Simple Gamma Rings With Involutions.

Transcription

1 IOSR Journl of Mthemtics (IOSR-JM) ISSN: Volume 4, Issue (Nov. - Dec. 2012), PP Simple Gmm Rings With Involutions. 1 A.C. Pul nd 2 Md. Sbur Uddin 1 Deprtment of Mthemtics University of Rjshhi, Rjshhi-6205 Bngldesh. 2 Associte Professor, Deprtment of Mthemtics, Crmichel College, Rngpur, Bngldesh Abstrct: Let M be simple gmm ring with n involution I. In this pper, we develop some chrcteriztions of these gmm rings with involution. We lso obtin some properties of Lie nd Jordn idels with involutions. Key words: Simple gmm rings, Involution, Symmetric elements, skew symmetric elements, Jordn idels, Lie idels AMS Subject Clssifiction : Primry 16 N 60, Secondry 16 W 25, 16 U 80. I. Introduction. The notion of gmm ring ws first introduced by N. Nobusw [6] s generliztion of the concept of clssicl ring. Brnes [1] generlized the concept of the Nobusw s gmm ring which is known s gmm ring nd Nobusw s gmm ring is known s N -ring (i.e. gmm ring in the sense of Nobusw) L. Luh [5] worked on simple gmm rings nd obtined some importnt properties. I. N. Herstein [, 4] obtined vrious chrcteriztions of simple rings with involution nd lso developed some structurl results of Lie nd Jordn rings. Pul nd Sbur Uddin [7, 8] worked on Lie nd Jordn structure in simple gmm rings nd obtined some remrkble results. In this pper, we introduce the concept of n involution of -ring. An exmple of the involution for -ring is given here. Some chrcteriztions of simple -rings re obtined by mens of the involution. Also, we develop some properties of Lie nd Jordn idels with involutions. II. Preliminries Definitions. Gmm Ring. [1] Let M nd be two dditive belin groups. Suppose tht there is mpping from M M M (sending (x,, y) into xy) such tht i) (x + y) z = xz + yz x ( + )z = xz + xz x(y + z) = xy + xz ii) (xy)z = x(yz), where x, y, zm nd,. Then M is clled -ring. Idel of -rings. A subset A of the -ring M is left (right) idel of M if A is n dditive subgroup of M nd MA = {c cm,, A}(AM) is contined in A. If A is both left nd right idel of M, then we sy tht A is n idel or two sided idel of M. If A nd B re both left (respectively right or two sided) idels of M, then A + B = { + b A, bb} is clerly left (respectively right or two sided) idel, clled the sum of A nd B. We cn sy every finite sum of left (respectively right or two sided) idel of -ring is lso left (respectively right or two sided) idel. Nilpotent element. Let M be -ring. An element x of M is clled nilpotent if for some, there exists positive integer n = n() such tht (x) n x = (xx...x)x = 0. Nilpotent idel. An idel A of -ring M is clled nilpotent if (A) n A = (AA...A)A = 0, where n is the lest positive integer. Simple -ring. A -ring M is clled simple -ring if MM 0 nd its idels re {0} nd M. Centre of -ring. Let M be -ring. The centre of M, written s is the set of those elements in M tht commute with every element in M, tht is, = {mmmx = xm for ll xm nd }. Jordn Structure. Let M be -ring. The Jordn structure is defined by- (x, y) = xy + yx for x, ym nd ll. We sy tht subset A of M is Jordn sub--ring of M if A is n dditive subgroup such tht for, ba nd, b + b must lso be in A. 40 Pge

2 Simple Gmm Rings With Involutions. Jordn Idel. Let A be Jordn sub--ring of M. The dditive subgroup UA is to sid to be Jordn idel of A if whenever uu, A, nd then (u, ) = u + u is in U. Lie Structure. Let M be -ring. The Lie structure is defined by- [x, y] = xy - yx for x, ym nd for ll. We sy tht subset A of M is Lie sub--ring of M if A is n dditive subgroup such tht for, ba nd, b - b must lso be in A. Lie Idel. Let A be Lie sub--ring of M. The dditive subgroup UA is sid to be Lie idel of A if whenever uu, A,, then [u, ] = u - u is in U. If A, B re subsets of M, then [A, B] is the dditive subgroup of M generted by ll b - b with A, bb nd. If M is non-commuttive simple -ring of chrcteristic 2, then the sub--ring generted by [M, M] in M. If U is Lie idel of M, let T(U)={xM[x, M] U}. We need the following theorems for obtining our results which re ppered in [7, 8]. 2.2 Theorem. Let M be simple -ring of chrcteristic 2. Then ny Lie idel of M which is lso sub- - ring of M must either be M itself or contined in the centre of M. 2. Theorem. Let M be -ring nd is centre of M. If M is qudrtic over, then M is t most 4- dimensionl over. 2.4 Theorem. If M is simple -ring nd if U is Lie idel of [. ] then either U or U [. ] except if M is of chrcteristic 2 nd is 4-dimensionl over. 2.5 Theorem. If M is simple non-commuttive -ring then the sub- -ring generted by [, ] is M. 2.6 Theorem. Let M be simple -ring of chrcteristic 2 nd let U be Lie idel of M. Then either U or U [, ]. 2.7 Theorem. If M is non-commuttive simple -ring of chrcteristic 2, then the sub- -ring generted by [, ] is M. 2.8 Theorem. Let M be -ring nd 0 N right idel of M. Suppose tht, given n N nd,( ) 0 for fixed integer n; then M hs non-nilpotent idel. 2.9 Theorem. Let M be -ring hving no-non-zero nilpotent idels in which 2x 0 implies tht x 0. If commutes with ll x x, x,, then is in the centre of M. III. Simple Gmm Rings with Involutions..1 Involution -ring. Let M be -ring. A mpping I: MM is clled n involution if ( i) ( ( ) ( ( ii) ( ( ( ) 2 ( iii) ( ) for ll, bm,. If ( ), then is clled symmetric element of M nd if ( ), then clled skew symmetric element of M..2 Exmple. Let R be n ssocitive ring with 1 hving n involution *. Let n1.1 M = M 1.2 (R) nd : n1, n Then M is -ring. Define I : M M by n I ((, ) = ( *, b * ). Then it is cler tht I is n involution on M.. Theorem. Let M be simple -ring with n involution I on M. Define S, the set of ll symmetric elements x ( x) x nd K, the set of ll skew symmetric elements of M by of M by S x ( x) x. Then S nd K re respectively Jordn sub- -ring nd Lie sub- -ring of M nd M = S. 41 Pge

3 Simple Gmm Rings With Involutions. Proof. We hve I (0) = 0 then 0 S. Let, b S, then I(- = I ()-I( = -b. So b S. Hence S in n dditive subgroup of M. Let, then ( b b ) ( ( b ) ( ( ) ( ) ( = b b b b. Thus b b S. Hence S is Jordn sub- -ring of M. We hve I (0) = 0 = -0, so 0. Let b,, then I(- = I ()-I( = - + b = -(-. Hence b. So K is n dditive subgroup of M. Let, then ( b b ) ( ( b ) ( ( ) ( ) ( ( ( ) ( ) ( b b ( b b ). Thus b b K. Hence K is Lie sub- -ring of M. Since 2M is n idel of M nd M is simple, 2M=M. So for every, x x ( x) x ( x) x mkes sense nd so we cn write x x ( x) x ( x) Now x ( x) ( x) ( x) ( x) x x ( x) x( x) Hence S. 2 x ( x) x ( x) Agin x ( x) ( x) ( x) ( x) x x ( x) x( x) Hence. 2 x ( x) x ( x) There fore x S K. Hence M = S + K. 2 2 Let xs K, then xs nd K. So ( x) x nd ( x) x. Therefore x x. This implies tht 2x 0. So x 0. Thus S K 0. Hence M S K. Now we shll determine the nture of S s Jordn -ring nd tht of K s Lie -ring. Also, if s S nd k K then sk ks S. In studying -rings with involution I two cses immeditely present themselves; these depend on the nture of the involution on certin prescribed subset. The definition we re bout to give should be mde using the centroid rther thn the centre, however in the mteril t hnd it is the centre, even if it is 0, tht plys the crucil role. Nottion. If A is subset of M then will denote the sub- -ring of M generted by A..4 Theorem. Let M be simple -ring with involution I of chrcteristic not 2 nd let S x ( x) x. Then S, the sub- -ring of M generted by S is M unless M is of dimension 4 or less (thus 4 or 1) over its centre. Proof. We clim tht S is Lie idel of M. To see this note first tht trivilly S, S S. If k K nd s S we wnt to show tht s, k S, ; to do so, since s is sum of monomils from S, we need merely do it for monomils s s1 s2... s, s S. But then n 1 2 n 1 2 n 1 i1 i i1 n i s s... s, k s, k s... s... s... s s, k s... s Pge

4 s1 s2... sn 1 sn, k which certinly in S. Thus Simple Gmm Rings With Involutions. S, S, S K S, S S, K S nd so S is Lie idel of M. By definition it is sub- -ring of M. There fore by Theorem 2.2 we conclude tht either S M or S. We consider the second possibility, nmely S. But then S. Given, s k, ss, k K hence s k. Then ( s) ( s) k k,. This implies tht s s s s k k. So s s s s k k. Consequently 2s s s k k 0, which is to sy, M is qudrtic over. By theorem 2., we get tht M is t most 4-dimensionl over. Relted to this theorem is the following remrk which holds for simple -rings of ny chrcteristic which hve involutions..5 Theorem. Let M be simple -ring with involution I whose centre =0 or for which M. Then the only element commuting with dim 4 S x ( x) x liein. Proof. Let commute with ll s S. If the chrcteristic of M is not 2, by theorem.4, S hence follows. Thus we my suppose tht M is of chrcteristic 2. m m s s m, ss,. is clerly sub- -ring of M. Given Let x, then ( x ( x)) ( x ( x). This implies tht x ( x) x ( x). So tht x 2 x ( x) x I( x) 2 ( x). Hence x x ( x) ( x). We wnt to show tht T is Lie idel of M. Given, y, s S then ( y y ) s y s y s y s y s y s 2 s( y) y s sin ce 2 s( y) 0 y s s( y) s( y) y s ( y s s( y)) s ( y) y s ( y s s( y)) s ( y) y s y s s( y) s ( y) y s 2 y s s( y) s ( y) s( y) s ( y) 2y s 0 s ( ( y) ( y)) s ( y y) s we hve just shown. In other words, T is both Lie idel nd sub- -ring of M. By our ssumption on dim M we get from Theorems 2.4 nd 2.5 tht T or T. If T = M then S which we hve seen forces dim R 4. Thus T, which is the ssertion of the theorem. We hve lredy seen in Theorem.4 tht S for most simple -rings. We now wish to estblish its compnion theorem nmely, tht K in generl. To do so we first show nother construction, in most - ring with involution I of Lie idel of the -ring..6 Definition. K K is the dditive group generted by ll k1 k2 with k1, k2k,..7 Lemm. Let M be ny -ring with involution I such tht M = S+K. Then K K is Lie idel of M. Proof. Let k1, k2 nd k K. Then ( k k ) k k( k k ) ( k k kk ) k k ( k k kk ) Pge

5 Simple Gmm Rings With Involutions. KK, so KK, K K K. On the other hnd, if s S then ( k1 k2) s s ( k1 k2) k1 ( k2 s s k2) ( k1 s s k1) k2 K K. ThusK K, S K K. K K, M K K, K S K K, K KK, S K K. Hence Now K K is Lie idel of M..8 Theorem. If M is simple -ring with involution I of chrcteristic not 2, then provided dim M 4. Proof. Then conditions of Lemm.7 hold in M hence K K, s Lie idel of M. By theorem 2.6 must either KK M, M or KK. Now if KK M, M then K KK M, M M by the theorem 2.7. Suppose then tht KK. If K is not invertible then since K nd ll the non-zero elements of re invertible we must hve K 0. In prticulr, 0,. If S then s s K. Hence 0 ( s s ) s s s. There fore S 0. Hence ( S K) S Consequently M is nilpotent left idel nd so = 0. Thus 0 in K forces to be invertible. If bk nd b, then we get 1 b ce n (sin ). Thus K. If ss commutes with then s K forcing s. Now if s S then s s t, t, in fct t S. Thus ( s s ts) ( s s ts). But since ss ts S nd commutes with, so ss ts. Given xm, x s p, ss, p. Hence x x ( s p ) ( s p ) s s s p p s ( p ) ( p ) s s p s p s p ( p) s s p s p s p p s s p ( s s) ( p p) ( ) s s p t p p n. Now x x t x s s pt p p n t x s s p t p p n t ( s p ) s s p t p p n t s t p s s p t p p n t s pt s s t s p p n. Since s we hve seen ss ts, we must hve xx tx. In this wy M hs been shown to be qudrtic over. By theorem 2., M must be t most 4-dimenionl over. This proves the theorem. We now prepre to study the Jordn structure of S. We begin with.9 Theorem. If 0 U is Jordn idel of S then for u U, m, s M, m ( u ) u s ( s) ( u ) u( m) U. Proof. Then proof will consist of breking m nd s into their symmetric nd skew symmetric prts nd verifying tht in these specil instnces the theorem holds. We do this in the sequence of three lemms..10 Lemm. If, x y S nd u U then xu y yu xu,. Proof. 2 x u x x ( xu u x) ( xu u x) x x xu u x x. Since x xs, x xu u x x U. Agin since xu ux U, 44 Pge

6 Simple Gmm Rings With Involutions. so is x ( xu u x) ( xu u x) x U. Thus 2 x u x U. But we hve 2S = S, so we get x u x U. Now Linerizing on x we get x u y U. Similrly we get y u xu. Thus xu y yu x U..11 Lemm. If ss, k K nd x U then su u k ku u su,. Proof. Since uk ku is in S, uu k ku u u ( u k ku) ( u k ku) u is in U. Being Jordn idel of S, s ( uu k ku u) ( uu k ku u) s U. Tht is, (1) s ( uu k ku u) ( uu k ku u) s su u k ku u s s ku u uu ks is in U. Consider (2) k( uu s su u) ( uu s su u) k ku u s su u k ks uu uu s k. Adding (1) nd (2) the right sides dd up to uu ( ks s k) ( ks s k) uu which, since ks sk K, we hve seen must be in U. Therefore the sum of the left sides must be in U; since the left side of (1) is lredy in U we get tht of (2) must lso be in U. Now subtrct (1) from (2); doing so we sty in U. The result on the right is 2( su u k ku u s) U. Since 2S = S this gives su u k ku u s U for ll ss, k K, uu,, which is the desired result..12 Lemm. If, b K nd u U then ( u ) ub b ( u ) u U,. Proof. Since bk, so b uu uu b U. Thus 2( b uu uu U. Since 2 K K( nd so 4 K K) this gives us ( b uu uu ( b uu uu U. But expnding we hve ( b uu uu ( b uu uu ( b uu) ( b uu) ( b uu) ( uu ( uu ( b uu) ( uu ( uu uu b uu b ( b uu ( uu) ( uu) ( b uu b uu b uu b ( u ) ub uu b b uu ( uu) ( b ( uu) b ( u ) ub. Now ( b uu ( uu) ( uu) ( b uu is in U, since 2 b u u bs nd uu U. By Lemm.10 since 2 uu U, 4 uu b b uu U, so ( uu) ( b ( uu) U. The upshot of ll this is tht b u u b U ( ). Linerizing on b we get u u b U ( ). Similrly we get b ( u ) u U. Thus ( u ) ub b ( u ) u U. Proof of theorem.9. Given u U, m, s M then m m0 m1, s s0 s1 with m, s, S, m, s K. Thus ( ) ( ) ( ) ( ) m u u s s u u m ( m m ) ( u ) u ( s s ) ( s s ) ( u ) u( m m ) ( m m ) ( u ) u ( s s ) ( ( s ) ( s )) ( u ) u ( ( m ) ( m )) ( m m ) ( u ) u ( s s ) ( s s ) ( u ) u ( m m ) m ( u ) u s s ( u ) u m ( m ( u ) u s s ( u ) u m ) ( m ( u ) u s s ( m ) u m ) ( m ( u ) u s s ( u ) u m ) Since 4( u ) u U combintion of the three Lmms.10,.11 nd.12. nd since we hve seen the fctor 4 cn be eliminted we obtin the desired theorem s 45 Pge

7 Simple Gmm Rings With Involutions. We re in position to prove the bsic.1 Theorem. The only Jordn idels of S re 0 nd S tht is, S is simple Jordn -ring. Proof. Let U 0 be Jordn idel of S. If u U then we hve seen tht m ( u ) ut ( t) ( u ) u( m) U for ll m, t M. If ( u ) u 0 then M ( u ) um M nd so, given xm then x mi ( u ) uti. But then ( x) ( ti) ( u ) u( mi). Hence x ( x) ( mi ( u ) uti ( ti) ( u ) u( mi )) is in U. Since x ( x) covers S s x runs over M we get tht U = S. Thus if U S we must ssume tht ( u ) u 0 for ll u U. Given u U, m m m, m S, m K we hve u u m m uu u u m m m m uu 2 ( ) 2 ( ) ( ) 2u u m 2u u m 2( ( m ) ( m )) uu 2u u m 2u u m 2( m m ) uu 2u u m 2u u m 2m uu 2muu 2( uu m m uu) 2( uu m m uu) is in U Since u u uu m m uu uu m m uu ( ) 0, 4 ( ( ) ) ( ) 0. We get 4 u u m u u m 0 for ll uu nd m M. By theorem 2.8 we conclude tht uu 0 for ll u, v U. Given ss, v us su U. for ll u U. Linerizing we get tht uv vu 0 Hence 0 uv vu u ( u s su) ( u s su) u uu s u su u su su u 2u su uu s su u 2 u s u, since uu 0. Thus u s u 0 ll u U nd s S. Given k K then k u k S. Hence u k u k u 0. mm, m m0 m1 with m0s, m1 K, then u m u m u u ( m m ) u ( m m ) u ( u m u u mu) ( m m ) u ( u m u u mu) ( m u mu) for For ny u m u m u u m u mu u mu m u u mu mu Hence 2 ( ) 0. There fore u m u m u m 0 m. Thus u m u m u m 0. u m u m By Theorem 2.8, we conclude tht u = 0. We hve prove tht U = 0 or U = S. Hence the theorem is proved. Hving determined the Jordn structure of S we now wnt to determine the Lie structure of K. We begin with the very esy.14 Lemm. If U is Lie idel of K nd if u U, s S then ( uu) s s ( uu) U,. Proof. To see the result merely note tht us su K nd ( uu) s s ( uu) u ( u s su) ( u s su) u U..15 Definition. If U is Lie idel of K then ( U) x K x, U. Clerly T (U) is Lie idel of K nd contins U. We wnt closer tie-in between U nd T(U)..16 Lemm. If U is Lie idel of K then u, v, wu implies u v u T( U) nd uv w w vu ( U),. Proof. Consider u v u, K ; for k K uv u k ku vu u ( vu k ku v) ( vu k ku v) u vu ku u ku v ( ) ( ( )) v ( u ku) ( u ku) v. u v u k v u k v u k v u k u 46 Pge

8 Simple Gmm Rings With Involutions. Since vu k ( vu k) is in K, so its commuttor with u is in U. Since, ( ) ( ). In ll we hve shown tht u v u, K u ku K v u ku u ku v U U nd so u v u ( U). Linerizing on U we obtin uv w w vu ( U). We proceed to prove.17 Theorem. If U is Lie idel of K then for ll u, vu, ( uu v vu u) m( m) ( uu v vu u) T( U) for ll mm,. Proof. We write m s k with ss, k K. Then 1. u u v s s v v s sv uu. u u s su uv v uu s su u gin uu v s s v v s sv uu uu v vu u s s uu v vu u uu s su u v v ( uu s su u) By Lemm.14, uu s su u U, so is in U. Also, since vs sv S, by Lemm.14 is in U. Being in U these re certinly in T(U). Hence ( uu v vu u) s s ( uu v vu u) is in T(U). 2. uu v vu uk k uu v v uu uu( k kv) ( vk k ) uu} ( kuu uuk) ( kuu uuk) v ( kuu uuk ) ( kuu uuk ) v(mod U)( by Lemm.14,sin ce vk kv S) v(( ku uk) u u( ku uk)) (( ku uk) u u( ku uk)) v(mod U) ( ) ( ) ( ) ( ) v u ku u k ku u k u v U v u ku u k ku u k u v U v ku u k u u ku u k ku u k u u ku uk v 2 ( ( ) ) mod 2( ( ) ( ) ) mod. But by Lemm.14, since ku u k U, vu ( u k ku) ( uk ku) uv is in T(U). Then upshot of ll this is tht uu v v uu k k uu v vu u U. Hence ( uu v vu u) m ( m) ( uu v vu u) uu v v uu ( s k) ( s k) ( uu v v uu) uu v v uu ( s k) ( s k) ( uu v v uu) uu v v uu s s uu v v uu uu v v uu k k uu v v uu { } { } is in ( U )..18 Theorem. If M is simple nd dim M 4 nd if U is Lie idel of K then either U K, K or uu v vu u for ll u, v U. Proof. Let uu v vu u, where u, v U. By theorem.17, m ( m) ( U) for ll m M,. If k1 K then b ( m ( m) ) k1 k1 ( m ( m) ) U ( U). Since b ( m k1) ( m k1) ( m) k1 k1 m, so ( m) k1 k1 m ( U) for ll mm, k1 K. We continue in this vein, let k2 K. Then ( ( m) k k m) k k ( ( m) k k m) ( U) ( m) k k ( ( m) k k ) k ( m k ) ( m k ) k ( U). Hence Since k1 ( m k2) ( m k2) k1 ( U), we obtin ( m) k k ( ( m) k k ) ( U). Continuing we get by induction tht for ll k K, ( m) k ( ( m) k) ( U). Since dim M 4, by theorem.8, K M. Then 47 Pge

9 m t ( m t ) ( U) for ll mt., Now Simple Gmm Rings With Involutions. is n idel of M, if 0, then i i i i ( mi ti) M M=M But for ny x, x mt, so ( x) ( mt ) i i i i ( t ) ( ) ( m ) ( t ) ( m ),sin ce ( ). Hence x ( x) mt ( mt ) ( mt ( mt )) ( U). i i i i i i i i Since x ( x) sweeps out K nd we hve tht if 0 then ( U) K. From the definition of T(U) this sys tht U, K K..19 Theorem. If M is simple, dim M 4 nd if U is Lie idel of K such tht [ K, K] uu, uu,. U then given Proof. Since uu v vu u by theorem.18, u u is in the centre of U, the sub- -ring generted by U. However uu s su u U for s S uu k ku u u ( u k ku) ( u k ku) u U for ks, thus u u uu ( s k) ( s k) uu, ss, k K, tht is with ll uu m m uu, for ll m M the chrcteristic is not 2, by theorem 2.9 tht this forces u u to be in..20 Corollry. If [ K, K] U, then uv vu for ll u, v U nd. nd commutes with ll. Since.21 Theorem. If M is simple, dim M 4 nd U is Lie idel of K such tht uu U, implies uu 0, then U = 0. Proof. On linerizing uu 0 we get uv vu 0 for ll u, v U. Thus uv vu. So uu v v uu 0. Given u U, k K then 2 u ku u ku kuu uu k u ku ( u k ku) u u ( u k ku) (sin ce uu 0). Hence u k u U. But then vu ku v 0. Since uv vu, we rrive t uv uv 0. Now ( uv) ( v) ( u) vu uv, tht is, uv, thus for s S, su v s nd so uv su v su vuv uv 0. Given m S, m s k, ss, k K, whence uv m uv m uv 0. The right idel uv is such tht every element in it hs cube 0. By theorem 2.8 this forces uv 0 for ll u, vu. But then for k, u ( u k ku) 0, leving us with uu 0. As bove we then get u su su 0 for s S nd so u is nil right idel, where every element hs cube 0. The outcome of this is tht u = 0 tht is, U = 0. Combining theorems.19 nd.21 we hve.22 Theorem. If M is simple nd = 0 then ny non-zero Lie idel U of K must contin[, ]. References [1] W.E. Brnes : On the gmm rings of Nobusw, Pcific J. Mth 18 (1966) [2] W. Bxter : Lie simplicity of specil clss of ssocitive rings. Proc. Amer. Mth. Soc. 7 (1958), [] I.N.Herstein : Topics in Ring Theory, The University of Chicgo Press, (1969). [4] I.N.Herstein : Lie nd Jordn System in Simple Rings with Involution, Amer. Journl Mth. 78(1956), [5] L. Luh : On the theory of simple Gmm rings, Michign Mth. J.,16(1969), [6] N.Nobusw: On generliztion of the ring theory Osk J. Mth. 1(1964), [7] A.C.Pul nd Sbur Uddin: Lie nd Jordn Structure in Simple Gmm Rings Journl of Physicl Sciences Vol.14 (2010), [8] A.C.Pul nd Sbur Uddin: Lie Structure in Simple Gmm Rings Interntionl Journl of Pure nd Applied Sciences nd Technology Vol.4(2) (2011), Pge