Let R be a prime ring with involution of the second kind and centre Z(R). Suppose R admits a generalized derivation F: R → R associated with a derivation d: R → R. The purpose of this paper is to study the commutativity of a prime ring R satisfying any one of the following identities: (i) F(x) ? x* ∈ Z(R) (ii) F([x, x*]) ± x ? x* ∈ Z(R) (iii) F(x?x*) ± [x, x*] ∈ Z(R) (iv) F(x)?d(x*) ± x?x* ∈ Z(R) (v) [F(x), d(x*)] ± x?x* ∈ Z(R) (vi) F(x) ± x ? x* ∈ Z(R) (vii) F(x) ± [x, x*] ∈ Z(R) (viii) [F(x), x*] ? F(x) ? x* ∈ Z(R) (ix) F(x ? x*) ∈ Z(R) for all x ∈ R.
Keywords: prime ring, generalized derivation, derivation, involution.
Throughout this paper R will represent a prime ring with center Z(R). 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 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. If char(R) ≠ 2, 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, S(R) ∩ Z(R) ≠ (0). A ring R is said to be normal if xx* = x*x for all x ∈ R. An example is the ring of quaternions. A description of such rings can be found in [12], 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. Following Bresar [9], an additive mapping F: R → R is called a generalized derivation if there exists a derivation d: R → R such that F(xy) = F(x)y + xd(y) for all x, y ∈ R. Basic examples are derivations and generalized inner derivations i.e., maps of type x ↦ ax + xb for some a, b ∈ R. A map f: R → R is said to be centralizing on R if [f(x), x] ∈ Z(R) for all x ∈ R. In a special case, when [f(x), x] = 0 holds for all x ∈ R, a map f is said to be commuting on R. The study of centralizing and commuting mappings on prime rings was initiated by the result of Posner [16], which states that the existence of a nonzero centralizing derivation on a prime ring implies that the ring has to be commutative. Through the years, a lot of work has been done in this context by a number of authors (see, for example, [1, 5, 7, 8, 13, 17] and references therein).
Very recently in many papers the additive mappings like derivations, generalized derivations have been studied in the setting of rings with involution and in fact it was seen that there is a close connection between these mappings and the commutativity of the ring R. For instance in [2], it is proved that let R be a prime ring with involution * such that char(R) ≠ 2. Let d be a nonzero derivation of R such that [d(x), x*] ∈ Z(R) for all x ∈ R and d(S(R) ∩ Z(R)) ≠ (0). Then R is commutative. Many other results in this direction can be found in [3, 4, 10, 15]. The goal of the present paper is to continue this line of investigation by considering certain identities involving Jordan product in the setting of generalized derivations.
Let R be a prime ring with involution ‘*’ of the second kind. Then x ∘ x* ∈ Z(R) for all x ∈ R if and only if R is commutative.
Theorem 2.3
Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: R → R associated with a derivation d: R → R, such that F(x) ∘ x* ∈ Z(R) for all x ∈ R. Then R is commutative.
and h ∈ H(R) ∩ Z(R). Using (2.2), we get [y ∘ x*, r]d(h) = 0 for all x, y, r ∈ R and h ∈ H(R) ∩ Z(R). Using the primeness of R we have either [y ∘ x*, r] = 0 for all x, y, r ∈ R or d(h) = 0 for all h ∈ H(R)∩Z(R). First if we consider [y ∘ x*, r] = 0 for all x, y, r ∈ R. Replacing y by x we get [x ∘ x*, r] = 0 for all x, r ∈ R. Thus in view of Lemma 2.2, we get R is commutative. Now consider the second case d(h) = 0 for all h ∈ H(R) ∩ Z(R). This intern implies d(k) = 0 for all k ∈ S(R) ∩ Z(R) and hence d(z) = 0 for all z ∈ Z(R). Replacing y by ky where k ∈ S(R) ∩ Z(R) in (2.2) and using d(z) = 0 for all z ∈ Z(R), We obtain
Using (2.2) in the previous equation, we get 2[F(y)∘x*, r]k = 0 for all x, y, r ∈ R and k ∈ S(R)∩Z(R). Since char(R) ≠ 2, we have [F(y)∘x*, r]k = 0for all x, y, r ∈ R and k ∈ S(R)∩Z(R). Now using the primeness and the fact that S(R)∩Z(R) ≠ (0), we have [F(y)∘x*, r] = 0 for all x, y, r ∈ R. That is, [F(y) ∘ x, r] = 0 for all x, y, r ∈ R. Taking x = z where z ∈ Z(R), we get [2F(x)z, r] = 0 for all x, r ∈ R and z ∈ Z(R). Since char(R) ≠ 2, we get [F(x)z, r] = 0 for all x, r ∈ R and z ∈ Z(R). Further since S(R) ∩ Z(R) ≠ (0) and using the primeness of R, we have [F(y), r] = 0 for all y, r ∈ R. Replace y by r, we get [F(r), r] = 0 for all r ∈ R. Thus in view of [18, Theorem 3.1], R is commutative.
Theorem 2.4
Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: R → R associated with a derivation d: R → R, such that F([x, x*]) ± x ∘ x* ∈ Z(R) for all x ∈ R. Then R is commutative.
Proof
We first consider the case
F([x,x*])+x∘x*∈Z(R)forallx∈R.
If F is zero, then we get x∘x* ∈ Z(R). Then by Lemma 2.2, we get R is commutative. Now consider F is non zero. Linearizing (2.3), we get
+[y ∘ x*, r]h = 0 for all x, y, r ∈ R and h ∈ H(R) ∩ Z(R). Using (2.4) in (2.5), we get ([[x, y*], r] + [[y, x*], r])d(h) = 0 for all x, y, r ∈ R and h ∈ H(R) ∩ Z(R). Using the primeness of R we have either ([[x, y*], r] + [[y, x*], r]) = 0 for all x, y, r ∈ R or d(h) = 0 for all h ∈ H(R) ∩ Z(R). If we consider ([[x, y*], r] + [[y, x*], r]) = 0 for all x, y, r ∈ R. This implies that [x, y*] + [y, x*] ∈ Z(R) for all x, y ∈ R. Replacing y by x we get [x, x*] ∈ Z(R) for all x ∈ R. Thus in view of Lemma 2.1, we get R is commutative. Now suppose that d(h) = 0 for all h ∈ H(R) ∩ Z(R). This further implies that d(z) = 0 for all z ∈ Z(R). Replacing y by ky in (2.4) where k ∈ S(R) ∩ Z(R), we get
for all x, y ∈ R and k ∈ S(R) ∩ Z(R). Using (2.4) and the fact that d(z) = 0 for all z ∈ Z(R), we get (F([y, x*]) + y ∘ x*)k ∈ Z(R) for all x, y ∈ R and k ∈ S(R) ∩ Z(R). Now using the primeness and the fact that S(R) ∩ Z(R) ≠ (0), we obtain (F([y, x*]) + y ∘ x*) ∈ Z(R) for all x, y ∈ R. Replace x by x*, we get (F([y, x]) + y ∘ x) ∈ Z(R) for all x, y ∈ R. Taking y = x, we get x2 ∈ Z(R) for all x ∈ R. Replacing x by x + y and using x2 ∈ Z(R) for all x ∈ R, we get xy + yx ∈ Z(R) for all x, y ∈ R. This can be further written as [xy + yx, r] = 0 for all x, y, r ∈ R. Replace y by z where z ∈ Z(R), we get 2[x, r]z = 0 for all x, r ∈ R and z ∈ Z(R). Finally using the facts that char(R) ≠ 2, S(R) ∩ Z(R) ≠ (0) and the primeness of R, we obtain R is commutative.
The second case can be proved in a similar manner with necessary variations.
Theorem 2.5
Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: R → R associated with a derivation d: R → R, such that F(x ∘ x*) ± [x, x*] ∈ Z(R) for all x ∈ R. Then R is commutative.
Proof
We first consider the positive sign case. If F is zero then we have [x, x*] ∈ Z(R) for all x ∈ R. Thus in view of Lemma 2.1, we get R is commutative. If F is nonzero then we have
F(x∘x*)+[x,x*]∈Z(R)forallx∈R.
Linearizing (2.6), we get
F(x∘y*)+F(y∘x*)+[x,y*]+[y,x*]∈Z(R)forallx,y∈R.
Replacing y by hy where h ∈ H(R) ∩ Z(R) and using (2.7), we get
(x∘y*+y∘x*)d(h)∈Z(R)forallx,y∈Z(R)andh∈H(R)∩Z(R).
Now using the primeness of R, we have (x ∘ y* + y ∘ x*) ∈ Z(R) for all x, y ∈ R or d(h) = 0 for all h ∈ H(R) ∩ Z(R). Replace y by x, we get x ∘ x* ∈ Z(R) for all x ∈ R. Thus by Lemma 2.2, we get R is commutative. Now consider d(h) = 0 for all h ∈ H(R) ∩ Z(R). This further implies that d(z) = 0 for all z ∈ Z(R). Replacing y by ky in (2.7) where k ∈ S(R) ∩ Z(R) and using (2.7), we have 2(F(y ∘ x*) + [y, x*])k ∈ Z(R) for all x, y ∈ R and k ∈ S(R) ∩ Z(R). This further implies that F(y ∘ x*) + [y, x*] ∈ Z(R) for all x, y ∈ R. Replacing x by x*, we get F(y ∘ x) + [y, x] ∈ Z(R) for all x, y ∈ R. Taking y = x, we get F(x2) ∈ Z(R) for all x ∈ R. Linearizing this and using F(x2) ∈ Z(R) for all x ∈ R, we get F(x ∘ y) ∈ Z(R) for all x, y ∈ R. Replacing y by z, where z ∈ Z(R) we get 2F(xz) ∈ Z(R) for all x ∈ R and z ∈ Z(R). This can be further written as 2[F(xz), r] = 0 for all x, r ∈ R and z ∈ Z(R). Since char(R) ≠ 2 this implies that [F(xz), r] = 0 for all x, r ∈ R and z ∈ Z(R). Since d(z) = 0 for all z ∈ Z(R), we finally arrive at [F(x), r]z = 0 for all x, r ∈ R and z ∈ Z(R). Now using the primeness and the fact that S(R)∩Z(R) ≠ (0) we have [F(x), r] = 0 for all x, r ∈ R. Replace r by x we get [F(x), x] = 0 for all x ∈ R. Then by [18, Theorem 3.1], we get R is commutative. Following the same steps, we get R is commutative in the second case as well.
Theorem 2.6
Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: R → R associated with a derivation d: R → R, such that F(x) ∘ d(x*) ± x ∘ x* ∈ Z(R) for all x ∈ R. Then R is commutative.
Proof
We first consider the case
F(x)∘d(x*)+x∘x*∈Z(R)forallx∈Z(R).
If either F or d or both are zero then we get x ∘ x* ∈ Z(R) for all x ∈ Z(R). Then by Lemma 2.2, we get R is commutative. Now consider the case in which both F and d are nonzero. Linearizing (2.8), we get
F(x)∘d(y*)+F(y)∘d(x*)+x∘y*+y∘x*∈Z(R)forallx,y∈R.
Replacing y by hy where h ∈ H(R) ∩ Z(R) and using (2.9), we get
Using the primeness condition, we get either F(x) ∘ y* + y ∘ d(x*) ∈ Z(R) for all x, y ∈ R or d(h) = 0 for all h ∈ H(R) ∩ Z(R). First consider F(x) ∘ y* +y ∘ d(x*) ∈ Z(R) for all x, y ∈ R. This can be further written as
[F(x)∘y*,r]+[y∘d(x*),r]=0forallx,y,r∈R.
Replacing y by ky where k ∈ S(R)∩Z(R) and using (2.10), we get 2[y∘d(x*), r]k = 0 for all x, y, r ∈ R and k ∈ S(R) ∩ Z(R). Using the primeness and the facts that char(R) ≠ 2 and S(R) ∩ Z(R) ≠ (0), we get [y ∘ d(x*), r] = 0 for all x, y, r ∈ R. Replacing x by x* we have [y∘d(x), r] = 0 for all x, y, r ∈ R. Replacing y by z where z ∈ Z(R), we get 2[d(x), r]z = 0 for all x, r ∈ R and z ∈ Z(R). Since char(R) ≠ 2 and S(R) ∩ Z(R) ≠ (0), this implies that [d(x), r] = 0 for all x, r ∈ R. Then by posner’s result [16], we get R is commutative. Now we consider the case d(h) = 0 for all h ∈ H(R) ∩ Z(R). This implies that d(z) = 0 for all z ∈ Z(R). Replacing y by ky in (2.9) where k ∈ S(R) ∩ Z(R) and using (2.9), we obtain
This further implies that F(y) ∘ d(x*) + y ∘ x* ∈ Z(R) for all x, y ∈ R. That is, F(y) ∘ d(x) + y ∘ x ∈ Z(R) for all x, y ∈ R. Taking x = z where z ∈ Z(R) and using d(z) = 0 for all z ∈ Z(R), we get 2yz ∈ Z(R) for all y ∈ R and z ∈ Z(R). Since char(R) ≠ 2, we get that yz ∈ Z(R) for all y ∈ R and z ∈ Z(R). This can be written as [yz, r] = 0 for all y, r ∈ R and z ∈ Z(R). This further implies that [y, r] = 0 for all y, r ∈ R. That is, R is commutative.
Similarly we can prove the second part.
Theorem 2.7
Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: R → R associated with a derivation d: R → R, such that [F(x), d(x*)] ± x ∘ x* ∈ Z(R) for all x ∈ R. Then R is commutative.
Proof
The proof is on the similar lines as in the above theorem.
Theorem 2.8
Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: R → R associated with a derivation d: R → R, such that F(x) ± x ∘ x* ∈ Z(R) for all x ∈ R. Then R is commutative.
Proof
we first consider the case
F(x)+x∘x*∈Z(R)forallx∈R.
If F is zero then we have x ∘ x* ∈ Z(R) for all x ∈ R. Then by Lemma 2.2, we get R is commutative. Consider F is nonzero. Linearizing (2.12), we get
x∘y*+y∘x*∈Z(R)forallx,y∈R.
Replacing y by ky where k ∈ S(R) ∩ Z(R) and using (2.13), we get (y ∘ x*)k ∈ Z(R) for all x, y ∈ R and k ∈ S(R) ∩ Z(R). Using primeness and the fact that S(R) ∩ Z(R) ≠ (0), we have y ∘ x* ∈ Z(R) for all x, y ∈ R. Taking y = x, we get x ∘ x* ∈ Z(R) for all x ∈ R. Thus in view of Lemma 2.2, we get R is commutative.
Similarly we can prove the other case.
On similar lines we can prove the following result.
Theorem 2.9
Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: R → R associated with a derivation d: R → R, such that F(x) ± [x, x*] ∈ Z(R) for all x ∈ R. Then R is commutative.
Theorem 2.10
Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: R → R associated with a derivation d: R → R, such that [F(x), x*] ∓ F(x) ∘ x* ∈ Z(R) for all x ∈ R. Then R is commutative.
Replacing y by hy where h ∈ H(R) ∩ Z(R) and using (2.15), we get
([y,x*]-y∘x*)d(h)∈Z(R)forallx,y∈Randh∈H(R)∩Z(R).
Using the primeness condition we have [y, x*] − y ∘ x* ∈ Z(R) for all x, y ∈ R or d(h) = 0 for all h ∈ H(R) ∩ Z(R). First suppose [y, x*] − y ∘ x* ∈ Z(R) for all x, y ∈ R. This further implies that [y, x] − y ∘ x ∈ Z(R) for all x, y ∈ R. Replacing y by x we get x2 ∈ Z(R) for all x ∈ R. Replacing x by x+y, we obtain x∘y ∈ Z(R) for all x, y ∈ R. That is, [x ∘ y, r] = 0 for all x, y, r ∈ R. Taking y = z, where z ∈ Z(R), we get 2[x, r]z = 0 for all x, r ∈ R and z ∈ Z(R). Since char(R) ≠ 2, S(R)∩Z(R) ≠ (0) and using the primeness of R, we have [x, r] = 0 for all x, r ∈ R. This implies that R is commutative. Now we consider the case d(h) = 0 for all h ∈ H(R) ∩ Z(R). This intern implies that d(z) = 0 for all z ∈ Z(R). Replacing y by ky where k ∈ S(R) ∩ Z(R) in (2.15) and using (2.15), we get
Since char(R) ≠ 2 and S(R) ∩ Z(R) ≠ (0), we obtain
[F(y),x*]-F(y)∘x*∈Z(R)forallx,y∈R.
Replacing x by x*, we get
[F(y),x]-F(y)∘x∈Z(R)forallx,y∈R.
This can be further written as F(y)x−xF(y)−F(y)x−xF(y) ∈ Z(R) for x, y ∈ R. This implies that −2xF(y) ∈ Z(R) for all x, y ∈ R. That is, xF(y) ∈ Z(R) for all x, y ∈ R, since char(R) ≠ 2. That is, [xF(y), r] = 0 for all x, y, r ∈ R. Replacing r by x, we get
x[F(y),x]=0forallx,y∈R.
Replacing y by yx we get x[F(y), x]x+xy[d(x), x]+x[y, x]d(x) = 0 for all x, y ∈ R. using (2.18), we get xy[d(x), x] + x[y, x]d(x) = 0 for all x, y ∈ R. Taking y = x, we have
x2[d(x),x]=0forallx∈R.
Replacing x by x + z where z ∈ Z(R) and using d(z) = 0 for all z ∈ Z(R), we obtain (z2 +2xz)[d(x), x] = 0 for all x ∈ R and z ∈ Z(R). Left multiplying by x we get z2x[d(x), x] + x2[d(x), x]2z = 0 for all x ∈ R and z ∈ Z(R). Using (2.19), we get z2x[d(x), x] = 0 for all x ∈ R and z ∈ Z(R). Now using the primeness and the fact that S(R) ∩ Z(R) ≠ (0), we get
x[d(x),x]=0forallx∈R.
Replacing x by x + z and using d(z) = 0 for all z ∈ Z(R), we get z[d(x), x] = 0 for all x ∈ R and z ∈ Z(R). Using primeness and the fact that S(R) ∩ Z(R) ≠ (0), we finally arrive at [d(x), x] = 0 for all x ∈ R. Thus by the result of Posner [16], we get R is commutative.
Now we consider the case
[F(x),x*]+x∘x*∈Z(R)forallx∈R.
Using the same steps as we did in the above case, we get
[F(y),x*]+F(y)∘x*∈Z(R)forallx,y∈R.
Replacing x by x*, we have
[F(y),x]+F(y)∘x∈Z(R)forallx,y∈R.
This can be further written as
F(y)x-xF(y)+F(y)x+xF(y)∈Z(R)forallx,y∈R.
That is, 2F(y)x ∈ Z(R) for all x, y ∈ R. Since char(R) ≠ 2, this implies that F(y)x ∈ Z(R) for all x, y ∈ R. That is, [F(y)x, r] = 0 for all x, y, r ∈ R. This intern implies that F(y)[x, r] + [F(y), r]x = 0 for all x, y, r ∈ R. Taking r = x, we get
[F(y),x]x=0forallx,y∈R.
Replacing y by yx, we arrive at
[F(y),x]x2+y[d(x),x]x+[y,x]d(x)x=0forallx,y∈R.
Using (2.22), we get
y[d(x),x]x+[y,x]d(x)x=0forallx,y∈R.
Replacing y by z where z ∈ Z(R). We obtain
z[d(x),x]x=0forallx∈Randz∈Z(R).
Using the primeness and the fact that S(R) ∩ Z(R) ≠ (0), we have [d(x), x]x = 0 for all x ∈ R. Replacing x by x+z, where z ∈ Z(R), using (2.23) and the fact that d(z) = 0 for all z ∈ Z(R), we get [d(x), x]z = 0 for all x ∈ R and z ∈ Z(R). This further implies that [d(x), x] = 0 for all x ∈ R. Hence in view of Posner’s result [16], we get R is commutative.
Theorem 2.11
Let R be a prime ring with involution of the second kind such that char(R) ≠ 2. If R admits a generalized derivation F: R → R associated with a derivation d: R → R, such that F(x ∘ x*) ∈ Z(R) for all x ∈ R. Then R is commutative.
Proof
Proceeding on the similar lines as we did in the previous result, we get F(x ∘ y*) ∈ Z(R) for all x, y ∈ R (where d(Z(R)) = (0)). Replacing y by z* where z ∈ Z(R), we get 2F(x)z ∈ Z(R) for all x ∈ R and z ∈ Z(R). Using the conditions that R is prime, char(R) ≠ 2 and S(R) ∩ Z(R) ≠ (0), we finally arrive at F(x) ∈ Z(R) for all x ∈ R. This implies that [F(x), r] = 0 for all x, r ∈ R. Then by the result [18], we get R is commutative.
At the end of paper, let us write an example which shows that the restriction of the second kind involution in our results is not superfluous
Example 2.21
Let R={(α1α2α3α4)|α1,α2,α3,α4∈Z2}. Of course R under matrix addition and matrix multiplication is a noncommutative prime ring. Define mappings F, D, *: R −→ R such that
Then x* = x for all x ∈ Z(R), and hence Z(R) ⊆ H(R), which shows that the involution * is of the first kind. Moreover, F, D are nonzero generalized derivation and derivation such that the following conditions hold
F(x) ∘ x* ∈ Z(R),
F([x, x*]) ± x ∘ x* ∈ Z(R),
F(x ∘ x*) ± [x, x*] ∈ Z(R),
F(x) ∘ D(x*) ± x ∘ x* ∈ Z(R),
[F(x), D(x*)] ± x ∘ x* ∈ Z(R),
F(x) ± x ∘ x* ∈ Z(R),
F(x) ± [x, x*] ∈ Z(R),
[F(x), x*] ∓ F(x) ∘ x* ∈ Z(R),
F(x ∘ x*) ∈ Z(R),
for all x ∈ R. However, R is not commutative. Hence, the hypothesis of second kind involution is crucial in our results.
