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

ISSN 2229-5518

# Abstract :

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)na = 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.

—————————— ——————————

# I. Introduction :

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)nk −1 a = 0 ,

k =0

n−1

d ( x)k (d 2 ( x) y + d ( x)d ( y))d ( x)nk −1 a = 0 ,

k =0

n−1

d ( x)k d 2 ( x) yd ( x)nk −1 a +

k =0

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

# II.Main Theorem :

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)nk −1 a = 0 .

k =0

Using the relation 1.1, this reduces to

n−1

d ( x)k d 2 ( x) yd ( x)nk −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 ( nk )−1 a = 0 .

k =0

Since bd(x)na = 0, we get d(x)n-1d2(x)ybd(x)n-1a = 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,yR,

n−1

where r ≥ 0 is any integer, which implies
d(x)rd2(x)ybd(x)n-1 = 0.

r

d ( x)k d ( y)d ( x)nk −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, d2(x)ybd(x)n-1a = 0.

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

k =0

Since bd(x)na = 0, this relation reduces to

International Journal of Scientific & Engineering Research, Volume 4, Issue 6, June-2013 2980

ISSN 2229-5518

d ( x)r d 2 ( x) ybd ( x)n−1 a +

n−1 .

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

k =r +1

Hence d2(z)ybd(x)n-1a = 0 by the semiprimeness of

R.

Lemma 1.4: For all x R, bd(x)2a = 0.
Hence if u is an arbitrary element in R, then
Proof: We replace z by x2 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(d2(x2)) yb d(x)n-1a = 0.

n-1

- d ( x)k d 2 ( x) ybd ( x)nk −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

(d2(x)x + xd(x)2

+ xd2d(x)) yb d(x)

n-1

a = 0,

n−1

= - d ( x)k d 2 ( x) ybd ( x)nk −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.

[d2(x)x + 2(d(x))2 + xd2(x)] yb d(x)n-1a = 0. By Lemma 1.3, this relation reduces to
2d(x)2 yb d(x)n-1a = 0. Let us assume that n ≥ 3.
Then R is 2-torsion free by assumption. So d(x)2 yb d(x)n-1a = 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 d2(x)ybd(x)n-1a = d2(z)ybd(x)n-1a = 0, we have
Since n is any integer larger than 2 we have
by induction bd(x)2a = 0.
Theorem 1.1: If R is a semiprime ring with a derivation d such that bd(x)na = 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)na = 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 d2(z)ybd(x)a = 0, for all x,y,z R.
Since y is arbitrary, we have bd2(z)aybd2(x)a = 0.

2 2

Since T(kx,z) = 0, for k = 1…….n-1, we have tn-1 =
0 by Lemma 1.1.
Note that γn-1 = d(x)n-1.
Thus 0 = tn-1 = d2(x)yb γn-2a+ d2(z)yb γn-1a
= d2(x)yb γn-2a + d2(z)ybd(x)n-1a.
Using this relation and Lemma 1.3, for every u R
we have
(d2(z)ybd(x)n-1a)ud2(z)ybd(x)n-1a =
In particular, bd (x)aybd (x)a = 0
and also bd2(z)d(x)aybd2(z)d(x)a = 0
which imply bd2(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 bd2(x)a = 0. Then we get

bd(x+y)2a = 0.

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

2 2

(-d2(x)ybγn-2a) u (d2(z)ybd(x)n-1a)=
-d2(x)(ybγn-2aud2(z)y)bd(x)n-1a = 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

International Journal of Scientific & Engineering Research, Volume 4, Issue 6, June-2013 2981

ISSN 2229-5518

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)ybd2(x)a+ bd(y)bd(x)2a+byd(b)d(x)2a+bybd2(x)d(x)a = 0.
Now, using the equations 1.4, 1.5 and bd(x)2a= 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)2a = 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,yR. 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) ybd2(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)ybd2(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)2d(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

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.