Right Ideals of Prime Rings with Left Generalized Derivations

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International Journal of Mathematics Trends and Technology- Volume25 Number1 – September 2015
ISSN: 2231-5373 http://www.ijmttjournal.org Page 47
Right Ideals of Prime Rings with Left Generalized
Derivations
Dr. C. Jaya Subba Reddy1, S.MallikarjunaRao2
1,2Department of Mathematics, S.V.University, Tirupati-517502, AndhraPradesh, India.
Abstract: In this paper we present some results concerning derivations of a prime ring to the left generalized
derivations associated with a derivation of and anonzero right ideal of which is semi prime as a ring.
We proved that is a left generalized derivation of a prime ring , is a non-commutative right ideal of
and or acts as a homomorphism or anti- homomorphism or )]=0 or
or , then there exists a Martindale ring of quotients i.e. such that
for all . And also proved and , for all , then
Key words: Prime ring, Semi prime ring, Derivation,Generalized derivation, Left generalized derivation.
Introduction:The study of thecommutativity of prime rings with derivations was initiated by E.C.Posner
[11].Recently, M.Bresar[4] defined generalized derivation of rings. Hvala[9] studied the properties of
generalized derivations in prime rings .Golbasi[7] extended some well known results concerning derivations of
prime rings to the generalized derivations and a nonzero left ideal of a prime ring which is semi-prime as a ring
. In this paper weextend some results concerning derivations of prime ring to the left generalized derivations
associated with a derivation of and a non zero right ideal of which is semi prime as a ring .
Throughout out this paper, will be prime ring with characteristic different from two and a nonzero right
ideal of which is semiprime as a ring, the multiplicative center of , the right Martindale ring of
quotients the extended centroid and the central closure.For any , the symbol will
represent the commutator . Recall that prime if implise that or and semi prime
if implies that .An additive map from to is called a derivation of if
holds for all Let is a right generalized derivation if there exists a derivation of
such that holds for all If is a left generalized derivation if there exists a
derivation of such that holds for all If is a generalized derivation of
associated with if it is both left and right generalized derivation of .
To prove the main results we require the following lemmas
Lemma 1: [6,lemma 1] Let be a prme ring and a nonzero right ideal of which is semiprime as a ring. If
for then
Lemma 2:[9,Lemma2]Left be an additive map satisfying for all .Then there
exists such that ,for all
Lemma 3: [11, lemma2.3]Let be prime ring and a nonzero right ideal of which is semiprime as a ring. If
is a derivation of such that then .
Theorem 1: Let be a prime ring, a nonzero right ideal of which is semiprime as a ring and a left
generalized derivation of .If is non commutative and for all , then there exists
such that , for all
Proof: Let for all (1)
Substitute for in , we get

International Journal of Mathematics Trends and Technology- Volume25 Number1 – September 2015
ISSN: 2231-5373 http://www.ijmttjournal.org Page 48
,
.
From (1), we get
for all .(2)
Substitute for in equation (2), we get
,
,
From (2) the first summand is zero, it is clear that
.
Writing , in place of in this equation, we get
, for all
Since is prime ring, we have
or ,for all
or,for all
By lemma 1, we get either or for all
Let and Then and are two additive subgroups of such
that However, a group cannot be the union of proper sub groups. Hence either or .
If then , and so is commutative, which contradicts the hypothesis.
So we must have ,for all . By lemma3, we get
Hence, there exists such that, ,for all by lemma2.
Theorem 2: Let be a prime ring, a nonzero right ideal of which is semiprime as a ring and a left
generalized derivation of . If is non commutative and for all , then there exists
such that , for all .
Proof:Assume that for all . (3)
Replacing by in above equation, we get
,
,

International Journal of Mathematics Trends and Technology- Volume25 Number1 – September 2015
ISSN: 2231-5373 http://www.ijmttjournal.org Page 49
From equation (3), we get
,for all . (4)
Substitute for in equation (4), we get
,
,
From (4) the first summand is zero, it is clear that
.
Writing , in place of in this equation, we get
, for all
Since is prime ring, we have
or ,for all
or ,for all
By lemma 1, we get either or for all
Let and Then and are two additive subgroups of such that
However, a group cannot be the union of proper sub groups. Hence either or .
If then , and so is commutative, which contradicts the hypothesis.
So we must have , for all . By lemma 3, we get
Hence, there exists such that, , for all by lemma 2.
Corollary 1:Let be a prime ring, a non zero right ideal of which is semi prime as a ring and a left
generalized derivation of .If is non commutative and , for all then there exists
such that, , for all .
Theorem 3: Let be a prime ring, a nonzero right ideal of which is semi prime as a ring and a left
generalized derivation of . If acts as a homomorphism or anti homomorphism on , then there exists
such that, , for all .
Proof:Assume that acts as a homomorphism on .
Then , for all . (5)
Replacing by , , in the second equality in (5), we have
(6)
Since is a homomorphism. On the other hand, we have
(7)

International Journal of Mathematics Trends and Technology- Volume25 Number1 – September 2015
ISSN: 2231-5373 http://www.ijmttjournal.org Page 50
From (6) & (7), we have
for all .
Replacing by , in the above equation, we arrive at
for all .
Since is prime ring, we have either is the identity map on or, .
Suppose that , for all
Then ,
,
And so, , for all .
Hence we conclude that by lemma 1. Thus, there exists such that, , for all
,by lemma 2.
Now assume that acts as an anti homomorphism on .
Then , for all (8)
Replacing by in the first equation (8), we get
(9)
The second equation (8), we get
(10)
From equation (9) & (10), we get
, for all .(11)
Replacing by , in (11), to get
.
That is,
for all , (12)
Again writing by , , we have either or , for all .
According to Brauer’s Trick and lemma1, we conclude that or
In the second case, the proof is complete. The first case gives that acts as a homomorphism on . Thus, there
exists such that, , for all .

International Journal of Mathematics Trends and Technology- Volume25 Number1 – September 2015
ISSN: 2231-5373 http://www.ijmttjournal.org Page 51
Theorem 4: Let be a prime ring with characteristic different from two, anonzero right ideal of which is
semiprime as a ring, and a leftgeneralized derivation of . If is noncommutative and for all
, then there exists such that, , for all .
Proof:Let for all (13)
Linearization of (13) gives that
Using (13) gives that
, for all . (14)
Writing instead of in equation (14), we get
,
,
From (14), we get
for all . (15)
Writing instead of in equation (15), and using this equation, we obtain that
,
,
,
, for all (16)
Replacing by , , in (16), we get
Since is prime, we get
or , for all .
By lemma 2.1, we have either or for all .
By a standard argument one of these must be held for all . The second result cannot hold since is non
commutative, so the first possibility gives and hence
The proof may be completed by using lemma 2.
Theorem5:Let be a prime ring with characteristic different from two, a nonzero left ideal of which is
semiprime as a ring, and a left generalized derivation of .If is noncommutative,
and ,for all then there exists such that, , for all .

International Journal of Mathematics Trends and Technology- Volume25 Number1 – September 2015
ISSN: 2231-5373 http://www.ijmttjournal.org Page 52
Proof:Let for all (17)
Taking instead of in the equation (17), we get
,
,
From (17), we get
for all (18)
Replacing by , where ,and using (18), we arrive at
,
Since so
for all
Since is a nonzero right ideal of , we have
, for all .
The proof is now completed using theorem 4
Theorem 6:Let be a prime ring with characteristic different from two, a nonzero left ideal of wich is
semiprime as a ring, and a left generalized derivation of . If is noncommutativeand , then there
exists such that, , for all .
Proof:Since then
, for all .
Taking instead of in the above equation, we have
,
,
.
Since then
.
Expanding this equation we conclude that

International Journal of Mathematics Trends and Technology- Volume25 Number1 – September 2015
ISSN: 2231-5373 http://www.ijmttjournal.org Page 53
for all (19)
Writing instead of in equation (19), and using (19) we get
,
, for all
Taking in the place of in the above equation and using the fact is prime, we conclude that
, for all . By the standard argument, we have either that is commutative or .
Since is not commutative, the proof is complete.
Theorem 7:Let be a prime ring with characteristic different from two, a non zero right ideal of wich is
semiprime as a ring, and a left generalized derivation of and . If is non commutative, and
for all , then .
Proof:Since , there exists such that .
Furthermore, since is derivation, it is clear that
Replacing by in the hypothesis, we have
,
.
The second term lies in , we get
, for all .
Thus we obtain that , for all , and so
for all . (20)
Taking instead of and using (20), we get
for all .since clear and for all
and so
or for all .
Let and . Then A and B are additive sub groups of such that
. By Brauer’s trick, either or .
Since is noncommutative, we have . Hence , and so
Corollary2: Let be a prime ring with characteristic different from two, a non zero right ideal of R which is
semiprime as a ring, and a left generalized derivation of . If is non commutative, and
, then there exists such that, , for all .

International Journal of Mathematics Trends and Technology- Volume25 Number1 – September 2015
ISSN: 2231-5373 http://www.ijmttjournal.org Page 54
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