = yxy = yz, z is central. Hence R is central symmetric. On the other hand, yx = yx1 = 0, but xy1 = z and so R is not symmetric. Recall that a ring R is semiprime if aRa = 0 implies a = 0 for a ∈ R. Our next aim is to find conditions under what a central symmetric ring is symmetric. Proposition 2.3. If R is a symmetric ring, then R is central symmetric. The converse holds if R satisfies any of the following conditions. (1) R is a semiprime ring. (2) R is a right (left) principally projective ring. (3) R is a right (left) principally quasi-Baer ring. Proof. Clearly every symmetric ring is central symmetric. Conversely, assume that R is a central symmetric ring and a, b, c ∈ R with abc = 0. It implies that bca is central due to 1 ∈ R. Now consider the following cases. (1) Let R be a semiprime ring. Since (bca)x(bca) = 0 for all x ∈ R, it follows that bca = 0. Therefore R is symmetric. (2) Let R be a right principally projective ring. Then there exists an idempotent e ∈ R such that rR(a) = eR. Thus ae = 0. Since bc ∈ rR(a) = eR, we have bc = ebc. It follows that bca = ebca = bcae = 0. Therefore R is symmetric. A similar proof may be given for left principally projective rings. (3) Let R be a right principally quasi-Baer ring. Then there exists an idempotent e ∈ R such that rR(aR) = eR. Hence ae = 0. On the other hand, axbca = abcax = 0 for all x ∈ R. This implies that bca ∈ rR(aR) = eR, and so bca = ebca = bcae = 0. Thus R is symmetric. In a similar way every left principally quasi-Baer central symmetric ring is symmetric. The following is an apparent consequence of Proposition 2.3. Corollary 2.4. If R is a central symmetric ring, then the following conditions are equivalent. (1) R is a right principally projective ring. (2) R is a left principally projective ring. (3) R is a right principally quasi-Baer ring. (4) R is a left principally quasi-Baer ring. Proof. (1) ⇒ (2) Let R be a right principally projective ring and a ∈ R. Then there exists an idempotent e ∈ R such that rR(a) = eR. Also by Proposition 2.3, R is symmetric. Let b ∈ lR(a). Being R symmetric, G. Kafkas, B. Ungor, S. Halicioglu, A. Harmanci 75 ab = 0 and so b = eb. Since e is central, we have b = be. Hence lR(a) ⊆ Re. Obviously, Re ⊆ lR(a). Thus R is a left principally projective ring. (2) ⇒ (3) Let R be a left principally projective ring and a ∈ R. Then there exists an idempotent e ∈ R such that lR(a) = Re. Since e is central, we have ea = ae = 0. Let b ∈ rR(aR). Then axb = 0 for all x ∈ R. Hence bax = 0, and ba = 0 due to 1 ∈ R. Thus b = be = eb, and so rR(aR) ⊆ eR. Hence eR = rR(aR) since eR ⊆ rR(aR) holds also. Therefore R is a right principally quasi-Baer ring. (3) ⇒ (4) and (4) ⇒ (1) are similar to the proofs of preceding cases. It is folklore that all reduced rings are symmetric. In the central case, we have the following result. Lemma 2.5. If R is a central reduced ring, then it is central symmetric. Proof. Let abc = 0 for some a, b, c ∈ R. Then (cab)2 = cabcab = 0 so cab is central. On the other hand, (bca)2 = bcabca = 0 so bca is central. For any r ∈ R, (arbc)2 = arbcarbc = abcar2bc = 0 since bca is central. So arbc is central. For any r ∈ R, (abrc)3 = abrcabrcabrc = abcabrcabrrc = 0 since cab is central, and (bcra)2 = bcrabcra = 0 implies bcra is central. (carb)2 = carbcarb = cabcarrb = 0. So carb is central. For any r, s ∈ R, (arbsc)2 = arbscarbsc = arbcarbs2c = abcar2bs2c = 0 since carb and then bca are central. Hence arbsc is central. (bac)4 = b(acbac)b(acbac) = 0 since acbac is central and (acbac)2 = 0. Hence bac is central. In the following we prove when a central symmetric ring is reduced. Theorem 2.6. Let R be a central symmetric ring. Then we have (1) If R is a semiprime ring, then R is reduced. (2) If R is a right (left) principally projective ring, then R is reduced. Proof. (1) Let a ∈ R with a2 = 0. For any r ∈ R, ra2 = 0. By hypothesis ara is central, (ara)2 = 0 and so aRa = 0. Since R is semiprime, a = 0. (2) Let R be a right principally projective ring and a ∈ R with a2 = 0. There exists e2 = e ∈ R such that rR(a) = eR. Then a ∈ rR(a) and ae = 0 and a = ea. By Proposition 2.3, e is central and so a = ea = ae = 0. This proof is left-right symmetric since idempotents are central. According to Cohn [9], a ring R is said to be reversible if for any a, b ∈ R, ab = 0 implies ba = 0. Similarly, a ring R is called central reversible if for any a, b ∈ R, ab = 0 implies ba is central in R. It is well known that every symmetric ring is reversible and the converse holds for semiprime rings. In this direction we have the following. 76 Generalized symmetric rings Lemma 2.7. Let R be a central symmetric ring. Then R is central reversible. The converse statement holds if R is a semiprime ring. Proof. Let a, b ∈ R with ab = 0. Then 1ab = 0. Hence 1ba = b1a = ba is central. Conversely, assume that R is a semiprime central reversible ring. Let a, b, c ∈ R with abc = 0. We may suppose that a, b and c are nonzero. For any r ∈ R, abcr = 0 implies crab is central, and (crab)2 = 0. By assumption crab = 0. For any s ∈ R, crab = 0 implies bscra is central and (bscra)2 = 0. By similar reason bscra = 0. Hence (bac)2 = bacbac = 0. For any t ∈ R, bactbac is central and (bactbac)2 = 0. Then bactbacRbactbac = 0. Hence bactbac = 0 for all t ∈ R. Being R semiprime we have bac = 0. The next example provides that there exists a central reversible ring which is not a central symmetric ring. Example 2.8. Let Q8 = {1, x−1, xi, x−i, xj , x−j , xk, x−k} be the quater- nion group and consider the group ring R = Z2Q8. The elements of Z2Q8 as Z2-linear combinations of {xg : g ∈ Q8}. By Courter’s result in [10, Corol- lary 2.3],R is reversible and so central reversible. But R is not symmetric as in the [18, Example 7] by taking a = 1+xj , b = 1+xi and c = 1+xi+xj+xk. Then abc = 0 but bac 6= 0. In fact bac = xi+xj +xk +x−i+x−k and it is easily checked that xi(bac) 6= (bac)xi. Hence R is not central symmetric. Now we show that the class of central symmetric rings is closed under finite direct sums. Proposition 2.9. Let I be a finite index set and {Ri}i∈I a class of rings. Then Ri is central symmetric for all i ∈ I if and only if ⊕ i∈I Ri is central symmetric. Proof. LetRi be central symmetric for all i ∈ I and (ai)i∈I , (bi)i∈I , (ci)i∈I ∈ ⊕ i∈I Ri with (ai)(bi)(ci) = 0. Then aibici = 0 and by hypothesis biaici is central in Ri for all i ∈ I. Hence (bi)(ai)(ci) is central in ⊕ i∈I Ri. Therefore ⊕ i∈I Ri is central symmetric. The sufficiency is clear since a subring of a central symmetric ring is central symmetric. The following result is a direct consequence of Proposition 2.9. Corollary 2.10. Let R be a ring. Then eR and (1 − e)R are central symmetric for some idempotent element e in R if and only if R is central symmetric. G. Kafkas, B. Ungor, S. Halicioglu, A. Harmanci 77 Note that the homomorphic image of a central symmetric ring need not be central symmetric. Consider the following example. Example 2.11. Let Z2 denote the field of integers modulo 2 and Z2(y) rational functions field of polynomial ring Z2[y] and R = Z2(y)[x] the ring of polynomials in x over Z2(y) subject to the relation xy + yx = 1. It is well known that R is a principal ideal domain and so is a non- commutative domain (see [14, p.30], [11, Note 3.9], [21, Example 5.3]). Let I = x2R. Then I is a maximal ideal of R. Consider the ring S = R/I. We write x̄ and ȳ for the images of x and y respectively under the natural epimorphism from R onto S. Let a, b, c ∈ R with abc = 0. Since R is a domain, at least one of a, b and c is zero. Therefore bac = 0 and so bac is central and R is central symmetric. For x̄, ȳ ∈ S, we have x̄2 = 0̄ and x̄ȳ + ȳx̄ = 1̄. Multiplying the last equality from the right by x̄ and using x̄2 = 0̄, we have x̄ȳx̄ = x̄. If S were central symmetric, (x̄ȳ − 1̄)x̄ = 0̄ would imply x̄(x̄ȳ − 1̄) = −x̄ is central in S and so x̄ is central in S. This is a contradiction since x̄ is not central. Our next endeavor is to find conditions when the homomorphic image of a ring is central symmetric. Recall that a ring R is called unit-central [15], if all unit elements are central in R. It is proven that every idempotent of a unit-central ring is central. Lemma 2.12. Let R be a unit-central ring. If I is a nil ideal of R, then R and R/I are central symmetric. Proof. Let a ∈ R with an = 0 for some positive integer n. Then (1 + a)(1−a+a2−a3+ ...+(−1)n−1an−1) = 1. Hence 1+a and therefore a is central. Let a, b and c ∈ R with āb̄c̄ = 0̄ in R/I. Then abc ∈ I. Hence abc is nilpotent. So 1 + abc, and therefore abc is central. Now (c̄āb̄)2 = 0̄ and (b̄c̄ā)2 = 0̄ imply (cab)2 ∈ I and (bca)2 ∈ I, and therefore cab and bca are central. āb̄c̄r̄ = 0̄ for all r ∈ R implies (c̄r̄āb̄)2 = 0̄. So (crab)2 ∈ I and (crab)2 is nilpotent and crab is central for all r ∈ R. Similarly, (b̄s̄c̄ā)2 = 0̄ implies bsca is central for all s ∈ R. Let s̄, r̄ ∈ R/I. Then (b̄s̄c̄r̄ā)2 = 0̄ since c̄r̄āb̄ is central nilpotent. Hence bscra is central for s, r ∈ R. Now (b̄āc̄)4 = 0̄ since b̄āc̄b̄ā is central nilpotent. Hence bac is central. Thus b̄āc̄ is central. The next example shows that for a ring R and an ideal I, if R/I is central symmetric, then R need not be central symmetric. Example 2.13. Let F be a field and R the ring of all 2 × 2 upper triangular matrices over F and eij matrix units with 1 at the entry (i, j) and zeros elsewhere. Let I = e12R. Then I is an ideal of R and R/I is a commutative ring, therefore central symmetric. Consider A = e22, 78 Generalized symmetric rings B = e11 + e12 and C = A + B. Then ABC = 0 but BAC = e12 is not central since e11e12 = e12 but e12e11 = 0. Hence R is not central symmetric. Let R be a ring with an ideal I. Then I is said to be prime if aRb ⊆ I implies a or b is in I, while I is called completely prime if ab ∈ I implies that a or b is in I. Completely prime ideals are prime ideals, but the converse is not true. For example, for any positive integer n, the zero ideal in the ring of all n× n matrices over a field is a prime ideal, but it is not completely prime. For a ring R and an ideal I, we show that if R/I is a central symmetric ring with a completely prime reduced ideal I, then R is a symmetric ring. Lemma 2.14. Let R be a ring. If R/I is a central symmetric ring with a completely prime reduced ideal I, then R is symmetric and so central symmetric. Proof. Let a, b, c ∈ R with abc = 0 and ā will be the image of a under natural epimorphism from R onto R/I. Then āb̄c̄ = 0̄. Since R/I is central symmetric and āb̄c̄r̄ = 0̄ for any r ∈ R, b̄āc̄r̄ is central in R/I () also, r̄āb̄c̄ = 0̄ for any r ∈ R implies b̄r̄āc̄ is central in R/I () By using () and (*) we prove (b̄āc̄)4 = 0̄. By (), we have (b̄āc̄)4 = (b̄āc̄b̄)āc̄b̄āc̄b̄āc̄ = āc̄b̄ā(b̄āc̄b̄)c̄b̄āc̄. Hence we have āc̄b̄ā(b̄āc̄b̄)c̄b̄āc̄ = āc̄(b̄āb̄āc̄)b̄c̄b̄āc̄ = āc̄b̄c̄b̄ā(b̄āb̄āc̄)c̄. Thus āc̄b̄c̄b̄ā(b̄āb̄āc̄)c̄ = āc̄(b̄c̄b̄āb̄āb̄āc̄)c̄ = ā(b̄c̄b̄āb̄āb̄āc̄)c̄c̄ = 0̄. It follows (bac)4 ∈ I and bac ∈ I and so one of a, b and c belongs to I since I is completely prime. So cab ∈ I and (cab)2 = 0 implies cab = 0 since I is reduced. Similarly, (bsca)2 = 0 implies bsca = 0, and (arbsc)2 = 0 implies arbsc = 0, and (cuarbs)2 = 0 implies cuarbs = 0. Hence bscuar = 0 for all r, s, u ∈ R. This implies (bac)2 = bacbac = 0. Thus bac = 0. This completes the proof. Symmetric rings are generalized by Ouyang and Chen to weak symmet- ric rings in [19]. A ring R is said to be weak symmetric, if for all a, b, c ∈ R, if abc is nilpotent, then acb is nilpotent. We now give an example to show that there exists a weak symmetric ring which is not a central symmetric ring. Example 2.15. Consider the ring R =   Z Z Z 0 Z Z 0 0 Z  , and the elements G. Kafkas, B. Ungor, S. Halicioglu, A. Harmanci 79 A =   0 1 2 0 0 0 0 0 0  , B =   1 1 1 0 0 −2 0 0 1  , C =   1 1 1 0 1 1 0 0 1   of R. Then ABC = 0. But BAC =   0 1 3 0 0 0 0 0 0   is not central in R. Hence R is not central symmetric. However R is weak symmetric by [19, Proposition 2.3]. A module M has the summand intersection property if the intersection of two direct summands is again a direct summand of M . A ring R is said to have the summand intersection property if the right R-module R has the summand intersection property. A module M has the summand sum property if the sum of two direct summands is a direct summand of M and a ring R is said to have the summand sum property if the right R-module R has the summand sum property. A ring R is said to be abelian if every idempotent is central. In this case we have the following. Proposition 2.16. Let R be a central symmetric ring. Then we have the following. (1) R is an abelian ring. (2) R has the summand intersection property. (3) R has the summand sum property. Proof. (1) Let e be an idempotent of R and x ∈ R. Since e(xe− exe) = 0 and (ex−exe)e = 0, being R central symmetric, (xe−exe)e and e(ex−exe) are central. Then (xe − exe)e = e(xe − exe) = 0 and e(ex − exe) = (ex − exe)e = 0. Hence we have ex = xe for all x ∈ R. Therefore R is abelian. (2) Let e and f be idempotents of R. By (1), e and f are central, we have eR ∩ fR = efR = feR and (ef)2 = ef . This completes the proof. (3) Let eR and fR be right ideals of R with e2 = e, f2 = f ∈ R. Then e+ f − ef is an idempotent of R by (1). Since R is abelian, it is easy to check that eR+ fR = (e+ f − ef)R. So eR+ fR is a direct summand of R. The converse of Proposition 2.16 (1) does not hold in general, that is, every abelian ring need not be central symmetric, as the following example shows. Example 2.17. Consider the ring R = {[ a b c d ] \| a, b, d, c ∈ Z, a ≡ d(mod2), b ≡ c ≡ 0(mod2) } 80 Generalized symmetric rings with the usual matrix operations. Since 0 and the identity matrices are the only idempotents of R, R is an abelian ring. Let A = [ 2 4 0 2 ] , B = [ 0 −4 0 4 ] , C = [ 2 2 2 2 ] ∈ R. Then ABC = 0 but BAC is not central. The next result shows that the converse statement of Proposition 2.16 (1) holds for right principally projective rings. Proposition 2.18. Let R be a right principally projective ring. If R is abelian, then it is central symmetric. Proof. Let a, b, c ∈ R with abc = 0. By hypothesis, there exists an idem- potent e of R such that rR(a) = eR. Hence ae = 0 and bc = ebc. It follows that bca = ebca = bcae = 0. Thus R is symmetric and so central symmetric. Recall that a ring R is called directly finite whenever a, b ∈ R, ab = 1 implies ba = 1. Then we have the following. Corollary 2.19. Every central symmetric ring is directly finite. Proof. It is clear from Proposition 2.16 since every abelian ring is directly finite. Recall that a ring R is called semicommutative if for any a, b ∈ R, ab = 0 implies aRb = 0. A ring R is called central semicommutative [3], if for any a, b ∈ R, ab = 0 implies arb is a central element of R for each r ∈ R, while R is weakly semicommutative [17], if for any a, b ∈ R, ab = 0 implies arb is a nilpotent element for each r ∈ R. In the next proposition, we prove that all central symmetric rings are central semicommutative and weakly semicommutative. Proposition 2.20. Let R be a central symmetric ring. Then the followings hold. (1) R is central semicommutative. (2) R is weakly semicommutative. Proof. (1) Let a, b ∈ R with ab = 0. For any r ∈ R, rab = 0 implies arb is central in R. Hence R is central semicommutative. (2) Let a, b ∈ R with ab = 0. Since R is central symmetric, ba is central in R. Hence for each r ∈ R, (arb)2 = arbarb = ar2bab = 0. Therefore R is weakly semicommutative. G. Kafkas, B. Ungor, S. Halicioglu, A. Harmanci 81 It is well known that a ring is a domain if and only if it is prime and symmetric. In addition to this fact, we have the following proposition when we deal with central case. Proposition 2.21. Let R be a ring. Then R is a domain if and only if R is a prime and central symmetric ring. Proof. First assume R is a domain. It is clear that R is prime and symmet- ric and so central symmetric. Conversely, assume R is a prime and central symmetric ring. Let a, b ∈ R with ab = 0. Then rab = 0 and abr = 0 for all r ∈ R. Being R central symmetric arb and bar are contained in the center of R. Hence we have (arb)R(arb) = 0 for any r ∈ R. Since R is prime, arb = 0 for any r ∈ R and so aRb = 0. This implies that a = 0 or b = 0. Therefore R is a domain. Let P (R) denote the prime radical and N(R) the set of all nilpotent elements of the ring R. The ring R is called 2-primal if P (R) = N(R) (see namely [12] and [13]). In this direction we obtain the following result. Theorem 2.22. If R is a central symmetric ring, then it is 2-primal. The converse holds for semiprime rings. Proof. To complete the proof it is enough to show that N(R) ≤ P (R) since P (R) is a nil ideal. Let a ∈ N(R). We first assume that a2 = 0. By hypothesis ara is central for any r ∈ R. Commuting ara with sa for any s ∈ R we have arasa = 0. It follows that a ∈ P for any prime ideal P and so a ∈ P (R). Assume now a3 = 0. Then a2ra and ara2 are central. Commuting a2ra by sa for any s ∈ R we have a2rasa = 0. By hypothesis arasata is central for any t ∈ R. Again commute arasata with az for any z ∈ R and use the centrality of ara2 for all r ∈ R to obtain (az)(arasata) = (arasata)(az) = aras(ata2)z = (ata2)arasz = 0. Since z, t, r and s are arbitrary in R, a ∈ P (R). By induction on the index of nilpotency we may conclude that P (R) consists of all nilpotent elements of R. Hence R is 2-primal. Conversely, let R be a semiprime and 2-primal ring. Then R is symmetric and so central symmetric. Corollary 2.23. Let R be a central symmetric ring. Then the ring R/P (R) is central symmetric. Recall that a ring R is said to be von Neumann regular if for every a ∈ R there exists b ∈ R with a = aba. A ring R is called strongly regular if for any a ∈ R there exists b ∈ R such that a = a2b. Now we give some relations between symmetric, central symmetric, regular, strongly regular and abelian rings. Also the following theorem provides some conditions for the converses of Proposition 2.3 and Proposition 2.16(1). 82 Generalized symmetric rings Theorem 2.24. Let R be a ring. Then the following conditions are equivalent. (1) R is strongly regular. (2) R is von Neumann regular and symmetric. (3) R is von Neumann regular and central symmetric. (4) R is von Neumann regular and abelian. Proof. (1) ⇒ (2) The first assertion is clear. Let a, b, c ∈ R with abc = 0. Since (bca)2 = 0 and bac = (bac)2r for some r ∈ R, we have bca = 0. Then R is symmetric. (2) ⇒ (3) Obvious. (3) ⇒ (4) By Proposition 2.16. (4) ⇒ (1) Let a ∈ R. By hypothesis, there exists b ∈ R such that a = aba. Since ab is an idempotent, ab is central. Hence a = a2b and therefore R is strongly regular. Let S denote a multiplicatively closed subset of a ring R consisting of central regular elements. Let S−1R be the localization of R at S. Then we have the following. Proposition 2.25. A ring R is central symmetric if and only if S−1R is central symmetric. Proof. Let R be a central symmetric ring and a/s1, b/s2, c/s3 ∈ S−1R, where a, b, c ∈ R, s1, s2, s3 ∈ S with (a/s1)(b/s2)(c/s3) = 0. Since abc/s1s2s3 = 0, we have abc = 0. So bac is central in R, then (b/s2)(a/s1)(c/s3) is also central in S−1R. Therefore S−1R is central symmetric. Conversely, assume that S−1R is a central symmetric ring. Since R may be embedded in S−1R, the rest is clear. Corollary 2.26. Let R be a ring. Then R[x] is central symmetric if and only if R[x, x−1] is central symmetric. Proof. Consider the subset S = {1, x, x2, x3, . . . } of R[x] consisting of central regular elements. Then it follows from Proposition 2.25. Let R be a ring and f(x) = n ∑ i=0 aix i, g(x) = m ∑ j=0 bjx j ∈ R[x]. Rege and Chhawchharia [20] introduce the notion of an Armendariz ring, that is, a ring R is called Armendariz, f(x)g(x) = 0 implies aibj = 0 for all i and j. The name of the ring was given due to Armendariz [6] who proved that reduced rings satisfied this condition. The interest of this notion lies in its natural and useful role in understanding the re- lation between the annihilators of the ring R and the annihilators of G. Kafkas, B. Ungor, S. Halicioglu, A. Harmanci 83 the polynomial ring R[x]. So far, Armendariz rings are generalized in different ways. A ring R is called nil-Armendariz [5], if f(x)g(x) has nilpotent coefficients, then aibj is nilpotent for 0 ≤ i ≤ n, 0 ≤ j ≤ s. According to Harmanci et al. [1], a ring R is called central Armendariz, f(x)g(x) = 0 implies that aibj is a central element of R for all i and j. In [4, Theorem 5], Anderson and Camillo proved that for a ring R and n ≥ 2 a natural number, R[x]/(xn) is Armendariz if and only if R is reduced. For central symmetric rings, we obtain the following result. Theorem 2.27. Let R be a right principally projective ring and n ≥ 2 a natural number. Then R is central symmetric if and only if R[x]/(xn) is central Armendariz. Proof. Suppose R is a central symmetric ring. By Theorem 2.6(2), R is a reduced ring. From [4, Theorem 5], R[x]/(xn) is Armendariz and so central Armendariz. Conversely, assume that R[x]/(xn) is central Armendariz. By hypothesis and [1, Theorem 2.5], R[x]/(xn) is Armendariz. It follows from [4, Theorem 5] that R is reduced and so central symmetric. We wind up the paper with some observations. Theorem 2.28. If R is a central symmetric ring, then R is nil-Armendariz. Proof. If R is central symmetric, then it is 2-primal by Theorem 2.22 and so N(R) is an ideal of R. Proposition 2.1 in [5] states that a ring in which the set of all nilpotent elements forms an ideal is nil-Armendariz. Corollary 2.29. If R is a central symmetric ring, then R[x]/(xn) is nil-Armendariz, where n ≥ 2 is a natural number and (xn) is the ideal generated by xn. Proof. If R is central symmetric, then it is nil-Armendariz by Theorem 2.28. From [5, Proposition 4.1], R[x]/(xn) is nil-Armendariz. References [1] N. Agayev, G. Gungoroglu, A. Harmanci and S. Halicioglu, Central Armendariz Rings, Bull. Malays. Math. Sci. Soc. (2) 34(1)(2011), 137-145. [2] N. Agayev, A. Harmanci and S. Halicioglu, Extended Armendariz Rings, Algebras Groups Geom. 26(4)(2009), 343-354. [3] N. Agayev, T. Ozen and A. Harmanci, On a Class of Semicommutative Rings, Kyungpook Math. J. 51(2011), 283-291. [4] D. D. Anderson and V. Camillo, Armendariz rings and Gaussian rings, Comm. Algebra 26(7)(1998), 2265-2272. [5] R. Antoine, Nilpotent elements and Armendariz rings, J. 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Jacobson, The theory of rings, Amer. Math. Soc. Math. Surveys II, New York, 1943. [15] D. Khurana, G. Marks and A. Srivastava, On unit-central rings, Advances in ring theory, 205-212, Trends Math., Birkhauser-Springer Basel AG, Basel, 2010. [16] J. Lambek , On the representation of modules by sheaves of factor modules, Canad. Math. Bull. 14 (1971), pp. 359-368 [17] L. Liang and L. Wang and Z. Liu, On a generalization of semicommutative rings, Taiwanese J. Math. 11(5)(2007), 1359-1368. [18] G. Marks, Reversible and symmetric rings, Journal of Pure and Applied Algebra 174(3), 2002, 311-318. [19] L. Ouyang and H. Chen, On weak symmetric rings, Comm. Algebra 38(2)(2010), 697-713. [20] M. B. Rege and S. Chhawchharia, Armendariz rings, Proc. Japan Acad. Ser.A, Math. Sci. 73(1997), 14-17. [21] G. Shin, Prime ideals and Sheaf Represantations of a Pseudo Symmetric ring, Trans. Amer. Math. Soc. 184(1973), 43-69. Contact information Gizem Kafkas, Burcu Ungor, Sait Halicioglu Department of Mathematics, Ankara University, Turkey E-Mail: giz.maths@gmail.com, burcuungor@gmail.com, halici@ankara.edu.tr Abdullah Harmanci Department of Mathematics, Hacettepe Univer- sity, Turkey E-Mail: harmanci@hacettepe.edu.tr Received by the editors: 11.07.2011 and in final form 18.12.2011.

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