Jordan triple (α,β)-higher ∗-derivations on semiprime rings

Definition

Let N 0 be the set of all nonnegative integers, α , β two endomorphisms of R , and let D = ( d i ) i ∈ N 0 be a family of additive mappings of R such that d 0 = i d R . Assume also that ∗ is an involution defined on R such that for all a ∈ R , a ∗ 0 = a , and a ∗ 2 = ( a ∗ ) ∗ = a . Hence, for nonnegative even powers s , a ∗ s = a , and for nonnegative odd powers t , a ∗ t = a ∗ . D is called:

1. an ( α , β )-higher ∗ -derivation of R if for each positive integer n ,

   d n ( a b ) = ∑ i + j = n d i ( β j ( a ) ) d j ( α i ( b ∗ i ) ) for all a , b ∈ R ;

2. a Jordan ( α , β )-higher ∗ -derivation of R if for each positive integer n ,

   d n ( a 2 ) = ∑ i + j = n d i ( β j ( a ) ) d j ( α i ( a ∗ i ) ) for all a ∈ R ;

3. a Jordan triple ( α , β )-higher ∗ -derivation of R if for each positive integer n ,

   d n ( a b a ) = ∑ i + j + k = n d i ( β j + k ( a ) ) d j ( β k ( α i ( b ∗ i ) ) ) d k ( α i + j ( a ∗ i + j ) ) for all a , b ∈ R .

Lemma 2.1

([7], Lemma 1.1) Let R be a 2-torsion free semiprime ring. If a , b ∈ R are such that a r b = 0 for all r ∈ R , then b r a = a b = b a = 0 .

Notation

Let D = ( d i ) i ∈ N 0 be a Jordan triple higher ∗ -derivation of R . For every fixed n ∈ N 0 and each a , b ∈ R , we denote by A n ( a ) and B n ( a , b ) the elements of R defined by:

Lemma 3.1

Let R be a 2-torsion free semiprime ∗ -ring and let D = ( d i ) i ∈ N 0 a Jordan triple ( α , β )-higher ∗ -derivation of R such that β is an automorphism of R and α β = β α . If A m ( a ) = 0 for all a ∈ R and for each m ≤ n , then β n ( a 2 ) A n ( a ) = A n ( a ) β n ( a 2 ) = 0 for all a ∈ R and for each n ∈ N 0 .

Proof

We can compute the value of L = d n ( a 2 b a 2 ) in the following two ways:

First by substitution of a b a for b in the definition of Jordan triple ( α , β )-higher ∗ -derivation. So we look at L as d n ( a ( a b a ) a ) and by using the definition twice, we obtain

Now the second way to compute L is the substitution of a 2 for a in the definition of Jordan triple ( α , β )-higher ∗ -derivation and using our assumption that A m ( a ) = 0 for m < n to obtain

Now, subtracting the two values so obtained for L and using our notation we find that

In case n is even (3.1) reduces to

Replacing b by r a 2 b , r ∈ R , gives

Using (3.2) for the value of A n ( a ) α n ( r a 2 ) yields that

Again, using (3.2) for the value of A n ( a ) α n ( b a 2 ) yields, in view of R is 2-torsion free that

Now put r = b β − n ( A n ( a ) ) r in the last expression, we reach to

Since β onto, this implies that

Hence, by the semiprimeness of R , we obtain β n ( a 2 b ) A n ( a ) = 0 for all a , b ∈ R . Again since β is onto we have β n ( a 2 ) R A n ( a ) = { 0 } for all a ∈ R , and by Lemma 2.1, we reach to β n ( a 2 ) A n ( a ) = A n ( a ) β n ( a 2 ) = 0 for all a ∈ R .

In case n is odd (3.1) reduces to

Putting b = r a 2 b gives for all a , b , r ∈ R that

Substituting the value of A n ( a ) α n ( b ∗ a 2 ∗ ) from (3.3) in the last relation gives for all a , b , r ∈ R that

Again by using (3.3) for the value of A n ( a ) α n ( r ∗ a 2 ∗ ) , we obtain for all a , b , r ∈ R

Taking r = b in (3.4) leads, in view of R is 2-torsion free, to

Now putting r = b β − n ( A n ( a ) ) r in (3.4) gives for all a , b , r ∈ R

But using (3.5), the first summand of the last equation is zero. Hence, we obtain β n ( a 2 b ) A n ( a ) β n ( r a 2 b ) A n ( a ) = 0 for all a , b , r ∈ R . Surjectiveness of β leads to β n ( a 2 b ) A n ( a ) R β n ( a 2 b ) A n ( a ) = { 0 } for all a , b ∈ R and since R is semiprime we obtain β n ( a 2 ) β n ( b ) A n ( a ) = 0 for all a , b ∈ R . Again by using the surjectiveness of β , we find β n ( a 2 ) R A n ( a ) = { 0 } for all a ∈ R . Thus, since R is semiprime we obtain by Lemma 2.1 that A n ( a ) β n ( a 2 ) = β n ( a 2 ) A n ( a ) = 0 for all a ∈ R .

So for either two cases the required result holds and our lemma is proved.□

Remark 3.1

Straightforward computations show that A n ( a + b ) = A n ( a ) + A n ( b ) + B n ( a , b ) , A n ( a ) = A n ( − a ) and B n ( − a , b ) = − B n ( a , b ) for each pair a , b ∈ R and for all nonnegative integer n .

Theorem 3.2

Let R be a 6-torsion free semiprime ∗ -ring and β an automorphism of R . Then every Jordan triple ( α , β )-higher ∗ -derivation D = ( d i ) i ∈ N 0 of R, with α β = β α , is a Jordan ( α , β )-higher ∗ -derivation of R.

Proof

We will use induction on n in our proof. We see trivially that A 0 ( a ) = 0 for all a ∈ R . In case n = 1 , we obtain from [11, Theorem 2.1] that A 1 ( a ) = 0 for all a ∈ R . So we suppose that A m ( a ) = 0 for all a ∈ R and m < n . Now we can use Lemma 3.1 to obtain

and

The replacement of a + b for a in (3.6) gives

By replacing a by − a in (3.8) we obtain

Adding (3.8) and (3.9) we obtain, since R is 2-torsion free that

Substituting 2 a for a in (3.10) gives in view of the fact that R is 2-torsion free that

Comparing (3.10) and (3.11) we have, since R is 3-torsion free, that

Multiply (3.12) by A n ( a ) β n ( a ) from the right and using (3.7) we obtain

Substituting b by b a in (3.13) and multiplying by β n ( a ) from the left we obtain using that β is onto ( β n ( a ) A n ( a ) β n ( a ) ) R ( β n ( a ) A n ( a ) β n ( a ) ) = { 0 } for all a , b ∈ R . But since R is semiprime β n ( a ) A n ( a ) β n ( a ) = 0 for all a ∈ R . So (3.13) reduces to A n ( a ) β n ( a ) β n ( b ) A n ( a ) β n ( a ) = 0 , for all a , b ∈ R . Since β is onto, we have A n ( a ) β n ( a ) R A n ( a ) β n ( a ) = { 0 } , for all a ∈ R . Again, since R is semiprime, we have