SEMI DERIVATIONS OF PRIME GAMMA RINGS

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1 GANIT J. Bangladesh Math. Soc. (ISSN ) xx (2011) 31 (2011) SEMI DERIVATIONS OF PRIME GAMMA RINGS Kalyan Kumar Dey 1 and Akhil Chandra Paul 2 1,2 Department of Mathematics Rajshahi University, Rajshahi-6205, Bangladesh 1 kkdmath@yahoo.com; 2 acpaulru_math@yahoo.com Received Accepted ABSTRACT Let M be a prime Γ-ring satisfying a certain assumption (*). An additive mapping f : M M is a semi-derivation if f(xαy) = f(x)αg(y) + xαf(y) = f(x)αy + g(x)αf(y) and f(g(x)) = g(f(x)) for all x, y M and α Γ, where g : M M is an associated function. In this paper, we generalize some properties of prime rings with semi-derivations to the prime Γ-rings with semi-derivations AMS Subject Classifications: 16A70, 16A72, 16A10. Keywords: Semi derivation, Prime gamma ring, Commuting mapping 1. Introduction J. C. Chang [6] worked on semi-derivations of prime rings. He obtained some results of derivations of prime rings into semi-derivations. H. E. Bell and W. S. Martindale III [1] investigated the commutativity property of a prime ring by means of semi-derivations. C. L. Chuang [7] studied on the structure of semi-derivations in prime rings. He obtained some remarkable results in connection with the semi-derivations. J. Bergen and P. Grzesczuk [3] obtained the commutativity properties of semiprime rings with the help of skew (semi)-derivations. A. Firat [8] generalized some results of prime rings with derivations to the prime rings with semi-derivations. In this paper, we generalize some results of prime rings with semi-derivations to the prime Γ-rings with semi-derivations. 2. Preliminaries Let M and Γ be additive abelian groups. M is called a Γ-ring if for all x, y, z M, α,β Γ the following conditions are satisfied: (i) (ii) xβy M, (x + y)αz = xαz + yαz, x(α + β)y = xαy + xβy, xα(y + z) = xαy + xαz, (iii) (xαy)βz = xα(yβz). Let M be a Γ-ring with center C(M). For any x, y M, the notation [x, y] α and (x, y) α will denote xαy yαx and xαy + yαx respectively. We know that [xβy, z] α = xβ[y,z] α + [x,z] α βy + x[β,α] z y and [x,yβz] α = yβ[x,z] α + [x,y] α βz + y[β,α] x z, for all x,y,z M and for

2 66 Dey and Paul all α,β Γ. We shall take an assumption (*) xαyβz = xβyαz for all x,y,z M, α,β Γ. Using the assumption (*) the identities [xβy, z] α = xβ[y,z] α + [x,z] α βy and [x,yβz] α =yβ[x,z] α +[x,y] α βz, for all x,y,z M and for all α,β Γ are used extensively in our results. So we make extensive use of the basic commutator identities: (xβy, z) α = (x, z) α βy + xβ[y, z] α = [x, z] α βy + xβ(y, z) α. A Γ-ring M is to be n-torsion free if nx = 0, x M implies x = 0. Recall that a Γ-ring M is prime if xγmγy = 0 implies that x = 0 or y = 0. A mapping D from M to M is said to be commuting on M if [D(x), x] α = 0 holds for all x M, α Γ, and is said to be centralizing on M if [D(x), x] α C(M) holds for all x M, α Γ. An additive mapping D from M to M is called a derivation if D(xαy) = D(x)αy + xαd(y) holds for all x, y M, α Γ. Let M be a Γ-ring. An additive mapping d: M M, is called a semi-derivation associated with a function g: M M, if, for all x, y M, α Γ, (i) d(xαy) = d(x)αg(y) + xαd(y) = d(x)αy + g(x)αd(y), (ii) d(g(x)) = g(d(x)). If g = I, i.e., an identity mapping of M, then all semi-derivations associated with g are merely ordinary derivations. If g is any endomorphism of M, then other examples of semi-derivations are of the form d(x) = x g(x). Example 2.1 Let M 1 be a Γ 1 ring and M 2 be a Γ 2 -ring. Consider M = M 1 M 2 and Γ = Γ 1 Γ 2. Define addition and multiplication on M and Γ by (m 1, m 2 ) + (m 3, m 4 ) = (m 1 + m 3, m 2 + m 4 ), (α 1, α 2 ) + (α 3, α 4 ) = (α 1 + α 3, α 2 + α 4 ), (m 1, m 2 )(α 1, α 2 )(m 3, m 4 ) = (m 1 α 1 m 3, m 2 α 2 m 4 ), for every (m 1, m 2 ), (m 3, m 4 ) M and (α 1, α 2 ), (α 3, α 4 ) Γ. Under these addition and multiplication M is a Γ-ring. Let δ: M 1 M 1 be an additive map and τ: M 2 M 2 be a left and right M 2 Γ -module which is not a derivation. Define d: M M such that d((m 1, m 2 )) = (0, τ(m 2 )) and g: M M such that g((m 1, m 2 )) = (δ(m 1 ), 0), m 1 M 1, m 2 M 2. Then it is clear that d is a semi-derivation of M (with associated map g) which is not a derivation. 3. Semi Derivations of Prime Γ-rings We obtain our results. Lemma 3.1 Let M be a prime Γ-ring satisfying the assumption (*) and let m M. If [[m, x] α, x] α = 0 for all x M, α Γ, then x C(M).

3 Semi Derivations of Prime Gamma Rings 67 Proof A linearization of [[m, x] α, x] α = 0 for all x M, α Γ, gives [[m, x] α, y] α + [[m, y] α, x] α = 0 for all x, y M, α Γ. (1) Replacing y by yβx in (1) and using [[m, x] α, x] α = 0 for all x M, α Γ, we obtain 0 = [[m, x] α, yβx] α + [[m, yβx] α, x] α = [[m, x] α, y] α βx + [[m, y] α βx + yβ[m, x] α = [[m, x] α, y] α βx + [[m, y]α, x] α βx + [y, x] α β[m,x] α, for all x M, α Γ, Applying (1), we then get [y, x] α β[m, x] α = 0, for all x M, α Γ. Taking yβz for y in this relation and using [yβz, x] α = [y, x] α βz + yβ[z, x] α, we see that [y, x] α βzβ[m, x] α = 0, for all x, y, z M, α Γ. In particular, [m, x] α βzβ[m,x] α = 0, for all x M, α Γ. Since M is prime, [m, x] α = 0. This implies x C(M). Theorem 3.2 Let M be a non-commutative 2-torsion free prime Γ-ring satisfying the condition (*) and d is a semi-derivation of M with g: M M is an onto endomorphism. If the mapping x [aβd(x), x] α for all α,β Γ, is commuting on M, then a = 0 or d = 0. Proof Firstly, we assume that a be a nonzero element of M. Then we know that the mapping x [aβd(x), x] α is commuting on M. Thus we have [[aβd(x), x] α, x] α = 0. By lemma 3.1, we have [aβd(x), x] α = 0, for all x M, α,β Γ. (2) By linearizing (2), we have [aβd(x), y] α + [aβd(y), x] α = 0, for all x, y M, α,β Γ. (3) From this relation it follows that aβ[d(x), y] α + [a, y] α βd(x) + aβ[d(y), x] α + [a, x] α βd(y) = 0, for all x, y M, α,β Γ. (4) We get Replacing y by yδx in (3) and using (2), we get [aβd(x), yδx] α + [aβd(yδx), x] α = 0, for all x, y M, α,β Γ. yδ[aβd(x), x] α + [aβd(x), y] α δx + [aβ(d(y)δx + g(y)δd(x)), x] α = [aβd(x), y] α δx + [aβd(y)δx, x] α + [aβg(y)δd(x), x] α = aβ[d(x), y] α δx + [a, y] α βd(x)δx+ aβd(y)δ[x, x] α + [aβd(y), x] α δx + aβg(y)δ[d(x), x] α + [aβg(y), x] α δd(x) = aβ[d(x), y] α δx + [a, y] α βd(x)δx + aβ[d(y), x] α δx + [a, x] α βd(y)δx + aβg(y)δ[d(x), x] α + aβ[g(y), x] α δd(x) + [a, x] α βg(y)δd(x) = 0 for all x,y M, α,β Γ. (5)

4 68 Dey and Paul Right multiplication of (3) by δx gives aβ[d(x), y] α δx + [a, y] α βd(x)δx + aβ[d(y), x] α δx + [a, x] α βd(y)δx = 0, for all x, y M, α,β Γ. (6) Subtracting (6) from (5), we obtain aβg(y)δ[d(x), x] α +aβ[g(y), x] α δd(x) + [a, x] α βg(y)δd(x) = 0 for all x, y M, α, β, δ Γ. Taking aλg(y) instead of g(y) in (7), we have aβaλg(y)δ[d(x), x] α + aβ[aλg(y), x] α δd(x) + [a, x] α βaλg(y)δd(x) = aβaλg(y)δ[d(x), x] α + aβaλ[g(y), x] α δd(x) + aβ[a, x] α λg(y)δd(x) + [a, x] α βaλg(y)δd(x) = aβaλg(y)δ[d(x), x] α + aβaλ[g(y), x] α δd(x) + aβ[a, x] α λg(y)δd(x) + [a, x] α βaλg(y)δd(x) =0 for all x, y M, α,β,λ,δ Γ. (8) Left multiplication of (6) by aλ leads to aλaβg(y)δ[d(x), x] α + aλaβ[g(y), x] α δd(x) + aλ[a, x] α βg(y)δd(x) =aβaλg(y)δ[d(x), x] α + aβaλ[g(y), x] α δd(x) + aβ[a, x] α λg(y)δd(x) = 0 for all x, y M, α,β,δ,λ Γ. (9) Subtracting (9) from (8), we get [a, x] α βaλg(y)δd(x) = 0 for all x, y M, α,β,λ,δ Γ. Since M is prime, we obtain that for any x M either d(x) = 0 or [a, x] α = 0. It means that M is the union of its additive subgroups P = {x M: d(x) = 0} and Q = {x M: [a, x] α βa = 0}. Since a group cannot be the union of two proper subgroups, we find that either P = M or Q = M. If P = M, then d = 0. If Q = M, then this implies that [a, x] α βa = 0, for all x M, α,β Γ. Let us take xδy instead of x in this relation. Then we get [a, xδy] α βa = 0, for all x M, α,β Γ. We get [a, xδy] α βa = xδ[a, y] α βa + [a, x] α δyβa = [a, x] α δyβa = 0, for all x, y M, α,β,δ Γ. Since a M is nonzero and M is prime, we obtain a C(M). Thus by this and (2), the relation (7) reduces to aβ[g(y), x] α δd(x) = 0, for all x, y M, α,β,δ Γ. Since g is onto, we see that aβzγ[u, x] α δd(x) = zβaγ[u, x] α δd(x) = 0, for all x, u, z M, α,β,δ,γ Γ. Now by primeness of M, we obtain that [u, x] α δd(x) = 0, for all x, u M, α,β,δ Γ. (7)

5 Semi Derivations of Prime Gamma Rings 69 Replacing u by uλw, we get [u, x] α λwδd(x) = 0, for all x, u, w M, α, β, δ, λ Γ. By the primeness of M, [u, x] α = 0 or d(x) = 0. Again using the fact that a group cannot be the union of two proper subgroups, it follows that d = 0, since M is non-commutative, i.e., [u, x] α. Hence we see that, in any case, d = 0. This completes the proof. Theorem 3.3 Let M be a prime 2-torsion free Γ-ring satisfying the condition (*), d is a nonzero semiderivation of M, with associated endomorphism g and a M. If g ± I (I is an identity map of M), then (d(m), a) α = 0 if and only if d((m, a) α ) = 0. Proof Suppose (d(m), a) α = 0. Firstly, we will prove that d(a) = 0. If a = 0 then d(a) = 0. So we assume that a 0. By our hypothesis, we have (d(x), a) α = 0, for all x M, α Γ. From this relation, we get 0 = (d(xβa), a) α = (d(x)βg(a) + xβd(a), a) α = d(x)β[g(a), a] α + (d(x), a) α βa + xβ(d(a), a) α + [x, a] α βd(a), and so, [x, a] α βd(a) = 0, for all x M, α,β Γ. (10) Now, replacing x by xδy in (10), we get [xδy, a] α βd(a) = 0, for all x M, α,β Γ. By calculation we get, xδ[y, a] α βd(a) + [x, a] α δyβd(a) = [x, a] α δyβd(a) = 0, for all x M, α,β,δ Γ (11) The primeness of M implies that [x, a] α = 0 or d(a) = 0 that is, a C(M) or d(a) = 0. Now suppose that a C(M). Since (d(a), a) α = 0, we have d(a)αa + aαd(a) = 2aαd(a) = 0. Since M is 2-torsion free, aαd(a) = 0. Since we assumed that 0 a and M is a prime Γ-ring, we get d(a) = 0. Hence we have d((x, a) α ) = d(xαa + aαx) = d(xαa)+ d(aαx) =d(x)αa + g(x)αd(a) + d(a)αg(x) + aαd(x) = (g(x), d(a)) α + (d(x), a) α = (d(x), a) α ) = 0, for all x M, α Γ. Hence (d(x), a) α = 0. Conversely, for all x M, 0 = d((aβx, a) α ) = d(aβ(x, a) α + [a, a] α βx) = d(aβ(x, a) α ) = d(a)β(x, a) α + g(a)βd((x), a) α. We have d(a)β(x, a) α = 0, for all x M, α, β Γ. (12) Replacing x by xδy in (12), we get 0 = d(a)β(xδy, a) α = d(a)βxδ[y, a] α + d(a)β(x, a) α δy = d(a)βxδ[y, a] α. This implies that d(a)βxδ[x, a] α = 0, for all x M, α, β, δ Γ.

6 70 Dey and Paul For the primeness of M, we have either d(a) = 0 or a C(M). If d(a) = 0, then we have 0 = d((x), a) α = (d(x), a) α + (d(a), g(x)) α = (d(x), a) α, for all x M, α Γ. This yields that (d(x), a) α = 0. If a C(M), then we have 0 = d((a, a) α ) = 2d(a)α(a + g(a)). Since M is 2-torsion free, we obtain d(a)α(a + g(a)) = 0. Since M is prime we have d(a) = 0 or a + g(a) = 0. But since g is different from ± I, we find that d(a) = 0. Finally, (d(x), a) α = 0 implies the required result. REFERENCES 1. Bell, H. E. and Martindale, W. S. III, Semiderivations and commutativity in prime rings, Canad. Math. Bull., 31(4), (1988), Bergen, J., Derivations in prime rings, Canad. Math. Bull., 26 (1983), Bergen, J. and Grzesczuk, P., Skew derivations with central invariants, J. London Math. Soc., 59(2), (1999), M. Bresar, On a generalization of the notation of centralizing mappings, Proc. Amer. Math. Soc., 114 (1992), Bresar, M., Semiderivations of prime rings, Proc. Amer. Math. Soc., 108 (4) (1990), Chang, J.-C., On semi-derivations on prime rings, Chinese J. Math., 12 (1984), Chuang, Chen-Lian, on the structure of semiderivations in prime rings, Proc. Amer. Math. Soc., 108 (4) (1990), Firat, A., Some results for semi derivations of prime rings, International J. of Pure and Applied Mathematics, 28(3), (2006),

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