International Journal of Scientific & Engineering Research, Volume 4, Issue 6, June-2013 2979

ISSN 2229-5518

Chung and Luh [1] studied semiprime rings with nilpotent derivatives and established the result for (*n*-

1)!-torsion free semiprime rings. Giambruno and Herstein [2] proved the same result without assuming that *R *is

(*n*-1)!-torsion free. Bresar [3] generalized the result of Chung and Luh. Herstein proved some related results in

[4] and [5]. In this paper we prove that if *R *is an (*n*-1)!-torsion free semiprime ring with a derivaiton *d *such that *bd*(*x*)n*a *= 0 for *a,b *∈ *R *and for all *x *∈ *R*, then *bd*(*x*)*a *= 0 for all *x *∈ *R*.

Key words : Prime ring, semiprime ring, derivative, 2-torsion free ring.

—————————— ——————————

We know that an additive map *d *from a ring R to R is called a derivation fon R if *d(xy) = d(x)y + xd(y) *for all *x,y *in R. A ring R is called prime if and only if *xay = 0 *for all *a *in R implies *x = 0 *or *y = 0 *and semiprime if and only if *xax = 0 *for all a in R implies *x = 0.***Proof:**** **Replacing *y *by *d*(*x*)*y *in 1.1, we obtain

n−1

∑ d ( x)k d (d ( x) y)d ( x)n−k −1 a = 0 ,

k =0

n−1

∑ *d *( *x*)k (*d *2 ( *x*) *y *+ *d *( *x*)*d *( *y*))*d *( *x*)n−k −1 *a *= 0 ,

k =0

n−1

∑ d ( x)k d 2 ( x) yd ( x)n−k −1 a +

k =0

Throughout this paper *R *denotes an (*n*-1)!- torsion free ring with a derivation *d *such that *bd*(*x*)n*a *= 0.

To prove the main Theorem we need the following

Lemmas.

**Lemma 1.1:**** **Let R be a m!-torsion free ring. Suppose that t1, t2…. tm ∈ R satisfy kt1 + k2t2

n−1

d ( x)∑ d ( x)k d ( y)d ( x)n−k −1 a = 0 .

k =0

Using the relation 1.1, this reduces to

n−1

∑ d ( x)k d 2 ( x) yd ( x)n−k −1 a = 0 , for all x,y ∈ R 1.2

k =0

Replacing *y = ybd*(*x*)n-1 in the relation 1.2, we get

n−1

∑d ( x) k d 2 ( x) ybd ( x) e ( n−k )−1 a = 0 .

k =0

Since *bd*(*x*)n*a *= 0, we get *d*(*x*)n-1*d*2(*x*)*ybd*(*x*)n-1*a *= 0. We will prove this Lemma by showing that

+…………… kmtm = 0 for k = 1,2……..m. Then ti =

d(x)

r+1

d2(x) ybd(x)

n-1

a = 0,

0 for all *i.***Lemma1.2:**** **For all *x,y*∈*R*,

n−1

where *r *≥ 0 is any integer, which implies*d*(*x*)r*d*2(*x*)*ybd*(*x*)n-1 = 0.

r

∑ d ( x)k d ( y)d ( x)n−k −1 a = 0 . 1.1

k =0

Taking *y *= *ybd*(*x*)

n−1

in the relation 1.2, we obtain

Using these, we prove the following.**Lemma 1.3:**** **For all *x,y *∈ *R*, *d*2(*x*)*ybd*(*x*)n-1*a *= 0.

∑ d ( x)k d 2 ( x) ybd ( x)n−k −1+r a = 0 .

k =0

Since *bd*(*x*)n*a *= 0, this relation reduces to

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d ( x)r d 2 ( x) ybd ( x)n−1 a +

n−1 .

∑ d ( x)k d 2 ( x) ybd ( x)n−k −1+r a = 0

k =r +1

Hence *d*2(*z*)*ybd*(*x*)n-1*a *= 0 by the semiprimeness of

R.

**Lemma 1.4:**** **For all *x *∈ *R*, *bd*(*x*)2*a *= 0.

Hence if *u *is an arbitrary element in *R*, then**Proof:**** **We replace *z *by *x*2 in the relation 1.3.

2 2 *n*-1

(*d *( *x*)r *d *2 ( *x*) *ybd *( *x*)n−1 *a*)*u*(*d *( *x*)r *d *2 ( *x*) *ybd *( *x*)n−1 *a*) =

Then *d *(*x *) *yb d*(*x*)

a = 0.

n−1

This implies *d*(*d*2(*x*2)) *yb d*(*x*)n-1*a *= 0.

n-1

- ∑ *d *( *x*)k *d *2 ( *x*) *ybd *( *x*)n−k −1+r *au *(*d *( *x*)r *d *2 ( *x*) *ybd *( *x*)n−1 *a*)

So *d*(*d*(*x*)*x + xd*(*x*)) *yb d*(*x*)

a = 0,

k =r +1

(*d*2(*x*)*x + xd*(*x*)2

+ xd2d(x)) yb d(x)

n-1

a = 0,

n−1

= - ∑ *d *( *x*)k *d *2 ( *x*) *ybd *( *x*)n−k −1+r *aud *( *x*)r *d *2 ( *x*) *ybd *( *x*)n−1 *a*

k =r +1

= 0. by hypothesis.

By semiprimeness of *R*, this relation implies that

d(x)rd2(x)ybd(x)n-1a= 0.

[*d*2(*x*)*x + *2(*d*(*x*))2 + *xd*2(*x*)] *yb d*(*x*)n-1*a *= 0. By Lemma 1.3, this relation reduces to

2*d*(*x*)2 *yb d*(*x*)n-1*a *= 0. Let us assume that *n *≥ 3.

Then *R *is 2-torsion free by assumption. So *d*(*x*)2 *yb d*(*x*)n-1*a *= 0.

Since *y *is arbitrary, we also have**Lemma 1.4:**** **For all *x,y,z *∈ *R*,

d(x)

n-1

n-1

aybd(x)

n-1

n-1

a = 0.

d2(z)ybd(x)n-1a = 0 1.3

Hence *bd*(*x*)

aybd(x)

a = 0.

Proof: By Lemma 1.3 we have

By semiprimeness of *R*, we obtain

n-1

d2(x)ybd(x)n-1a = 0.

bd(x)

a = 0.

Linearizing, we obtain

T(x,z) = d2(x+z)ybd(x+z)n-1a = 0. That is, (d2(x)+d2(z))yb (d(x) + d(z))n-1a = 0.

Let us take (*d*(*x*) + *d*(*z*))n-1 as γo + γ1 +…………+

γn-1 where γj denotes the sum of these terms in

which *d*(*x*) appears as a factor in the product *j *times. Since *d*2(*x*)*ybd*(*x*)n-1*a *= *d*2(*z*)*ybd*(*x*)n-1*a = *0, we have

Since *n *is any integer larger than 2 we have

by induction *bd*(*x*)2*a *= 0.**Theorem 1.1:**** **If *R *is a semiprime ring with a derivation *d *such that *bd*(*x*)n*a *= 0 for all *a,b,x *∈ *R*

and *n *is a positive integer, then *bd*(*x*)*a *= 0 for all

a,b,x ∈ R. Moreover, if R is prime, then either a= 0

or *b *= 0 or *d *= 0.

n−2

T ( x, z) = ∑d 2 ( x) ybγ

k =0

n−1

a +∑d 2 ( z) ybγ

j =1

j a = 0 .

**Proof:**** **Let us assume that *bd*(*x*)n*a *= 0 for all

Thus if tk = d2(x)ybγk-1a + d2(z)ybγka,

then we can write

x,a,b ∈ R. By lemma 1.4, we may assume that

n = 2.

T(x,z) = t1 + ………… + tn-1.

Clearly T(kx,z) = kt1 + k2t2 +………..+ kn-1tn-1 for every integer k.

Hence by the relation 1.3, we have *d*2(*z*)*ybd*(*x*)*a *= 0, for all *x,y,z *∈ *R*.

Since *y *is arbitrary, we have *bd*2(*z*)*aybd*2(*x*)*a *= 0.

2 2

Since *T*(*kx,z*) = 0, for *k *= 1…….*n-*1, we have *t*n-1 =

0 by Lemma 1.1.

Note that γn-1 = *d*(*x*)n-1.

Thus 0 = *t*n-1 = *d*2(*x*)*yb *γn-2*a*+ *d*2(*z*)*yb *γn-1*a*

= *d*2(*x*)*yb *γn-2*a *+ *d*2(*z*)*ybd*(*x*)n-1*a*.

Using this relation and Lemma 1.3, for every *u *∈ *R*

we have

(*d*2(*z*)*ybd*(*x*)n-1*a*)*ud*2(*z*)*ybd*(*x*)n-1*a =*

In particular, *bd *(*x*)*aybd *(*x*)*a *= 0

and also *bd*2(*z*)*d*(*x*)*aybd*2(*z*)*d*(*x*)*a *= 0

which imply *bd*2(*x*)*a *= 0, for all *x *∈ *R *and 1.4

bd2(z)d(x)a = 0, for all x,z ∈ R 1.5

by the semiprimeness of *R*.

We linearize *bd*2(*x*)*a *= 0. Then we get

bd(x+y)2a = 0.

That is,*b*[*d*(*x*) + *d(y*)]2*a *= 0 which implies

2 2

(-*d*2(*x*)*yb*γn-2*a*) *u *(*d*2(*z*)*ybd*(*x*)n-1*a*)=

-*d*2(*x*)(*yb*γn-2*aud*2(*z*)*y*)*bd*(*x*)n-1*a *= 0.

bd(x) a + bd(y) a + bd(x)d(y)a + bd(y)d(x)a = 0.

Using the equation 1.4, we obtain

bd(x)d(y)a + bd(y)d(x)a = 0, for all x,y ∈ R. 1.6

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By replacing *y *by *ybd*(*x*) in the equaion 1.6, we get

bd(x)d(ybd(x))a + bd(ybd(x))d(x)a = 0.

This implies *bd*(*x*)*d*(*y*)*bd*(*x*)*a+bd*(*x*)*yd*(*b*)*d*(*x*)*a+bd*(*x*)*ybd*2(*x*)*a+ bd*(*y*)*bd*(*x*)2*a+byd*(*b*)*d*(*x*)2*a*+*bybd*2(*x*)*d*(*x*)*a *= 0.

Now, using the equations 1.4, 1.5 and *bd*(*x*)2*a*= 0, this relation reduces to *bd*(*x*)*d*(*y*)*bd*(*x*)*a+*

bd(x)yd(b)d(x)a + byd(b)d(x)2a = 0.

Replacing *b *by *byd*(*b*) in *bd*(*x*)2*a *= 0, we get

byd(b)d(x)2a = 0. 1.7

Hence *bd*(*x*)[*d*(*y*)*b *+ *yd*(*b*)]*d*(*x*)*a *= 0 implies

bd(x)d(yb)d(x)a=0, for all x,y∈R. 1.8

Linearizing the equation 1.8, we obtain

bd(x+z) d(yb) d(x+z)a = 0,

bd(x)d(yb) d(x)a + bd(z)d(yb)d(x)a + bd(x)d(yb)d(z)a

+ bd(z)d(yb)d(z)a = 0.

Using the equation 1.8, we get,

bd(x)d(yb) d(z)a + bd(z)d(yb)d(x)a = 0. 1.9

By taking *yb *= *ybd*(*z*) in the equation 1.9, we get *bd*(*x*)*d*(*ybd*(*z*))*d*(*z*)*a + bd*(*z*)*d*(*ybd*(*z*))*d*(*x*)*a = *0. This implies

bd(x)d(y)bd(z)2a+bd(x)yd(b)d(z)2a+bd(x)yb

d2(z)d(x)a+bd(z)d(y)bd(z)d(x)a+bd(z)yd(b)d(z)d(x)a

+ *bd*(*z*) *ybd*2(*z*)*d*(*x*)*a *= 0.

Using the equation 1.5 and *bd*(*z*)2 *a*= 0, we obtain

bd(z)d(y)bd(z)d(x)a + bd(z)yd(b)d(z)d(x)a

+ bd(x)yd(b)d(z)2a = 0. Replacing y by d(x)y in the relation 1.7, we get

bd(x)yd(b)d(z)2a = 0.

Therefore bd(z)(d(yb + yd(b))d(z)d(x)a = 0.

Hence *bd*(*z*)*d*(*yb*)*d*(*z*)*d*(*x*)*a = *0. Put *yb *= *ybd*(*x*)*u *in this equation. Then we have

bd(z) d(ybd(x)u) d(z) d(x)a = 0.

That is,

bd(z)[d(y)bd(x)u+yd(b)d(x)u+ybd2(x)d(u)+ +

ybd(x)d(u)]d(z)d(x)a = 0,

bd(z)d(y)bd(x)ud(z)d(x)a+bd(z)yd(b)d(x)ud(z)d(x)a

+*bd*(*z*)*ybd*2(*x*)*d*(*u)d*(*z*)*d*(*x*)*a*+

bd(z)ybd(x)d(u)d(z)d(x)a = 0. 1.10

By replacing *y *by *d*(*u*)*z *in the equation 1.8, we obtain

bd(x)d(d(u)zb) d(x)a = 0

bd(x)d2(u)zbd(x)a + bd(x)d(u)d(zb)d(x)a = 0. Using the equation 1.3, it reduces to

bd(x)d(u)d(zb)d(x)a = 0.

The equation 1.10 reduces to

bd(z)d(yb)d(x)ud(z)d(x)a = 0, 1.11

for all *x,y,z,u *∈ *Z*.

By replacing *b *by *bd*(*yb*) in the equation 1.6

bd(yb)d(x)d(yb)a + bd(y)2d(x)a = 0.

By the relation 1.7, it follows that *bd*(*yb*)2*d*(*x*)*a *= 0 for all *x,y *∈ *R*.

On linearizing we get *bd*(*yb*+*z*)2 *d*(*x*)*a *= 0,

bd(yb)2d(x)a + bd(z)2d(x)a + bd(yb)d(z)d(x)a

+ bd(z) d(yb)d(x)a = 0.

Using the equation 1.5, it reduces to

bd(yb)d(z)d(x)a + bd(z)d(yb)d(x)a = 0.

Since the element *u *the equation 1.11 is arbitrary,

we also have

bd(z)d(yb)d(x)au bd(y)d(z)d(x)a = 0.

Combining these two relations,

bd(z)d(yb)d(x)aubd(z)d(yb)d(x)a = 0, for all

x,y,z ∈ R.

Since *R *is semiprime this relation implies

bd(z)d(yb)d(x)a = 0, for all x,y,z ∈ R. 1.12

By replacing *d(z) *by *xd(z), *we get

bxd(z)d(yb)d(x)a=0. 1.13

By substituting *xz *for *z *in the equation 1.12, we obtain

b d(xz) d(yb) d(x)a = 0. This implies

bd(x)zd(yb)d(x)a +b xd(z)d(yb)d(z)a = 0.

Hence *bd*(*x*)*zd*(*yb*)*d*(*x*)*a = *0, for all *x,y,z *∈ *R *by

using the equation 1.12 which yields

bd(yb)d(x)a = 0, since R is semiprime. Now, by replacing yb by xyb, we get

bd(xyb) d(x)a = 0,

bd(x)ybd(x)a + bxd(yb) d(x)a = 0. 1.14

By replacing *d(yb) *by *xd(yb) *in the equation 1.14, we get *bxd(yb)d(x)a = 0, *hence the above equation reduces to

bd(x)ybd(x)a = 0.

Since *y *is arbitrary, we have

bd(x)aybd(x)a = 0.

Hence *bd*(*x*)*a *= 0.

If *R *is prime then either *bd*(*x*) = 0 or *a *= 0 for all *x*

∈ *R*. Again by primeness of *R *we get either *a*= 0 or

b = 0 or d(x) = 0.

The proof of Theorem 1.1 is thus completed.

REFERENCES

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International Journal of Scientific & Engineering Research, Volume 4, Issue 6, June-2013 2982

ISSN 2229-5518

[1] Chung,L.O., and Luh,J., Semiprime rings with nilpotent derivations Canad. Math. Bull. 24(4) (1981), 415-421.

[2] Giambruno,A., Derivations with nilpotent values, Rend. And Herstein, I.N.Circ. Mat. Palermo, 30(1981), 199-206.

[3] Bresar,M., A note on derivations, Math. J. Okayama Univ. 2(1990), 83-88.

[4] Herstein, I.N., Center-like elements in prime rings, J. Algebra, 60(1979), 569-574.

[5] Herstein, I.N., Derivations of prime rings having power central values, Algebraist’s homage : Vol

13(1982), 163-171.

Asst. Professor, Ananthalakshmi Institute of

Technology & Sciences, Anantapur (A.P.), India.

Research Supervisor, Department of Mathematics, S. K. University, Anantapur (A.P.), India.

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