Commutativity conditions on derivations and Lie ideals σ -prime rings
aa r X i v : . [ m a t h . R A ] J u l Commutativity conditions on derivations and Lie idealsin σ -prime rings L. Oukhtite, S. Salhi & L. Taoufiq
Universit´e Moulay Isma¨ıl, Facult´e des Sciences et TechniquesD´epartement de Math´ematiques, Groupe d’Alg`ebre et ApplicationsB. P. 509 Boutalamine, Errachidia; [email protected], [email protected], lahcentaoufi[email protected]
Abstract
Let R be a 2-torsion free σ -prime ring, U a nonzero square closed σ -Lieideal of R and let d be a derivation of R. In this paper it is shown that:1) If d is centralizing on U, then d = 0 or U ⊆ Z ( R ) .
2) If either d ([ x, y ]) = 0 for all x, y ∈ U, or [ d ( x ) , d ( y )] = 0 for all x, y ∈ U and d commutes with σ on U, then d = 0 or U ⊆ Z ( R ) . : 16W10, 16W25, 16U80. Key words and phrases : σ -prime ring, derivation, commutativity. Throughout this paper, R will represent an associative ring with center Z ( R ) . Recall that R is said to be 2-torsion free if whenever 2 x = 0 , with x ∈ R, then x = 0 . R is prime if aRb = 0 implies that a = 0 or b = 0 for all a and b in R. If σ is an involution in R, then R is said to be σ -prime if aRb = aRσ ( b ) = 0 implies that a = 0 or b = 0. It is obvious that everyprime ring equipped with an involution σ is also σ -prime, but the converseneed not be true in general. An additive mapping d : R → R is said to be aderivation if d ( xy ) = d ( x ) y + xd ( y ) for all x, y in R. A mapping F : R → R issaid to be centralizing on a subset S of R if [ F ( s ) , s ] ∈ Z ( R ) for all s ∈ S. Inparticular, if [ F ( s ) , s ] = 0 for all s ∈ S , then F is commuting on S. In all thatfollows Sa σ ( R ) will denote the set of symmetric and skew-symmetric elementsof R ; i.e., Sa σ ( R ) = { x ∈ R/σ ( x ) = ± x } . For any x, y ∈ R, the commutator xy − yx will be denoted by [ x, y ] . An additive subgroup U of R is said to bea Lie ideal of R if [ u, r ] ∈ U for all u ∈ U and r ∈ R. A Lie ideal U whichsatisfies σ ( U ) ⊆ U is called a σ -Lie ideal. If U is a Lie (resp. σ -Lie) idealof R, then U is called a square closed Lie (resp. σ -Lie) ideal if u ∈ U for1ll u ∈ U. Since ( u + v ) ∈ U and [ u, v ] ∈ U , we see that 2 uv ∈ U for all u, v ∈ U. Therefore, for all r ∈ R we get 2 r [ u, v ] = 2[ u, rv ] − u, r ] v ∈ U and2[ u, v ] r = 2[ u, vr ] − v [ u, r ] ∈ U, so that 2 R [ U, U ] ⊆ U and 2[ U, U ] R ⊆ U. Thisremark will be freely used in the whole paper.Many works concerning the relationship between commutativity of a ring andthe behavior of derivations defined on this ring have been studied. The firstimportant result in this subject is Posner’s Theorem, which states that theexistence of a nonzero centralizing derivation on a prime ring forces this ringto be commutative ([9]). This result has been generalized by many authors inseveral ways.In [3], I. N. Herstein proved that if R is a prime ring of characteristic not 2which has a nonzero derivation d such that [ d ( x ) , d ( y )] = 0 for all x, y ∈ R, then R is commutative. Motivated by this result, H. E. Bell, in [1], studiedderivations d satisfying d ([ x, y ]) = 0 for all x, y ∈ R . In [4] and [7], L. Oukhtiteand S. Salhi generalized these results to σ -prime rings. In particular, theyproved that if R is a 2-torsion free σ -prime ring equipped with a nonzeroderivation which is centralizing on R, then R is necessarily commutative.Our purpose in this paper is to extend these results to square closed σ -Lieideals. In order to prove our main theorems, we shall need the following lemmas.
Lemma 1 ([8], Lemma 4) If U Z ( R ) is a σ -Lie ideal of a 2-torsion free σ -prime ring R and a, b ∈ R such that aU b = σ ( a ) U b = 0 or aU b = aU σ ( b ) = 0 , then a = 0 or b = 0 . Lemma 2 ([5], Lemma 2.3) Let 0 = U be a σ -Lie ideal of a 2-torsion free σ -prime ring R. If [
U, U ] = 0 , then U ⊆ Z ( R ) . Lemma 3 ([6], Lemma 2.2) Let R be a 2-torsion free σ -prime ring and U anonzero σ -Lie ideal of R. If d is a derivation of R which commutes with σ andsatisfies d ( U ) = 0 , then either d = 0 or U ⊆ Z ( R ) . Remark.
One can easily verify that Lemma 3 is still valid if the conditionthat d commutes with σ is replaced by d ◦ σ = − σ ◦ d. Theorem 1
Let R be a 2-torsion free σ -prime ring and U a square closed σ -Lie ideal of R. If d is a derivation of R satisfying [ d ( u ) , u ] ∈ Z ( R ) for all u ∈ U, then U ⊆ Z ( R ) or d = 0 . roof. Suppose that U Z ( R ) . As [ d ( x ) , x ] ∈ Z ( R ) for all x ∈ U, bylinearization [ d ( x ) , y ] + [ d ( y ) , x ] ∈ Z ( R ) for all x, y ∈ U. Since char R = 2 , thefact that [ d ( x ) , x ] + [ d ( x ) , x ] ∈ Z ( R ) yields x [ d ( x ) , x ] ∈ Z ( R ) for all x ∈ U ;hence [ r, x ][ d ( x ) , x ] = 0 for all x ∈ U, r ∈ R, and therefore [ d ( x ) , x ] = 0 for all x ∈ U. Since [ d ( x ) , x ] ∈ Z ( R ),[ d ( x ) , x ] R [ d ( x ) , x ] σ ([ d ( x ) , x ]) = 0 for all x ∈ U and the σ -primeness of R yields [ d ( x ) , x ] = 0 or [ d ( x ) , x ] σ ([ d ( x ) , x ]) = 0 . If [ d ( x ) , x ] σ ([ d ( x ) , x ]) = 0, then [ d ( x ) , x ] Rσ ([ d ( x ) , x ]) = 0; and the fact that[ d ( x ) , x ] = 0 gives[ d ( x ) , x ] Rσ ([ d ( x ) , x ]) = [ d ( x ) , x ] R [ d ( x ) , x ] = 0 . Since R is σ -prime, we obtain[ d ( x ) , x ] = 0 for all x ∈ U. Let us consider the map δ : R R defined by δ ( x ) = d ( x ) + σ ◦ d ◦ σ ( x ) . One can easily verify that δ is a derivation of R which commutes with σ andsatisfies [ δ ( x ) , x ] = 0 for all x ∈ U. Linearizing this equality, we obtain[ δ ( x ) , y ] + [ δ ( y ) , x ] = 0 for all x, y ∈ U. Writing 2 xz instead of y and using char R = 2 , we find that δ ( x )[ x, z ] = 0 for all x, z ∈ U. Replacing z by 2 zy in this equality, we conclude that δ ( x ) z [ x, y ] = 0 , so that δ ( x ) U [ x, y ] = 0 for all x, y ∈ U. (1)By virtue of Lemma 1, it then follows that δ ( x ) = 0 or [ x, U ] = 0 , for all x ∈ U ∩ Sa σ ( R ) . Let u ∈ U . Since u − σ ( u ) ∈ U ∩ Sa σ ( R ) , it follows that δ ( u − σ ( u )) = 0 or [ u − σ ( u ) , U ] = 0 . δ ( u − σ ( u )) = 0 , then δ ( u ) ∈ Sa σ ( R ) and (1) yields δ ( u ) = 0 or [ u, U ] = 0 . If [ u − σ ( u ) , U ] = 0 , then [ u, y ] = [ σ ( u ) , y ] for all y ∈ U and (1) assures that δ ( u ) U [ u, y ] = 0 = δ ( u ) U σ ([ u, y ]) , for all y ∈ U. Applying Lemma 1, we find that δ ( u ) = 0 or [ u, U ] = 0 . Hence, U is a unionof two additive subgroups G and G , where G = { u ∈ U such that δ ( u ) = 0 } and G = { u ∈ U such that [ u, U ] = 0 } . Since a group cannot be a union of two of its proper subgroups, we are forcedto U = G or U = G . Since U Z ( R ) , Lemma 2 assures that U = G and therefore δ ( U ) = 0 . Now applying Lemma 3, we get δ = 0 and therefore d ◦ σ = − σ ◦ d. As [ d ( x ) , x ] = 0 for all x ∈ U, in view of the above Remark,similar reasoning leads to d = 0 . Corollary 1 ([7], Theorem 1.1) Let R be a 2-torsion free σ -prime ring and d a nonzero derivation of R. If d is centralizing on R, then R is commutative. Theorem 2
Let U be a square closed σ -Lie ideal of a -torsion free σ -primering R and d a derivation of R which commutes with σ on U. If [ d ( x ) , d ( y )] = d ([ y, x ]) for all x, y ∈ U , then d = 0 or U ⊆ Z ( R ) . Proof.
Suppose that U Z ( R ) . We have[ d ( x ) , d ( y )] = d ([ y, x ]) for all x, y ∈ U. (2)Substituting 2 xy for y in (2) and using char R = 2 , we get d ( x )[ y, x ] = [ d ( x ) , x ] d ( y ) + d ( x )[ d ( x ) , y ] for all x, y ∈ U. (3)Replacing y by 2[ y, z ] x and using (3), we find that[ d ( x ) , x ][ y, z ] d ( x ) + d ( x )[ y, z ][ d ( x ) , x ] = 0 for all x, y, z ∈ U. (4)Replace y by 2[ y, z ] d ( x ) in (3) to get d ( x )[ y, z ][ d ( x ) , x ] − [ d ( x ) , x ][ y, z ] d ( x ) = 0 for all x, y, z ∈ U. (5)From (4) and (5) we obtain[ d ( x ) , x ][ y, z ]( d ( x ) + d ( x )) = 0 for all x, y, z ∈ U. (6)4riting 2[ u, v ]( d ( x ) + d ( x )) y instead of y in (6) , where u, v ∈ U, we obtain[ d ( x ) , x ][ u, v ] z ( d ( x ) + d ( x )) y ( d ( x ) + d ( x )) = 0 , so that[ d ( x ) , x ][ u, v ] z ( d ( x ) + d ( x )) U ( d ( x ) + d ( x )) = 0 for all x, u, v, z ∈ U. (7)If x ∈ U ∩ Sa σ ( R ), then Lemma 1 together with (7) assures that d ( x ) + d ( x ) = 0 or [ d ( x ) , x ][ u, v ] z ( d ( x ) + d ( x )) = 0 for all u, v, z ∈ U. Suppose that [ d ( x ) , x ][ u, v ] z ( d ( x ) + d ( x )) = 0 . Then[ d ( x ) , x ][ u, v ] U ( d ( x ) + d ( x )) = 0 . (8)Since d commutes with σ and x ∈ Sa σ ( R ) , in view of (8), Lemma 1 gives d ( x ) + d ( x ) = 0 or [ d ( x ) , x ][ u, v ] = 0 for all u, v ∈ U. (9)If [ d ( x ) , x ][ u, v ] = 0, then replacing u by 2 uw in (9) where w ∈ U, we obtain[ d ( x ) , x ] U [ u, v ] = 0 . (10)As σ ( U ) = U and [ U, U ] = 0 , by (10), Lemma 2 yields that [ d ( x ) , x ] = 0.Thus, in any event,either [ d ( x ) , x ] = 0 or d ( x ) + d ( x ) = 0 for all x ∈ U ∩ Sa σ ( R ) . Let x ∈ U. Since x + σ ( x ) ∈ U ∩ Sa σ ( R ) , either d ( x + σ ( x )) + d ( x + σ ( x )) = 0or [ d ( x + σ ( x )) , x + σ ( x )] = 0 . If d ( x + σ ( x )) + d ( x + σ ( x )) = 0 , then d ( x ) + d ( x ) ∈ Sa σ ( R ) and (7) yieldsthat d ( x ) + d ( x ) = 0 or [ d ( x ) , x ][ u, v ] U ( d ( x ) + d ( x )) = 0 . If [ d ( x ) , x ][ u, v ] U ( d ( x ) + d ( x )) = 0 , once again using d ( x ) + d ( x ) ∈ Sa σ ( R ) , we find that d ( x ) + d ( x ) = 0 , or [ d ( x ) , x ][ u, v ] for all u, v ∈ U, in which case[ d ( x ) , x ] = 0 . Now suppose that [ d ( x + σ ( x )) , x + σ ( x )] = 0 . As x − σ ( x ) ∈ U ∩ Sa σ ( R ) wehave to distinguish two cases:1) If d ( x − σ ( x )) + d ( x − σ ( x )) = 0 , then d ( x ) + d ( x ) ∈ Sa σ ( R ). Reasoningas above we get d ( x ) + d ( x ) = 0 or [ d ( x ) , x ] = 0 .
2) If [ d ( x − σ ( x )) , x − σ ( x )] = 0, then [ d ( x ) , x ] ∈ Sa σ ( R ) . Replace u by 2 yu in(7), with y ∈ U , to get [ d ( x ) , x ] y [ u, v ] z ( d ( x ) + d ( x )) U ( d ( x ) + d ( x )) = 0 , sothat[ d ( x ) , x ] U [ u, v ] z ( d ( x ) + d ( x )) U ( d ( x ) + d ( x )) = 0 for all x, u, v, z ∈ U. (11)Since [ d ( x ) , x ] ∈ Sa σ ( R ) , from (11) it follows that[ d ( x ) , x ] = 0 or [ u, v ] U ( d ( x ) + d ( x )) U ( d ( x ) + d ( x )) = 0 for all u, v ∈ U. u, v ] U ( d ( x )+ d ( x )) U ( d ( x )+ d ( x )) = 0 . As σ ( U ) = U and [ U, U ] = 0 , then ( d ( x ) + d ( x )) U ( d ( x ) + d ( x )) = 0 . (12)In (6), write 2[ u, v ]( d ( x ) + d ( x )) r instead of z , where u, v ∈ U and r ∈ R, toobtain[ d ( x ) , x ][ u, v ] y ( d ( x )+ d ( x )) r ( d ( x )+ d ( x )) = 0 , for all u, v, y ∈ U, r ∈ R. (13)Replacing r by rσ ( d ( x ) + d ( x )) z in (13), where z ∈ U, we find that[ d ( x ) , x ][ u, v ] y ( d ( x ) + d ( x )) rσ ( d ( x ) + d ( x )) z ( d ( x ) + d ( x )) = 0 , which leads us to[ d ( x ) , x ][ u, v ] y ( d ( x ) + d ( x )) U σ ( d ( x ) + d ( x )) U ( d ( x ) + d ( x )) = 0 . (14)Since σ ( d ( x ) + d ( x )) U ( d ( x ) + d ( x )) is invariant under σ , by virtue of (14) , Lemma 1 yields σ ( d ( x ) + d ( x )) U ( d ( x ) + d ( x )) = 0 or [ d ( x ) , x ][ u, v ] y ( d ( x ) + d ( x )) = 0 . If σ ( d ( x ) + d ( x )) U ( d ( x ) + d ( x )) = 0 , then (12) implies that d ( x ) + d ( x ) = 0 . Now assume that[ d ( x ) , x ][ u, v ] y ( d ( x ) + d ( x )) = 0 for all u, v, y ∈ U. (15)Replace v by 2 wv in (15), where w ∈ U , and use (15) to get[ d ( x ) , x ] w [ u, v ] y ( d ( x ) + d ( x )) = 0 , so that [ d ( x ) , x ] U [ u, v ] y ( d ( x ) + d ( x )) = 0 for all u, v, y ∈ U. (16)As [ d ( x ) , x ] ∈ Sa σ ( R ) , (16) yields [ u, v ] U ( d ( x ) + d ( x )) = 0 , in which case d ( x ) + d ( x ) = 0 , or [ d ( x ) , x ] = 0.In conclusion, for all x ∈ U we have either [ d ( x ) , x ] = 0 or d ( x ) + d ( x ) = 0 . Now let x ∈ U such that d ( x ) + d ( x ) = 0 . In (2), put y = 2[ y, z ] d ( x ) to get d ([ y, z ])[ d ( x ) , x ] − [[ y, z ] , x ] d ( x ) + [ d ( x ) , [ y, z ]] d ( x ) = [ y, z ][ d ( x ) , x ] . (17)If in (2) we put y = 2[ y, z ] x, we get[[ y, z ] , x ] d ( x ) = [ d ( x ) , [ y, z ]] d ( x ) + d ([ y, z ])[ d ( x ) , x ] = 0 . (18)6rom (17) and (18) it then follows that[ y, z ][ d ( x ) , x ] = 0 for all y, z ∈ U, hence [ y, z ] U [ d ( x ) , x ] = 0 for all y, z ∈ U. Applying Lemma 1, this leads to[ d ( x ) , x ] = 0 , for all x ∈ U. By virtue of Theorem 1, this yields that d = 0 . Note that if d is a derivation of R which acts as an anti-homomorphism on U ,then d satisfies the condition [ d ( x ) , d ( y )] = d ([ y, x ]) for all x, y ∈ U. Thus wehave the following corollary.
Corollary 2 ([6], Theorem 1.1) Let d be a derivation of a 2-torsion free σ -prime ring R which acts as an anti-homomorphism on a nonzero square closed σ -Lie ideal U of R. If d commutes with σ , then either d = 0 or U ⊆ Z ( R ) . Theorem 3
Let U be a square closed σ -Lie ideal of a -torsion free σ -primering R and d a derivation of R. If either d ([ x, y ]) = 0 for all x, y ∈ U, or [ d ( x ) , d ( y )] = 0 for all x, y ∈ U and d commutes with σ on U, then d = 0 or U ⊆ Z ( R ) . Proof.
Suppose that U Z ( R ) . Assume that d ([ x, y ]) = 0 for all x, y ∈ U. Let δ be the derivation of R defined by δ ( x ) = d ( x ) + σ ◦ d ◦ σ ( x ) . Clearly, δ commutes with σ and δ ([ x, y ]) = 0 for all x, y ∈ U, so that[ δ ( x ) , y ] = [ δ ( y ) , x ] for all x, y ∈ U. (19)Writing [ x, y ] instead of y in (19) , we find that[ δ ( x ) , [ x, y ]] = 0 for all x, y ∈ U. (20)Replacing x by x in (19), we conclude that δ ( x )[ x, y ] + [ x, y ] δ ( x ) = 0 for all x, y ∈ U. (21)As char R = 2, from (20) and (21) it follows that δ ( x )[ x, y ] = 0 for all x, y ∈ U. (22)Replacing y by 2 zy in (22) , we get δ ( x ) z [ x, y ] = 0 , so that δ ( x ) U [ x, y ] = 0 for all x, y ∈ U. δ = 0 and thus d ◦ σ = − σ ◦ d. Since d satisfies d ([ x, y ]) = 0 for all x, y ∈ U, by similar reasoning, we areforced to d = 0 . Now assume that d commutes with σ and satisfies [ d ( x ) , d ( y )] = 0 for all x, y ∈ U. The fact that [ d ( x ) , d (2 xy )] = 0 implies that d ( x )[ d ( x ) , y ] + [ d ( x ) , x ] d ( y ) = 0 for all x, y ∈ U. (23)Replace y by 2[ y, z ] d ( u ) in (23), where z, u ∈ U, to find that[ d ( x ) , x ][ y, z ] d ( u ) = 0 for all x, y, u ∈ U. (24)Write 2[ s, t ] d ( w ) y instead of y in (24), where s, t, w ∈ U, thereby concludingthat [ d ( x ) , x ] z [ s, t ] d ( w ) yd ( u ) = 0 . Accordingly,[ d ( x ) , x ] z [ s, t ] d ( w ) U d ( u ) = 0 for all s, t, u, w, x ∈ U. (25)Since d commutes with σ and σ ( U ) = U, using (25) we find that d ( U ) = 0 or [ d ( x ) , x ] U [ s, t ] d ( w ) = 0 . Suppose that [ d ( x ) , x ] U [ s, t ] d ( w ) = 0 for all s, t, w, x ∈ U. (26)Replacing t by 2 tv in (26), where v ∈ U, we are forced to[ d ( x ) , x ][ s, t ] vd ( w ) = 0and hence [ d ( x ) , x ][ s, t ] U d ( w ) = 0 for all s, t, w, x ∈ U. (27)Since σ ( U ) = U and d commutes with σ , then (27) implies that either d ( U ) =0 , or [ d ( x ) , x ][ s, t ] = 0 for all s, t, x ∈ U, in which case [ d ( x ) , x ] = 0 for all x ∈ U. Thus, in any event, we find that d ( U ) = 0 or [ d ( x ) , x ] = 0 for all x ∈ U. If d ( U ) = 0 , then [[5], Theorem 1.1] assures that d = 0 . If [ d ( x ) , x ] = 0 for all x ∈ U , then Theorem 1 yields d = 0 . Corollary 3 ([4], Theorem 3.3) Let d be a nonzero derivation of a 2-torsionfree σ -prime ring R. If d ([ x, y ]) = 0 for all x, y ∈ R, then R is commutative.8 eferences [1] H. E. Bell, M. N. Daif, On derivations and commutativity in prime rings,
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