Let R be a prime ring of characteristic not 2 and U be a noncentral square closed Lie ideal of R. An additive mapping Hon R is called a homoderivation if H(xy) =H(x)H(y)+H(x)y+xH(y)for all x, y∈R. In this paper we investigate homoderivations satisfying certain differential identitieson square closed Lie ideals of prime rings

A. Sarikaya

O. Gölbasi

Extracta Mathematicae

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 EXTRACTAMATHEMATICAEVolumen 33, Número 2, 2018instituto de investigación de matemáticas de launiversidad de extremaduraEXTRACTA MATHEMATICAEVol. 38, Num. 2 (2023), 125 – 137doi:10.17398/2605-5686.38.2.125Available online December 8, 2023Results on Lie ideals of prime ringswith homoderivationsA. Sarikaya , Ö. Gölbaşi @Department of Mathematics, Faculty of Science, Sivas Cumhuriyet UniversitySivas, Turkeyeraysen345@gmail.com , ogolbasi@cumhuriyet.edu.trReceived May 26, 2023 Presented by C. Mart́ınezAccepted November 14, 2023Abstract : Let R be a prime ring of characteristic not 2 and U be a noncentral square closed Lie idealof R. An additive mapping H on R is called a homoderivation if H(xy) = H(x)H(y)+H (x) y+xH(y)for all x, y ∈ R. In this paper we investigate homoderivations satisfying certain differential identitieson square closed Lie ideals of prime rings.Key words: prime ring, Lie ideal, homoderivation, commutativity.MSC (2020): 16W25, 16N60.1. IntroductionLet R be an associative ring with center Z. For any x, y ∈ R the symbol[x, y] represents the Lie commutator xy − yx and the Jordan product xoy =xy+yx. An additive subgroup U of R is said to be a Lie ideal of R if [u, r] ∈ U ,for all u ∈ U , r ∈ R and U is called a square closed Lie ideal of R if U is aLie ideal and u2 ∈ U for all u ∈ U . If u2 ∈ U for all u ∈ U , then we have(u + v)2 ∈ U and so (u + v)2 − u2 − v2 = uv + vu ∈ U , for all u, v ∈ U . Alsouv − vu ∈ U , for all u, v ∈ U . Hence we find 2uv ∈ U , for all u, v ∈ U .Recall that a ring R is prime if for x, y ∈ R, xRy = 0 implies eitherx = 0 or y = 0. An additive mapping d : R → R is called a derivation ifd(xy) = d(x)y + xd(y) holds for all x, y ∈ R. E.C. Posner studied conditionsabout derivations of a ring that imply commutativity of the ring in [8]. Overthe last several years, a number of authors studied commutativity theorems forprime rings admitting automorphisms or derivations on appropriate subsetsof R.In 2000, El Sofy [5] defined a homoderivation on R as an additive mappingH : R → R satisfying H(xy) = H(x)H(y) + H(x)y + xH(y) for all x, y ∈ R.An example of such mapping is to let H(x) = f(x)− x, for all x, y ∈ R where@ Corresponding authorISSN: 0213-8743 (print), 2605-5686 (online)c©The author(s) - Released under a Creative Commons Attribution License (CC BY-NC 3.0)126 a. sarikaya, ö. gölbaşif is an endomorphism on R. It is clear that a homoderivation H is also aderivation if H(x)H(y) = 0 for all x, y ∈ R. In this case, H(x)RH(y) = 0for all x, y ∈ R. Hence if R is a prime ring, then the only additive mappingwhich is both a derivation and a homoderivation is the zero mapping.If S ⊆ R, then a mapping f : R→ R preserves S if f(S) ⊆ S. A mappingf : R → R is zero-power valued on S if f preserves S and if for each x ∈ S,there exists a positive integer n(x) > 1 such that fn(x)(x) = 0.In [3], Daif and Bell proved that if R is semiprime ring, U is a nonzeroideal of R and d is a derivation of R such that d([x, y]) = ±[x, y], for allx, y ∈ U , then R contains a nonzero central ideal. There is also a growingliterature on strong commutativity preserving (SCP) maps and derivations ofoperator algebras, prime rings and semiprime rings. In [7], this condition wasexamined by Melaibari et al. for homoderivations. On the other hand, Ashrafand Rehman showed that if R is a prime ring with a nonzero ideal U of Rand d is a derivation of R such that d(xy) ± xy ∈ Z, d(xy) ± yx ∈ Z, for allx, y ∈ U , then R is commutative in [2]. Alharfie and Muthana proved similarresults regarding homoderivations in [1].The purpose of this paper is to study homoderivations satisfying certaindifferential identities on Lie ideals of prime rings. We will give new results inthis field and generalize some theorems found in the literature.2. ResultsThroughout this paper, R will be a prime ring of characteristic not 2, U asquare closed Lie ideal of R and H a homoderivation of R. For any x, y ∈ R,we make extensive use of basic commutator identities:[x, yz] = y[x, z] + [x, y]z,[xy, z] = [x, z]y + x[y, z],xo(yz) = (xoy)z − y[x, z] = y(xoz) + [x, y]z,(xy)oz = x(yoz)− [x, z]y = (xoz)y + x[y, z].Remark 1. For all x, y ∈ R, we getH([x, y]) = H(xy − yx) = H(xy)−H(yx)= H(x)H(y) + H(x)y + xH(y)−H(y)H(x)−H(y)x− yH(x)= [H(x), H(y)] + [H(x), y] + [x,H(y)].results on lie ideals of prime rings 127Lemma 1. ([4, Lemma 2]) Let R be a prime ring of characteristic not 2and U be a noncentral Lie ideal of R. Then CR(U) = Z.Lemma 2. ([4, Lemma 4]) Let R be a prime ring of characteristic not2 and U be a noncentral Lie ideal of R, a, b ∈ R. If aUb = (0) thena = 0 or b = 0.Lemma 3. Let R be a prime ring of characteristic not 2 and U be anoncentral Lie ideal of R, a ∈ R. If Ua = (0) (aU = (0)), then a = 0.Proof. Take an arbitrary nonzero b in R. Ua = (0) implies bUa = (0).Then a = 0 immediatly follows from Lemma 2.Lemma 4. ([6, Lemma 9]) Let R be a prime ring of characteristic not 2and U be a noncentral Lie ideal of R, a ∈ R. If aoU = (0), then a ∈ Z.Lemma 5. ([6, Theorem 2]) Let R be a prime ring of characteristic not2, U a noncentral Lie ideal of R and H a homoderivation of R. If H(U) ⊆ Z,then H = 0.Theorem 1. Let R be a prime ring of characteristic not 2, U a noncentralsquare closed Lie ideal of R and H a homoderivation which is zero-powervalued on U . If [H(u), H(v)] = [u, v] for all u, v ∈ U , then H = 0.Proof. By the hypothesis, we get[H(u), H(v)] = [u, v] for all u, v ∈ U.Replacing v by 2vw in the last equation, we have2[H(u), H(v)]H(w) + 2H(v)[H(u), H(w)]+ 2[H(u), H(v)]w + 2H(v)[H(u), w]+ 2[H(u), v]H(w) + 2v[H(u), H(w)] = 2[u, v]w + 2v[u,w].Using the hypothesis and charR 6= 2, we obtain that[u, v]H(w) + H(v)[u,w] + H(v)[H(u), w] + [H(u), v]H(w) = 0.That is,[u + H(u), v]H(w) + H(v)[H(u) + u,w] = 0.128 a. sarikaya, ö. gölbaşiSince H is zero-power valued on U, there exists an integer n = n(u) > 1 suchthat Hn(u) = 0. Replacing u by u −H(u) + H2(u) + · · · + (−1)n−1Hn−1(u)in this equation, we find[u, v]H(w) + H(v)[u,w] = 0.Replacing w by u in the last equation, we get[u, v]H(u) = 0 for all u, v ∈ U.Taking 2vw instead of v in the last equation and using this, we have2[u, v]wH(u) = 0 for all u, v ∈ U.Since charR 6= 2, we get[u, v]UH(u) = (0) for all u, v ∈ U.Now Lemma 2 implies that either H(u) = 0 or u belongs to the centralizer ofU in R for all u ∈ U . Hence we have H(u) = 0 or u ∈ Z for any u ∈ U byLemma 1.Let L = {u ∈ U : u ∈ Z } and K = {u ∈ U : H(u) = 0}. Clearly each ofL and K is additive subgroup of U such that U = L∪K. But a group can notbe union of two proper subgroups. Hence L = U or K = U . In the first case,U ⊆ Z, which forces our hypothesis. In the last case, we get H(U) = (0), andso H(U) ⊆ Z. By Lemma 5, we obtain the required result.Theorem 2. Let R be a prime ring of characteristic not 2, U a noncentralsquare closed Lie ideal of R and H a homoderivation which is zero-powervalued on U . If one of the following conditions is satisfied then H = 0.(i) H (u)H(v) = [u, v], for all u, v ∈ U .(ii) H (u)H(v) = u ◦ v, for all u, v ∈ U .Proof. (i) By the hypothesis, we getH(u)H(v) = [u, v] for all u, v ∈ U.Replacing u by 2uw in this equation, we have2H(u)H(w)H(v) + 2H(u)wH(v) + 2uH(w)H(v) = 2 [u, v]w + 2u[w, v].results on lie ideals of prime rings 129Using the hypothesis, we see that2[u,w]H(v) + 2H(u)wH(v) = 2 [u, v]w.Taking v = u in the above equation and using charR 6= 2, we find that[u,w]H(u) + H(u)wH(u) = 0.By the hypothesis, we have(2.1) H(u)H(w)H(u) + H(u)wH(u) = 0.That is,H(u)(H(w) + w)H(u) = 0.Since H is zero-power valued on U , there exists an integer n = n(w) > 1 suchthat Hn(w) = 0. Replacing w by w−H(w) +H2(w) + · · ·+ (−1)n−1Hn−1(w)in this equation, we getH(u)wH(u) = 0.That is, H(u)UH(u) = (0), for all u ∈ U . By Lemma 2, we get H(u) = 0, forall u ∈ U , and so we conclude that H = 0 by Lemma 5.(ii) We getH(u)H(v) = u ◦ v for all u, v ∈ U.Replacing u by 2uw,w ∈ U in this equation, we get2H(u)H(w)H(v) + 2H(u)wH(v) + 2uH(w)H(v) = 2u(w ◦ v)− 2[u, v]w.Using the hypothesis, we obtain that2H(u)H(w)H(v) + 2H(u)wH(v) = −2[u, v]w.Replacing v by u in this equation and using charR 6= 2, we arrive atH(u)H(w)H(u) + H(u)wH(u) = 0.The rest of the proof is the same as equation (2.1). This completes theproof.Theorem 3. Let R be a prime ring of characteristic not 2, U a noncentralsquare closed Lie ideal of R and H a homoderivation which is zero-powervalued on U . If uH(v)± uv ∈ Z, for all u, v ∈ U , then H = 0.130 a. sarikaya, ö. gölbaşiProof. By the hypothesis, we getuH(v) + uv ∈ Z, for all u, v ∈ U.Replacing u by 2vu in this equation, we have2v(uH(v) + uv) ∈ Z, for all u, v ∈ U.Since R is prime ring, uH(v) + uv ∈ Z and charR 6= 2, we find thatv ∈ Z or uH(v) + uv = 0.Let L = {v ∈ U : v ∈ Z } and K = {v ∈ U : uH(v) + uv = 0, for all u ∈ U}.Clearly each of L and K is additive subgroup of U such that U = L∪K. Buta group can not be union of two proper subgroups. Hence L = U or K = U .In the first case, U ⊆ Z, which forces our hypothesis. So, we must haveuH(v) + uv = 0, for all u, v ∈ U.Henceu(H(v) + v)H(u) = 0, for all u, v ∈ U.Since H is zero-power valued on U , there exists an integer n = n(v) > 1 suchthat Hn(v) = 0. Replacing v by v−H(v) +H2(v) + · · ·+ (−1)n−1Hn−1(v) inthis equation, we getuvH(u) = 0, for all u, v ∈ U.Lemma 2 yields that u = 0 or H(u) = 0 for each u ∈ U . If u = 0, then wehave H(u) = 0. Hence we get H(u) = 0, for all v ∈ U for any cases. ByLemma 5, we obtain the required result.Using the same arguments, we obtain a similar result for the case uH(v)−uv ∈ Z, for all u, v ∈ U .Theorem 4. Let R be a prime ring of characteristic not 2, U a noncentralsquare closed Lie ideal of R and H a homoderivation which is zero-powervalued on U . If uH(v)± vu ∈ Z, for all u, v ∈ U , then H = 0.Proof. By the hypothesis, we getuH(v) + vu ∈ Z, for all u, v ∈ U.results on lie ideals of prime rings 131Replacing u by 2vu in this equation, we have2v(uH(v) + vu) ∈ Z, for all u, v ∈ U.Since R is prime ring, uH(v) + vu ∈ Z and charR 6= 2, we find thatv ∈ Z or uH(v) + vu = 0.If we apply Brauer’s Trick as we had done above, then we haveuH(v) + vu = 0, for all u, v ∈ U.Writing v by 2uv in this equation, we obtain that0 = 2uH(u)H(v) + 2uH(u)v + 2u2H(v) + 2uvu= 2uH(u)(H(v) + v) + 2u(uH(v) + vu)and souH(u)(H(v) + v) = 0, for all u, v ∈ U.Since H is zero-power valued on U , there exists an integer n = n(v) > 1 suchthat Hn(v) = 0. Replacing v by v−H(v) +H2(v) + · · ·+ (−1)n−1Hn−1(v) inthis equation, we getuH(u)v = 0, for all u, v ∈ U.Lemma 3 yields that uH(u) = 0 for all u ∈ U .By our hypothesis, we get uH(u) + u2 ∈ Z. Using uH(u) = 0, we getu2 ∈ Z, for all u ∈ U . Hence we have(u + v)2 = u2 + v2 + uv + vu ∈ Zand souv + vu ∈ Z, for all u, v ∈ U.Subtracting this equation from our hypothesis, we arrive atuH(v) + uv ∈ Z, for all u, v ∈ U.Hence we get H = 0 by Theorem 3.By the same tecniques, we get the required result for the case uH(v)−uv ∈Z, for all u, v ∈ U .132 a. sarikaya, ö. gölbaşiTheorem 5. Let R be a prime ring of characteristic not 2, U a noncentralsquare closed Lie ideal of R and H a homoderivation which is zero-powervalued on U . If uH(v)± [u, v] ∈ Z, for all u, v ∈ U , then H = 0.Proof. Let us assumeuH(v)± [u, v] ∈ Z, for all u, v ∈ U.Replacing u by 2vu in the above equation and using this, we arrive at2v(uH(v)± [u, v]) ∈ Z, for all u, v ∈ U.Since R is prime ring, charR 6= 2 and uH(v)± [u, v] ∈ Z, we getv ∈ Z or uH(v)± [u, v] = 0.Define L = {v ∈ U : v ∈ Z } and K = {v ∈ U : uH(v) ± [u, v] = 0, for allu ∈ U}. Applying Brauer’s Trick, we have L = U or K = U . In the first case,U ⊆ Z, which forces our hypothesis. So, we must haveuH(v)± [u, v] = 0, for all u, v ∈ U.Writing v by 2vu in this equation, we obtain that0 = 2uH(v)H(u) + 2uH(v)u + 2uvH(u)± 2[u, v]u= 2u(H(v) + v)H(u) + 2(uH(v)± [u, v])u.Using charR 6= 2 and uH(v)± [u, v] = 0, we haveu(H(v) + v)H(u) = 0, for all u, v ∈ U.Since H is zero-power valued on U , there exists an integer n = n(v) > 1 suchthat Hn(v) = 0. Replacing v by v−H(v) +H2(v) + · · ·+ (−1)n−1Hn−1(v) inthis equation, we getuvH(u) = 0, for all u, v ∈ U.Lemma 2 yields that u = 0 or H(u) = 0 for each u ∈ U . If u = 0, thenH(u) = 0. Hence we get H(u) = 0 for all u ∈ U . By Lemma 5, we obtain therequired result.Theorem 6. Let R be a prime ring of characteristic not 2, U a noncentralsquare closed Lie ideal of R and H a homoderivation which is zero-powervalued on U . If H(v)u± [u, v] ∈ Z, for all u, v ∈ U , then H = 0.results on lie ideals of prime rings 133Proof. By the hypothesis, we getH(v)u± [u, v] ∈ Z, for all u, v ∈ U.Replacing u by 2uv in the this equation and using this, we find that2(H(v)u± [u, v])v ∈ Z, for all u, v ∈ U.Since R is prime ring, charR 6= 2 and H(v)u± [u, v] ∈ Z, we havev ∈ Z or H(v)u± [u, v] = 0.Again using Brauer’s Trick as we had done above, we arrive atH(v)u± [u, v] = 0, for all u, v ∈ U.Substituing v by 2uv in this equation, we obtain that0 = 2H(u)H(v)u + 2H(u)vu + 2uH(v)u± u[u, v]= 2H(u)(H(v) + v)u + 2u(H(v)u± [u, v]).Since H(v)u± [u, v] = 0 and charR 6= 2, we have2H(u)(H(v)± v)u = 0, for all u, v ∈ U.Using H is zero-power valued on U , we get Hn(v) = 0, for an integer n =n(v) > 1. Replacing v by v − H(v) + H2(v) + · · · + (−1)n−1Hn−1(v) in thelast equation, we haveH(u)vu = 0, for all u, v ∈ U.Lemma 2 yields that u = 0 or H(u) = 0 for each u ∈ U . If u = 0, thenH(u) = 0. Hence we take H(u) = 0 for all u ∈ U . We get the required resultby Lemma 5.Theorem 7. Let R be a prime ring of characteristic not 2, U a noncentralsquare closed Lie ideal of R, H a homoderivation which is zero-power valuedon U and H(U) ⊆ U . If [H(u), v]± uv ∈ Z, for all u, v ∈ U , then H = 0.Proof. Let us assume[H(u), v]± uv ∈ Z, for all u, v ∈ U.134 a. sarikaya, ö. gölbaşiReplacing v by 2vH(u) in the above equation and using this, we get2([H(u), v]± uv)H(u) ∈ Z, for all u, v ∈ U.Since R is prime ring, charR 6= 2 and [H(u), v]± uv ∈ Z, we haveH(u) ∈ Z or [H(u), v]± uv = 0.Let L = {u ∈ U : H(u) ∈ Z } and K = {u ∈ U : [H(u), v] ± uv = 0, forall v ∈ U}. Clearly each of L and K is additive subgroup of U such thatU = L ∪K. But a group can not be union of two proper subgroups. HenceL = U or K = U .Now, we assume that K = U . Then we have[H(u), v]± uv = 0, for all u, v ∈ U.Substituing v by 2uv in this equation, we obtain that0 = 2[H(u), uv]± 2u2v= 2u([H(u), v]± uv) + 2[H(u), u]vand so2[H(u), u]v = 0, for all u, v ∈ U.Since charR 6= 2, we have[H(u), u]v = 0, for all u, v ∈ U.Lemma 3 yields that(2.2) [H(u), u] = 0, for all u, v ∈ U.On the other hand, we get [H(u), u] ± u2 ∈ Z, by our hypothesis. Using[H(u), u] = 0, for all u, v ∈ U , we have u2 ∈ Z, for all u ∈ U . Hence we get(u + v)2 = u2 + v2 + uv + vu ∈ Zand souv + vu ∈ Z, for all u, v ∈ U.Replacing u by 2uv in the above equation and using this, we have2(uv + vu)v ∈ Z, for all u, v ∈ U.results on lie ideals of prime rings 135By the primeness of R, charR 6= 2 and uv + vu ∈ Z, we find thatv ∈ Z or uv + vu = 0.Define A = {v ∈ U : v ∈ Z } and B = {v ∈ U : uv + vu = 0, for all v ∈ U}.Applying Brauer’s Trick, we have A = U or B = U . In the first case, U ⊆ Z,which forces our hypothesis. So, we must haveuv + vu = 0, for all u, v ∈ U.Hence we arrive at UoU = 0, and so U ⊆ Z by Lemma 4. This contradictsU * Z. So, we must have L = U , and so H(u) ∈ Z, for all u ∈ U . ByLemma 5, we get H = 0. This completes the proof.Theorem 8. Let R be a prime ring of characteristic not 2, U a noncentralsquare closed Lie ideal of R, H a homoderivation which is zero-power valuedon U and H(U) ⊆ U . If [H(u), v]± vu ∈ Z, for all u, v ∈ U , then H = 0.Proof. Let us assume[H(u), v]± vu ∈ Z, for all u, v ∈ U.Replacing v by 2H(u)v in the above equation and using this, we get2H(u)([H(u), v]± vu) ∈ Z, for all u, v ∈ U.Since R is prime ring, charR 6= 2 and [H(u), v]± uv ∈ Z, we haveH(u) ∈ Z or [H(u), v]± vu = 0.Define L = {u ∈ U : H(u) ∈ Z } and K = {u ∈ U : [H(u), v] ± vu = 0,for all v ∈ U}. Clearly each of L and K is additive subgroup of U such thatU = L ∪K. But a group can not be union of two proper subgroups. HenceL = U or K = U .Now, we assume that K = U . Then we have[H(u), v]± vu = 0, for all u, v ∈ U.Replacing v by 2uv in this equation, we obtain that0 = 2[H(u), uv]± 2u2v= 2u([H(u), v]± uv) + 2[H(u), u]v136 a. sarikaya, ö. gölbaşiand so[H(u), u]v = 0, for all u, v ∈ U.Lemma 3 yields that[H(u), u] = 0, for all u, v ∈ U.Using the same arguments after the equation (2.2) in the proof of Theorem 7,we get the required result.Example 1. SupposeR = a b 00 0 00 0 0 : a, b ∈ RandU = 0 u 00 0 00 0 0 : u ∈ RR is a prime ring of characteristic not 2 and U a noncentral Lie ideal of R.Let us defineH : R −→ U, H a b 00 0 00 0 0 = 0 a 00 0 00 0 0 .It is clear that H is a zero-power valued nonzero homoderivation on R. Somecommutativity conditions given above are satisfied. However, R is not com-mutative and U is noncentral.Conclusion. Let R be a ring. An additive mapping d : R→ R is called aderivation if d(xy) = d(x)y+xd(y) holds for all x, y ∈ R. The study of deriva-tions in prime rings was initiated by E.C. Posner in [8]. Over the last severalyears, a number of authors studied commutativity theorems for prime ringsadmitting automorphisms or derivations on appropriate subsets of R. In 2000,El Sofy [5] defined a homoderivation on R as an additive mapping H : R→ Rsatisfying H(xy) = H(x)H(y) + H(x)y + xH(y) for all x, y ∈ R. After thisdefinition several results appeared where the authors proved commutativityresults for the domain of these mappings. The goal of this paper is to provethe commutativity of Lie ideals of prime rings with homoderivations satisfyingsome algebraic conditions and collect the information about the commutativestructure of these rings.results on lie ideals of prime rings 137References[1] E.F. Alharfie, N.M. Muthana, The commutativity of prime rings withhomoderivations, International Journal of Advanced and Applied Sciences5 (5) (2018), 79 – 81.[2] M. Ashraf, N. Rehman, On derivations and commutativity in prime rings,East-West J. Math. 3 (1) (2001), 87 – 91.[3] H.E. Bell, M.N. Daif, On commutativity and strong commutativity-preserving maps, Canad. Math. Bull. 37 (1994), 443 – 447.[4] J. Bergen, I.N. Herstein, J.W. Kerr, Lie ideals and derivation of primerings, J. Algebra 71 (1981), 259 – 267.[5] M.M. El Sofy, “Rings with some kinds of mappings”, M.Sc. Thesis, CairoUniversity, Branch of Fayoum, 2000.[6] Ö. Gölbaşi, Homoderivations on Lie ideals of prime rings, to appear in Int.J. Math. Stat.[7] A. Melaibari, N. Muthana, A. Al-Kenani, Homoderivations on Rings,Gen. Math. Notes 35 (1) (2016), 1 – 8.[8] E.C. Posner, Derivations in prime rings, Proc. Amer. Math. Soc. 8 (1957),1093 – 1100.

Results on Lie ideals of prime ringswith homoderivations

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Results on Lie ideals of prime ringswith homoderivations

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