EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 12, No. 3, 2019, 1138-1148 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global A characterization of derivations in prime rings with involution Shakir Ali1,∗, M. Rahman Mozumder1, Adnan Abbasi1, M. Salahuddin Khan2 1 Department of Mathematics, Faculty of Science, Aligarh Muslim University, Aligarh-202002, India 2 Department of Applied Mathematics, Faculty of Engineering, Aligarh Muslim University, Aligarh-202002, India Abstract. The purpose of this paper is to investigate ∗-differential identities satisfied by pair of derivations on prime rings with involution. In particular, we prove that if a 2-torsion free noncommutative ring R admit nonzero derivations d1, d2 such that [d1(x), d2(x∗)] = 0 for all x ∈ R, then d1 = λd2, where λ ∈ C. Finally, we provide an example to show that the condition imposed in the hypothesis of our results are necessary. 2010 Mathematics Subject Classifications: 16W10, 16N60, 16W25 Key Words and Phrases: Prime ring, derivation, involution, ∗-differential identity 1. Introduction In all that follows, R will represent an associative ring with center Z(R). We de- note by Q and C the maximal ring of quotient and the extended centroid of a prime ring, respectively. For the explanation of Q and C we refer the reader to [4]. We denote [x, y] = xy − yx, the commutator of x and y and x ◦ y = xy + yx, the anti-commutator of x and y. A ring is said to 2-torsion free if 2x = 0 (where x ∈ R) implies x = 0. A ring R is said to be prime if aRb = (0) (where a, b ∈ R) implies either a = 0 or b = 0, and is called semiprime ring if aRa = (0) (where a ∈ R) implies a = 0. An additive mapping ∗ : R → R is called an involution if ∗ is an anti-automorphism of order 2; that is, (x∗)∗ = x for all x ∈ R. An element x in a ring with involution is said to be hermitian if x∗ = x and skew-hermitian if x∗ = −x. The sets of all hermitian and skew-hermitian elements of R will be denoted by H(R) and S(R), respectively. A ring equipped with an involution is known as ring with involution or ∗-ring. The involution is said to be of the first kind if Z(R) ⊆ H(R), otherwise it is said to be of the second kind. In the later case, ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v12i3.3496 Email addresses: shakir.ali.mm@amu.ac.in (S. Ali), muzibamu81@gmail.com (M. R. Mozumder), adnan.abbasi001@gmail.com (A. Abbasi), salahuddinkhan50@gmail.com (M. S. Khan) http://www.ejpam.com 1138 c© 2019 EJPAM All rights reserved. S. Ali et al. / Eur. J. Pure Appl. Math, 12 (3) (2019), 1138-1148 1139 S(R)∩Z(R) 6= (0). If R is 2-torsion free then every x ∈ R can be uniquely represented in the form 2x = h + k, where h ∈ H(R) and k ∈ S(R). Note that in this case x is normal i.e., xx∗ = x∗x, if and only if h and k commute. If all elements in R are normal, then R is called a normal ring. An example is the ring of quaternions. A description of such rings can be found in [14], where further references can be found. A derivation on R is an additive mapping d : R→ R such that d(xy) = d(x)y+xd(y) for all x, y ∈ R. A derivation d is said to be inner if there exists a ∈ R such that d(x) = ax−xa for all x ∈ R. Over the last 30 years, several authors have investigated the relationship between commutativity of the ring R and certain special types of maps on R. The first result in this direction is due to Divinsky [12], who proved that a simple artinian ring is commutative if it has a commuting non-trivial automorphism. Two years later, Posner [18] proved that the existence of a nonzero centralizing derivation on a prime ring forces the ring to be commutative. Over the last few decades, many authors have refined and extended these results in various directions (see for example [3, 5–7, 9] where further references can be looked). In [13], Herstein proved that if R is a prime ring of characteristic not two admitting a nonzero derivation d such that [d(x), d(y)] = 0 for all x, y ∈ R, then R is commutative. Further, Daif [10] showed that a 2-torsion free semiprime ring R admits a nonzero deriva- tion d such that [d(x), d(y)] = 0 for all x, y ∈ I, where I is a nonzero ideal of R, then R contains a nonzero central ideal. In [15], Lanski prove that if L is a noncommutative Lie ideal of a 2-torsion free prime ring R and d, h are nonzero derivations of R such that [d(x), h(x)] ∈ C for all x ∈ L, then h = λd, where λ ∈ C. Very recently, the first author together with Dar [11] proved the following result: Let R be a prime ring with involution ∗ of the second kind such that char(R) 6= 2. If R admits a nonzero derivation d such that [d(x), d(x∗)] = 0 for all x ∈ R, then R is commutative. In the last three decades many au- thors have generalized the above mention result in several ways (viz.; [1, 2, 8, 11, 15, 17, 19] where further references can be found). Motivated by the above results, here we continue this line of investigation by consid- ering more general situations. Besides proving some other results, the main result is the following theorem. Main Theorem. Let R be a 2-torsion free noncommutative prime ring with involution ∗ of the second kind and d1, d2 be two nonzero derivations on R such that [d1(x), d2(x ∗)] = 0 for all x ∈ R. Then d1 = λd2, where λ ∈ C. 2. Main results In order to prove our results, we need the following lemma. S. Ali et al. / Eur. J. Pure Appl. Math, 12 (3) (2019), 1138-1148 1140 Lemma 1. Let R be a 2-torsion free noncommutative prime ring with involution ∗ of the second kind and d1, d2 be two nonzero derivations on R. If one of the following conditions holds: (i) [d1(x), d2(y)] = x ◦ y for all x, y ∈ R, (ii) [d1(x), d2(y)] = −x ◦ y for all x, y ∈ R, then d1 = λd2, where λ ∈ C. Proof. (i) We consider the case [d1(x), d2(y)] = x ◦ y for all x, y ∈ R. (1) Substituting yh for y, where h ∈ H(R) ∩ Z(R), we get [d1(x), y]d2(h) = 0 for all x, y ∈ R. (2) Using the primeness of R we have either d2(h) = 0 for all h ∈ H(R)∩Z(R) or [d1(x), y] = 0 for all x, y ∈ R. If [d1(x), y] = 0 for all x, y ∈ R, then by Posner’s result [18] R is commutative, a contradiction. Therefore we are left with the case d2(h) = 0 for all h ∈ H(R) ∩ Z(R). Replacing y by yx in (1), we obtain d2(y)[d1(x), x] + [d1(x), d2(y)]x+ y[d1(x), d2(x)] + [d1(x), y]d2(x) = xyx+ yx2 (3) for all x, y ∈ R. Multiplying (1) by x from right side and subtracting it from (3), we arrive at d2(y)[d1(x), x] + y[d1(x), d2(x)] + [d1(x), y]d2(x) = 0 for all x, y ∈ R. Now taking h for y where h ∈ H(R) ∩ Z(R), we get h[d1(x), d2(x)] = 0 for all x ∈ R. Now using the primeness of R and the fact that S(R) ∩ Z(R) 6= (0), we finally arrive at [d1(x), d2(x)] = 0 for all x ∈ R. Thus in view of [15, Theorem 4] we get d1 = λd2, where λ ∈ C. (ii) Using a similar approach with necessary variations, we can prove that the same con- clusion holds for the case [d1(x), d2(y)] = −x ◦ y for all x, y ∈ R. Proof of Main Theorem. By the given assumption, we have [d1(x), d2(x ∗)] = 0 (4) for all x ∈ R. A linearization of (4) yields that [d1(x), d2(y ∗)] + [d1(y), d2(x ∗)] = 0 (5) for all x, y ∈ R. Replacing y by hy in (5), where y ∈ R and h ∈ H(R) ∩ Z(R), we get h([d1(x), d2(y ∗)] + [d1(y), d2(x ∗)]) + d2(h)[d1(x), y∗] + d1(h)[y, d2(x ∗)] = 0. S. Ali et al. / Eur. J. Pure Appl. Math, 12 (3) (2019), 1138-1148 1141 Using (5), we get d2(h)[d1(x), y∗] + d1(h)[y, d2(x ∗)] = 0 (6) for all x, y ∈ R and h ∈ H(R)∩Z(R). Substituting ky for y in (6), where k ∈ S(R)∩Z(R), we have −d2(h)[d1(x), y∗k] + d1(h)[ky, d2(x ∗)] = 0. This further implies that −d2(h)k[d1(x), y∗] + d1(h)k[y, d2(x ∗)] = 0. (7) Multiplying (6) by k and comparing with (7), we obtain 2d1(h)k[y, d2(x ∗)] = 0. Since char(R) 6= 2 and S(R) ∩ Z(R) 6= (0), the above expression gives d1(h)[y, d2(x ∗)] = 0 (8) for all x, y ∈ R and h ∈ H(R) ∩ Z(R). Invoking the primeness of R, we get d1(h) = 0 for all h ∈ H(R) ∩ Z(R) or [y, d2(x ∗)] = 0 for all x, y ∈ R. Suppose [y, d2(x ∗)] = 0 for all x, y ∈ R. Replacing x by x∗ we get [y, d2(x)] = 0 for all x, y ∈ R. Thus in view of Posner’s result [18], R is commutative, which is a contradiction. Now suppose d1(h) = 0 for all h ∈ H(R)∩Z(R). This further implies that 0 = d1(k 2) = 2d1(k)k. Since char(R) 6= 2 and S(R)∩Z(R) 6= (0), we have d1(k) = 0 for all k ∈ S(R)∩Z(R). Now since every z ∈ Z(R) can be represented as 2z = h + k where h ∈ H(R) ∩ Z(R) and k ∈ S(R) ∩ Z(R), we get d1(Z(R)) = (0). Now in view of (7), we have d2(h)k[d1(x), y∗] = 0 for all x, y ∈ R, h ∈ H(R) ∩ Z(R) and k ∈ S(R) ∩ Z(R). Using primeness, we get either d2(h) = 0 for all h ∈ H(R)∩Z(R) or [d1(x), y∗] = 0 for all x, y ∈ R. Replacing y by y∗, we get [d1(x), y] = 0 for all x, y ∈ R. Again using Posner’s result [18], we get a contradiction. Now suppose d2(h) = 0 for all h ∈ H(R)∩Z(R). This intern implies that d2(Z(R)) = (0). Replacing y by −ky in (5), we have k([d1(x), d2(y ∗)]− [d1(y), d2(x ∗)]) = 0. This further implies that [d1(x), d2(y ∗)]− [d1(y), d2(x ∗)] = 0 (9) for all x, y ∈ R, since S(R) ∩ Z(R) 6= (0). On comparing (9) with (5), we get 2[d1(x), d2(y ∗)] = 0 for all x, y ∈ R. The last relation gives, [d1(x), d2(y)] = 0 for all x, y ∈ R. This implies that [d1(x), d2(x)] = 0 for all x ∈ R. Hence in view of [15, Theorem 4], we conclude that d1 = λd2, where λ ∈ C. This completes the proof of the theorem. � S. Ali et al. / Eur. J. Pure Appl. Math, 12 (3) (2019), 1138-1148 1142 Corollary 1. Let R be a 2-torsion free noncommutative prime ring with involution ∗ of the second kind and d1, d2 be two nonzero derivations on R such that [d1(x), d2(y ∗)] = 0 for all x, y ∈ R. Then d1 = λd2, where λ ∈ C. Theorem 1. Let R be a 2-torsion free noncommutative prime ring with involution ∗ of the second kind and d1, d2 be two nonzero derivations on R. If one of the following conditions holds: (i) [d1(x), d2(x ∗)] = [x, x∗] for all x ∈ R, (ii) [d1(x), d2(x ∗)] = −[x, x∗] for all x ∈ R, then d1 = λd2, where λ ∈ C. Proof. By the given assumption, we have [d1(x), d2(x ∗)] = [x, x∗] for all x ∈ R. (10) A linearization of (10) yields that [d1(x), d2(y ∗)] + [d1(y), d2(x ∗)] = [x, y∗] + [y, x∗] for all x, y ∈ R. (11) Replace y by hy in (11), where h ∈ H(R) ∩ Z(R), we get [d1(x), d2((hy)∗)] + [d1(hy), d2(x ∗)] = [x, (hy)∗] + [hy, x∗] (12) for all x, y ∈ R and h ∈ H(R) ∩ Z(R). On solving, we obtain [d1(x), y∗]d2(h) + [y, d2(x ∗)]d1(h)+ (13) h([d1(x), d2(y ∗)] + [d1(y), d2(x ∗)]) = ([x, y∗] + [y, x∗])h for all x, y ∈ R and h ∈ H(R) ∩ Z(R). Multiplying (11) by h and adding with (13), we arrive at [d1(x), y∗]d2(h) + [y, d2(x ∗)]d1(h) = 0 (14) for all x, y ∈ R and h ∈ H(R)∩Z(R). Replacing y by ky, where k ∈ S(R)∩Z(R), we get −[d1(x), y∗]kd2(h) + d1(h)k[y, d2(x ∗)] = 0 for all x, y ∈ R. (15) Multiplying (14) by k and adding with (15), we obtain 2d1(h)k[y, d2(x ∗)] = 0 for all x, y ∈ R. This implies that d1(h)k[y, d2(x ∗)] = 0 for all x, y ∈ R. Using the primeness of the ring R and the fact that S(R) ∩ Z(R) 6= (0), we arrive at either d1(h) = 0 or [y, d2(x ∗)] = 0 for all x, y ∈ R. (16) S. Ali et al. / Eur. J. Pure Appl. Math, 12 (3) (2019), 1138-1148 1143 [y, d2(x ∗)] = 0 for all x, y ∈ R implies that R is commutative, a contradiction. Therefore we are left with d1(h) = 0 for all h ∈ H(R) ∩ Z(R). Using this in (15), we get −[d2(x), y∗]kd2(h) = 0 for all x, y ∈ R. The primeness of R yields that d2(h) = 0 for all h ∈ H(R) ∩ Z(R). (17) or [d1(x), y∗] = 0 for all x, y ∈ R. Again if [d1(x), y∗] = 0 for all x, y ∈ R, we get a contradiction. Therefore we are left with d2(h) = 0. This implies that d2(k) = 0 and hence d2(Z(R)) = (0). Similarly in view of (16) we get d1(Z(R)) = (0). Now on replacing y by ky in (11), where k ∈ S(R) ∩ Z(R), we get [d1(x), d2((ky)∗)] + [d1(ky), d2(x ∗)] = [x, (ky)∗] + [ky, x∗] (18) for all x, y ∈ R and k ∈ S(R) ∩ Z(R). On solving, we have −[d1(x), d2(y ∗)]k + k[d1(y), d2(x ∗)] = −[x, y∗]k + k[y, x∗] (19) for all x, y ∈ R and k ∈ S(R) ∩ Z(R). Multiplying (11) by k and adding with (19), we obtain 2k[d1(y), d2(x ∗)] = 2[y, x∗]. Since char(R) 6= 2 and invoking primeness of R, we get [d1(y), d2(x ∗)] = [y, x∗] for all x, y ∈ R. Hence [d1(y), d2(x)] = [y, x] for all x, y ∈ R. Taking y for x, we finally arrive at [d1(x), d2(x)] = 0 for all x ∈ R. Thus in view of [15, Theorem 4], we get d1 = λd2, where λ ∈ C. (ii) This can be proved by similar manner with necessary variations. Theorem 2. Let R be a 2-torsion free noncommutative prime ring with involution ∗ of the second kind and d1, d2 be two nonzero derivations on R such that [d1(x), x∗d2(x)] = 0, for all x ∈ R. Then d1 = λd2, where λ ∈ C. Proof. By the assumption, we have [d1(x), x∗d2(x)] = 0 for all x ∈ R. (20) Linearization of (20) give us [d1(x), x∗d2(x)] + [d1(x), x∗d2(y)] + [d1(x), y∗d2(x)] + [d1(x), y∗d2(y)] +[d1(y), x∗d2(x)] + [d1(y), x∗d2(y)] + [d1(y), y∗d2(x)] + [d1(y), y∗d2(y)] = 0 for all x, y ∈ R. Using (20), we get [d1(x), x∗d2(y)] + [d1(x), y∗d2(x)] + [d1(x), y∗d2(y)] (21) S. Ali et al. / Eur. J. Pure Appl. Math, 12 (3) (2019), 1138-1148 1144 +[d1(y), x∗d2(x)] + [d1(y), x∗d2(y)] + [d1(y), y∗d2(x)] = 0 for all x, y ∈ R. Replacing y by h where h ∈ H(R) ∩ Z(R), we get [d1(x), x∗]d2(h) + h[d1(x), d2(x)] = 0 for all x ∈ R. (22) Substituting x+ y for x where x, y ∈ R and combining it with (22), we have ([d1(x), y∗] + [d1(y), x∗])d2(h) + h([d1(x), d2(y)] + [d1(y), d2(x)]) = 0 (23) for all x, y ∈ R. Now replacing y by hy where y ∈ R and h ∈ H(R) ∩ Z(R), we obtain [d1(x), y∗]hd2(h) + [d1(y), x∗]hd2(h) + [y, x∗]d1(h)d2(h) + h2[d1(x), d2(y)] (24) +h[d1(x), y]d2(h) + h2[d1(y), d2(x)] + hd1(h)[y, d2(x)] = 0 for all x, y ∈ R. Multiplying (23) by h where h ∈ H(R) ∩ Z(R) and using in (24) we get [y, x∗]d1(h)d2(h) + h[d1(x), y]d2(h) + hd1(h)[y, d2(x)] = 0 (25) for all x, y ∈ R. Replacing x by kx where x ∈ R and k ∈ S(R) ∩ Z(R), we arrive at −[y, x∗]kd1(h)d2(h) + hk[d1(x), y]d2(h) + h[x, y]d1(k)d2(h) (26) +hd1(h)[y, d2(x)]k + hd1(h)[y, x]d2(k) = 0 for all x, y ∈ R. Multiplying (25) by k where k ∈ S(R) ∩ Z(R) and adding it with (26), we get 2hk([d1(x), y]d2(h) + [y, d2(x)]d1(h)) + h[x, y]d1(k)d2(h) + hd1(h)[y, x]d2(k) = 0 for all x, y ∈ R. Taking y = x, we obtain 2hk([d1(x), x]d2(h) + [x, d2(x)]d1(h)) = 0 for all x ∈ R. Since R is 2-torsion free prime ring and S(R) ∩ Z(R) 6= (0), the above relation implies that [d1(x), x]d2(h) + [x, d2(x)]d1(h) = 0 for all x ∈ R. (27) Replacing y by x in (25), we get [x, x∗]d1(h)d2(h) + h[d1(x), x]d2(h) + hd1(h)[x, d2(x)] = 0 (28) for x ∈ R and h ∈ H(R) ∩ Z(R). Using (27) in (28), we get [x, x∗]d1(h)d2(h) = 0 for all x ∈ R and h ∈ H(R) ∩ Z(R). Now use the primeness condition we get either [x, x∗] = 0 for all x ∈ R or d1(h)d2(h) = 0 for all h ∈ H(R) ∩ Z(R). If we consider [x, x∗] = 0, then in view of [16, Lemma 2.1] R is commutative, which is a contradiction to our assumption, now we have d1(h)d2(h) = 0 for all h ∈ H(R) ∩ Z(R). Using the primeness of the ring R we get either d1(h) = 0 or d2(h) = 0 for all h ∈ H(R) ∩ Z(R). If consider d1(h) = 0. S. Ali et al. / Eur. J. Pure Appl. Math, 12 (3) (2019), 1138-1148 1145 Then by (25) we get h[d1(x), y]d2(h) = 0 for all x, y ∈ R. Now using the primeness of the ring R, we obtain [d1(x), y]d2(h) = 0 for all x, y ∈ R. Again by the primeness of the ring R, we have either d2(h) = 0 for all h ∈ H(R) ∩ Z(R) or [d1(x), y] = 0 for all x, y ∈ R. If we consider [d1(x), y] = 0 for all x, y ∈ R. This gives R is commutative by Posner’s result [18], a contradiction. Therefore we are left with d2(h) = 0 for all h ∈ H(R)∩Z(R). Similarly in view of (25) we get d1(h) = 0 for all h ∈ H(R)∩Z(R). Replacing y by h where h ∈ H(R) ∩ Z(R) in (21) and using d1(h) = 0 and d2(h) = 0, we get h[d1(x), d2(x)] = 0 for all x ∈ R. Now using the primeness and S(R) ∩ Z(R) 6= (0) conditions, we get [d1(x), d2(x)] = 0 for all x ∈ R. Thus by the result of Lanski [15, Theorem 4], we get d1 = λd2, where λ ∈ C. Corollary 2. Let R be a 2-torsion free noncommutative prime ring with involution ∗ of the second kind and d1, d2 be two nonzero derivations on R such that [d1(x), y∗d2(y)] = 0 for all x, y ∈ R. Then d1 = λd2, where λ ∈ C. Theorem 3. Let R be a 2-torsion free noncommutative prime ring with involution ∗ of the second kind and d1, d2 be two nonzero derivations on R. If one of the following conditions holds: (i) [d1(x), d2(x ∗)] = x ◦ x∗ for all x ∈ R, (ii) [d1(x), d2(x ∗)] = −x ◦ x∗ for all x ∈ R, then d1 = λd2, where λ ∈ C. Proof. By the given hypothesis, we have [d1(x), d2(x ∗)] = x ◦ x∗ for all x ∈ R. (29) Replacing x by x+ y in (29), we get [d1(x), d2(x ∗)] + [d1(y), d2(y ∗)] + [d1(x), d2(y ∗)] + [d1(y), d2(x ∗)] = x ◦ x∗ + y ◦ y∗ + xy∗ + yx∗ + x∗y + y∗x for all x, y ∈ R. Using (29), we get [d1(x), d2(y ∗)] + [d1(y), d2(x ∗)] = xy∗ + yx∗ + x∗y + y∗x (30) for all x, y ∈ R. Substituting hy for y in (30) where h ∈ H(R) ∩ Z(R), we have h([d1(x), d2(y ∗)] + [d1(y), d2(x ∗)]) + [d1(x), y∗]d2(h) + d1(h)[y, d2(x ∗)] (31) = h(xy∗ + yx∗ + x∗y + y∗x) for all x, y ∈ R. Using (30), (31) reduces to [d1(x), y∗]d2(h) + d1(h)[y, d2(x ∗)] = 0 for all x, y ∈ R. (32) S. Ali et al. / Eur. J. Pure Appl. Math, 12 (3) (2019), 1138-1148 1146 Now (32) is same as (14) and thus following the same techniques we get d1(Z(R)) = (0) and d2(Z(R)) = (0). Now replace y by ky in (30) where k ∈ S(R) ∩ Z(R), we get −[d1(x), d2(y ∗)]k − [d1(x), y∗]d2(k) + d1(k)[y, d2(x ∗)] + k[d1(y), d2(x ∗)] (33) = −xy∗k + kyx∗ + x∗ky − ky∗x for all x, y ∈ R. Now multiplying (30) by k ∈ S(R) ∩ Z(R) and adding with (33), we get 2[d1(y), d2(x ∗)]k = 2k(yx∗ + x∗y) for all x, y ∈ R. This implies that k([d1(y), d2(x ∗)]− (y ◦ x∗)) = 0 for all x, y ∈ R. Invoking the primeness of R, we get [d1(y), d2(x ∗)]− (y ◦ x∗) = 0 for all x, y ∈ R. Now replace x by x∗, we obtain [d1(y), d2(x)]− (y ◦ x) = 0 for all x, y ∈ R. Hence application of Lemma 1 gives that d1 = λd2, where λ ∈ C. (ii) Similarly we can prove the second part. The following example shows that the primeness hypothesis in main theorem and Theorem 2 is not superfluous. Example 1. Let R = {( a1 + ib1 a2 + ib2 a3 + ib3 a4 + ib4 ) ∣∣∣ a1, a2, a3, a4, b1, b2, b3, b4 ∈ R } , where R is a ring of real numbers. Of course, R with matrix addition and matrix multiplication is a noncommutative prime ring. Define mappings ∗, d1 : R −→ R such that( a1 + ib1 a2 + ib2 a3 + ib3 a4 + ib4 )∗ = ( a1 − ib1 a3 − ib3 a2 − ib2 a4 − ib4 ) and, d1 ( a1 + ib1 a2 + ib2 a3 + ib3 a4 + ib4 ) = ( 0 −(a2 + ib2) (a3 + ib3) 0 ) . It can be easily checked that ∗ and d1 are respectively involution and derivation on R. Let H be a ring of real quaternions. Define involution − and derivation d2 = di (where di is an inner derivation on H determined by i ∈ H) as follows q = α − iβ − jγ − kδ and di(q) = [i, q] for all q ∈ H. Let S = R×H, where R is same as defined above with involution ∗ and derivation d1 and H is the ring of real quaternions with involution − and derivation d2 as above. Clearly, S is a 2-torsion free noncommutative semiprime ring. 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