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)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.
—————————— ——————————
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)na = 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)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,y∈R,
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)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, d2(x)ybd(x)n-1a = 0.
∑ d ( x)k d 2 ( x) ybd ( x)n−k −1+r a = 0 .
k =0
Since bd(x)na = 0, this relation reduces to
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ISSN 2229-5518
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 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)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
(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)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.
[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
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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)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,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) 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
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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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