Let $R$ be a ring with unity and $U(R)$ its group of units. Let $\Delta
U=\{a\in U(R)\mid [U(R):C_{U(R)}(a)]<\infty\}$ be the $FC$-radical of $U(R)$
and let $\nabla(R)=\{a\in R\mid [U(R):C_{U(R)}(a)]<\infty\}$ be the
$FC$-subring of $R$.
  An infinite subgroup $H$ of $U(R)$ is said to be an $\omega$-subgroup if the
left annihilator of each nonzero Lie commmutator $[x,y]$ in $R$ contains only
finite number of elements of the form $1-h$, where $x,y \in R$ and $h\in H$. In
the case when $R$ is an algebra over a field $F$, and $U(R)$ contains an
$\omega$-subgroup, we describe its $FC$-subalgebra and the $FC$-radical. This
paper is an extension of [1].Comment: 8 pages, AMS-Te

Bovdi, Victor

English

arXiv

Let R be a ring with unity and U(R) its group of units. Let Delta U = {a is an element of U(R) \textbackslash [U(R) : C-U(R)(a)] < infinity} be the FC-radical of U(R) and let del(R) = {a is an element of R \textbackslash [U(R) : C-U(R)(a)] < infinity} be the FC-subring of R. An infinite subgroup H of U(R) is said to be an omega-subgroup if the left annihilator of each nonzero Lie commmutator [x, y] in R contains only finite number of elements of the form 1 - h, where x, y is an element of R and h is an element of H. In the case when R is an algebra over a field F, and U(R) contains an omega-subgroup, we describe its FC-subalgebra and the FC-radical. This paper is an extension of [1]

Bódi, Viktor

Repository of the Academy's Library

Publ. Math. Debrecen57 / 1-2 (2000), 231–239On elements in algebrashaving finite number of conjugatesBy VICTOR BOVDI (Debrecen)Dedicated to Professor Ka´lma´n Gyo˝ry on his 60th birthdayAbstract. Let R be a ring with unity and U(R) its group of units. Let ∆U ={a ∈ U(R) | [U(R) : CU(R)(a)] < ∞} be the FC-radical of U(R) and let ∇(R) = {a ∈R | [U(R) : CU(R)(a)] <∞} be the FC-subring of R.An infinite subgroup H of U(R) is said to be an ω-subgroup if the left annihilatorof each nonzero Lie commmutator [x, y] in R contains only finite number of elementsof the form 1 − h, where x, y ∈ R and h ∈ H. In the case when R is an algebra overa field F , and U(R) contains an ω-subgroup, we describe its FC-subalgebra and theFC-radical. This paper is an extension of [1].1. IntroductionLet R be a ring with unity and U(R) its group of units. Let∆U = {a ∈ U(R) | [U(R) : CU(R)(a)] <∞},and∇(R) = {a ∈ R | [U(R) : CU(R)(a)] <∞},which are called the FC-radical of U(R) and FC-subring ofR, respectively.The FC-subring ∇(R) is invariant under the automorphisms of R andcontains the center of R.Mathematics Subject Classification: Primary: 16U50, 16U60, 20C05; Secondary: 16N99.Key words and phrases: units, FC-elements, finite conjugacy.The research was supported by OTKA T 025029 and T 029132.232 Victor BovdiThe investigation of the FC-radical ∆U and the FC-subring ∇(R)was proposed by S. K. Sehgal and H. Zassenhaus [8]. They describedthe FC-subring of a Z-order as a unital ring with a finite Z-basis and asemisimple quotient ring.Definition. An infinite subgroupH of U(R) is said to be an ω-subgroupif the left annihilator of each nonzero Lie commmutator [x, y] = xy − yxin R contains only a finite number of elements of the form 1 − h, whereh ∈ H and x, y ∈ R.The groups of units of the following infinite rings R contain ω-sub-groups, of course:1. Let A be an algebra over an infinite field F . Then the subgroup U(F )is an ω-subgroup.2. Let R = KG be the group ring of an infinite group G over the ring K.It is well-known (see [6], Lemma 3.1.2, p. 68 ) that the left annihilatorof any z ∈ KG contains only a finite number of elements of the formg − 1, where g ∈ G. Thus G is an ω-subgroup.3. Let R = FλG be an infinite twisted group algebra over the field F withan F -basis {ug | g ∈ G}. Then the subgroup G = {λug | λ ∈ U(F ),g ∈ G} is an ω-subgroup.4. If A is an algebra over a field F , and A contains a subalgebra D suchthat 1 ∈ D and D is either an infinite field or a skewfield, then everyinfinite subgroup of U(D) is an ω-subgroup.2. ResultsIn this paper we study the properties of the FC-subring ∇(R) whenR is an algebra over a field F and U(R) contains an ω-subgroup. We showthat the set of algebraic elements A of ∇(R) is a locally finite algebra,the Jacobson radical J(A) is a central locally nilpotent ideal in ∇(R) andA/J(A) is commutative. As a consequence, we describe the FC-radical∆U , which is a solvable group of length at most 3, and the subgroup t(∆U)is nilpotent of class at most 2. If F is an infinite field then any algebraicunit over F belongs to the centralizer of ∇(R), and, as a consequence,we obtain that t(∆U) is abelian and ∆U is nilpotent of class at most 2.These results are extensions of the results obtained by the author in [1]for groups of units of twisted group algebras.On elements in algebras having finite number of conjugates 233By the Theorem of B. H. Neumann [5], elements of finite order in∆U form a normal subgroup, which we denote by t(∆U), and the factorgroup ∆U/t(∆U) is a torsion free abelian group. If x is a nilpotent elementof the ring R, then the element y = 1 + x is a unit in R, which is calledthe unipotent element of U(R).Let ζ(G) be the center of G and (g, h) = g−1h−1gh, where g, h ∈ G.Lemma 1. Assume that U(R) has an ω-subgroup. Then all nilpotentelements of the subring ∇(R) are central in ∇(R).Proof. Let x be a nilpotent element of ∇(R). Then xk = 0, and byinduction on k we shall prove that vx = xv for all v ∈ ∇(R).Choose an infinite ω-subgroup H of U(R). By Poincare’s Theoremthe centralizer S of the subset {v, x} in H is a subgroup of finite indexin H. Since H is infinite, S is infinite and fx = xf for all f ∈ S. Thenxf is nilpotent and 1 + xf is a unit in U(R). Since v ∈ ∇(R), the set{(1 + xf)−1v(1 + xf) | f ∈ S} is finite. Let v1, . . . , vt be all the elementsof this set andWi = {f ∈ S | (1 + xf)−1v(1 + xf) = vi}.Obviously, S =⋃Wi and there exists an index j such that Wj is infinite.Fix an element f ∈Wj . Then any element q ∈Wj such that q 6= f satisfies(1 + xf)−1v(1 + xf) = (1 + xq)−1v(1 + xq)andv(1 + xf)(1 + xq)−1 = (1 + xf)(1 + xq)−1v.Thenv{(1 + xq) + (xf − xq)}(1 + xq)−1 = {(1 + xq) + (xf − xq)}(1 + xq)−1v,v(1 + x(f − q)(1 + xq)−1) = (1 + x(f − q)(1 + xq)−1)vand(1) vx(f − q)(1 + xq)−1 = x(f − q)(1 + xq)−1v.Let xv 6= vx and k = 2. Then x2 = 0 and (1 + xq)−1 = 1− xq. Sincef and q belong to the centralizer of the subset {x, v}, from (1) we have(f − q)vx(1− xq) = (f − q)x(1− xq)v,234 Victor Bovdiwhence (f−q)(vx−vx2q−xv+x2qv) = 0 and evidently (f−q)(vx−xv) = 0.Therefore, (q−1f − 1)(vx − xv) = 0 for any q ∈ Wj . Since q−1Wj is aninfinite subset of the ω-subgroup H, we obtain a contradiction, and thusvx = xv.Let k > 2. If i ≥ 1 then xi+1 is nilpotent of index less than k, thusapplying an induction on k, first we obtain that xi+1v = vxi+1 and thenx(f − q)xiqiv = (f − q)xi+1qiv = (f − q)vxi+1qi = vx(f − q)xiqi.Hencevx(f − q)(1− xq + x2q2 + · · ·+ (−1)k−1xk−1qk−1)= x(f − q)((1− xq)v + (x2q2 + · · ·+ (−1)k−1xk−1qk−1)v).and (f − q)(vx− xv) = 0. As before, we have a contradiction in the casexv 6= xv.Thus nilpotent elements of ∇(R) are central in ∇(R). ¤Lemma 2. Let R be an algebra over a field F such that the group ofunits U(R) contains an ω-subgroup. Then the radical J(A) of every locallyfinite subalgebra A of ∇(R) consists of central nilpotent elements of thesubalgebra ∇(R), and A/J(A) is a commutative algebra.Proof. Let x ∈ J(A), then x ∈ L for some finite dimensional sub-algebra L of A. Since L is left Artinian, Proposition 2.5.17 in [7] (p. 185)ensures that L ∩ J(A) ⊆ J(L), moreover J(L) is nilpotent. Now x ∈ J(L)implies that x is nilpotent and the application of Lemma 1 gives that xbelongs to the center of ∇(R). Then Theorem 48.3 in [4] (p. 209) willenable us to verify the existence of the decomposition into the direct sumL = Le1 ⊕ · · · ⊕ Len ⊕N,where Lei is a finite dimensional local F -algebra (i.e. Lei/J(Lei) is a divi-sion ring), N is a commutative artinian radical algebra, and e1, . . . , en arepairwise orthogonal idempotents. Since nilpotent elements of ∇(R) be-long to the center of ∇(R), by Lemma 13.2 of [4] (p. 57) any idempotentei is central in L and the subring Lei of ∇(R) is also an FC-ring, whenceJ(Lei) is a central nilpotent ideal.On elements in algebras having finite number of conjugates 235Suppose that Lei/J(Lei) is a noncommutative division ring. Then1 + J(Lei) is a central subgroup andU(Lei)/(1 + J(Lei)) ∼= U(Lei/J(Lei)).Applying Herstein’s Theorem [2] we establish that a noncentral unitof Lei/J(Lei) has an infinite number of conjugates, which is impossible.Therefore, L/J(L) is a commutative algebra and from J(L) ⊆ J(A) andJ(L) is nil (actualy nilpotent) in A, it follows that A/J(A) is a commutativealgebra. ¤Theorem 1. Let R be an algebra over a field F such that the group ofunits U(R) contains an ω-subgroup, and let ∇(R) be the FC-subalgebraof R. Then the set of algebraic elements A of ∇(R) is a locally finitealgebra, the Jacobson radical J(A) is a central locally nilpotent ideal in∇(R) and A/J(A) is commutative.Proof. Since any nilpotent element of ∇(R) is central in ∇(R) byLemma 1, one can see immediately that the set of all nilpotent elements of∇(R) form an ideal I, and the factor algebra ∇(R)/I contains no nilpotentelements. Obviously, I is a locally finite subalgebra in ∇(R), and allidempotents of ∇(R)/I are central in ∇(R)/I.Let x1, x2, . . . , xs be algebraic elements of ∇(R)/I. We shall provethat the subalgebra generated by x1, x2, . . . , xs is finite dimension.For every xi the subalgebra 〈xi〉F of the factor algebra ∇(R)/I is adirect sum of fields〈xi〉F = Fi1 ⊕ Fi2 ⊕ · · · ⊕ Firi ,where Fij is a field and is finite dimensional over F . Choose F -basiselements uijk (i = 1, . . . , s, j = 1, . . . , ri, k = 1, . . . , [Fij : F ]) in Fij over Fand denote by wijk = 1− eij + uijk, where eij is the unit element of Fij .Obviously, wijk is a unit in ∇(R)/I. We collect in the direct summandall these units wijk for each field Fij (i = 1, . . . , s, j = 1, . . . , ri) and thisfinite subset in the group U(∇(R)/I) is denoted by W .Let H be the subgroup of U(∇(R)/I) generated byW . The subgroupH of ∇(R)/I is a finitely generated FC-group, and as it is well-known,a natural number m can be assigned to H such that for any u, v ∈ Hthe elements um, vm are in the center ζ(H), and (uv)m = umvm (see [5]).236 Victor BovdiSince H is a finitely generated group, the subgroup S = {vm | v ∈ H} hasa finite index in H and {wm | w ∈ W} is a finite generated system for S.Let t1, t2, . . . , tl be a transversal to S in H.Let HF be the subalgebra of ∇(R)/I spanned by the elements of Hover F . Clearly, the commutative subalgebra SF of HF , generated bycentral algebraic elements wm (w ∈ W ), is finite dimensional over F andany u ∈ HF can be written asu = u1t1 + u2t2 + · · ·+ ultl,where ui ∈ SF . Since titj = αijtr(ij) and αij ∈ SF , it yields that thesubalgebra HF is finite dimensional over F . Recall thatxi =∑j,kβjkwijk −∑j,kβjk(1− eij),where βjk ∈ F and eij are central idempotents of ∇(R)/I. The subalgebraT generated by eij (i = 1, 2, . . . , s, j = 1, 2, . . . , ri) is finite dimensionalover F and T is contained in the center of ∇(R)/I. Therefore, xi belongsto the sum of two subspaces HF and T and the subalgebra of ∇(R)/Igenerated by HF and T is finite dimensional over F . Since 〈x1, . . . , xs〉F isa subalgebra of 〈HF , T 〉F , is also finite dimensional over F . We establishedthat the set of algebraic elements of ∇(R)/I is a locally finite algebra.One can see that all the algebraic elements of ∇(R) form a locally finitealgebra A (see [3], Lemma 6.4.1, p. 162). Since the radical of an algebraicalgebra is a nil ideal, according to Lemma 1 we have that J(A) is a centrallocally nilpotent ideal in ∇(R), and A/J(A) is commutative by Lemma 2.¤Recall that by Neumann’s Theorem [5] the set t(∆U) of ∆U contain-ing all elements of finite order of ∆U is a subgroup.Theorem 2. Let R be an algebra over a field F such that the groupof units U(R) contains an ω-subgroup. Then1. the elements of the commutator subgroup of t(∆U) are unipotent andcentral in ∆U ;2. if all elements of ∇(R) are algebraic then ∆U is nilpotent of class 2;On elements in algebras having finite number of conjugates 2373. ∆U is a solvable group of length at most 3, and the subgroup t(∆U)is nilpotent of class at most 2.Proof. It is easy to see that ∆U ⊆ ∇(R), and any element of t(∆U)is algebraic. According to Theorem 1 the set A of algebraic elementsof ∇(R) is a subalgebra, the Jacobson radical J(A) is a central locallynilpotent ideal in ∇(R), and A/J(A) is commutative. The isomorphismU(A)/(1 + J(A)) ∼= U(A/J(A)),implies that(t(∆U)(1 + J(A)))/(1 + J(A)) is abelian, the commutatorsubgroup of t(∆U) is contained in 1 + J(A) and consists of unipotentelements.By Neumann’s Theorem ∆U/t(∆U) is abelian, therefore ∆U is a solv-able group of length at most 3. ¤Let R be an algebra over a field F . Let m be the order of the elementg ∈ U(R) and assume that the element 1 − αm is a unit in F for someα ∈ F . It is well-known that g − α ∈ U(R) and(g − α)−1 = (1− αm)−1m−1∑i=0αm−1−igi.We know that the number of solutions of the equation xm − 1 = 0 in Fdoes not exceed m. If F is an infinite field, then it follows that, thereexists an infinite set of elements α ∈ F such that g − α is a unit. We willshow that this is true for any algebraic unit.Lemma 3. Let g ∈ U(R) be an algebraic element over an infinitefield F . Then there are infinitely many elements α of the field F such thatg − α is a unit.Proof. Since g is an algebraic element over F , F [g] is a finite di-mensional subalgebra over F . Let T be the radical of F [g]. There existsan orthogonal system of idempotents e1, e2, . . . , es such thatF [g] = F [g]e1 ⊕ F [g]e2 ⊕ · · · ⊕ F [g]es,and Tei is a nilpotent ideal such that F [g]ei/Tei is a field. It is well-knownthat F [g]ei is a local ring, and all elements of F [g]ei, which do not belongto Tei, are units. Moreover, if α ∈ F , then(2) g − α = (ge1 − αe1) + (ge2 − αe2) + · · ·+ (ges − αes).238 Victor BovdiClearly, gei is a unit and gei /∈ Tei for every i. PutLi = {gei − αei | α ∈ F}.Suppose that gei − βei and gei − γei belong to Tei for some α, β ∈ F .Then(gei − βei)− (gei − γei) = (γ − β)ei ∈ Tei,which is impossible for β 6= γ. Therefore, Tei contains at most one elementfrom Li. Since F [g]ei is a local ring, all elements of the form gei − αeiwith gei − αei /∈ Tei are units, and there are infinitely many units of theform (2). ¤Lemma 4. Let g ∈ U(R) and a ∈ R. If g − α, g − β are units forsome α, β ∈ F and ag 6= ga, then(g − α)−1a(g − α) 6= (g − β)−1a(g − β).Proof. Suppose that (g − α)−1a(g − α) = (g − β)−1a(g − β). Then(g − α− (β − α))(g − α)−1a = a(g − α− (β − α))(g − α)−1and (1− (β − α)(g − α)−1)a = a(1− (β − α)(g − α)−1).Hence(β − α)(g − α)−1a = a(β − α)(g − α)−1and (g − α)−1a = a(g − α)−1, which provides the contradiction ag = ga.¤Theorem 3. Let R be an algebra over an infinite field F . Then1. any algebraic unit over F belongs to the centralizer of ∇(R);2. if R is generated by algebraic units over F , then ∇(R) belongs to thecenter of R.Proof. Let a ∈ ∇(R), and g ∈ U(R) be an algebraic element over F .Then by Lemma 3 there are infinitely many elements α ∈ F such that g−αis a unit for every α. If [a, g] 6= 0, then by Lemma 4 the elements of theform (g − α)−1a(g − α) are different, and a has an infinite number ofconjugates, which is impossible. Therefore, g belongs to the centralizerof ∇(R).Now, suppose that R is generated by algebraic units {aj} over F .Since every w ∈ U(R) can be written as a sum of elements of the formαiaγi1i1. . . aγisis, where αj ∈ F , γij ∈ Z, by the first part of this theorem wcommute with elements of ∇(R). Hence ∇(R) is central in R. ¤On elements in algebras having finite number of conjugates 239Corollary. Let R be an algebra over an infinite field F . Then1. t(∆U) is abelian and ∆U is a nilpotent group of class at most 2;2. if every unit of R is an algebraic element over F , then ∆U is centralin U(R).Proof. Clearly, all elements from t(∆U) are algebraic and by The-orem 3 every algebraic unit belongs to the centralizer of ∇(R). Sincet(∆U) ⊆ ∇(R), it follows that t(∆U) is central in ∇(R). Since ∆U/t(∆U)is abelian, by Neumann’s Theorem ∆U is a nilpotent group of class atmost 2.Let a ∈ ∆U and g ∈ U(R) be an algebraic element over F . Thenby Theorem 3 we get [a, g] = 0. Hence, if every unit of R is an algebraicelement over F , then ∆U is central in U(R). ¤References[1] V. Bovdi, Twisted group rings whose units form an FC-group, Canad. J. Math.47 (1995), 247–289.[2] I. N. Herstein, Conjugates in division ring, Proc. Amer. Soc. 7 (1956), 1021–1022.[3] I. N. Herstein, Noncommutative rings, Math. Association of America, John Wi-ley and Sons, 1968.[4] A. Kertesz, Lectures on artinian rings, Akade´miai Kiado´, Budapest, 1987.[5] B. H. Neumann, Groups with finite clasess of conjugate elements, Proc. LondonMath. Soc. 1 (1951), 178–187.[6] D. S. Passman, The algebraic structure of group rings, A Wiley-Interscience,New York – Sydney – Toronto, 1977.[7] L. H. Rowen, Ring Theory, vol. 1, Academic Press, Boston, MA, 1988, 538.[8] S. K. Sehgal and H. J. Zassenhaus, On the supercentre of a group and its ringtheoretical generalization., Integral Represantations and Applications. Proc. Conf.,Oberwolfach, 1980, Lect. Notes Math. 882 (1981), 117–144.VICTOR BOVDIINSTITUTE OF MATHEMATICS AND INFORMATICSUNIVERSITY OF DEBRECENH–4010 DEBRECEN, P.O. BOX 12HUNGARYE-mail: vbovdi@math.klte.hu(Received February 23, 2000; revised May 23, 2000)

On elements in algebras having finite number of conjugates

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On elements in algebras having finite number of conjugates

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