DIVISION RINGS RELATED TO THE KUROSH PROBLEM

Both the Burnside problem in Group Theory and its ringtheoretic analogue, the Kurosh problem for algebras over fields, were answered negatively by Golod and Shafarevich. However, the restricted case when algebras are division rings, the Kurosh problem for division rings, is still open. This paper is devoted mainly to the construction of the new class of division rings for which the Kurosh problem has the affirmative answer

DIVISION RINGS RELATED TO THE KUROSH PROBLEM BUI XUAN HAI, MAI HOANG BIEN, AND TRINH THANH DEO Abstract. Both the Burnside problem in Group Theory and its ring-theoretic analogue, the Kurosh problem for algebras over fields, were answered negatively by Golod and Shafarevich. However, the restricted case when algebras are division rings, the Kurosh problem for division rings, is still open. This paper is devoted mainly to the construction of the new class of division rings for which the Kurosh problem has the affirmative answer. 1. Introduction In 1941, Kurosch [10, Problem R] asked if a finitely generated algebraic algebra is necessarily a finite dimensional vector space over a base field. This is a ringtheoretic analogue of the famous Burnside problem in Group Theory: whether a finitely generated group whose elements all have finite order is necessarily finite. Both the Kurosh problem and the Burnside problem were solved negatively by Golod and Shafarevich [6]-[7]. In fact, they constructed an example of an infinite finitely generated group in which every element has finite order as well as one of an infinite dimensional finitely generated algebraic algebra. Concerning the Kurosh problem, as he remarked, the particular case [10, Problem K] of division rings is of special interest. Moreover, Rowen [16] pointed out that, in general, there are two special cases we have to consider: the case of nil rings and the case of division rings. The problem has the negative answer for nil rings: there are several valuable examples [18]-[20], [12], [2] by Smoktunovicz and others of infinite dimensional finitely generated algebraic nil rings. However, for division rings, the problem remains without definite answer and this case is usually refered as the Kurosh problem for division rings. For an additional information about this problem we refer to [9] and [21]. Let D be a division ring with center F . We say that D is centrally finite if the vector space F D is finite dimensional over F ; D is locally finite if the division subring F (S) of D generated by S ∪ F for every finite subset S of D, is finite dimensional over F . Note that, there are vast numbers of locally finite division rings which are not centrally finite. An element a ∈ D is algebraic over F iff the field extension F ⊆ F (a) is finite. A division ring D is algebraic over F (briefly, D is algebraic), if every element of D is algebraic over F . Clearly, a locally finite division ring is algebraic, and the converse is equivalent to the Kurosh problem. More exactly, the Kurosh problem for division rings will be answered in the affirmative iff any algebraic division ring is locally finite. At the present, this problem remains still unsolved in general: there are no similar examples as in the case of nil rings on one side, and on the other side, it is answered in the affirmative Key words and phrases. division ring; centrally finite; locally finite; weakly locally finite; linear groups. 2010 Mathematics Subject Classification. 16K20. 1 2 BUI XUAN HAI, MAI HOANG BIEN, AND TRINH THANH DEO for all known special cases. In particular, it is the case: for F uncountable [16], for F finite [11], and for F having only finite algebraic field extensions (in particular for F algebraically closed). The last case follows from the Levitzki-Shirshov theorem which states that any algebraic algebra of bounded degree is locally finite (see e.g. [4], [9]). The answer for the case of finite F is due to Jacobson who proved that an algebraic division ring D is commutative provided its center is finite (see, for example, [11]). In this work, we introduce the notion of weakly locally finite division rings, and we prove that the class of weakly locally finite division rings strictly contains the class of locally finite division rings by giving an example of a weakly locally finite division ring which is not even algebraic over its center. Further, we prove that the Kurosh problem has the affirmative answer for this class of division rings. We devote Section 2 to do this. In Section 3, we study Herstein’s conjecture [8, Conjecture 3], and we show that this conjecture is true for weakly locally finite division rings. The symbols and notation we use in this paper are standard. In particular, for a division ring D we denote by D∗ and D the multiplicative group and the derived subgroup of D∗ respectively. If A is a ring or a group, then Z(A) denotes the center of A. 2. Kurosh problem for weakly locally finite division rings Let us begin with the observation that in a centrally finite division ring, every division subring is itself centrally finite. Using this fact, it is easy to show that in a locally finite division ring, every finite subset generates the centrally finite division subring. Motivating by this observation, we introduce the following notion. Definition 1. We say that a division ring D is weakly locally finite if for every finite subset S of D, the division subring generated by S in D is centrally finite. From this definition, the following is obvious. Proposition 2. Every locally finite division ring is weakly locally finite. Now, it is natural to ask, whether the class of weakly locally finite division rings is really different from the class of locally finite division rings. Our purpose in this paragraph is to construct an example showing that the class of weakly locally finite division rings strictly contains the class of locally finite division rings, and we then prove that the Kurosh problem has the affirmative answer for this class of rings. In order to do so, following the general Mal’cev-Neumann construction of Laurent series rings (see [15], and also [11]), in the following theorem, firstly, we construct a Laurent series ring with the base ring which is an extension of the field Q of rational numbers. Further, we construct its division subring which is weakly locally finite but it is not even algebraic. Theorem 3. There exists a weakly locally finite division ring which is not algebraic. ∞ Z the direct sum of infinitely many copies of the additive Proof. Denote by G = i=1 group Z. For any positive integer i, denote by xi = (0, . . . , 0, 1, 0, . . .) the element of G with 1 in the i-th position and 0 elsewhere. Then G is a free abelian group generated by all xi and every element x ∈ G is written uniquely in the form x= ni xi , i∈I (1) DIVISION RINGS RELATED TO THE KUROSH PROBLEM 3 with ni ∈ Z and some finite set I. Now, we define an order in G as follows: For elements x = (n1 , n2 , n3 , . . .) and y = (m1 , m2 , m3 , . . .) in G, define x < y if either n1 < m1 or there exists k ∈ N such that n1 = m1 , . . . , nk = mk and nk+1 < mk+1 . Clearly, with this order G is a totally ordered set. Suppose that p1 < p2 < . . . < pn < . . . is a sequence of prime numbers and √ √ K = Q( p1 , p2 , . . .) is the subfield of the field R of real numbers generated by Q √ √ and p1 , p2 , . . ., where Q is the field of rational numbers. For any i ∈ N, suppose that fi : K → K is Q-isomorphism satisfying the following condition: √ √ √ √ fi ( pi ) = − pi ; and fi ( pj ) = pj for any j = i. It is easy to verify that fi fj = fj fi for any i, j ∈ N. • Step 1. Proving that, for x ∈ K, fi (x) = x for any i ∈ N if and only if x ∈ Q: The converse is obvious. Now, suppose that x ∈ K such that fi (x) = x for √ √ any i ∈ N. By setting K0 = Q and Ki = Q( p1 , . . . , pi ) for i ≥ 1, we have the following ascending series: K0 ⊂ K1 ⊂ . . . ⊂ Ki ⊂ . . . √ If x ∈ Q, then there exists i ≥ 1 such that x ∈ Ki \ Ki−1 . So, we have x = a + b pi , √ with a, b ∈ Ki−1 and b = 0. Since fi (x) = x, 0 = x − fi (x) = 2b pi , a contradiction. • Step 2. Constructing a Laurent series ring: For any x = (n1 , n2 , ...) = ni xi ∈ G, define Φx := i∈I i∈I fini . Clearly Φx ∈ Gal(K/Q) and the map Φ : G → Gal(K/Q), defined by Φ(x) = Φx is a group homomorphism. The following conditions hold. i) Φ(xi ) = fi for any i ∈ N. √ √ ii) If x = (n1 , n2 , . . .) ∈ G, then Φx ( pi ) = (−1)ni pi . For the convenience, from now on we write the operation in G multiplicatively. For G and K as above, consider formal sums of the form α= ax x, ax ∈ K. x∈G For such an α, define the support of α by supp(α) = {x ∈ G : ax = 0}. Put ax x, ax ∈ K | supp(α) is well-ordered . D = K((G, Φ)) := α = x∈G For α = ax x and β = x∈G bx x from D, define x∈G α+β = (ax + bx )x; x∈G and αβ = ax Φx (by ) z. z∈G xy=z With operations defined as above, D = K((G, Φ)) is a division ring (we refer to [6, pp. 243-244]). Moreover, the following conditions hold. iii) For any x ∈ G, a ∈ K, xa = Φx (a)x. √ √ √ √ iv) For any i = j, xi pi = − pi xi and xj pi = pi xj . √ √ √ √ v) For any i = j and n ∈ N, xni pi = (−1)n pi xni and xnj pi = pi xnj . 4 BUI XUAN HAI, MAI HOANG BIEN, AND TRINH THANH DEO • Step 3. Finding the center of D: Put H = {x2 | x ∈ G} and Q((H)) = α = ax x, ax ∈ Q | supp(α) is well-ordered . x∈H It is easy to check that H is a subgroup of G and for every x ∈ H, Φx = IdK . Denote by F the center of D. We claim that F = Q((H)). Suppose that α = ax x ∈ Q((H)). Then, for every β = by y ∈ D, we have Φx (by ) = by and x∈H y∈G Φy (ax ) = ax . Hence αβ βα = ax Φx (by ) z = z∈G xy=z z∈G xy=z = ax by z, z∈G xy=z z∈G xy=z by Φy (ax ) z = ax by z. Thus, αβ = βα for every β ∈ D, so α ∈ F . ax x ∈ F. Denote by S the set of all elements x Conversely, suppose that α = x∈G appeared in the expression of α. Then, it suffices to prove that x ∈ H and ax ∈ Q √ √ for any x ∈ S. In fact, since α ∈ F , we have pi α = α pi and αxi = xi α for √ √ any i ≥ 1, i.e. pi ax x = Φx ( pi )ax x and ax (xxi ) = Φxi (ax )(xi x). x∈S x∈S x∈S x∈S √ Therefore, by conditions mentioned in the beginning of Step 2, we have pi ax = √ √ Φx ( pi )ax = (−1)ni pi ax and ax = Φxi (ax ) = fi (ax ) for any x = (n1 , n2 , . . .) ∈ S. From the first equality it follows that ni is even for any i ≥ 1. Therefore x ∈ H. From the second equality it follows that ax = fi (ax ) for any i ≥ 1. So by Step 1, we have ax ∈ Q. Therefore α ∈ Q((H)). Thus, F = Q((H)). • Step 4. Proving that D is not algebraic over F : −1 −1 −1 Suppose that γ = x−1 1 + x2 + . . . is an infinite formal sum. Since x1 < x2 < . . . , supp(γ) is well-ordered. Hence γ ∈ D. Consider the equality a0 + a1 γ + a2 γ 2 + . . . + an γ n = 0, ai ∈ F. (2) −1 −1 2 n−1 Note that X = x−1 1 x2 ...xn does not appear in the expressions of γ, γ , . . . , γ n and the coefficient of X in the expression of γ is n!. Therefore, the coefficient of X in the expression on the left side of the equality (2) is an .n!. It follows that an = 0. By induction, it is easy to see that a0 = a1 = . . . = an = 0. Hence, for any n ∈ N, the set {1, γ, γ 2 , . . . , γ n } is independent over F . Consequently, γ is not algebraic over F . • Step 5. Constructing a division subring of D which is a weakly locally finite: Consider the element γ from Step 4. For any n ≥ 1, put √ √ √ Rn = F ( p1 , p2 , . . . , pn , x1 , x2 , . . . , xn , γ), ∞ and R∞ = Rn . First, we prove that Rn is centrally finite for each positive n=1 integer n. Consider the element −1 γn = x−1 n+1 + xn+2 + . . . (infinite formal sum). −1 −1 Since γn = γ − (x−1 1 + x2 + . . . + xn ), we conclude that γn ∈ Rn and √ √ √ √ √ √ F ( p1 , p2 , . . . , pn , x1 , x2 , . . . , xn , γ) = F ( p1 , p2 , . . . , pn , x1 , x2 , . . . , xn , γn ). DIVISION RINGS RELATED TO THE KUROSH PROBLEM 5 √ Note that γn commutes with all pi and all xi (for i = 1, 2, ..., n). Therefore √ √ √ Rn = F ( p1 , p2 , . . . , pn , x1 , x2 , . . . , xn , γn ) √ √ √ = F (γn )( p1 , p2 , . . . , pn , x1 , x2 , . . . , xn ). √ √ √ In combination with the equalities ( pi )2 = pi , x2i ∈ F , pi xj = xj pi , i = j, √ √ pi xi = −xi pi , it follows that every element β from Rn can be written in the form √ √ β= a(ε1 ,...,εn ,µ1 ,...,µn ) ( p1 )ε1 . . . ( pn )εn xµ1 1 . . . xµnn , 0≤εi ,µi ≤1 where a(ε1 ,...,εn µ1 ,...,µn ) ∈ F (γn ). Hence Rn is a vector space over F (γn ) having the finite set Bn which consists of the products √ √ ( p1 )ε1 . . . ( pn )εn xµ1 1 . . . xµnn , 0 ≤ εi , µi ≤ 1 as a base. Thus, Rn is a finite dimensional vector space over F (γn ). Since γn √ commutes with all pi and all xi , F (γn ) ⊆ Z(Rn ). It follows that dimZ(Rn ) Rn ≤ dimF (γn ) Rn < ∞ and consequently, Rn is centrally finite. For any finite subset S ⊆ R∞ , there exists n such that S ⊆ Rn . Therefore, the division subring of R∞ , generated by S over F is contained in Rn , which is centrally finite. Thus, R∞ is weakly locally finite. • Step 6. Finding the center of R∞ : √ √ We claim that Z(R∞ ) = F . Put Sn = { p1 , ..., pn , x1 , ..., xn }. Since for any √ √ √ √ √ √ √ 2 √ 2 i = j, xi , ( pi ) ∈ F , xi xj = xj xi , pi pj = pj pi , xi pj = pj xi , xi pi = √ − pi xi , every element from F [Sn ] can be expressed in the form √ √ α= a(ε1 ,...,εn ,µ1 ,...,µn ) ( p1 )ε1 . . . ( pn )εn xµ1 1 . . . xµnn , a(ε1 ,...,εn ,µ1 ,...,µn ) ∈ F. 0≤εi ,µi ≤1 (3) Moreover, the set Bn consists of products √ √ ( p1 )ε1 . . . ( pn )εn xµ1 1 . . . xµnn , 0 ≤ εi , µi ≤ 1 is finite of 22n elements. Hence, F [Sn ] is a finite dimensional vector space over F . So, it follows that F [Sn ] = F (Sn ). Therefore, every element from F (Sn ) can be expressed in the form (3). In the first, we show that Z(F (S1 )) = F . Thus, suppose that α ∈ Z(F (S1 )). √ √ √ Since x21 , ( p1 )2 = p1 ∈ F and x1 p1 = − p1 x1 , every element α ∈ F (S1 ) = √ F ( p1 , x1 ) can be expressed in the following form: √ √ α = a + b p1 + cx1 + d p1 x1 , a, b, c, d ∈ F. √ Since α commutes with x1 and p1 , we have √ √ √ √ ax1 + b p1 x1 + cx21 + d p1 x21 = ax1 − b p1 x1 + cx21 − d p1 x21 , and √ √ √ √ a p1 + bp1 − c p1 x1 − dp1 x1 = a p1 + bp1 + c p1 x1 + dp1 x1 . From the first equality it follows that b = d = 0, while from the second equality we get c = 0. Hence, α = a ∈ F , and consequently, Z(F (S1 )) = F . Suppose that n ≥ 1 and α ∈ Z(F (Sn )). By (3), α can be expressed in the form √ √ α = a1 + a2 pn + a3 xn + a4 pn xn , with a1 , a2 , a3 , a4 ∈ F (Sn−1 ). 6 BUI XUAN HAI, MAI HOANG BIEN, AND TRINH THANH DEO From the equality αxn = xn α, it follows that √ √ √ √ a1 xn + a2 pn xn + a3 x2n + a4 pn x2n = a1 xn − a2 pn xn + a3 x2n − a4 pn x2n . Therefore, a2 + a4 xn = 0, and consequently we have a2 = a4 = 0. Now, from √ √ √ √ √ √ the equality α pn = pn α, we have a1 pn − a3 pn xn = a1 pn + a3 pn xn , and it follows that a3 = 0. Therefore, α = a1 ∈ F (Sn−1 ), and this means that α ∈ Z(F (Sn−1 )). Thus, we have proved that Z(F (Sn )) ⊆ Z(F (Sn−1 )). By induction we can conclude that Z(F (Sn )) ⊆ Z(F (S1 )) for any positive integer n. Since F ⊆ Z(F (Sn )) ⊆ Z(F (S1 )) = F , it follows that Z(F (Sn )) = F for any positive integer n. Now, suppose that α ∈ Z(R∞ ). Then, there exists some n such that α ∈ Rn , and clearly α ∈ Z(F (Sn )) = F . Hence Z(R∞ ) = F . • Step 7. Proving that R∞ is not algebraic over F : It was shown in Step 4 that γ ∈ R∞ is not algebraic over F . The theorem above shows, in particular that the class of weakly locally finite division rings strictly contains the class of locally finite division rings. By the following theorem, we prove that the Kurosh problem is true for the class of weakly locally finite division rings. Theorem 4. A division ring D is locally finite if and only if D is weakly locally finite and algebraic. Proof. If D is locally finite, then clearly D is both weakly locally finite and algebraic. Conversely, assume that D is both weakly locally finite and algebraic. Let F = Z(D) and S be a finite subset of D. Since D is weakly locally finite, the division subring L of D generated by S is centrally finite. Let B = {x1 , x2 , . . . , xn } be the basis of [L : Z(L)]. For any 1 ≤ i, j ≤ n, write xi xj = aij1 x1 + aij2 x2 + . . . + aijn xn , where aijk ∈ Z(L). Let K be the division subring of D generated by F and all aijk . One has K is a subfield of D. By D is algebraic over F and set of all aijk is finite, K/F is a finite field extension. Let H = {a1 x1 + . . . + an xn |ai ∈ K}. Then H is a finite dimensional vector space over K, and it is clear that H is a subring of D. Now, for any x ∈ H, the n set {1, x, x2 , . . . , xn+1 } is linearly dependent over K, hence i=0 ci xi = 0 for some ci ∈ K not all zero. It follows that x−1 ∈ H, so H is a division subring of D. Moreover dimF H < ∞, since dimK H < ∞ and K/F is a finite field extension. It is easy to see that H = F (S), and the proof is now complete. 3. Herstein’s conjecture for weakly locally finite division rings Let K D be division rings. Recall that an element x ∈ D is radical over K if there exists some positive integer n(x) depending on x such that xn(x) ∈ K. A subset S of D is radical over K if every element from S is radical over K. In 1978, I.N. Herstein [8, Conjecture 3] conjectured that given a subnormal subgroup N of D∗ , if N is radical over center F of D, then N is central, i. e. N is contained in F . Herstein, himself in the cited above paper proved this fact for the special case, when N is torsion group. However, the problem remains still open in general. In [5], it was proved that this conjecture is true in the finite dimensional case. In this section, we shall prove that this conjecture is also true for weakly locally finite division rings. First, we note the following two lemmas we need for our further purpose. DIVISION RINGS RELATED TO THE KUROSH PROBLEM 7 Lemma 5. Let D be a division ring with center F . If N is a subnormal subgroup of D∗ , then Z(N ) = N ∩ F . Proof. If N is contained in F , then there is nothing to prove. Thus, suppose that N is non-central. By [9, 14.4.2, p. 439], CD (N ) = F . Hence Z(N ) ⊆ N ∩ F . Since the inclusion N ∩ F ⊆ Z(N ) is obvious, Z(N ) = N ∩ F . Lemma 6. If D is a weakly locally finite division ring, then Z(D ) is a torsion group. Proof. By Lemma 5, Z(D ) = D ∩F . For any x ∈ Z(D ), there exists some positive integer n and some ai , bi ∈ D∗ , 1 ≤ i ≤ n, such that −1 −1 −1 −1 −1 x = a1 b1 a−1 1 b1 a2 b2 a2 b2 . . . an bn an bn . Set S := {ai , bi | 1 ≤ i ≤ n}. Since D is weakly locally finite, the division subring L of D generated by S is centrally finite. Put n = [L : Z(L)]. Since x ∈ F , x commutes with every element of S. Therefore, x commutes with every element of L, and consequently, x ∈ Z(L). So, −1 −1 −1 −1 −1 xn = NL/Z(L) (x) = NL/Z(L) (a1 b1 a−1 1 b1 a2 b2 a2 b2 . . . an bn an bn ) = 1. Thus, x is torsion. In [4, Theorem 1], Herstein proved that, if in a division ring D every multiplicative commutator aba−1 b−1 is torsion, then D is commutative. Further, with the assumption that D is a finite dimensional vector space over its center F , he proved [4, Theorem 2] that, if every multiplicative commutator in D is radical over F , then D is commutative. Now, using Lemma 6, we can carry over the last fact for weakly locally finite division rings. Theorem 7. Let D be a weakly locally finite division ring with center F . If every multiplicative commutator in D is radical over F , then D is commutative. Proof. For any a, b ∈ D∗ , there exists a positive integer n = nab depending on a and b such that (aba−1 b−1 )n ∈ F. Hence, by Lemma 6, it follows that aba−1 b−1 is torsion. Now, by [4, Theorem 1], D is commutative. The following theorem gives the affirmative answer to Conjecture 3 in [8] for weakly locally finite division rings. Theorem 8. Let D be a weakly locally finite division ring with center F and N be a subnormal subgroup of D∗ . If N is radical over F , then N is central, i.e. N is contained in F . Proof. Consider the subgroup N = [N, N ] ⊆ D and suppose that x ∈ N . Since N is radical over F , there exists some positive integer n such that xn ∈ F . Hence xn ∈ F ∩ D = Z(D ). By Lemma 6, xn is torsion, and consequently, x is torsion too. Moreover, since N is subnormal in D∗ , so is N . Hence, by [4, Theorem 8], N ⊆ F . Thus, N is solvable, and by [9, 14.4.4, p. 440], N ⊆ F . In Herstein’s conjecture a subgroup N is required to be radical over center F of D. What happen if N is required to be radical over some proper division subring of D (which not necessarily coincides with F )? In the other words, the following question should be interesting: “Let D be a division ring and K be a proper division subring of D and given a subnormal subgroup N of D∗ . If N is radical over K, 8 BUI XUAN HAI, MAI HOANG BIEN, AND TRINH THANH DEO then is it contained in center F of D?” In the following we give the affirmative answer to this question for a weakly locally finite ring D and a normal subgroup N. Lemma 9. Let D be a weakly locally finite division ring with center F and N be a subnormal subgroup of D∗ . If for every elements x, y ∈ N , there exists some positive integer nxy such that xnxy y = yxnxy , then N ⊆ F . Proof. Since N is subnormal in D∗ , there exists the following series of subgroups N = N1 N2 ... Nr = D ∗ . Suppose that x, y ∈ N . Let K be the division subring of D generated by x and y. Then, K is centrally finite. By putting Mi = K ∩ Ni , ∀i ∈ {1, . . . , r} we obtain the following series of subgroups M1 M2 ... Mr = K ∗ . For any a ∈ M1 ≤ N1 = N , suppose that nax and nay are positive integers such that anax x = xanax and anay y = yanay . Then, for n := nax nay we have an = (anax )nay = (xanax x−1 )nay = xanax nay x−1 = xan x−1 , and an = (anay )nax = (yanay y −1 )nax = yanay nay y −1 = yan y −1 . Therefore an ∈ Z(K). Hence M1 is radical over Z(K). By Theorem 8, M1 ⊆ Z(K). In particular, x and y commute with each other. Consequently, N is abelian group. By [9, 14.4.4, p. 440], N ⊆ F . Theorem 10. Let D be a weakly locally finite division ring with center F and K be a proper division subring of D. Then, every normal subgroup of D∗ which is radical over K is contained in F . Proof. Assume that N is a normal subgroup of D∗ which is radical over K, and N is not contained in the center F . If N \ K = ∅, then N ⊆ K. By [9, p. 433], either K ⊆ F or K = D. Since K = D by the assertion, it follows that K ⊆ F . Hence N ⊆ F , that contradicts to the assertion. Thus, we have N \ K = ∅. Now, to complete the proof of our theorem we shall show that the elements of N satisfy the requirements of Lemma 9. Thus, suppose that a, b ∈ N . We examine the following cases: Case 1: a ∈ K. Subcase 1.1: b ∈ K. We shall prove that there exists some positive integer n such that an b = ban . Thus, suppose that an b = ban for any positive integer n. Then, a + b = 0, a = ±1 and b = ±1. So we have x = (a + b)a(a + b)−1 , y = (b + 1)a(b + 1)−1 ∈ N. Since N is radical over K, we can find some positive integers mx and my such that xmx = (a + b)amx (a + b)−1 , y my = (b + 1)amy (b + 1)−1 ∈ K. Putting m = mx my , we have xm = (a + b)am (a + b)−1 , y m = (b + 1)am (b + 1)−1 ∈ K. Direct calculations give the equalities xm b − y m b + xm a − y m = xm (a + b) − y m (b + 1) = (a + b)am − (b + 1)am = am (a − 1), from that we get the following equality (xm − y m )b = am (a − 1) + y m − xm a. DIVISION RINGS RELATED TO THE KUROSH PROBLEM 9 If (xm − y m ) = 0, then b = (xm − y m )−1 [a(am − 1) + y m − xm a] ∈ K, that is a contradiction to the choice of b. Therefore (xm − y m ) = 0 and consequently, am (a − 1) = y m (a − 1). Since a = 1, am = y m = (b + 1)am (b + 1)−1 and it follows that am b = bam , a contradiction. Subcase 1.2: b ∈ K. Consider an element x ∈ N \ K. Since xb ∈ K, by Subcase 1.1, there exist some positive integers r, s such that ar xb = xbar and as x = xas . From these equalities it follows that ars = (xb)−1 ars (xb) = b−1 (x−1 ars x)b = b−1 ars b, and consequently, ars b = bars . Case 2: a ∈ K. Since N is radical over K, there exists some positive integer m such that am ∈ K. By Case 1, there exists some positive integer n such that amn b = bamn . Acknowledgments The authors thank Professor Pham Ngoc Anh for his comments and useful suggestions. The authors are funded by Vietnam National Foundation for Science and Technology Development (NAFOSTED) under Grant No. 101.04-2013.01. A part of this work was done when the first and the third authors were working as researchers at the Vietnam Institute for Advanced Study in Mathematics (VIASM). They would like to express their sincere thanks to VIASM for providing a fruitful research environment and hospitality. References [1] S. AKBARI and M. MAHDAVI-HEZAVEHI, ‘Normal subgroups of GLn (D) are not finitely generated’, Proc. of the Amer. Math. Soc., 128 (6) (2000), 1627-1632. [2] J.B. BELL, L.W. 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Bui Xuan Hai, Faculty of Mathematics and Computer Science, University of Science, VNU-HCM 227 Nguyen Van Cu Str., Dist. 5, Ho Chi Minh City, Vietnam e-mail: bxhai@hcmus.edu.vn Trinh Thanh Deo, Faculty of Mathematics and Computer Science, University of Science, VNU-HCM 227 Nguyen Van Cu Str., Dist. 5, Ho Chi Minh City, Vietnam e-mail: ttdeo@hcmus.edu.vn Mai Hoang Bien Department of Basic Sciences, University of Architecture, 196 Pasteur Str., Dist. 1, HCM-City, Vietnam e-mail: maihoangbien012@yahoo.com  ... paragraph is to construct an example showing that the class of weakly locally finite division rings strictly contains the class of locally finite division rings, and we then prove that the Kurosh problem. .. algebraic over F The theorem above shows, in particular that the class of weakly locally finite division rings strictly contains the class of locally finite division rings By the following theorem, we... lemmas we need for our further purpose DIVISION RINGS RELATED TO THE KUROSH PROBLEM Lemma Let D be a division ring with center F If N is a subnormal subgroup of D∗ , then Z(N ) = N ∩ F Proof