ISSN 2148-838Xhttp://dx.doi.org/10.13069/jacodesmath.284972 J. Algebra Comb. Discrete Appl. 4(2) • 219–225 Received: 14 October 2016 Accepted: 28 November 2016 Journal of Algebra Combinatorics Discrete Structures and Applications Reversible elements in rings∗ Research Article Dilshad Alghazzawi Abstract: We study some properties related to zero divisors and reversibility in noncommutative rings. 2010 MSC: 16U10, 16U99 Keywords: Reversible rings, Zero divisors, Right and left annihilators, Strongly π-regular rings 1. Introduction All our rings will have a unity. A ring R is reversible if, for any a,b ∈ R, ab = 0 if and only if ba = 0. These rings are natural generalizations of commutative rings. Reversible rings were studied, in particular, by P.M. Cohn [1], Gutan and Kisielewicz [3], Kim and Lee [4] and many others. Our aim, in this short note, is to introduce elementwise definitions that are directly connected to reversibility but can be applied in a more flexible way. For a ∈ R, we will write r(a) = {x ∈ R | ax = 0} and l(a) = {x ∈ R | xa = 0}. An element a ∈ R is right reversible if r(a) ⊆ l(a). The set of right reversible elements will be denoted by rRev(R). A ring R is semi-commutative if for any a,b ∈ R, ab = 0 implies that aRb = 0. In other words a ring R is semi-commutative when the annihilator of an element is a two sided-ideal. The subsets N(R) and U(R) will respectively stand for the set of nilpotent elements and invertible elements of the ring R. A ring R is 2-primal if the set N(R) coincides with the prime radical. Other notions will be defined when and where needed. The second section is devoted to the definition, characterizations and properties of the reversible set of a ring and its behavior relative to some ring constructions. Reversible nilpotent and idempotent elements are characterized and connections with 2-primal rings are established. In the third section we study some connections with other, more classical, notions such as strongly regular rings and McCoy rings. The paper ends with considerations related to elements having the property that r(a) 6= 0 implies l(a) 6= 0. ∗ Supported by a Ph.D. grant from King Abdulaziz University (Rabigh), Saudi Arabia. Dilshad Alghazzawi; Department of Mathematics, KAU University (Saudi Arabia) and Université d’Artois, France (email: Dalghazzawi@kau.edu.sa). 219 D. Alghazzawi / J. Algebra Comb. Discrete Appl. 4(2) (2017) 219–225 2. Reversible set of a ring Definition 2.1. Let R be a ring with 1 ∈ R. For an element a in R we write rR(a) or just r(a) (resp. lR(a) or l(a)) the right (resp. left) annihilator of a. The element a is right (resp. left) reversible if r(a) ⊆ l(a) (resp. l(a) ⊆ r(a)). An element which is both left and right reversible is a reversible element. The set of right (resp. left) reversible elements of R will be denoted by rRev(R) (resp. lRev(R)). The set of reversible elements is denoted by Rev(R). Observe that for any ring R, Rev(R) = rRev(R)∩lRev(R). A ring is reversible (cf. [1]) if rRev(R) = R (equivalently lRev(R) = R). Example 2.2. 1. Observe that 0 ∈ rRev(R). Also if r(a) = 0 than a ∈ rRev(R). In particular, left invertible elements, right regular elements in a ring are right reversible. 2. If R is commutative or if R is domain then rRev(R) = lRev(R) = R. 3. Of course, if R is reduced then rRev(R) = R = lRev(R). This is easily checked as follows: for any a ∈ R, if ab = 0 then (ba)2 = 0 and hence ba = 0, showing that rRev(R) = R. 4. A ring R is semi-commutative if, for any a ∈ R, r(a) is a 2-sided ideal. In other words R is semi- commutative if for any elements a,b ∈ R we have ab = 0 implies that aRb = 0. In general, if an element a ∈ rRev(R) is such that r(a) ⊆ rRev(R), then r(a) is a 2-sided ideal of R. In particular, a reversible ring is always semi-commutative. 5. Let k be a field. The set of 2 × 2 lower triangular matrices over k will be denoted L2(k). We have rRev(L2(k)) = { ( α 0 β γ ) | αγ 6= 0}∪{ ( 0 0 β γ ) | β ∈ k,γ ∈ k \ {0}}. Indeed, if αγ 6= 0 then the corresponding matrix is invertible and hence it belongs to rRev(R). It is easy to check that, for any α,β,γ ∈ k \{0} and any δ ∈ k, we have( 0 0 β 0 ) /∈ rRev(R), ( 0 0 δ γ ) ∈ rRev(R), ( α 0 δ 0 ) /∈ rRev(R). 6. Let R be any ring. Observe that for ( 1 0 0 0 ) and ( 0 0 1 0 ) in M2(R) we have ( 1 0 0 0 )( 0 0 1 0 ) = ( 0 0 0 0 ) and ( 0 0 1 0 )( 1 0 0 0 ) = ( 0 0 1 0 ) This shows that for any ring R, rRev(M2(R)) 6= M2(R). 7. It is easy to check that if two rings R,S are such that R ⊆ S then rRev(S) ∩R ⊆ rRev(R). The reverse inclusion generally does not hold. Indeed, let R be a domain and σ an endomorphism of R with nonzero kernel. Consider the Ore extension S = R[t;σ] with polynomials of the form ∑ Xiai and commutation rule aX = Xσ(a). If a ∈ ker(σ) then aX = Xσ(a) = 0 but Xa 6= 0 this shows that a ∈ rRev(R) but a /∈ rRev(S)∩R. The following proposition provides a characterization of right reversible elements. Proposition 2.3. Let R be a ring. For an element a ∈ R, the following are equivalent: (i) a ∈ rRev(R), (ii) r(a) ⊆ C(a), where C(a) = {x ∈ R | ax = xa} is the centralizer of a in R, (iii) The correspondence ϕ : aR → Ra defined by ϕ(ar) = ra is a well-defined additive map, 220 D. Alghazzawi / J. Algebra Comb. Discrete Appl. 4(2) (2017) 219–225 (iv) a ∈ r(r(a)), (v) For every b ∈ R we have that (ab)2 = ab implies that (ba)2 = ba. Proof. (i) ⇔ (ii): This is clear. (i) ⇒ (iii): Notice that if ar = ar′ then a(r−r′) = 0 and hence (r−r′)a = 0 so that ra = r′a. Now we have ϕ(ar) = ra = r′a = ϕ(ar′). This shows that ϕ is well defined. The fact that the map is additive is clear. The converse implication is obvious. (iii) ⇒ (iv): For any x ∈ r(a) we have ax = 0 and 0 = ϕ(ax) = xa. This shows that a ∈ r(r(a)). (iv) ⇒ (v) Suppose b ∈ R is such that (ab)2 = ab, this gives that bab − b ∈ r(a) hence by (iv) we (bab− b)a = 0, i.e., (ba)2 = ba. (v) ⇒ (i): Suppose ab = 0. Then (ab)2 = ab = 0 and the hypothesis shows that 0 = (ba)2 = ba and so ba = 0. Corollary 2.4. If a ∈ rRev(R) is such that (ab)2 = ab then (ab)R ∼= (ba)R and R(ba) ∼= R(ab) and the idempotents ab and ba are isomorphic. Proof. This is a direct consequence of Proposition 2.3 above and Proposition 21.20 in [5]. Let us now show how the reversible notion behaves. Theorem 2.5. Let R be any ring. The following hold: (a) The set rRev(R) (respectively lRev(R)) is closed under product (,i.e., if a,b ∈ rRev(R), then ab ∈ rRev(R)). (b) If R and S are two rings and ϕ : R −→ S is an isomorphism of rings then ϕ(rRev(R)) = rRev(S). In particular, if u ∈ U(R) is a unit in R, then a ∈ rRev(R) if and only if uau−1 ∈ rRev(R). (c) If a ∈ rRev(R) then, for invertible elements u,v ∈ U(R), we have uav ∈ rRev(R). (d) If R is a prime ring we have {a ∈ R | r(a) = 0} = rRev(R). (e) If a is right invertible then a is right reversible if and only if a is left invertible (and hence invertible). (f) If R and S are two rings then rRev(R×S) = rRev(R)×rRev(S). (g) If R is a semisimple ring then rRev(R) = U(R), the set of invertible elements of R. (h) An idempotent e ∈ R is right reversible if and only if (e−1)Re = 0. (i) Let {x1, . . . ,xn} be a subset of rRev(R). If b ∈ r(x1x2 · · ·xn), then RbRx1Rx2R...RxnR = 0. (j) If a ∈ R is right reversible and nilpotent, then the ideal RaR is nilpotent. (k) If a ∈ R is such that the descending chain of left ideals Rai stabilizes then l(a) = 0 if and only if a ∈ U(R). Proof. (a) Let a,b ∈ rRev(R) and let c ∈ r(ab). Then we have abc = 0 and since a ∈ rRev(R), this gives bca = 0. Since b ∈ rRev(R), we get cab = 0, and hence c ∈ l(ab). (b) and (c) These are left to the reader. (d) If R is prime and 0 6= a ∈ rRev(R) we have, for any b ∈ r(a) and any r ∈ R, we have abr = 0 and hence bra = 0. This gives bRa = 0 and the primeness of R leads to b = 0, as required. (e) If a is right invertible then there exists b ∈ R such that ab = 1 and hence a(ba−1) = 0. Since a is also right reversible we get that (ba− 1)a = 0. This gives ba2 = a and, right multiplying by b, we get ba = 1. The converse is clear. 221 D. Alghazzawi / J. Algebra Comb. Discrete Appl. 4(2) (2017) 219–225 (f) This is obvious. (g) In the light of Wedderburn-Artin Theorem and Part (f) of this theorem, we may assume that R is a matrix ring over a division ring. In particular, R is a prime ring and so the point (d) shows that the right reversible elements are nonzero divisors. Now, a nonzero divisor matrix with coefficients in a division ring must be invertible, this easily yields the statement. (h) This is clear. (i) We, then, have x1x2 · · ·xnbR = 0, and since x1 ∈ rRev(R), we get x2x3 · · ·xnbRx1 = 0. So x2x3 · · ·xnbRx1R = 0. But x2 ∈ rRev(R), hence x3x4 · · ·xnbRx1Rx2 = 0. Continuing this process we get the desired result. (j) Let us suppose that a ∈ rRev(R) is such that an = 0, for some n ∈ N . We then have a ∈ r(an−1) and the above statement (i) yields the result. (k) There exists n ∈ N and x ∈ R such that an = xan+1. Since l(a) = 0, this leads to 1 = xa and hence to a = axa and also 1 = ax, showing that a ∈ U(R). Let us recall that N(R) = {x ∈ R | ∃n ∈ N : xn = 0}. Corollary 2.6. (1) For any ring R, rRev(R)∩N(R) is contained in the prime radical of R. (2) If all nilpotent elements of a ring are right reversible then the ring is 2-primal. (3) In a semiprime ring a nilpotent element cannot be right or left reversible. (4) If a ∈ N(R)∩rRev(R) and b ∈ N(R) then a + b ∈ N(R). (5) If a ∈ rRev(R), Rr(a) is a proper ( ,i.e., different from R ) two-sided ideal. Proof. 1) This is clear from Theorem 2.5 (i). 2) This is an obvious consequence of Corollary 2.6 (1). 3) It is enough to use the fact that a semiprime ring does not have nonzero nilpotent two sided ideal 4) Let l ∈ N, be such that bl = 0. When we develop (a + b)l all monomials will be in the prime radical. 5) This is due to the fact that Rr(a) is contained in l(a). We observe that the ring R of upper triangular 2 × 2 matrices over a field is 2-primal but not reversible. Example 2.7. (a) Observe that rRev(R) is in general not closed under addition. To give a concrete example let us consider the ring of 2 × 2 lower matrices over a field k. It is easy to check that the matrix( 1 0 1 0 ) = ( 1 0 0 1 ) + ( 0 0 1 −1 ) is not right reversible although the two matrices on the right hand side are, indeed, right reversible. (b) In connection with Theorem 2.5 (j) we remark that we might have RaR nilpotent even if a /∈ rRev(R). This is the case of the element a = e12 of the strictly upper 2×2 matrices over a field. 3. Connections with other notions Proposition 3.1. Let R be a semiprime ring and a ∈ rRev(R). Then: 222 D. Alghazzawi / J. Algebra Comb. Discrete Appl. 4(2) (2017) 219–225 (a) The right annihilator r(a) is a two-sided ideal of R. (b) for any n ∈ N, we have r(a) = r(an). Proof. (a) If b ∈ r(a) we have bRa = 0 and hence (RaRb)2 = 0. Since R is semiprime this leads to RaRb = 0, which, in turn, implies that aRb = 0. (b)For n ∈ N, n ≥ 1, and a ∈ rRev(R),b ∈ r(an) we have an−1(an−1b) = 0. Since an−1 is also in rRev(R) we have that an−1bRan−1b = 0. The fact that R is semiprime leads to an−1b = 0,i.e., b ∈ r(an−1). The same method leads to b ∈ r(an−2) and the desired result follows by iterations. We remark that the above statement admits a partial converse: if a ring R is such that for any a ∈ R, there exists l > 1 such that r(a) = r(al) then R is reduced and hence semiprime. We also recall that a strongly regular ring is a ring R such that for every a ∈ R there exists x ∈ R such that a = a2x. The following proposition is based on Exercise 12. 6A in Lam’s book [6]. Proposition 3.2. The following are equivalent: (i) The ring R is strongly regular, (ii) The ring R is regular and reduced, (iii) The ring R is regular and reversible. Let us now give some applications to McCoy condition on polynomials. Let us first define Rev(R) = rRev(R)∩ lRev(R) and say that a polynomal f(x) ∈ R[x] is right McCoy if rR[x](f(x)) 6= 0 implies that there exists a nonzero c ∈ R such that f(x)c = 0. We denote the set of right McCoy polynomials by rMC(R[x]). Proposition 3.3. For any ring R we have Rev(R)[x] ⊆ rMC(R[x]). Proof. Let f(x) = ∑n i=0 aix i ∈ Rev(R)[x]. If rR[x](f(x)) = 0 then clearly f(x) ∈ rMC(R(x)). So let us suppose that 0 6= g(x) = ∑m j=0 bjx j ∈ R[x] is of minimal degree such that f(x)g(x) = 0. We, then have anbm = 0 and since an ∈ Rev(R), we get bman = 0, this leads to deg(g(x)an) < deg(g(x)) and since f(x)g(x)an = 0, the minimality of deg(g(x)) shows that we have g(x)an = 0 and hence also ang(x) = 0. We now have an−1bm = 0 which leads to bman−1 = 0 and hence deg(g(x)an−1) < deg(g(x)). Since we have f(x)g(x)an−1 = 0 the minimality of deg(g(x)) implies that g(x)an−1 = 0. Since an−1 ∈ Rev(R) we thus conclude that an−1g(x) = 0. Continuing this process we will finally obtain that for all i ∈{0, . . . ,n}, aig(x) = 0. In particular, we obtain f(x)bm = 0, as desired. We also have the following properties also connected with the McCoy condition. Proposition 3.4. Let f(x) = ∑n i=0 aix i,g(x) = ∑m j=0 bjx j ∈ R[x] be such that f(x)g(x) = 0. Then: (a) If a0 ∈ rRev(R) then g(x)am+10 = 0. In particular if a m+1 0 6= 0, then r(g(x))∩R 6= 0. (b) If b0 ∈ lRev(R), then bn+10 f(x) = 0. In particular if b n+1 0 6= 0, then l(f(x))∩R 6= 0. Proof. (a) With the notation as in the statement of the theorem, it is enough to prove that, for any 0 ≤ i ≤ m, biai+10 = 0. If n = deg(f(x)) < deg(g(x)) = m, we put al = 0 for any n < l ≤ m. With this notation the equality f(x)g(x) = 0 gives, for any 0 ≤ k ≤ m, a0bk +a1bk−1 +· · ·+akb0 = 0. In particular, a0b0 = 0. Since a0 ∈ rRev(R) we also have b0a0 = 0. This shows that the required equality mentioned above is valid for i = 0. Let l < m and assume we have proved that bia i+1 0 = 0 for any 0 ≤ i ≤ l < m. Multiplying the equation a0bl+1 +a1bl + · · ·+al+1b0 = 0 on the right by al+10 we then get a0bl+1a l+1 0 = 0 and hence, since a0 ∈ rRev(R), bl+1al+20 = 0. This yields the required equalities. (b) The second part of the theorem is proved similarly. 223 D. Alghazzawi / J. Algebra Comb. Discrete Appl. 4(2) (2017) 219–225 We now consider relations between reversible elements and other kind of classical elements. Let us recall that an element a ∈ R is strongly π-regular if there exists n ∈ N such that an ∈ Ran+1 ∩ an+1R. This is equivalent to asking that both chains aR ⊃ a2R ⊃ ... ⊃ anR ⊃ . . . and Ra ⊃ Ra2 ⊃ ... ⊃ Ran ⊃ . . . stabilize. The set of strongly π-regular elements is denoted by sregπ(R). R is a π regular ring if sregπ(R) = R. Let us recall that Dischinger [2] showed that a ring R is strongly π-regular if and only if any descending chain condition Ra ⊃ Ra2 ⊃ ... stabilizes, i.e., only one of the above chain conditions is required for a ring to be strongly π-regular. Let us mention that using the above, we can show that a right (resp. left) artinian ring is such that every element a ∈ R with r(a) = 0 (resp. l(a) = 0) must be invertible. In particular any left or right artinian rings is strongly π-regular. This short discussion leads quickly to the following classical result. Proposition 3.5. If R is π-strongly regular then every left or right nonzero divisor is invertible. Prompted by this proposition, we introduce another elementwize condition. This concept is more general than the right reversible one. For this we define the following two sets: Sr(R) = {a ∈ R | r(a) 6= 0 , if l(a) 6= 0} Sl(R) = {a ∈ R | l(a) 6= 0 , if r(a) 6= 0}. We say that the ring R satisfies the R (resp. L) property if Sr(R) = R (resp.Sl(R) = R). Corollary 3.6. Let R be any ring. Both rRev(R) and sregπ(R) are contained in Sr(R). Example 3.7. Consider the upper triangular matrix ring of the form( Z Z/2Z 0 Z ) It is easy to check that the element a = ( 2 1 0 1 ) is such that r(a) 6= 0 but l(a) = 0. We have seen that it was not possible to pass the right reversible property from a ring to the matrix ring. In the next proposition we show that in some cases the property Sr(R) = R goes up to the matrix ring Mn(R). Proposition 3.8. (a) Let a,u,v ∈ R such that u,v are invertible. Then a ∈ Sr(R) if and only if uav ∈ Sr(R). A similar result is true for Sl(R). (b) Let R be such that Sr(R) = R and suppose that every square matrix A ∈ Mn(R) is diagonalizable. Then Sr(Mn(R)) = Mn(R). (c) Let R be a ring with a total left ring of quotient S. If Sr(S) = S then Sr(R) = R. (d) Let R ⊆ S be rings such that RR is essential in RS. If Sr(S) = S then Sr(R) = R. (e) Let a ∈ R be a unit regular element (i.e., there exists an invertible element u ∈ U(R) such that a = aua). Then a ∈ Sr(R)∩Sl(R). Proof. (a) It easy to check that r(uav) = v−1r(a) and l(uav) = u−1l(a). So if we assume that r(a) 6= 0 implies l(a) 6= 0, then r(uav) 6= 0 implies l(uav) 6= 0. (b) By Part (a) above, it is enough to show that a diagonal matrix A is such that r(A) 6= 0 also satisfies l(A) 6= 0. This is easy and left to the reader. (c) This is easy as follows: let a ∈ R be such that rR(a) 6= 0. Hence rS(a) 6= 0 and since Sr(S) = S, we have that lS(a) 6= 0, so there exists elements x,y ∈ R with lS(x) = 0 and s = x−1y ∈ lS(a). We then get that 0 6= y ∈ lR(a). (d) Suppose a ∈ R is such that rR(a) 6= 0 then rS(a) 6= 0. The fact that a ∈ Sr(S) implies that lS(a) 6= 0 and since R is essential in S, we obtain that lR(a) 6= 0. 224 D. Alghazzawi / J. Algebra Comb. Discrete Appl. 4(2) (2017) 219–225 (e) Suppose that a ∈ R is unit regular, i.e., there exists an invertible element u such that a = aua. Suppose that l(a) = 0, then since (1 − au)a = 0, we have that 1 = au, and if b ∈ r(a), we get that u−1b = auu−1b = ab = 0, this gives that b = 0. So that a ∈ Sr(R). The fact that a ∈ Sl(R) is obtained similarly. Acknowledgment: This work will be part of my Ph.D. thesis under the supervision of Prof. A. Leroy whom I thank for his continuous guidance and help. I would like also to thank King Abdulaziz University in Saudi Arabia for the financial supports received during the preparation of this work. References [1] P. M. Cohn, Reversible rings, Bull. London Math. Soc. 31(6) (1999) 641–648. [2] F. Dischinger, Sur les anneaux fortement π-réguliers, C. R. Acad. Sci. Paris Sér. A–B 283(8) (1976) Aii A571–A573. [3] M. Gutan, A. Kisielewicz, Reversible group rings, J. Algebra 279 (2004) 280–271. [4] N. K. Kim, Y. Lee, Extensions of reversible rings, J. Pure Appl. Algebra, 185(1–3) (2003) 207–223. [5] T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics, Springer Verlag, New York, Berlin, Heidelberg, 1990. [6] T. Y. Lam, Exercises in Classical Ring Theory, Problem Books in Mathematics, Springer Verlag, New York, Berlin, Heidelberg, 1994. 225 http://dx.doi.org/10.1112/S0024609399006116 http://www.ams.org/mathscinet-getitem?mr=422330 http://www.ams.org/mathscinet-getitem?mr=422330 http://dx.doi.org/10.1016/j.jalgebra.2004.02.011 http://dx.doi.org/10.1016/S0022-4049(03)00109-9 Introduction Reversible set of a ring Connections with other notions References