An identity related to centralizers in semiprime rings .. 449 \ hspacetheorem∗{\ thatf i l D l } =An 0 comma identity .. since related we have to assumed centralizers that R is noncommutative in semiprime period rings .. In\quad 449 other words in this case we have a in Z open parenthesis R closing parenthesis period .. But we have the weaker assumption \noindentthat R is atheoremthat semiprime ring and $D in semiprime = 0 rings ,$ Posner\quad quoterightsince s second we have theorem assumed in that $ R $ is noncommutative . \quad In othergeneral words does not in hold this as the case following we have simple $ example a \ in showsZ period ( .. Take R R ) sub 1 . to $ be\ aquad But we have the weaker assumption thatnoncommutative $ R $ is prime a semiprime ring and R sub ring 2 to and be a in commutative semiprime prime rings ring that Posner admits ’ s second theorem in generala nonzero does derivation not d hold : R sub as 2 the right following arrow R sub simple 2 period Thenexample R sub shows 1 times . R\ subquad 2 isTake a noncommutative $ R { 1 semiprime}$ to be a noncommutativering and the mapping prime D : R ring sub 1 and times $ R R sub{ 2 right2 }$ arrow to R be sub a 1 commutative times R sub 2 comma prime D ring open parenthesis that admits r sub 1 comma a nonzero derivation $ d : R { 2 }\rightarrow R { 2 } . $ Then $ R { 1 }\times r sub 2 closing parenthesis = open parenthesis 0 comma dAn open identity parenthesis related to rcentralizers sub 2 closing in semiprime parenthesis rings closing449 parenthesis is a Rnonzero{ 2 }$theorem is a that noncommutativeD = 0, since semiprime we have assumed that R is noncommutative . In other words in ringderivation andthis thewhich case mapping mapswe have R suba ∈ $D 1Z( timesR). : RBut sub R we 2{ intohave1 its}\ the center weakertimes period assumptionR In the{ 2 sequelthat}\R weisrightarrow a shall semiprime prove that ringR and{ in1 }\times R { 2 } ,in D case we (semiprime have r an{ rings inner1 } Posner derivation,’ r s second D{ comma2 } theorem) which = in is general commuting ( does 0 on not , a hold noncommutative d as ( the following r { 2 simple} ) example )$ is anonzero derivation which maps $ R { 1 }\times R { 2 }$ into its center . In the sequel we shall prove that semiprimeshows ring . comma Take thenR1 to D be = 0 a period noncommutative For this purpose prime we ring put and x plusR2 to y forbe a x commutative in open parenthesis prime 3ring closing that parenthesis period Wein have caseadmits we have a nonzero an inner derivation derivationd : R → R . $Then DR ,× $R whichis a noncommutative is commuting semiprime on a ring noncommutative and the semiprime ring , then $D =2 02 . $ For1 this2 purposeweput $x + y$ for $x$ in ( 3 ) .Wehave brackleftmapping D open parenthesisD : R1 × R2 x→ closingR1 × R parenthesis2,D(r1, r2) comma = (0, d( yr2 brackright)) is a nonzero plusbrackleft derivation D which open mapsparenthesisR1 × R y2 closing parenthesis comma x brackrightinto its center = 0 comma . In the x sequelcomma we y in shall R comma prove that and in in case particular we have for an y = inner a comma derivation sinceD, Dwhich open parenthesis is a closing parenthesis\noindentcommuting = 0$[D(x),y]+[D(y),x]=0,x comma on a noncommutative semiprime ring , then D = 0. For this purpose we put x + y for x ,we y obtain\inin brackleft ( 3 )R . We D have,$ open parenthesisandinparticularfor x closing parenthesis $y comma = a brackright a ,$ = 0 since comma x $D in R period ( In a other ) words = we 0 have, $ [D(x), y] + [D(y), x] = 0, x, y ∈ R, and in particular for y = a, since D(a) = 0, Equation:we open obtain parenthesis [D(x), a] = 4 closing 0, x ∈ R. parenthesisIn other words .. D to we the have power of 2 open parenthesis x closing parenthesis = 0 comma x in R\noindent period weobtain$[ D ( x ) , a ] = 0 , x \ in R . $ In other words we have Posner quoteright s first theorem brackleft 6 brackright .. asserts that if R is a prime ring of characteristic \ begindifferent{ a l ifrom g n ∗} two and D comma G are nonzeroD derivations2(x) = 0, on x R∈ commaR. then DG cannot be (4) \ taga derivation∗{$ ( period 4 ) .. $ Let} D us ˆ point{ 2 out} that( Posner x ) quoteright = 0 s first , theorem x \ doesin notR. hold for \endsemiprime{ a l i gPosner n ∗} rings ’ period s first .. theorem However [ 6 comma ] asserts it is wellthat known if R is that a prime if D ring and ofDto characteristic the power of different 2 are derivations from two in 2 hyphenand torsionD,G freeare semiprime nonzero derivations ring comma on thenR, then D = 0DG opencannot parenthesis be a derivation see the proof . of Let Lemma us point 1 period out that 1 period 9 in brackleft 5\noindent brackrightPosner closingPosner ’ parenthesiss first ’ s theorem first period does theorem not hold [ for 6 ] \quad asserts that if $ R $ is a prime ring of characteristic 2 differentHence wesemiprime have from D = two rings 0 period and . However .. $D In other , it , words is well G$ a knownin Z are open that nonzero parenthesis if D and derivationsD R closingare derivations parenthesis on in $R period , .. $ Now then it follows $DG$ from open cannot be parenthesisa derivation2 2 - closing torsion . parenthesis\ freequad semiprimeLet that us ring point , then outD = that 0( see Posner the proof ’ of s Lemma first 1 . theorem 1 . 9 in [ 5 does ] ) . not hold for T open parenthesisHence we have x closingD = 0 parenthesis. In other = words ax commaa ∈ Z( andR). T openNow itparenthesis follows from x closing ( 2 ) that parenthesis = xa for all x in R comma which\noindent completesT (semiprimex) =theax, proofand of ringsT ( thex) = theoremxa . \forquad all xHowever∈ R, which , completes it is well the proof known of the that theorem if in $D$ case R has and an $Dˆ{ 2 }$ are derivations in in case Ridentity has an element identity . element period \noindentIn the sequelIn2 the− we sequeltorsion present we present the free proof the semiprime of proof Theorem of Theorem 1 ring without 1, without then assuming assuming $D that R that = hasR 0has an ( identity $ see element the proof . The of Lemma1 . 1 . 9 in [ 5 ] ) . an identityproof element is , as we period shall The see , proof rather is long comma but itas is we elementary shall see commaP parenleft rather− longifive but−nt it−oo is− elementaryhe sense h−te− e e−brackleftT −ccparenleft−ia−fi−x \noindentP parenleft-ihatHencewehave−r five-nit −− t-orequires o-h subno e$Dsp sense h-t = e-h 0 sub a .$ t-r it\ toquad the powerInparenright other of e− endashc k− wordsperiod requires− $xbrackright a no sp\ toin−noequal the powerZ−w ( of e-brackleft R ) T-c . $ c\quad parenleft-iNowl a-fi-xe it− zerocomma follows sub parenright-c− fromdge concerning k-period-x( 2 ) that semiprime sub brackright-n rings in o ord equal-wW − le e-zeror t n − comma-do − it − f sube−on ge− concerningl l o − dw − semiprime rings in ord W-ett r− t n-o-ioh − t-fpr− subeo − e-op − n-lv lr o-do − w-teo − t-oth h-p− f. sub r-e o-p-v r o-e o-t h-f sub period dertoa \noindentt sub olrtolr o o$To r o aa r aa sub (x x in∈ xR. R periodI )n o =I n ochie to ax the power ,$and$T of derto a chie ( x ) = xa$ forall $x \ in Ru .. , t $ .. a which .. T s .. completes h r l t the proof ofu the t a theoremT s h r l t inopen case parenthesis $ R $ 6 closing has an parenthesis identity element . Sin .. r to the power of i-t ten n .. th e sub f (6) \noindenti P o sub uIn tin the gin t sequel e el .. io we n open present parenthesis the 1 proof closing of parenthesis Theorem x 1 r sub without plus y f assuming o sub or to that the power $ R of $n x nhas i to the an identity element . The proof is , as we shall see , rather long but it is elementary power of n sub e btaii−t s .. o mm u .. n n n $P{ p a rSin e n l er f t −iten} nf i th v ee−fni P o tu−tino gin o− th e el{ ioe n}$ (1)xr sense+y f oorx $n h−ietbtai e s−h o mm{ a u tn−r } i t ˆ{ e } −−{ r e q u i r e s }$ t .. h .. r .. t t h r t noEquation: $ sp ˆ{ opene−b parenthesis r a c k l e f t to the T−c power c{ ofp a7 r closing e n l e f t parenthesis−i } a−f .. i − xx to{ theparenright power of closing−c }} parenthesisk−period y plus−x T{ subbrackright open −n parenthesiso equal plus−w } closing$ l parenthesis $ e−zero plus comma−d { ge }$ concerning semiprime rings in ord $ W−e $ r t $ n−o−i t−f { e−o } n−l $ l $ o−d w−t t−o h−p { r−e } o−p−v $ x)y + T + )+ (7) r $ o−e o−t h−f { . }$ (

\noindent $ t { o l r }$ o o r $ aa { x }\ in R . $ I n $ o ˆ{ derto a }$ c h i e

\ centerline {u \quad t \quad a \quad $ T $ s \quad h r l t }

\ begin { a l i g n ∗} ( 6 ) \end{ a l i g n ∗}

\noindent Sin \quad $ r ˆ{ i−t }$ ten n \quad th $ e { f }$ i P $ o { u }$ tin gin t e el \quad i o n $ ( 1 ) x r { + } y $ f $ o ˆ{ n } { or } x $ n $ i ˆ{ n } { e }$ b t a i s \quad o mm u \quad n

\ centerline { t \quad h \quad r \quad t }

\ begin { a l i g n ∗} \ tag ∗{$ ( ˆ{ 7 } ) $} x ˆ{ ) } y + T { ( } + ) + \end{ a l i g n ∗} 450 .. J period Vukman \noindentOur next step450 is\ toquad proveJ the . Vukman relation For .. h i-s p u r .. Row 1 e Row 2 s . .. p u .. t n t sub h to the power of e r l-e t o to the power of n open parenthesis 7 sub closing\ centerline parenthesis{Our 2 to next the power step of is parenleft-x to prove y plus the relation } T x plus y closing parenthesis x 2 T to the power of open parenthesis x y plus y plus T x 2 closing parenthesis plus open parenthesis\ centerline y plus{For y Equation:\quad h closing $ i− parenthesiss $ p u T r =\quad closing parenthesis$\ l e f t . p\ Tbegin open{ parenthesisarray }{ c x} pluse \\ T xs to\ theend power{ array of} closingwe\ right . $ parenthesis\quad p u x\ 3quad 2 x closingt n parenthesis$ t { h plus ˆ{ e x}} open$ parenthesis r $ l−e x $ plus t to $ the o powerˆ{ n of} plus( open 7 { parenthesis) } 2 2ˆ{ x plusp a r e y n lopen e f t −x } parenthesisy + $ x} T x to the power of closing parenthesis .. closing parenthesis 4 T open parenthesis x plus x to the power of y plus y closing parenthesis450 J plus . Vukman plus T y to the power of closing parenthesis plus x y \ begine h a{ va e l i plus g n ∗} x T to the power of openOur parenthesis next step x is closing to prove parenthesis the relation y y T open parenthesis x closing parenthesis x plus 2 T x + y ) x 2 T ˆ{ ( } xy+y+Tx2)+(y+y \\ ) x to the power of T open parenthesis y x pluse x T open parenthesis plus sub Tn openparenleft parenthesis−x y x comma x comma y For h i − s p u r p we p u t n t e r l − e t o (7 2 y + 4h-e T r h an ( d comma x + u n-i x g ˆ ..{ openy } parenthesis+s y 7 to ) the + power + ofh closing T y parenthesis ˆ{ ) } a-n+ d open x parenthesis y \ tag ∗{$ 1 closing ) T parenthesis = ) Tw .. e ( b t a x n + T x ˆ{ ) } x32x)+x(x+ˆ{ + } ( 2 x + y ( x T x ˆ{ ) }$} Line 1 4 T x open parenthesisT x y+ plusy ) yx Line2T 2( 2x 2 T y open+ y parenthesis+ T x2 y )plus + 2 ( toy the+ powery of T open parenthesis y x plus y to\end the{ powera l i g n of∗} T open parenthesis x closing parenthesis plus 8 T open parenthesis x to the power of y x closing parenthesis = T y ) sub open parenthesis x x plus x T x sub closing parenthesis) to 4 theT ( powerx + x of+ y) plus + +T 2 y x+ 2 Linexy 3 plus 2 sub T open parenthesis y ) + ) closing\noindent parenthesise h ax plus v e y to $ the + power x of T T ˆ open{ ( parenthesis} x)yyT(x)x+2xˆ x closing)T =)T parenthesis(x + T x x32 xx) plus + x y(x x+ T open(2x + parenthesisy(xT x x to the{ T power} ( ofy closing x parenthesis + x T8 open ( parenthesis + { T( x x Line} 4y W sub x e h sub, a x e t sub , he y e sub $ or e ( T $ h−e$e h rhand,u a v e + xT x)y $n yT (−xi)x $+ g 2x\quad(yx + x$ T (( 7 + ˆT{( )yx,} x,a− yn h $− e dr h( an 1 d) , w u \nquad− i g e b t a n (7)a − n d ( 1 ) w e b t a n \ [ \ begin { a l i g n e d } 4 T x ( y + y \\ 2 2 T ( y + 2 ˆ{ T } ( y x + y ˆ{ T } 4(T x x(y + )y + 8 T ( x ˆ{ y } x ) = T { ( } x xT + xT T x { y) ˆ{ y }} + 2 x 2 \\ 22T ( y + 2 (yx + y ( x) + 8T (x x) = T( xx + x T x)y + 2x2 + 2 { T } ( y ) x + y ˆ{ T } (x)x+yxT(xˆ{ ) } 8 T ) ( x x \\ +2T (y)x + y (x)x + yxT (x 8(xx W { e } h { a } e t { he } e { or } e \end{ a l i g nW e de }\h]aetheeore An identity related to centralizers in semiprime rings .. 45 1 AnWe identity have therefore related to centralizers in semiprime rings \quad 45 1 Weplus have x plus therefore x plus open parenthesis y open parenthesis sub x plus 3 3 x to the power of T open parenthesis closing parenthesis e h .. and comma we o tain fi stu i ng open parenthesis 8 closing parenthesis an d h n to the power of closing parenthesis f terc l\ ctings[ + to x the power + x of x + ( y ( { x } + 3 3 x ˆ{ T } () \ ] o Line 1 1 6 T open parenthesis x 2 yx plus xyx sub 2 closing parenthesis = 16 T open parenthesis x open parenthesis xy closing parenthesis\noindent closinge h \ parenthesisquad and plus ,weotain 1 6 T open parenthesis $ fi $ x parenleft-ystu i ng(8)andh x closing parenthesis $nˆ x closing{ ) parenthesis}$ f terc Line l 2 = $ Line ctings ˆ{ x }$ o 3 2 2 to the powerAn identity of y plus related x closing to centralizers parenthesis in semiprime plus x rings x 2 x y45 T 1 2 We closing have parenthesistherefore plus Line 4 2 open parenthesis y closing parenthesis x open parenthesis 2 plus 2 to the power of plus open parenthesis x plus 2 closing parenthesis x T y 2 Line 5 T 2 x \ [ \ begin { a l i g n e d } 1 6 T ( x 2 yx + xyx { T2 } ) = 16 T ( x ( xy closing parenthesis y x closing parenthesis+ x open+ parenthesisx + (y sub(x 2+ 2 plus 3 y 3 yx closing() parenthesis plus plus 2 Line 6 x y 2 T 4 2 T x) x x )2 2 to + the power1 6 of plus T 2 2 ( x Line x 7 2 parenleft x 2 y plus− plusy 2 plusx plus ) sub x 2 y ) 2\\ x open parenthesis x 3 open parenthesis closing e h and , we o tain fi stu i ng ( 8 ) an d h n) f terc l ctingsx o parenthesis= \\ Line 8 plus x sub 2 plus 2 x plus minus x y minus x Line 9 open parenthesis closing parenthesis T x minus 2 2 ˆ{ y } +x)+xx2xyT2)+ \\ 2 ( y ) x16 (T (x 2yx ++ xyx 22) ˆ{ =+ 16T}(x(xy()) + x 16 +T (xparenleft 2 )− xyx)x T) y 2 \\ T 2 x ) y x ) ( { 2 } 2 + y y ) + += 2 \\ xy2T42Txxx22ˆ2 2y + x) +{ +x} x22xy 2 T x2)\\ + 2 x 2 y + + 2 + + { 2 y } 2 x ( x 3 ( ) \\ 2( y) x (2 + 2+ ( x + 2 ) x T y 2 + x { 2 } + 2 x + − x y − x \\ ( ) T x − \end{ a l i g n e dT}\2 ] x ) y x)(2 2 + y y ) + + 2 x y 2T 42 T x x x 2 2+ 2 2 x

2 x2y + + 2 + +2y 2 x (x 3 ( )

+ x2 + 2 x + − x y − x () T x − 452 .. J period Vukman \noindentSubstituting452 T open\quad parenthesisJ . Vukman x closing parenthesis y for y in the above relation gives open parenthesis 1 7 closing parenthesis xT open parenthesis x closing parenthesis y brackleft x to the power of 2 comma T open\ centerline parenthesis{ Substituting x closing parenthesis $T brackright ( x plus )2 x to y$ the power for of $y$ 2 T open in parenthesis the above x closing relation parenthesis gives y brackleft} T open parenthesis x closing parenthesis comma x brackright plus 2 T open parenthesis x closing parenthesis yx to the power of 2 brackleft\ begin { Ta l open i g n ∗} parenthesis x closing parenthesis comma x brackright Equation: x comma y in R period .. plus T open parenthesis x(17)xT(x)y[xˆ closing parenthesis yx brackleft x comma T open parenthesis{ x2 closing} , parenthesis T ( brackright x ) x plus ] T + open 2parenthesis x ˆ{ x2 closing} T parenthesis(x)y[T(x),x]+2T(x)yxˆ y brackleft x comma T open parenthesis x closing parenthesis brackright x to the power of 2{ =2 0 comma} [ T ( x ) , x ] \\ +T(x)yx[x,T(x)]x+T(x) Left multiplication452 J . Vukman of open parenthesis 1 6 closing parenthesis by T open parenthesis x closing parenthesis leads to y[x,T(x)]xˆopen parenthesis 1 8 closing parenthesisSubstituting T openT (x) parenthesisy{ for2 y} in= the xclosing above 0 relation , parenthesis\ tag gives∗{$ xy x brackleft , y x to\ thein powerR of . 2 comma$} T open\end{ parenthesisa l i g n ∗} x closing parenthesis brackright plus 2 T open parenthesis x closing parenthesis x to the power of 2 y brackleft T open parenthesis x closing parenthesis comma x brackright plus 2 T open parenthesis x closing parenthesis yx to the power of 2 brackleft\ centerline T open{ Left parenthesis multiplication x(17) closingxT parenthesis(x)y[x of2,T ((x comma 16)] + 2 )x x2T by brackright(x)y $T[T (x), Equation: x] ( + 2T x(x) xyx comma )2[T $(x) yleads, x in] R period to } .. plus T open parenthesis x closing parenthesis yx brackleft x comma T open+ parenthesisT (x)yx[x, T x( closingx)]x + T parenthesis(x)y[x, T (x brackright)]x2 = 0, x plus T x, open y ∈ R. parenthesis x closing parenthesis\ begin { a l y i g brackleft n ∗} x comma T open parenthesis x closing parenthesis brackright x to the power of 2 = 0 comma (18)T(x)xy[xˆ{ 2 } ,T(x)]+2T(x Subtracting open parenthesis 1 8 closingLeft multiplication parenthesis from of ( open 1 6 ) parenthesis by T (x) leads 1 7 to closing parenthesis we arrive at )brackleft x ˆ{ 2 T} openy[T(x),x]+2T(x)yxˆ parenthesis x closing parenthesis comma x brackright y brackleft T open parenthesis{ x2 closing} [T( parenthesis commax ) x to , the x power ] of\\ 2+T(x)yx[x,T(x)]x+T(x brackright minus 2 brackleft T open parenthesis x closing parenthesis comma x to the power of 2 brackright)y[x,T(x)]xˆ y brackleft T open(18) parenthesisT (x)xy x[x closing2,T (x)] parenthesis + 2T ({x)x22 commay}[T (x=), x] brackright + 0 2T (x , )yx\ tag2 =[T 0(∗{x comma)$, x] x x comma , y y in\ in R periodR . $} \end{ a l i g n ∗} We set +T (x)yx[x, T (x)]x + T (x)y[x, T (x)]x2 = 0, x, y ∈ R. a = brackleft T open parenthesis x closing parenthesis comma x brackright comma b-b = to the power of plus brackleft to \ centerline { Subtracting ( 1 8 ) from ( 1 7 ) we arrive at } the power of a-T-y open parenthesis x-equalSubtracting closing ( 1 parenthesis 8 ) from ( comma1 7 ) we comma-x-zero arrive at to the power of 2 sub brackright comma element-c = to the power of R minus 2 brackleft T open parenthesis x closing parenthesis comma x to the power of 2 brackright \ [[T(x),x]y[T(x),xˆ{ 2 } ] − 2 [ T ( period [T (x), x]y[T (x), x2] − 2[T (x), x2]y[T (x), x] = 0, x, y ∈ R. xT-P ) h-u , sub xt-e ˆ i-n-t{ 2 n-t} h-g]y[T(x),x]=0,x,y sub e-i a parenleft-b sub one-o v-nine parenright-e r-y a-e z-l sub a t-f sub i-o o-r n\ bin e-w c-eR o We set b-m-o. \ ] e-t a-s open parenthesis 2 0 closing parenthesis zb plus c yaz = a 0 comma y in Let .. mu tia = l i [T at(x on), x ..], f open b − b parenthesis=+ [a−T −y 9(x closing− equal parenthesis), comma − bx y− ayzero gi 2 element − c =R −2[T (x), x2]. \ centerlineparenleft-parenleft{We s e to t the} power of two-two two-one parenright-parenright], a-parenleft y-a to the power of y-a z-c b-minus c-plus y-a y-a to the power of parenright-c z-z a-a = sub = zero-zero sub comma z-z sub comma y-y element-element R-R sub period T − P h − ut−ei−n−tn − th − ge−i a parenleft − bone−ov−nineparenright−er − ya − ez − lat−fi−oo−rn b e − wc − e o b − m − oe − ta − s \ [a=[T(x),x],bS-L sub e-u tb r-n-a parenleft-c two-t sub two-i sub n parenright-g−b z open = ˆ{ parenthesis+ } [ b-two ˆ{ a− one-eT−y sub} closing( x− parenthesisequal z ) r-c-f , comma−x−zero ˆ{ 2 } { ], } element−c = ˆ{ R } − 2 [ T ( x ) , x ˆ{ 2 } ] o m-w e open parenthesis zero-o-two b t-a-w sub(20) i nozb tain+ cyaz = a 0 , y ∈ . \open] parenthesis 2 3 closing parenthesis ayc minus c-y a closing parenthesis zcy = a 0 z comma in R Right m uli li c a iono .. open parenthesisLet 22 mu by ti .. l yc i at giv on f ( 9 ) b y ay gi open parenthesis 24 closing parenthesis ayc minus c ya closing parenthesis z ay c = 0 z comma in R \ centerlineSub tr ctin ..{ 23$ closingT−P parenthesis h−u { t− f r-oe m i ..−n 24−t closing} n− parenthesist h−g ..{ w e−i }$ a $ parenleft −b { one−o v−nine parenleft−parenlefttwo−twotwo−oneparenright−parenright a−parenlefty−ay−az−cb−minusc−plusy−ay−aparenright−cz−za−a = zero−zero z−z y−yelement−elementR−R parenrightw to the power−e } of hr− ey a−e z−l { a t−f { i−o o−r n }}$ b $ e−w c−e $ o $ b−m−o = , , . e−t a−s $ }

S − Le−u tb r − n − aparenleft − ctwo − ttwo−inparenright−g z(b − twoone − e)zr−c−f o m − w e (zero − o − two b t − a − wi no tain \ [( 2 0 ) zb+c yaz=a 0 , y \ in \ ] (23) ayc − c − ya)zcy = a 0 z, ∈ R Right m uli li c a iono ( 22 by yc giv \ centerline { Let \quad mu t i l i at on \quad f(9)by $ay$ gi } (24) ayc − cya)zayc = 0 z, ∈ R \ [ p a r e n l e f t −p a r e n l e f t ˆ{ two−two two−one } parenright −parenright a−p a r e n l e f t y−a ˆ{ y−a z−c b−minus } c−plus y−aSub y− tra ctin ˆ{ parenright 23 ) f r − o−mc 24 z− )z w a−a } = { = } zero−zero { , } z−z { , } y−y element−element R−R { . }\ ] wh e

\ centerline { $ S−L { e−u }$ tb $ r−n−a p a r e n l e f t −c two−t { two−i { n parenright −g }} z ( b−two one−e { ) z r−c−f }$ o $ m−w $ e $ ( zero−o−two $ b $ t−a−w { i }$ no t a i n }

\ [ ( 2 3 ) ayc − c−ya ) zcy=a0 z , \ in R \ ]

\ centerline { Right m uli li c a iono \quad ( 22 by \quad $ yc $ giv }

\ [ ( 24 ) ayc − cya ) zayc=0z , \ in R \ ]

\ centerline {Sub t r c t i n \quad 23 ) f $ r−o $ m \quad 24 ) \quad w }

\ [ w ˆ{ h } e \ ] An identity related to centralizers in semiprime rings .. 453 AnCombining identity open related parenthesis to centralizers 1 9 closing parenthesis in semiprime with open parenthesis rings \quad 25 closing453 parenthesis we arrive at Combininga-x brackright-y ( 1 y-parenleft9 ) with to ( the 25 power ) we of arrive brackleft-b at plus-parenleft c-x parenright-parenright = sub x brackright-zero sub comma zero-y-comma in comma-R-x sub y In F-o r-t h-o sub e-m r t-w-h e-o r d-a sub b-s \ beginx to the{ a l power i g n ∗} of brackleft T open parenthesis = 0 x R .. W .. h .. e .. r a−7x brackleft brackright T open parenthesis−y y−p sub a r e x n comma l e f t ˆ xT{ b open r a c parenthesisk l e f t −b } x subplus T− subp a x r to e n the l e f powert c of−x x sub parenright closing parenthesis−p a r e comma n r i g h t = brackright{ x } 0brackright = comma 0− inzero R x comma{ , } in Rzero−y−comma \ in comma−R−x { y }\\ In F−o r−t h−o { e−m r }S ubt− c-sw− h i sub e− i-to ut p-o r n-i-l d−a e x plus{ b− ys f o} x r i n .. open parenthesis 7 closing parenthesis i-v sub e s \end{ a l i g n ∗} x to the powerAn identity of 2 brackright related to centralizers plus T open in semiprime parenthesis rings x closing453 Combining parenthesis ( 1 comma 9 ) with xy ( 25plus ) we y x arrive brackright at plus brackleft T closing parenthesis comma x y plus y \noindentPutti gin t h$ x ˆ{ [T } ( = 0 x R $ \quad W \quad h \quad e \quad r brackleft−b w .. i h ..a − bxbrackright to the power− ofyy t− a ..parenleft n .. n sub c .. w ..plus e a − parenleftc − xparenright − parenright =x brackright − zero, zero − y − comma ∈ comma − R − xy \ [ 7 [ T{ ( } { x } , xT ( x { T { x }}ˆ{ x } { ), } = ] 0 ={ , } 0 \ in Equation: open parenthesis .. 8 brackleft plus y x to the power of brackright InF − or − th − oe−mrt − w − he − ord − ab−s Rt to x the , power\ in of i nR .. i\ ..] t sub h e b to the power of o v .. r .. a sub t i n to the power of 2 open parenthesis y x .. r y w to the power ofx[ eT ( o .. = b .. 0 a ..x o ..R n ..W a n h e r 0 = 2 T 2 x closing parenthesis y plus 2 2 plus plus plus T y closing parenthesis 2 y T to the power of open parenthesis 2 = \noindent S ub $ c−s $ h $ i { i−t }$ ut $ p−o n−i−l$e$x + y$ fo$x$rin \quad sub 2 x sub T open parenthesis x closing7 [T parenthesis( , xT (xx comma=]0 = y, brackright0 ∈ Rx, plus∈ 2R T open parenthesis x parenright-comma y sub x Tx ), brackright$ ( 7 x plus ) 4 i sub−v brackleft{ e }$ T x s closing parenthesis comma sub y x plus T open parenthesis x brackleft-parenright comma x S ub c − s h i ut p − on − i − l e x + y f o x r i n (7)i − v s brackright plus x bracklefti− Tt sub open parenthesis y sub closing parenthesise comma x brackright Equation: open parenthesis to the power\ [ x of ˆ{ 292 closing} ]+T(x),xy+yx]+[T),xy+y parenthesis .. plus 2 T y plus 4 brackleft to the power of T open parenthesis x x sub y 2 plus y 2 \ ] x2] + T (x), xy + yx] + [T ), xy + y T .. hu to the power of s .. a v x Putti gin t h w i h bt a n n w e a \noindent Putti gin t h c w \quad i h \quad $ b ˆ{ t }$ a \quad n \quad $ n { c }$ \quad w \quad e a 8 [ +yx] ( \ begin { a l i g n ∗} i o 2 e \ tag ∗{$ ( $} 8 [t n + i yth e b xv ˆ{ r] }at i n ( y x r yw o b a o n a n \end{ a l i g n ∗}

\ centerline { $ t ˆ{ i }$ n \quad i \quad $ t {0 =h } 2$T 2x e) y $ b+ ˆ{ 2o 2} +$ v +\quad + Tr y)\quad 2yT ( 2$ a { t }$ i $ n ˆ{ 2 } ( y x $ \quad r $ y w ˆ{ e }$ o \quad b \quad a \quad o \quad n \quad a n } =2 xT (x), y] + 2T (xparenright − commay]x + 4[T x),y x + T (xbrackleft − parenright, x] + x[T(y),x ] + 2 T y + 4[T ( x x 2 + y2 \ begin { a l i g n ∗} y (29) 0 = 2 T{ 2 } x)y+22+++Ty)2yTˆ{ ( } 2 \\ = { 2 } x { TT (hu xs } a v) , y ] + 2 T ( x parenright −comma y { ] } x +{ 4 } { [ T } x ) , { y } x + T ( x brackleft −parenright , x ] + x{ [ } T { ( } y { ), } x ] \\\ tag ∗{$ ( ˆ{ 29 } ) $}x + 2 T y + 4 [ ˆ{ T } ( x x { y } 2 + y 2 \end{ a l i g n ∗}

\noindent T \quad $ hu ˆ{ s }$ \quad a v

\ [ x \ ] 454 .. J period Vukman \noindentAccording454 to the\quad relationJ brackleft . Vukman T open parenthesis x closing parenthesis comma x brackright x plus x brackleft T open parenthesis x closing parenthesis comma x brackright = 0 open parenthesis see open parenthesis 27 closing parenthesis closing parenthesis\ hspace ∗{\ onef i can l l } replaceAccordingtotherelation in $[ T ( x ) , x ] x + x [ T (the x above ) relation , x x to the ] power = of0 2 brackleft ($ see(27))onecanreplacein T open parenthesis x closing parenthesis comma x brackright by brackleft T open parenthesis x closing parenthesis comma x brackright x to the power of 2 comma which gives \ centerlineLine 1 parenleft-parenleft-parenleft{ the above relation to the $ power x ˆ{ of2 three-three-three} [T( zero-one-two x ) parenright-parenright-parenright , x ]$by$[T( x-x-brackleft x to) the , power x of ] T-two-brackleft-T x ˆ{ 2 } , open $ parenthesis which gives T-parenleft-x} parenright-parenleft-x sub x-parenright-comma parenright-x- comma x-comma-brackright454 J . Vukman brackright-x-x to the power of 2 = = zero-zero-zero sub comma-comma x-x-x element-element-element R-R-R\ [ \ begin Line{ 2a la-W i g n en d c-d-e} p aLine rAccording e n 3 l e 0 f yt − brackrightp to a rthe e n relationl e fx t to−p the a[T r( epowerx n), l xe] fx t of+ ˆ2x{ plus[Tthree(x) 4, xbrackleft]− =three 0( see T− ( openthre 27 ) parenthesise ) one zero can replace− xone comma−two in x} brackrightparenright plus −parenright −p a r e n r i g h t Tx− openx−b r parenthesis a c k l e f t ˆ Line{ T− 4two minus−bthe r x a brackleft c above k l e f t relation−T T Line} x(T 52[T open(x)−, parenthesisxp] a by r e n[T l( ex f) t, closing− x]xx2, which parenrightparenthesis gives to− thep a r power e n l e f tof− commax { x y− brackrightparenright x −comma } plusparenright 4 x y T comma−x−comma x plus Line x−comma 6 open−brackright parenthesis 33 closing brackright parenthesis−x−x x 2 ˆ{ brackleft2 } = T open = parenthesis zero−zero x closing−zero parenthesis{ comma−comma } commax−x−x y plus element brackleft−element T open−element parenthesis x R− closingR−R \\ parenthesis y brackright 2 plus 3 brackleft T open parenthesis x closing three−three−threezero−one−two T −two−brackleft−T 2 parenthesisa−W nparenleft comma c−d x-brackleft−e− parenleft\\ y-brackright-T− parenleft open parenthesis sub x x toparenright the power− ofparenright plus y 3 x-y− parenright minus open parenthesisx − x − brackleft T (T − parenleft − xparenright − parenleft − xx−parenright−commaparenright − x − commax − comma − brackrightbrackright − x − x == zero − zero − zerocomma−comma x − x − xelement − element − elementR − R − R parenright-parenleft0 y ] x x ˆ parenright-x{ 2 } +4[T(x,x]+T( sub comma y x-plus to the power of 2 sub minus y brackleft\\ T open parenthesis x closing a − W nc − d − e − x [ T \\ parenthesis comma brackright-x period 0 y]x2 + 4[T (x, x ] + T ( (e ah vet) ˆ he{ ref, } o rey]x+4xyT,x+ \\ (33)x2[T(x),y+[T(x)y]2+3[ −x[T T ( x ) , x−b r a c k l e f t y−brackright −T( { x } x ˆ{ + } y 3 x−y − (T (),y]x + 4xyT , x+ parenright −parenleft x parenright −x { , } y x−plus ˆ{ 2 } { − } y [ T ( x ) (33) x2[T (x), y + [T (x)y]2 + 3[T (x), x − bracklefty − brackright − T ( x+y3x − y − (T parenright − parenleftxparenright − x yx − plus2 y[T (x), brackright − x. , brackright −x . \end{ a l i g n e d }\ ] x , − e ah vet he ref o re

\ centerline {e ah vet he ref o re } An identity related to centralizers in semiprime rings .. 455 AnRight identity multiplication related of open to centralizers parenthesis 33 closing in semiprime parenthesis by rings x gives\quad 455 RightEquation: multiplication open parenthesis of 35 ( closing 33 ) parenthesis by $ x $ .. x gives to the power of 2 brackleft T open parenthesis x closing parenthesis comma y brackright x plus brackleft T open parenthesis x closing parenthesis comma y brackright x to the power of 3 plus 3 brackleft\ begin { Ta l open i g n ∗} parenthesis x closing parenthesis comma x brackright yx to the power of 2 plus 3 xy brackleft T open parenthesis x\ tag closing∗{$ parenthesis ( 35 comma ) $} xx brackright ˆ{ 2 } x[T(x),y]x+[T(x), plus 2 x brackleft T open parenthesis x closing parenthesis comma y brackright x to they power ] ofx 2ˆ{ plus3 } T open+3[T(x) parenthesis x closing parenthesis brackleft ,x]yxˆ y comma x to{ the2 } power+ of 3 2 brackright xy [ x plus T brackleft ( x y),x]x+2x[T(x),y]xˆ comma x to the power of 2 brackright T open parenthesis x closing parenthesis x plus x{ brackleft2 }\\ T+ open T parenthesis ( x x closing ) [ y , x ˆ{ 2 } ] x + [ y , x ˆ{ 2 } ]T(x)x+x[T( parenthesis commaAn identity x brackright related to yx centralizers = 0 comma in semiprime x comma rings y in R455 period Right multiplication of ( 33 ) by x gives x),x]yx=0,x,ySubtracting open parenthesis 35 closing parenthesis from open parenthesis\ in R. 34 closing parenthesis we obtain \endx{ toa thel i g n power∗} of 2 y brackleft T open parenthesis x closing parenthesis comma x brackright plus 3 xy brackleft x comma brackleft T open parenthesisx2[T x(x closing), y]x + parenthesis [T (x), y]x3 comma+ 3[T (x x), brackright x]yx2 + 3xy brackright[T (x), x]x plus+ 2x 2[T xy(x) brackleft, y]x2 T open(35) parenthesis x closing \noindent Subtracting ( 35 ) from ( 34 ) we obtain parenthesis comma x brackright x plus+ brackleftT (x)[y, x y2] commax + [y, x x2] toT (x the)x + powerx[T (x of), 2 x] brackrightyx = 0, x, brackleft y ∈ R. x comma T open parenthesis x closing parenthesis brackright = 0 comma \ [ x ˆ{ 2 } y[T(x),x]+3xy[x,[T(x), which reducesSubtracting because ( 35 of ) open from parenthesis ( 34 ) we obtain 3 1 closing parenthesis to x]]+2xy[T(x),x]x+[y,xˆ2 x to the power of 2 y brackleft T open parenthesis x closing parenthesis comma x brackright plus 3 xyx{ 2 brackleft} ] T [ open x , T ( x ) ] = 0 , \ ] parenthesis x closing parenthesisx2y[T comma(x), x]x + brackright 3xy[x, [T (x minus), x]] + xy 2 brackleftxy[T (x), xT]x open+ [y, parenthesis x2][x, T (x)] x =closing 0, parenthesis comma x brackright x = 0 comma x comma y in R period which reduces because of ( 3 1 ) to Replacing in the above relation minus brackleft T open parenthesis x closing parenthesis comma x brackright x by x brackleft T\noindent open parenthesiswhich x closing reduces parenthesis because comma of ( x 3brackright 1 ) to we obtain 2x2y[T (x), x] + 3xyx[T (x), x] − xy[T (x), x]x = 0, x, y ∈ R. x to the power of 2 y brackleft T open parenthesis x closing parenthesis comma x brackright plus 2 xyx brackleft T open \ [ 2 x ˆ{ 2 } y[T(x),x]+3xyx[T(x),x] parenthesisReplacing x closing parenthesis in the above comma relation x brackright−[T (x), x]x =by 0 commax[T (x), x x] comma we obtain y in R period − Becausexy[T(x),x]x=0,x,y of open parenthesis 27 closing parenthesis comma open parenthesis 30 closing parenthesis\ in commaR. open\ ] parenthesis 3 1 closing parenthesis and open parenthesisx2 32y[T closing(x), x] + parenthesis 2xyx[T (x) the, x] =relation 0, x, open y ∈ R. parenthesis 1 6 closing parenthesis reduces to x to the power of 2 y brackleft T open parenthesis x closing parenthesis comma x brackright = 0 comma Because of ( 27 ) , ( 30 ) , ( 3 1 ) and ( 32 ) the relation ( 1 6 ) reduces to x2y[T (x), x] = 0, \noindentx comma yReplacing in R comma in which the gives above together relation with the $ relation− [ above T( xyx brackleft xT ) open , parenthesis x ] x x$by$x closing parenthesis x, y ∈ R, which gives together with the relation above xyx[T (x), x] = 0, x, y ∈ R, whence it follows comma[ T x brackright ( x = ) 0 comma , x comma ]$weobtain y in R comma whence it follows \ [ x ˆ{ 2 } y[T(x),x]+2xyx[T(x),x]= x brackleftx[T T(x open),T − parenthesisx − brackleftbrackright x closing parenthesis− parenleft comma− T T-x-brackleft y − x − parenleft brackright-parenleft-Tx−parenright−xbrackleft y-x-parenleft− parenright to thecomma power−x of−T x − parenleftbrackright − x − x), x − zero − zero] = x − x − zero R − R − x y ∈ R. x-parenright-x0 , x brackleft-parenright , y \ in R. to the power\ ] of comma-x-T x-parenleft brackright-x-x closing parenthesis comma x-zero-zero , , brackright = x-x-zero sub comma R-R-x sub comma y in R period F-T-O o-f-h-r sub u-c-m o-s u-w-parenleft-r s-e-three h-parenrightF − T − Oo − f − h − ru−c−mo−su − w − parenleft − rs − e − threeh − parenrighto−a−wv−e−ne sub o-a-w v-e-n e brackleft T open parenthesis x closing parenthesis comma y brackright x plus brackleft T parenright-comma x = [T (x), y] x + [T parenright − commax = \noindentLe t mult p-lBecause i aton o of ft ( 27 ) , ( 30 ) , ( 3 1 ) and ( 32 ) the relation ( 1 6 ) reduces to $ x ˆ{ 2 } y[T(x),x]=0,$Equation:Le open t mult parenthesisp − l i aton 38 o closing ft parenthesis .. comma to the power of brackleft-x-T brackright x-y-parenleft brackleft sub T-parenright open parenthesis x closing parenthesis x = comma x comma y \noindent $ x , y \ in R , $ which gives together with the relation above $ xyx [ when e .. i-t .. o l-l o ws brackleft−x−T T(x),x]=0,x,y, ]x − y − parenleft[T −parenright\ in (xR)x = ,, x,$ y (38) whence it follows when e i − t o l − l o ws \ begin { a l i g n ∗} x [ T ( x ) , T−x−brackleft brackright −p a r e n l e f t −T y−x−p a r e n l e f t ˆ{ x−parenright −x } b r a c k l e f t −parenright ˆ{ comma−x−T } x−parenleft brackright −x−x ) , x−zero−zero ] = x−x−zero { , } R−R−x { , } y \ in R. \\ F−T−O o−f−h−r { u−c−m o−s } u−w−p a r e n l e f t −r s−e−thre e h−p a r e n r i g h t { o−a−w v−e−n e }\\ [T(x),y]x+[T parenright −comma x = \end{ a l i g n ∗}

\noindent Le t mult $ p−l $ i aton o f t

\ begin { a l i g n ∗} \ tag ∗{$ ( 38 ) $} , ˆ{ b r a c k l e f t −x−T } ] x−y−p a r e n l e f t [ { T−p a r e n r i g h t } ( x ) x = , x , y \end{ a l i g n ∗}

\noindent when e \quad $ i−t $ \quad o $ l−l $ o ws