GANIT J. Bangladesh Math. Soc. (ISSN 1606-3694) 34 (2014) 35-46
JORDAN DERIVATIONS ON LIE IDEALS OF
SEMIPRIME -RINGS
Md. Mizanor Rahman1 and Akhil Chandra Paul2
Department of Mathematics, Jagannath University, Dhaka-1100, Bangladesh
2
Department of Mathematics, University of Rajshahi, Rajshahi-6205, Bangladesh
1
E-mail: [email protected]; 2E-mail: [email protected]
1
Received 05.05.2014
Accepted 07.12.2014
ABSTRACT
Let M be a 2-torsion free semiprime -ring satisfying the condition abc = abc,  a, b, c  M
and ,   . Let U be an admissible Lie ideal of M that is, uu  U, u  U,    and
U ̷ Z(M), the centre of M. If d : M  M is an additive mapping such that d is a Jordan derivation
on U of M, then d is a derivation on U.
Keywords: Derivation, Jordan derivation, Lie ideal, admissible Lie ideal, -ring,
semiprime -ring
1. Introduction
Let M and  be additive abelian groups. If there is a mapping M    M  M
(sending ( x,  , y ) into xy ) such that
• ( x  y )z = xz  yz , x(   ) y = xy  xy, x ( y  z ) = xy  xz
• ( xy )  z = x ( y z ), x, y , z  M and  ,   
then M is called a -ring. This concept is more general than a ring and was introduced
by Barnes [3]. A -ring M is called a prime -ring if a, b  M , aMb = 0 implies
a = 0 or b = 0 and M is called semiprime if aMa = 0 (with a  M ) implies a = 0 .
A -ring M is 2-torsion free if 2a = 0 implies a = 0, a  M . For any x, y  M and
   , we denote the commutator x y  y x by [ x, y ] An additive subgroup U  M
is said to be a Lie ideal of M if whenever u  U , m  M and    , then [u , m]  U .
In the main results of this article we assume that the Lie ideal U verifies
uu  U , u  U ,    . A Lie ideal of this type is called a square closed Lie ideal.
Furthermore, if the Lie ideal U is square closed and ⊈ ( ), where Z (M ) denotes the
center of M then U is called an admissible Lie ideal of M . Let M be a -ring. An
additive mapping d : M  M is a derivation if d (ab) = d (a )b  ad (b) and a Jordan
derivation if d (aa) = d (a)a  ad (a), a, b  M and    . Throughout the article,
we use the condition ab c = a bc, a , b, c  M and  ,    and this is denoted
by (*). The relationship between usual derivations and Lie ideals of prime rings has been
extensively studied in the last forty years. In particular, when this relationship involves
the action of the derivations on Lie ideals. In 1984, R. Awtar [2] extended to Lie ideals a
36
Rahman and Paul
well known result proved by I. N. Herstein [10] which states that ’every Jordan derivation
on a 2-torsion free prime ring is a derivation’. In fact, R. Awtar [2] proved that if ⊈
( ) is a square closed Lie ideal of a 2-torsion free prime ring R and d : R  R is an
additive mapping such that d (u 2 ) = d (u )u  ud (u ), u  U , then d (uv) = d (u )v  ud (v),
u , v  U . M. Ashraf and N. Rehman [1] studied on Lie ideals and Jordan left derivations
of prime rings. They proved that if d : R  R is an additive mapping on a 2-torsion free
prime ring R satisfying d (u 2 ) = 2ud (u ), u  U , where U is a Lie ideal of R such that
u 2 U , u U , then d (uv) = d (u )v  ud (v), u , v  U . A. K. Halder and A. C. Paul [9]
extended the results of Y. Ceven [6] in Lie ideals. We have generalized the R. Awtar’s [2]
result in semiprime -rings.
2. Jordan Derivation on a Lie Ideal
Definition 2.1 Let M be a -ring and U be a Lie ideal of M . An additive mapping
d : M  M is said to be a Jordan derivation on a Lie ideal U of M if
d (uu ) = d (u )u  ud (u ), u  U and    .
Example 2.1 Let R be a ring of characteristic 2 having a unity element 1.
.1
Let M = M1,2 ( R) and Γ =
: ∈ ℤ, 2 ∤ .
.1
Then M is a  -ring. Let N = {( x, x) : x  R}  M .
n
Now ( x, x)  N , (a, b)  M and     , we have
n
n
n
( x, x)  (a, b)  (a, b) ( x, x) = ( xna  bnx, xnb  anx )
n
n
= ( xna  2bnx  bnx, bnx  2anx  xna = ( xna  bnx, bnx  xna)  N .
Therefore, N is a Lie ideal of M .
Example 2.2 Let M be a -ring satisfying the condition (*) and let U be a Lie ideal of
M . Let a  M and    be fixed elements. Define d : M  M by
d ( x) = ax  xa, x  U . Now y  U and    , we have d ( xy ) = axy  xya ,
 ( ax  xa )  y  for every x, y  U and    . Therefore d : M  M is a derivation
on U .
Example 2.3 Let M be a  -ring and U be a Lie ideal of M . Let d : M  M be a
derivation on U . Suppose M 1 = {( x, x) : x  M } and 1 = {( ,  ) :   }. Define
addition and multiplication on M 1 as ( x, x)  ( y, y ) = ( x  y, x  y ) and
( x, x )( ,  )( y , y ) = ( xy , xy ). Then M 1 is a 1 -ring. Suppose U1 = {(u, u ) : u  U } .
Now (u, u)(, )(x, x)  (x, x)(, )(u, u) = (ux, ux)  (xu, xu) = (ux  xu, ux  xu) U1
for ux  xu U . Hence U1 is a Lie ideal of M 1. If we define a mapping
Jordan Derivations on Lie Ideals of Semiprime -rings
37
D : M 1  M 1 on U1 by D((u , u )) = (d (u ), d (u )) . Then it is clear that D is a Jordan
derivation on U1 which is not a derivation on U1 .
Lemma 2.1 Let M be a -ring and U be a Lie ideal of M such that uu  U , u  U
and    . If d is a Jordan derivation on U of M, then a , b, c  U and  ,    , the
following statements hold:
(i)
d (ab  ba ) = d (a )b  d (b)a  ad (b)  bd (a ) .
(ii) d (aba  aba) = d (a)ba  d (a) ba  ad (b) a  ad (b)a
 ab d ( a )  a bd ( a ) In particular, if M is 2-torsion free and satisfies the
condition (*), then
(iii) d (aba ) = d (a )ba  ad (b) a  abd (a ) .
(iv) d (abc  cba) = d (a)bc  d (c)ba  ad (b) c  cd (b) a  abd (c)
 cbd (a) .
Proof. Since U is a Lie ideal satisfying the condition aa  U , a  U ,    . For
a, b  U ,   , (ab  ba ) = (a  b) (a  b)  (aa  bb) and so (ab  ba )  U .
Also, [a, b] = ab  ba  U and it follows that 2ab U .
Hence 4abc = 2(2ab) c  U , a, b, c  U ,  ,    .
Thus
d ( a b  b a ) = d (( a  b ) ( a  b )  ( a a  b b ))
 d ( a  b ) ( a  b )  ( a  b ) d ( a  b )  d ( a ) a  a d ( a )  d (b ) b  b d (b )
= d ( a ) a  d ( a ) b  d (b ) a  d (b ) b  a d ( a )  a d (b )  b d ( a )  b d (b ) 
d ( a ) a  a d ( a )  d (b ) b  b d (b ) = d ( a ) b  a d (b )  d (b ) a  b d ( a ).
Replacing a b  b a for b in (i) we get
d (a (ab  ba )  (ab  ba )a )
 d (a ) (ab  ba )  ad (ab  ba )  d (ab  ba )a  (ab  ba )d (a ).
This implies that
d ( a  a )  b  ( a  a )  d (b )  d (b )  ( a  a )  b  d ( a  a )  d ( a  b  a  a  b  a )
 d ( a ) a  b  d ( a ) b  a  a  d ( a )  b  a  a  d (b )  a  d (b )  a  a  b  d ( a )
 d ( a )  b  a  a  d (b ) a  d (b )  a  a  b  d ( a ) a  a  b  d ( a )  b  a  d ( a ).
This implies that
d (a)ab  ad (a)b  aad (b)  d (b)aa  bd (a)a  bad (a)  d (aba  aba)
 d (a)ab  d (a)ba  ad (a)b  aad (b)  ad (b)a  abd (a)  d (a)ba
 ad (b)a  d (b)aa  bd (a)a  abd (a)  bad (a).
38
Rahman and Paul
Now canceling the like terms from both sides we get the required result. Using the
condition (*) and since M is 2-torsion free, (iii) follows from (ii). And finally (iv) is
obtained by replacing a  c for a in (iii).
Lemma 2.2 Let U be a Lie ideal of a 2-torsion free  -ring M satisfying the condition(*)
then T (U ) = {x  M : [ x, M ]  U } is both a subring and a Lie ideal of M such that
U  T (U ) .
Proof. We have U is a Lie ideal of M , so [U , M ]  U . Thus U  T (U ). Also we
have [T (U ), M ]  U  T (U ) . Hence T (U ) is a Lie ideal of M . Now suppose that
x, y  T (U ) then [ x, m]  U and [ y, m] U , m  M and    . Now
[ xy , m ] = x [ y , m ]  [ x, m ] y  U . Therefore, [ x y , m ]  U , x, y  T (U ), m  M
and  ,    . Hence xy  T (U ).
Lemma 2.3 Let ⊈ ( ) be a Lie ideal of a 2-torsion free semiprime -ring M
satisfying the condition (*), then there exists a nonzero ideal K = M[U , U ] M of M
generated by [U , U ] such that [ K , M ]  U .
Proof. First we prove that if [U ,U ] = 0 then U  Z (M ) , so let [U ,U ] = 0 for u U
and    , we have [u, [u, x] ] = 0, x  M . For all z  M and    , we replace x
by xz in [u, [u, x] ] = 0 and obtain
0 = [u , [u , x z ] ]
= [u , x [u , z ]  [u , x ]  z ]
= [u , x [u , z ] ]  [u , [u , x ]  z ]
= x [u , [u , z ] ]  [u , x ]  [u , z ]  [u , [u , x ] ]  z  [u , x ]  [u , z ]
= 2[u , x ]  [u , z ]
By the 2-torsion freeness of M , we obtain [u, x]  [u, z ] = 0. Now replacing z by zx
, we obtain
0 = [u , x]  [u , zx]
= [u , x]  z [u , x]  [u , x]  [u , z ] x
= [u , x]  z [u , x]
That is, [u, x] M [u, x] = 0. Since M is semiprime, [u, x] = 0 . This implies that
u  Z (M ) and therefore, U  Z (M ) is a contradiction. So let [U , U ]  0 . Then
K = M[U , U ] M is a nonzero ideal of M generated by [U , U ] . Let x, y  U , m  M
and  ,    , we have [ x, ym] , y, [ x, m]  U  T (U ).
Hence [ x, y ] m = [ x, ym]  y [ x, m]  T (U ).
Jordan Derivations on Lie Ideals of Semiprime -rings
39
Also we can show that, m [ x, y ]  T (U ) and therefore, we obtain [[U , U ] , M ]  U .
That is, [[[[x, y ] , m] , s] , t ]  U , m, s, t  M and   .
Hence [ x, y ] ms  m [ x, y ] s  [ s, m]  [ x, y ]  [ s [ x, y ] , m] , m] t ]  T (U ).
Since [ x, y ] ms , s [ x, y ] , [ s , m ]  [ x, y ]  T (U ) .
Thus we have, [ m [ x, y ] s, t ]  U , m, s, t  M and   . Hence [K, M ]  U.
Lemma 2.4 Let ⊈ ( ) be a Lie ideal of a 2-torsion free semiprime  -ring M
satisfying the condition (*) and aU . If aUa = {0},  ,    then aa = 0 and
there exists a nonzero ideal K = M[U ,U ] M of M generated by [U ,U ] such that
[K , M ]  U and Ka = aK = {0}.
Proof. If aU a = {0},  ,   , then a [ a , am ]  a = 0, m  M and   . Thus,
by using (*)
0 = a (aam  ama) a
= aaama  aamaa
= aaam  aamaa.
Since aaa = 0 , we have ( aa )m ( aa ) = 0. Since M is semiprime, aa = 0. Now
we obtain a [ ka , m ]  u a = 0, k  K , m  M , u  U and  ,  ,  , . Therefore, by
using (*) and aUa = {0}.
0 = a (kam  mka)ua
= akamua  amkaua
= akamua.
So we obtain, akam [ k , a ] a = 0. This implies that akam ( ka  ak )a = 0
and hence akamkaa  akamaka = 0. By using (*) and aa = 0 , we
obtain, ( aka ) m ( aka ) = 0. Since M is semiprime, aka = 0. Thus we find that
(ak )M (ak ) = 0. Hence ak = 0, k  K , that is aK = {0}. Similarly we obtain
Ka = {0} .
Lemma 2.5 Let ⊈ ( ) be a Lie ideal of a 2-torsion free semiprime  -ring M
satisfying the condition (*). Let a, b  U and ,    .
(i)
If aUa = {0} , then a = 0 .
(ii) If aU = {0} (or Ua = {0}) , then a = 0 .
(iii) If uu  U , u  U and aUb = {0} then ab = 0 and ba = 0,    .
40
Rahman and Paul
Proof. (i) By Lemma 2.4, we have Ka = M[U ,U ] Ma = {0} and aa = 0  .
Therefore, x, y  M and  ,   , we obtain:
0 = [[ a, x ] , a ] ] ya
= [ ax  xa, a ]  ya
= a [ x, a ]  ya  [ x, a ] a ya
= axa ya  aax ya  xaa ya  axa ya
= axa ya  axa ya
= 2axa ya.
By the 2-torsion freeness of M , we have axa ya = 0 . Thus we obtain,
axayaxa = 0. By using (*), we have (axa ) y (axa ) = 0. This implies that
(axa ) M (axa ) = 0. Since M is semiprime axa = 0, x  M and  ,    .
Again using the semiprimeness of M ,we obtain a = 0.
(ii)If aU = {0} , then aUa = {0}),    , therefore by(i), we obtain a = 0 .
Similarly, if Ua = {0} , then a = 0 .
(iii) If aUb = {0} , then we have (ba)U (ba) = {0} and hence by (i),
ba = 0,    . Also (ab)U (ab) = {0} if aUb = {0} and hence ab = 0.
Definition 2.2 Let M be a 2-torsion free semiprime  -ring satisfying the condition(*)
and U be a Lie ideal of M. Let D be a Jordan derivation on U of M. Then a, b  U and
   , we define G ( a, b) = d ( ab)  d ( a )b  ad (b) .
Lemma 2.6 Let M be 2-torsion free semiprime  -ring satisfying the condition (*) and U
be a square closed Lie ideal of M. Let d be a Jordan derivation on U of M . Then
a , b, c  U and ,    , the following statements hold.
(i)
G ( a , b )  G (b, a ) = 0;
(ii) G ( a  b, c ) = G ( a , c )  G (b, c ) ;
(iii) G ( a , b  c ) = G ( a , b )  G ( a , c ) ;
(iv) G   ( a , b ) = G ( a , b )  G  ( a , b ).
Remark 2.1 d is a derivation on U of M if and only if G (a, b) = 0, a, b U and    .
Remark 2.2 If U is a commutative Lie ideal of M, then by Lemma 2.3 U  Z (M ). In this
situation, we have seen that G ( a , b ) = 0, a , b  U and    which is obtained from
Lemma 2.1(i). So, we proved our main result for admissible Lie ideal of M.
Lemma 2.7 Let M be 2-torsion free  -ring satisfying the condition(*) and U be a square
closed Lie ideal of M. If d is a Jordan derivation on U of M. Then
Jordan Derivations on Lie Ideals of Semiprime -rings
(i )G ( a, b)  u [ a, b]  [ a, b]  uG ( a, b) = 0, a, b, u  U
41
and ,   
(ii )G ( a , b)u [ a , b]  [ a , b] uG ( a , b) = 0, a , b, u  U and ,   
(iii )G ( a, b)  u [ a, b]  [ a, b]  u G ( a, b) = 0, a, b, u  U and ,    .
Proof. (i) For any a, b, u  U and  ,  ,    , we have by using Lemma 2.1(iv)
W
=
=
=

d (4 a  b  u  b  a  4 b  a  u  a  b )
d ((2 a  b )  u  (2 b  a )  (2 b  a )  u  (2 a  b ))
4 d ( a  b )  u  b  a  4 a  b  d ( u ) b  a  4 a  b  u  d ( b  a )
4 d ( b  a )  u  a  b  4 b  a  d ( u )  a  b  4 b  a  u  d ( a  b ).
On the other hand by using Lemma 2.1(iii)
W
= d ( a  (4 b  u  b ) a  b  (4 a  u  a ) b )
= d ( a  (4 b  u  b ) a )  d ( b  (4 a  u  a ) b )
= 4 d ( a ) b  u  b  a  4 a  d ( b  u  b ) a  4 a  b  u  b  d ( a )  4 d ( b ) a  u  a  b
 4 b  d ( a  u  a ) b  4 b  a  u  a  d ( b )
= 4( d ( a ) b  u  b  a  a  d ( b )  u  b  a  a  b  d ( u ) b  a  a  b  u  d ( b ) a
 a  b  u  b  d ( a )  d ( b ) a  u  a  b  b  d ( a )  u  a  b  b  a  d ( u ) a  b
 b  a  u  d ( a ) b  b  a  u  a  d ( b )).
Equating two expressions for W and canceling the like terms from both sides, we get
4( d ( a  b )  u  b  a  a  b  u  d ( b  a )  d ( b  a )  u  a  b  b  a  u  d ( a  b ))
 4( d ( a ) b  u  b  a  a  d ( b )  u  b  a  a  b  u  d ( b ) a  a  b  u  b  d ( a )
 d ( b ) a  u  a  b  b  d ( a )  u  a  b  b  a  u  d ( a ) b  b  a  u  a  d ( b )).
This gives
4( d ( a  b )  u  b  a  d ( a ) b  u  b  a  a  d ( b )  u  b  a  d ( b  a )  u  a  b
 d ( b ) a  u  a  b  b  d ( a )  u  a  b  a  b  u  d ( b  a )  a  b  u  d ( b ) a
 a  b  u  b  d ( a )  b  a  u  d ( a  b )  b  a  u  d ( a ) b  b  a  u  a  d ( b )) = 0.
This implies that
4((d (ab)  d (a )b  ad (b))  uba  (d (ba )  d (b)a  bd (a ))  uab 
ab u (d (ba )  d (b)a  bd (a ))  ba u (d (ab)  d (a )b  ad (b))) = 0.
Now using the Definition 2.2 and 2-torsion freeness of M , we obtain
G ( a , b )  ub a  G (b, a )  ua b  a b uG (b, a )  b a uG ( a , b ) = 0.
This implies that
G ( a , b )  u [ a , b ]  [ a , b ]  uG ( a , b ) = 0, a , b, u  U ,  ,  ,   .
42
Rahman and Paul
If we consider
W = d (4abuba  4bauab) and W = d (4abuba  4bauab)
for (ii) and (iii) respectively and proceeding in the same way as in the proof of (i) by the
similar arguments, we get (ii) and (iii).
Lemma 2.8 Let U be an admissible Lie ideal of a 2-torsion free semiprime  -ring M
and let a, b  U . If au b  bu a = 0, u  U ,  ,    , then aub = 0 = bua .
xU and    be any elements.
Proof. Let
au b  bu a = 0, u  U ,  ,    repeatedly, we get
Using
the
relation
4( au b)x ( au b) = 4(bu a )x ( au b)
= (b (4u ax)a )u b
= ( a (4u ax)b)u b
= au (4axb)u b
=  au (4bxa )u b
= 4( au b)x ( au b).
This implies, 8((aub)x (aub)) = 0 . Since
is 2-torsion free,
M
( au b)x ( au b) = 0. Therefore, ( au b)U ( au b) = 0. Thus by Lemma 2.5, we
get aub = 0. Similarly, it can be shown that bua = 0.
Corollary 2.1 Let M be a 2-torsion free semiprime  -ring satisfying the condition(*)
and U be an admissible Lie ideal of M. If d is a Jordan derivation on U of M. Then
a, b, u  U ,  ,  ,   
(i )G ( a, b)  u [ a, b] = 0; (ii )[ a, b]  uG ( a, b) = 0; (iii )G ( a, b)u [ a, b] = 0;
(iv )[ a, b] uG ( a, b) = 0; (v )G ( a, b)  u [ a, b] = 0; (vi )[ a, b]  u G ( a, b) = 0.
Proof. Applying the result of Lemma 2.8 in that of Lemma 2.7, we obtain these results.
Lemma 2.9 Let M be a 2-torsion free semiprime  -ring satisfying the condition (*) and
U be an admissible Lie ideal of M. Let d be a Jordan derivation on U of M. Then
a, b, u  U ,  ,  ,   
(i )G ( a , b )  u [ x , y ] = 0; (ii )[ x , y ]  u G ( a , b ) = 0
(iii )G ( a , b )  u [ x , y ] = 0; (iv )[ x , y ]  u G ( a , b ) = 0
Proof. (i) If we substitute a  x for a in the Corollary 2.1 (v), we get
G ( a  x, b )  u [ a  x, b ] = 0
This implies
G ( a, b)  u [ a, b]  G ( a, b)  u [ x, b]  G ( x, b)  u [ a, b]  G ( x, b)  u [ x, b] = 0.
Jordan Derivations on Lie Ideals of Semiprime -rings
43
Thus by using Corollary 2.1 (v), we have
G ( a, b)  u [ x, b]  G ( x, b)  u [ a, b] = 0 .
Thus, we obtain
(G ( a , b )  u [ x, b ] )  u (G ( a , b )  u [ x, b ] )
= G ( a , b )  u [ x, b ]  u G ( x, b )  u [ a , b ] = 0.
Hence, by Lemma 2.5(i), we get
G ( a, b)  u [ x, b] = 0
Similarly, by replacing b  y for b in this result, we get
G ( a , b )  u [ x, y ] = 0
(ii) Proceeding in the same way as described above by the similar replacements
successively in Corollary 2.1 (vi), we obtain
[ x, y ]  u G ( a, b) = 0, a, b, x, y , u  U ,  ,   
(iii) Replacing    for  in (i), we get
G  ( a , b )  u [ x, y ]  = 0
This implies
(G ( a , b )  G ( a , b ))  u ([ x, y ]  [ x, y ] ) = 0
Therefore
G ( a, b)  u [ x, y ]  G ( a, b)  u [ x, y ]  G ( a, b)  u [ x, y ]  G ( a, b)  u [ x, y ] = 0.
Thus by using Corollary 2.1 (vi), we get
G ( a , b )  u [ x, y ]  G ( a , b )  u [ x, y ] = 0.
Thus, we obtain
(G ( a, b)  u [ x, y ] )  u (G ( a, b)  u [ x, y ] )
= G ( a, b)  u [ x, y ]  u G ( a, b)  u [ x, y ] = 0.
Hence, by Lemma 2.5(i), we obtain G ( a , b )  u [ x, y ] = 0.
(iv) As in the proof of (iii), the similar replacement in (ii) produces (iv).
Lemma 2.10 ([8], Lemma 3.6.1) Every semiprime  -ring contains no nonzero nilpotent
ideal.
Corollary 2.2 ([8], Corollary 3.6.2) Semiprime  -ring has no nonzero nilpotent element.
Lemma 2.11 ([8], Lemma 3.6.2) The center of a semiprime  -ring does not contain any
nonzero nilpotent element.
44
Rahman and Paul
Theorem 2.1 Let M be a 2-torsion free semiprime  -ring satisfying the condition(*)
and U be an admissible Lie ideal of M . If d : M  M is a Jordan derivation on U ,
then d is a derivation on U of M .
Proof. Let a, b, y, u  U and ,    . Then by Lemma 2.9 (iii) , we get
2[G ( a , b ), y ]   u [G ( a , b ), y ]  = 2(G ( a , b )  y  y G ( a , b ))  u [G ( a , b ), y ] 
= G ( a , b )  (2 y u )  [G ( a , b ), y ]   2 y G ( a , b )  u [G ( a , b ), y ]  = 0.
Since 2 y u  U and G ( a , b )  M , a , b, y , u  U and  ,   . Hence, by Lemma
2.5(i), [G ( a , b ), y ] = 0 , where G ( a, b)  M , a, b, y  U and  ,   . Therefore,
G ( a , b )  Z ( M ) , the center of M . Now, let xU and  ,   . Then, we have
G ( a , b ) [ x, y ] uG ( a , b ) [ x, y ] = 0, by using Lemma 2.9(ii).
Thus by using Lemma 2.5(i)
G ( a, b) [ x, y ] = 0.
(1)
Similarly, we get [ x, y ] G ( a , b )u [ x, y ] G ( a , b ) = 0, by using Lemma 2.9(i).
Now by Lemma 2.5(i), we get
[ x, y ] G ( a , b ) = 0.
(2)
Similarly, by Lemma 2.9(iv), we have G ( a , b ) [ x, y ] uG ( a , b ) [ x, y ] = 0 . By
Lemma 2.5(i), it follows that
G ( a , b ) [ x, y ] = 0.
(3)
Also, by Lemma 2.9(iii), we have [ x, y ] G ( a , b )u [ x, y ] G ( a , b ) = 0. Hence by
Lemma 2.5(i), we get
[ x, y ] G ( a , b ) = 0.
(4)
Thus, we have
2G (a, b)G (a, b) = G (a, b) (G (a, b)  G (a, b))
= G (a, b) (G (a, b)  G (b, a))
= G (a, b) (d (ab)  d (a)b  ad (b)  d (ba)  d (b)a  bd (a))
= G (a, b) (d (ab  ba)  (bd (a)  d (a)b)  (d (b)a  ad (b)))
= G (a, b) (d ([a, b] )  [b, d (a)]  [d (b), a] )
= G (a, b)d ([a, b] )  G (a, b) [d (a), b]  G (a, b) [a, d (b)] .
Since d (a), d (b)  M , so by using Lemma 2.9(i), we get
G ( a, b) [ d ( a ), b] = G ( a, b) [ a, d (b)] = 0.
Jordan Derivations on Lie Ideals of Semiprime -rings
45
Thus
2G ( a, b)G ( a, b) = G ( a, b)d ([ a, b] ).
(5)
Adding (3) and (4), we obtain G ( a , b ) [ x, y ]  [ x, y ] G ( a , b ) = 0. Then by Lemma
2.1 (i ) with the use of (3), we have
0 = d (G ( a, b) [ x, y ]  [ x, y ] G ( a, b))
= d (G ( a, b)) [ x, y ]  d ([ x, y ] )G ( a, b)  G ( a, b)d ([ x, y ] )  [ x, y ] d (G ( a, b))
= d (G ( a, b)) [ x, y ]  2G ( a, b)d ([ x, y ] )  [ x, y ] d (G ( a, b)).
Since G ( a , b )  Z ( M ) and therefore d ([ x, y ]  )G ( a , b ) = G ( a , b )d ([ x, y ]  ).
Hence, we get
2G ( a, b)d ([ x, y ] ) =  d (G ( a, b)) [ x, y ]  [ x, y ] d (G ( a, b)).
(6)
Then from (5) and (6) we have
4G ( a, b)G ( a, b) = 2G ( a, b)d ([ a, b] )
=  d (G ( a, b)) [ a, b]  [ a, b] d (G ( a, b)).
Thus we obtain
4G ( a, b)G ( a, b)G ( a, b)G ( a, b) =
 G ( a, b)d (G ( a, b)) [ a, b] G ( a, b)  G ( a, b) [ a, b] d (G ( a, b))G ( a, b).
Since by (1) and (2), G ( a, b) [ a, b] d (G ( a, b))G ( a, b) = 0 and
G ( a, b)d (G ( a, b)) [ a, b] G ( a, b) = 0.
So, we get
4G ( a , b )G ( a , b )G ( a , b )G ( a , b ) = 0.
Therefore, 4(G ( a , b ) ) 3 G ( a , b ) = 0. Since M is 2-torsion free, so we have
(G ( a , b ) ) 3 G ( a , b ) = 0
But, it follows that G ( a, b) is a nilpotent element of the  -ring M . Since by Lemma
2.11, the center of a semiprime  -ring does not contain any nonzero nilpotent element, so
we get G ( a, b) = 0, a, b  U and   . It means that, d is a derivation on U of M .
REFERENCES
[1]
[2]
Ashraf , M. and Rehman, N., On Lie ideals and Jordan left derivations of prime rings, Arch.
Math.(Brno), 36 (2000), 201-206.
Awtar, R., Lie ideals and Jordan derivations of prime rings, Amer.Math.Soc. 90(1) (1984), 9-14.
46
Rahman and Paul
[3]
[4]
Barnes, W. E., On the -rings of Nobusawa, Pacific J. Math., 18(1966), 411-422.
Bergen, J., Herstein, I. N. and Kerr ,J. W., Lie ideals and derivations of prime rings, J. Algebra,
71(1981), 259-267.
Bresar, M., Jordan derivations on semiprime rings, Proc. Amer. Math. Soc., 104(4), (1988) 10031004.
Ceven, Y., Jordan left derivations on completely prime gamma rings, C. U. Fen-Edebiyat Fakultesi
Fen Bilimleri Dergisi 23(2002), 39-43.
[5]
[6]
[7]
Chakraborty, S. and Paul, A. C., On Jordan k-derivations of 2-torsion free Prime N -rings, Punjab
Univ. J. Math., 40(2008), 97-101.
[8]
Chakraborty, S. and Paul, A. C., k-derivations and k-homomorphisms of -rings, Lap Lambert
Academic Publishing GmbH & Co. KG, Geometry, 2012.
Haetinger, C., Higher derivations on Lie ideals, Tenclencias em Matematica Aplicada e.
computacional, 3(1) (2002), 141-145.
[9]
[10] Halder, A.K. and Paul, A. C., Jordan left derivations on Lie ideals of prime -rings, Punjab University
J.Math.,(2011), 01-07.
[11] Herstein, I. N., Jordan derivations of prime rings, Proc. Amer. Math. Soc., 8(1957), 1104-1110.
[12] Herstein, I. N., Topics in Ring Theory, (Ed. The University of Chicago Press, Chicago) ,(1969).
[13] Hongan, M., Rehman, N. and Al Omary, R. M., Lie ideals and Jordan triple derivations of rings,
Rend. Sem. Mat. Univ. Padova., 125, (2011), 147-156.
[14] Lanski, C. and Montgomery, S., Lie structure of prime rings of characteristic 2, Pacific J. Math., 42,
No. 1 (1972), 117-136.
[15] Nobusawa, N., On the generalizeation of the ring theory, Osaka J. Math. 1(1964), 81-89.
[16] Sapanci, M. and Nakajima, A., Jordan Derivations on Completely Prime -Rings, Math. Japonica 46,
(1997), 47-51.