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JOURNAL OF 184, 570᎐574Ž. 1996 ARTICLE NO. 0275

Free Subgroups and Free Subsemigroups of Rings

Katsuo Chiba

Niihama National College of Technology, Yagumo-Cho 7-1, Niihama 792, Japan

Communicated by Susan Montgomery View metadata, citation and similar papers at core.ac.uk brought to you by CORE

Received July 31, 1995 provided by Elsevier - Publisher Connector

Inwx 4 Lichtman raised a question whether the multiplicative of a noncommutative division D contains a noncyclic free subgroup. The answer is positive when D is finite-dimensional over its centerw 2, Theorem 2.1x . Recently Reichstein and Vonessen showed that the multiplicative group of D contains a noncyclic free subgroup, if the K of D is uncountable and there exists a non-central element a g D which is algebraic over K wx8, Theorem 1 . In this note we prove the above result without the assumption of the existence of a non-central element a g D which is algebraic over K Ž.Theorem 2 . On the other hand Klein raised a question whether the multiplicative of a noncommutative do- main contains a noncommutative free subsemigroup and he proved that the multiplicative semigroup of the ring of polynomials in two central indeterminates over a noncommutative contains a noncommuta- tive free subsemigroupwx 3 . Both of these questions are still open. Our main results are the following three theorems. THEOREM 1. Let D be a noncommutati¨edi¨ision ring with center K, Duwx,¨ the in two central indeterminates u, ¨ o¨er D, and DuŽ.,¨ the quotient di¨ision ring of Dwx u, ¨ . Let a and b be elements of D Ž.y1Ž. 2 such that ab y ba / 0 and write d12 s ab y ba a y u , d21 syb Ž.y1Žy1. ÄŽ. ab y ba bab y u g Duwx,¨.Then the elements of S s 1 y f¨d12 Ž.Ž.Ž. Ä44 1yg¨d21 1qf¨d12 1qg¨d21 Nf,ggKwx¨_0 freely generates a free subgroup of the multiplicati¨e group of DŽ. u, ¨ . THEOREM 2. Let D be a noncommutati¨edi¨ision ring with uncountable center K. Then the multiplicati¨e group of D contains a noncyclic free subgroup whose cardinal number is the same as that of K.

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0021-8693r96 $18.00 Copyright ᮊ 1996 by Academic Press, Inc. All rights of reproduction in any form reserved. FREE SUBGROUPS OF DIVISION RINGS 571

THEOREM 3. Let D be a noncommutati¨edi¨ision ring with center K, Dwx u the polynomial ring in a central indeterminate u o¨er D, and<< K the cardinal number of K. Then the multiplicati¨e semigroup of Dwx u contains a noncommutati¨e free subsemigroup on<< K free generators. Theorem 2 sharpens results of Makar᎐Limanovwx 6, Theorem , and of Makar᎐Limanov and Malcolmsonwxwx 7, Theorem , and of 8, Theorem 1 . In wx3 Klein asked whether the multiplicative semigroup of the polynomial ring in one central indeterminate over a noncommutative domain contains a noncommutative free subsemigroup. Theorem 3 gives an affirmative answer to the question for a special case.

Let R be a ring. We denote by MRnŽ.the n = n ring over R,by GLnnŽ. R the group of units of MRŽ., and by eijthe matrix with 1 in the Ž.i,j-entry and 0 elsewhere. We denote by Rwx¨ the polynomial ring in a central indeterminate ¨ over R and by Rww¨ xx the ring Ž .Ž. over R. By the natural isomorphisms MRnnwx¨ (MRwx¨ and Ž.Ž. Ž .Ž. MRnnww¨ xx (MRww¨ xx, we shall identify MRnwx¨ with MR nwx¨and Ž.Ž. MRnnww¨ xx with MRww¨ xx, respectively. Let V denote the of Rww¨ xx generated by ¨ and let X be a subset of Rww¨ xx. We denote by X the closure of X in Rww¨ xxwith respect to V-adic topology on R ww¨ xx.If Ris a division ring, we denote by RŽ¨ .and R ŽŽ¨ .. the quotient division rings of Rw¨ xand R ww¨ xx, respectively. There is a natural embedding RŽ.¨ into RŽŽ¨ .., so we shall think of R Ž¨ .as a of R ŽŽ¨ ... Let G be a group, and Y a subset of G.If Y is a free set of generators of G, we also say that Y freely generates G or that G is free on Y. Let S be a set. We denote by <

LEMMA 1. Let k be a commutati¨e and kwx¨ the polynomial ring in a Ä 4 commutati¨e indeterminate ¨ o¨er k. Let A s 1 q f¨e12 N f g kwx¨ and Ä 4 Ž . Bs1qg¨e21 Nggkwx¨,the subgroups of GL2 kwx¨ . Then the subgroup Ž . Äy1 y1 4 of GL2 kwx¨ generated by S s ababNagA,a/1, b g B, b / 1 is free on S. Ž. Proof. Bywx 1, Lemma 5 the subgroup of GL2 kwx¨ generated by A j B is the free product A) B of the groups A and B. Byw 5, Sect. 4.1, Ž . Problem 24, p. 196x the subgroup of GL2 kwx¨ generated by S is free on S. This completes the proof.

Proof of Theorem 1. We may regard Dwx¨ as a right Ku w,¨ x- by putting x и fuŽ.,¨ sxf Ž. a, ¨ for x g Dwx¨ and fuŽ.,¨ gKuwx,¨. Since Duwx,¨ sDmK Ku wxwx,¨,D¨ becomes a left Du wx,¨-module in the natu- ²: ral way. Let R s Kd12,d 21,¨ be the K-subalgebra of Duwx,¨ generated by d12, d 21, and ¨ and M s 1 и Kwx¨ q b и K wx¨ a right K wx¨ -submodule of Dwx¨ generated by Kwx¨ -linearly independent elements 1 and b. Since 572 KATSUO CHIBA

Ž .Ž. rM ; M for all r g R, the map ␸: R ª MK2 wx¨ defined by r1, rb s Ž .Ž. Ž. Ž. 1, b ␸ r is a such that ␸ d12s e 12, ␸ d 21s e 21, Ž. and ␸ ¨ s ¨. Let V12be the ideal of Duwx,¨ generated by ¨, V the ideal Ž. of R generated by ¨, V32the ideal of MKwx¨ generated by ¨, and V the nn ideal of Duw xww¨ xx generated by ¨. Then V21s V l R for all positive Ž. n and ␸ V23s V ; hence the V2-adic completion of R is naturally isomorphic to R, the closure of R in Duw xww¨ xx with respect to V-adic Ž . topology on Duw xww¨ xx, and the V32-adic completion of MKwx¨ s Ž. Ž. Ž . MK222wx¨ is MKww¨ xx sMKww¨ xx , therefore ␸ extends to a ring Ž.Ž. Ž. homomorphism ␸: R ª MK22ww¨ xx sMKww¨ xx. By Lemma 1, ␸ S s ÄŽ.Ž.Ž.Ž. Ä44 1yf¨e12 1yg¨e21 1qf¨e12 1qg¨e21 Nf,ggKwx¨_0 freely gen- Ž . erates a free subgroup of GL2 Kwx¨ . Since the elements of S are invertible in R, S freely generates a free subgroup of the group of R. Since DuŽ,¨ .;Du Ž .ŽŽ¨ .. and R ; Duw xww¨ xx ;DuŽ .ŽŽ¨ .., the inverses of the elements of S in R are contained in DuŽ.,¨. Thus the subgroup of the multiplicative group of DuŽ.,¨ generated by S is free on S. This completes the proof. Proof of Theorem 2. By Theorem 1, it suffices to show that D contains a noncommutative division ring E and two central elements u, ¨ such that Ž.1 uand ¨ are algebraically independent over E, and Ž. 2 the cardinality of the center of E is equal to the cardinality of the center of D.By hypothesis D / K, there are two elements a, b in D such that ab y ba / 0. Let D1 be the subdivision ring of D generated by a and b, K1 the center of D11, KKŽ.the subfield of D generated by K1and K, and Ä4Ä 4 Ž. u,¨j¨i NigIa transcendental basis of KK11over K , where I is the index set. Since K1 is finite or countable and K is uncountable, we Ä pqr1r2 rn have <

LEMMA 2. Let D be a noncommutati¨edi¨ision ring with center K, Duwx the polynomial ring in a central indeterminate u o¨er D, and DŽ. u the quotient FREE SUBGROUPS OF DIVISION RINGS 573 di¨ision ring of Dwx u and<< K the cardinal number of K. If there is a non-central element a of D which is algebraic o¨er K, then the multiplicati¨e group of DŽ. u contains a noncyclic free subgroup on<< K free generators which are contained in Dwx u . Proof. If K is uncountable, the results follow from Theorem 2, since the multiplicative group of D contains a free group on a subset X of D such that <

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