Commentationes Mathematicae Universitatis Carolinae

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1 Commentationes Mathematicae Universitatis Carolinae Borut Zalar On centralizers of semiprime rings Commentationes Mathematicae Universitatis Carolinae, Vol. 32 (1991), No. 4, Persistent URL: Terms of use: Charles University in Prague, Faculty of Mathematics and Physics, 1991 Institute of Mathematics of the Academy of Sciences of the Czech Republic provides access to digitized documents strictly for personal use. Each copy of any part of this document must contain these Terms of use. This paper has been digitized, optimized for electronic delivery and stamped with digital signature within the project DML-CZ: The Czech Digital Mathematics Library

2 Comment.Math.Univ.Carolin. 32,4(1991) On centralizers of semiprime rings Borut Zalar Abstract. Let K be a semiprime ring and T : K K an additive mapping such that T(x 2 )=T(x)xholdsforall x K.Then Tisaleftcentralizerof K.Itisalsoprovedthat Jordan centralizers and centralizers of K coincide. Keywords: semiprime ring, left centralizer, centralizer, Jordan centralizer Classification: 16N60, 16W10, 16W25 Throughout this paper, K will represent an associative ring with the center Z. Kiscalledprimeif akb=(0)implies a=0or b=0andsemiprimeif aka=(0) implies a=0.amapping D:K Kiscalledderivationif D(xy)=D(x)y+xD(y) holdsforall x, y K. Aleft(right)centralizerof Kisanadditivemapping T : K Kwhichsatisfies T(xy)=T(x)y(T(xy)=xT(y))forall x, y K. If a K, then L a (x)=axisaleftcentralizerand R a (x)=xaisarightcentralizer. If Kisaringwithinvolution,theneveryadditivemapping E: K Kwhich satisfies E(x 2 )=E(x)x +xe(x)forall x KiscalledJordan -derivation.these mappings are closely connected with a question of representability of quadratic forms by bilinear forms. Some algebraic properties of Jordan -derivations are considered in[1], where further references can be found. For quadratic forms see[6]. In[2] M. Brešar and the author obtained a representation of Jordan -derivations in terms of left and right centralizerson the algebra of compact operators on ahilbertspace. Wearrivedataproblemwhetheranadditivemapping T which satisfiesaweakercondition T(x 2 )=T(x)xisautomaticallyaleftcentralizer. We provedin[2]thatthisisinfactsoif Kisaprimering(generallywithoutinvolution). In the present paper, we generalize this result on semiprime rings. Our second result is motivated by the study of Jordan mappings in associative rings. Ifweintroduceanewproductin Kgivenby x y=xy+ yx,thenjordan derivationisanadditivemapping Dwhichsatisfies D(x y)=d(x) y+ x D(y) forall x, y KandJordanhomomorphismisanadditivemapping Awhichsatisfies A(x y)=a(x) A(y)forall x, y K.ThereforewecandefineaJordancentralizer tobeanadditivemapping Twhichsatisfies T(x y)= T(x) y=x T(y).Sincethe product is commutative, there is no difference between the left and right Jordan centralizers. Acentralizerof Kisanadditivemappingwhichisbothleftandrightcentralizer. An easy computation shows that every centralizer is also a Jordan centralizer. We prove as our second result that every Jordan centralizer of a semiprime ring is a centralizer. ThisresearchwaspartiallysupportedbytheResearchCouncilofSlovenija.

3 610 B. Zalar 1. The first result. Toproveourfirstresult,weneedthreesimplelemmaswhichwenowstate. Lemma1.1. Let Kbeasemiprimering.If a, b Karesuchthat axb=0forall x K,then ab=ba=0. Proof:Takeany x K. (ab)x(ab)=a(bxa)b=0, (ba)x(ba)=b(axb)b=0. Bythesemiprimenessof K,itfollows ab=ba=0. Lemma 1.2. Let Kbeasemiprimeringand A:K K Kbiadditivemappings. If A(x, y)wb(x, y)=0forall x, y, w K,then A(x, y)wb(u, v)=0forall x, y, u, v, w K. Proof:Firstweshallreplace xwith x+u. Weusedthebiadditivityof Aand B. A(x+u, y)wb(x+u, y)=0, A(x, y)wb(u, y)= A(u, y)wb(x, y). (A(x, y)wb(u, y))z(a(x, y)wb(u, y)) = = A(u, y)wb(u, y)za(x, y)wb(x, y)=0. Hence A(x, y)wb(u, y)=0bysemiprimenessof K.Nowwereplace yby y+ vand obtain the assertion of the lemma with a similar approach as above. Lemma1.3. Let Kbeasemiprimeringand a Ksomefixedelement. If a[x, y] =0forall x, y K,thenthereexistsanideal Uof Ksuchthat a U Zholds. Proof: [z, a]x[z, a]=zax[z, a] azx[z, a]= = za[z, xa] za[z, x]a a[z, zxa]+a[z, zx]a=0. Hence a Z. Since zaw[x, y]=0forall z, w, x, y Kwecanrepeattheabove argumentwith zawinsteadof atoobtain KaK Zandnowitisobviousthatthe ideal generated by a is central. Proposition1.4. Let Kbeasemiprimeringofcharacteristicnottwoand T : K Kanadditivemappingwhichsatisfies T(x 2 )=T(x)xforall x K. Then T is a left centralizer. Proof: (1) T(x 2 )=T(x)x.

4 On centralizers of semiprime rings 611 Ifwereplace xby x+y,weget (2) T(xy+ yx)=t(x)y+ T(y)x. Byreplacing ywith xy+ yxandusing(2),wearriveat (3) T(x(xy+ yx)+(xy+ yx)x)=t(x)xy+ T(x)yx+T(x)yx+T(y)x 2. Butthiscanalsobecalculatedinadifferentway. (4) T(x 2 y+ yx 2 )+2T(xyx)=T(x)xy+ T(y)x 2 +2T(xyx). Comparing(3) and(4) we obtain (5) T(xyx)=T(x)yx. Ifwelinearize(5),weget (6) T(xyz+ zyx)=t(x)yz+ T(z)yx. Nowweshallcompute j= T(xyzyx+ yxzxy)intwodifferentways.using(5)we have (7) j= T(x)yzyx+ T(y)xzxy. Using(6)wehave (8) j= T(xy)zyx+ T(yx)zxy. Comparing(7)and(8)andintroducingabiadditivemapping B(x, y)=t(xy) T(x)ywearriveat (9) B(x, y)zyx+b(y, x)zxy=0. Equality(2)canberewritteninthisnotationas B(x, y)= B(y, x). Usingthis fact and equality(9) we obtain (10) B(x, y)z[x, y]=0. UsingfirstLemma1.2andthenLemma1.1wehave (11) B(x, y)z[u, v]=0. Nowfixsome x, y Kandwrite Binsteadof B(x, y)tosimplifyfurtherwriting. UsingLemma1.3wegettheexistenceofanideal Usuchthat B U Zholds. Inparticular, bb, Bb Zforall b K.Thisgivesus x B 2 y= B 2 y x=yb 2 x=y B 2 x.

5 612 B. Zalar Thisgivesus4T(x B 2 y)=4t(y B 2 x).bothsidesofthisequalitywillbecomputed infewstepsusing(2)andtheaboveremarks. Since we obtain 2T(xB 2 y+ B 2 yx)=2t(yb 2 x+b 2 xy), 2T(x)B 2 y+2t(b 2 y)x=2t(y)b 2 x+2t(b 2 x)y, 2T(x)B 2 y+ T(B 2 y+ yb 2 )x=2t(y)b 2 x+t(b 2 x+xb 2 )y, 2T(x)B 2 y+t(b)byx+t(y)b 2 x=2t(y)b 2 x+t(b)bxy+ T(x)B 2 y, T(x)B 2 y+ T(B)Byx=T(y)B 2 x+t(b)bxy. Byx=By x=x By= xby= Bxy, (12) T(x)B 2 y=t(y)b 2 x. Ontheotherhand,wealsohave 4T(xyB 2 )=4T(xB yb), 2T(xyB 2 + B 2 xy)=2t(xbyb+ ybxb), 2T(xy)B 2 +2T(B)Bxy=2T(Bx)By+2T(By)Bx, 2T(xy)B 2 +2T(B)Bxy= T(xB+ Bx)By+ T(yB+ By)Bx, 2T(xy)B 2 +2T(B)Bxy=T(x)B 2 y+ T(B)Bxy+ T(y)B 2 x+t(b)bxy, 2T(xy)B 2 = T(x)yB 2 + T(y)xB 2. Using(12)wefinallyarriveat T(xy)B 2 = T(x)yB 2. Butthismeansthat B 3 =0 so that B 2 KB 2 = B 4 K=(0), BKB= B 2 K=(0), whichimplies B=0andtheproofiscomplete. It was proved in[5] that left centralizers of semisimple Banach algebras are automatically continuous. Corollary1.5. Let AbeasemisimpleBanachalgebraand T: A Aanadditive mappingsuchthat T(x 2 )=T(x)xholdsforall x A. Then T isacontinuous linear operator. Proof: Every semisimple Banach algebra is a semiprime ring. Linearity follows from (T(λx) λt(x))y= T(λx)y T(x)λy= T(λxy) T(xλy)=0. The concept of a left and right centralizer are symmetric, therefore it is obvious thateveryadditivemapping Twhichsatisfies T(x 2 )=xt(x)isarightcentralizer if K is semiprime ring of characteristic not two.

6 On centralizers of semiprime rings The second result. Weagaindividetheproofinfewlemmas. Lemma2.1. Let Kbeasemiprimering, Daderivationof Kand a Ksome fixed element. (i) D(x)D(y)=0forall x, y Kimplies D=0. (ii) ax xa Zforall x Kimplies a Z. Proof:(i) D(x)yD(x) = D(x)D(yx) D(x)D(y)x = 0. (ii) Define D(x)=ax xa. Itiseasytoseethat Disaderivation. Since D(x) Z forall x K,wehave D(y)x=xD(y)andalso D(yz)x=xD(yz). Hence D(y)zx+yD(z)x=xD(y)z+ xyd(z), D(y)[z, x]=d(z)[x, y]. Nowtake z= a.obviously D(a)=0,soweobtain From(i)weget D=0andhence a Z. 0=D(y)[a, x]=d(y)d(x). Lemma2.2. Let Kbeasemiprimeringand a Ksomefixedelement.If T(x)= ax+xaisajordancentralizer,then a Z. Proof: gives us T(xy+ yx)=t(x)y+ yt(x) axy+ ayx+xya+yxa=(ax+xa)y+ y(ax+xa), ayx+xya xay yax=0=(ay ya)x x(ay ya). ThesecondpartofLemma2.1nowgivesus a Z. Lemma2.3. Let Kbeasemiprimering.TheneveryJordancentralizerof Kmaps Zinto Z. Proof:Takeany c Zanddenote a=t(c). 2T(cx)=T(cx+xc)=T(c)x+xT(c)=ax+xa. A straightforward verification shows that S(x) = 2T(cx) is also a Jordan centralizer. ByLemma2.2,wehave T(c) Z.

7 614 B. Zalar Lemma2.4. Let Kbeasemiprimeringand a, b Ktwofixedelements.If ax=xb forall x K,then a=b Z. Proof: xyb=axy= xbyimplies x[b, y]=0forall x, y K.Hence[b, y]x[b, y]=0 andbysemiprimenessof K,wehave b Z. Therefore ax=bxandthisclearly implies a=b. Proposition 2.5. Every Jordan centralizer of semiprime ring K of characteristic not two is a centralizer. Proof:Let TbeaJordancentralizer,i.e. Ifwereplace yby xy+ yx,weget T(xy+ yx)=t(x)y+ yt(x)=xt(y)+t(y)x. T(x)(xy+ yx)+(xy+ yx)t(x)=t(xy+ yx)x+xt(xy+ yx)= =(T(x)y+ yt(x))x+x(t(x)y+ yt(x)). Nowitfollowsthat[T(x), x]y= y[t(x), x]holdsforall x, y Kandso[T(x), x] Z. Thenextgoalistoshowthat[T(x), x]=0holds.takeany c Z. UsingLemma2.3weget 2T(cx)=T(cx+xc)=T(c)x+xT(c)=2T(x)c. T(cx)=T(x)c=T(c)x, [T(x), x]c=t(x)xc xt(x)c=t(c)x 2 xt(c)x=0. Since[T(x), x]itselfiscentralelement,ourgoalisachieved. 2T(x 2 )=T(xx+xx)=T(x)x+xT(x)=2T(x)x=2xT(x). Proposition 1.4 now concludes the proof. References [1] Brešar M., Vukman J., On some additive mapping in rings with involution, Aequationes Math. 38(1989), [2] Brešar M., Zalar B., On the structure of Jordan -derivations, Colloquium Math., to appear. [3] Herstein I.N., Topics in ring theory, University of Chicago Press, [4], Theory of rings, University of Chicago Press, [5] Johnson B.E., Sinclair A.M., Continuity of derivations and a problem of Kaplansky, Amer. J. Math. 90(1968), [6] ŠemrlP.,QuadraticfunctionalsandJordan -derivations,studiamath.97(1991), Institute of Mathematics, University of Ljubljana, Jadranska 21, Ljubljana, Slovenija (Received May 21, 1991)