Free Group Algebras in Certain Division Rings

JOURNAL OF ALGEBRA ARTICLE NO.

185, 298]313 Ž1996.

0326

Free Group Algebras in Certain Division Rings L. M. V. FigueiredoU Department of Mathematics, Uni¨ ersity of Sao Paulo, CP 20570 Ag Iguatemi, Sao Paulo SP, Brazil

J. Z. Gonc¸alves† Department of Mathematics, Uni¨ ersity of Sao Paulo, CP 20570 Ag Iguatemi, Sao Paulo SP, Brazil

and M. Shirvani ‡ Department of Mathematical Sciences, Uni¨ ersity of Alberta, Edmonton, Alberta, Canada T6G 2G1 Communicated by Susan Montgomery Received March 29, 1995

Let D be a division ring with centre k. We show that D contains the k-group algebra of the free group on two generators when D is the ring of fractions of a suitable skew polynomial ring, or it is generated by a polycyclic-by-finite group which is not abelian-by-finite, or it is the ring of fractions of the universal enveloping algebra of a finite-dimensional Lie algebra of characteristic zero. Q 1996 Academic Press, Inc.

1. INTRODUCTION Let A1 be the first Weyl algebra over the field of complex numbers, and let D be the field of fractions of A1. In w9x, in the course of computing the Gel’fand]Kirillov dimension of certain subalgebras of D, Makar-Limanov U

E-mail: [email protected]. Research partially supported by CNPq-Brazil. E-mail: [email protected]. ‡ Research partially supported by NSERC, Canada, and FAPESP-Brazil. E-mail: [email protected]. †

298 0021-8693r96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.

FREE GROUP ALGEBRAS

299

showed that D contains the complex group algebra of a Žfrom now on always noncyclic. free group. That this may be a general phenomenon, at least when the centre of a division ring is uncountable, is indicated by the following: THEOREM w1x. If the centre k of a di¨ ision ring D is uncountable and if D contains a free k-algebra, then D contains the group algebra of a free group o¨ er k. In this paper we address the question of the existence of free group algebras in certain families of division rings. More precisely, we prove THEOREM A. Let K be a field, u g AutŽ K . of infinite order, and let k be the fixed field of u . Define d by c d s c y cu for all c g K. Let a, b g K U be such that a has infinite orbit under u and K d l bkw ax s  04 . Let N be an integer. Then  a, bx N Ž1 y x .y1 4 is k-free. THEOREM B. Let K s k Ž t . be the rational function field o¨ er k and R s k Ž t .w x; u , c x, where u is a k-automorphism of K and c is a u-deri¨ ation of K. Let D be the field of fractions of R. Ži. If u has infinite order, then D contains a k-free group algebra. Žii. If u s 1, d / 0, and char K s 0, then D contains a k-free group algebra. Žiii. In all other cases, D satisfies a polynomial identity. THEOREM C. Let D s k Ž G . be the di¨ ision k-algebra generated by a polycyclic-by-finite group G F DU . Then D contains a k-free group algebra if and only if G is not abelian-by-finite. THEOREM D. The field of fractions of the uni¨ ersal en¨ eloping algebra of a nonabelian, soluble, or finite-dimensional Lie algebra o¨ er a field k of characteristic zero contains a k-free group algebra. To prove the above results, we begin by producing k-free sets in the field of fractions of skew polynomial rings. If the free generators have valuation at least 1 Žrelative to whatever valuation happens to be present., then the following result of Lichtman may be applied to produce elements that generate a free group algebra: PROPOSITION ŽLichtman w4x..

Let U be an Ore domain with a discrete

¨ aluation ¨ . Let D be the ring of fractions of U and k the centre of D. Then for any k-free set x 1 , x 2 such that ¨ Ž x i . G 1 for i s 1, 2, the elements 1 q x i

generate a free group ring kF : D.

To consider the existence of free algebras in the field of fractions of skew polynomial rings, we introduce a polynomial ring with the property

300

FIGUEIREDO, GONC ¸ALVES, AND SHIRVANI

that the k-freeness of two elements is equivalent to the question of whether a certain family of polynomials is contained in an ideal in this new ring. We then use a suitable functional equation for a generating function of the family of polynomials to show that the family is not contained in the ideal. In Section 4, we give a more combinatorial criterion ŽCorollary 4.1., deduce Theorem D, and indicate how the main freeness arguments of w7x, w8x, and w10x also follow from the criterion. The reader should also compare our approach with that of Lorenz w5x. Both ultimately rely on MakarLimanov’s ideas w7, 9x, but ours seem to be simpler and more suitable for generalization.

2. FREE GROUP ALGEBRAS IN THE RING OF FRACTIONS OF SKEW POLYNOMIAL RINGS Let K, u , k, and N be as in the statement of Theorem A. Let p s x N Ý nis0 x ia i g k w x; u x, where n G 1, a i g k for all i, and a0 a n / 0. Viewing K w x; u x as a subset of the ring of fractions K Ž x; u . and the latter as a subset of the ring of skew Laurent series K ŽŽ x; u .., we may write `

py1 s xyN

Ý

x ma m ,

ms0

where a0 a 0 s 1 and Ý iqmsj a i a m s 0 for all j G 1. It is more convenient to study the question of whether the set  a, bpy1 4 is k-free in a more general ring. Let A s k w X i , Yi , i g Z x be the ring of polynomials in commuting indeterminates X i and Yi over k, and let s be the automorphism of A defined by X i s s X iq1 and Yi s s Yiq1 for all i. In the ring of skew Laurent series AŽŽ X; s .., consider the element z s XyN Ý`ms 0 X ma m . Given a, b in K, define w s wa, b : AŽŽ X; s .. ª K ŽŽ x; u .. by X w s x, X i w s au i, and Yi w s bu i for all i. Then z w s py1 . Next, let W be the free k-algebra on  u, ¨ 4 . For w s w Ž u, ¨ . g W, the element w Ž X 0 , Y0 z . g AŽŽ X; s .. may be written as w Ž X 0 , Y0 z . s

Ý X i Pi Ž w . igZ

with all Pi Ž w . g A Žand the series has only a finite number of nonzero terms with i - 0.. The idea, of course, is that w Ž a, bpy1 . s w Ž X 0 , Y0 z . wa, b , so the polynomials Pi Ž w . contain the information on the k-freeness of

301

FREE GROUP ALGEBRAS

 a, bpy1 4 . Next, define the monomials m I g W by mB s 1, and for a tuple I s Ž i 0 , . . . , i s . of nonnegative integers, with s G 0, let

½

mI s

if s s 0, if s G 1.

ui0 , s u i 0 Ł rs1 Ž ¨uir . ,

Clearly, every element of W is uniquely of the form w s Ýg I m I , with g I g k. Also, Pi Ž w . s Ýg I Pi Ž m I .. We need the following result: LEMMA 2.1.

Let I s Ž i 0 , . . . , i s . with s G 1. Then for all integers t, Pt Ž m I . s

Ptyd Ž m I X . s Yd a dqN X 0i s . d

ž

Ý

/

dGyN

Proof. By the definition of m I , we have m I s m I X Ž ¨ u i s ., and so

Ý X t Pt Ž m I . s m I Ž X 0 , Y0 z . Y0 zX0i X

s

t

s s

ž

Ý X j Pj Ž m I . X

j

X ma m X 0i s

X jyN Pj Ž m I X . s

/

YyN X ma m X 0i s

mG0

Ý Ý j

Ý ms0

yN

Ý Ý j

s

/ž

`

Y0 XyN

X jyNqm Pj Ž m I X . s

my N

mG0

=Ymy N a m X 0i s . Now put t s j q m y N, d s m y N, and observe that d G yN since m G 0. The result follows upon rewriting the last double summation and equating the coefficients of the powers of X. As usual, for I s Ž i 0 , . . . , i s ., we write t Ž I . s i s , I X s Ž i 0 , . . . , i sy1 ., and l Ž I . s s q 1. Let AŽŽ z .. be the ring of Laurent series in a central indeterminate z over A, and let s act on AŽŽ z .. through the coefficients. For nonempty I, define H Ž IX . s

ŽI. i z g AŽ Ž z . . . Ý Pi Ž m I . Xyt 0

igZ

We claim that the right-hand side only depends on I X . Indeed, if I X s J X , ŽI. ŽJ. then m I Ž X 0 , Y0 z . Xyt s m I X Ž X 0 , Y0 z .Y0 z s m J Ž X 0 , Y0 z . Xyt , whence 0 0 yt Ž I . yt Ž J . Pi Ž m I . X 0 s Pi Ž m J . X 0 for all i. Define an equivalence relation ; on the set of tuples by I ; J if and only if I X s J X , and let I be a complete set of representatives of the equivalence classes of ; . Write L a, b s ker wa, b < AŽŽ z .., a prime ideal of AŽŽ z ... We then have the following:

FIGUEIREDO, GONC ¸ALVES, AND SHIRVANI

302

PROPOSITION 2.1. Let w s Ýg I m I g W, where almost all g I s 0. Then the following are equi¨ alent: Ža. w Ž a, bpy1 . s 0. Žb. Ý J g I ŽÝ I ; J g I X 0tŽ I . . H Ž J X . g L a, b . Proof. Of course 0 s w Ž a, bpy1 . s w Ž X 0 , Y0 z . wa, b if and only if Ý i X i Pi Ž w . s w Ž X 0 , Y0 z . g ker wa, b , if and only if each Pi Ž w . g ker wa, b Žby the definition of w ., if and only if Ý i Pi Ž w . z i g L a, b Žagain by definition.. However,

Ý Pi Ž w . z i s Ý Ý g I Pi Ž m I . z i i

i

s

Ý Ý Pi Ž m I . z i I

s

I

ž

i

/

Ý g I H Ž I X . X 0tŽ I . I

s

Ý Ý Ž g I X 0tŽ I . . H Ž J X . , Jg I I;J

as required. The point is that condition Žb. above is quite tractable, especially in view of a recurrence relation satisfied by the H. We also observe that H ŽB. s 1, as follows immediately from the various definitions. THEOREM 2.1. H Ž I X . X 0tŽ I . Y0 .

For all I with l Ž I . G 2 we ha¨ e z N Ý nrs0 a r z r H Ž I . s

Proof. Given J, Lemma 2.1 implies that H Ž JX . s

ŽJ. t z Ý Pt Ž m J . Xyt 0

t

s

t

s

Ptyd Ž m J X . s Yd a dqN z d

Ý Ý dGyN

Ý

Yd a dqN z d

dGyN

s

t

ž

Ý Ptyd Ž m J X . z t X

Ý

Yd a dqN z d Ž H Ž J Y . X 0tŽ J . .

sd

dGyN

s

dGyN

X

H Ž J Y . s X dtŽ J . Yd a dqN z d . d

Ý

tyd

sd

/

rq N

s

303

FREE GROUP ALGEBRAS

In particular, replacing J X by I gives H Ž I . s Ý d G N H Ž I X . s X dtŽ I . Yd a dqN z d, whence d

n

zN

Ý ar z r H Ž I . s

rq N

rs0 n

s

Ý ar z

rqN

rs0 n

s

ž

HŽ I .

Ý

d

X dtŽ I . Yd a dqN z d

dGyN

H Ž IX . s

Ý Ý

X s

rq Nq d

s rq N

/

tŽ I . z rqNqd X rqNqd YrqNqd a r a dqN .

rs0 dGyN

Letting N q d s m and r q N q d s j, the last double summation beX j comes Ý jG 0 Ý rqmsj a r a m H Ž I X . s z j X jtŽ I . Yj . Since Ý rqmsj a r a m is 0 if j G 1 and is 1 when j s 0, the result follows. Proof of Theorem A. In this case p s X N Ž1 y X .y1 , so n s 1, a0 s 1, and a1 s y1. Suppose first that N ) 0. The formula in Theorem 2.1 becomes

z N H Ž J . s y z Nq1 H Ž J . s N

Nq 1

s Y0 X 0tŽ J . H Ž J X . ,

Ž 1.

i.e., y1

HŽ J . s

tŽ J . s z H Ž J . q zyN YyN y1 XyNy1 H Ž JX . s

yN y 1

.

Ž 2.

Now H ŽB. s 1 is invariant under s . Therefore, it can be proved by induction on l Ž J . that y1

HŽ J . s

lŽ J .

s z HŽ J . q

Ý l i Ž J . H Ž J Ž i. . ,

Ž 3.

is1

where l i Ž J . g Aw z , zy1 x and J Ž i. s Ž J Ž iy1. .X . To do this, use Ž2. and the yN y 1 inductive hypothesis to get a formula for H Ž J X . s , and then use Ž2. again. From Ž1. we also have that s

tŽ J . z H Ž J . s H Ž J . y zyN YyN XyN H Ž JX . s

yN

,

and so using Ž2. with J X in place of J, it follows that s

tŽ J . z H Ž J . s H Ž J . y YyN XyN H Ž JX . q

lŽ J .

Ý t i Ž J Ž i. . H Ž J Ž i. . ,

Ž 4.

is2

where t i Ž J Ž i. . g Aw z , zy1 x. Now, by Proposition 2.1Žb., it is enough to prove that if ŽÝ J g I Ý I ; J g I X 0tŽ I . . H Ž J . g L a, b , then Ý I ; J g I X 0tŽ I . g ker w

FIGUEIREDO, GONC ¸ALVES, AND SHIRVANI

304

for each J, for then the transcendence of a over k implies that g I s 0 for all I. We will work with a more general expression, proving Claim. If g J g Aw z , zy1 x are such that l s Ýtis0 Ý lŽ J .si g J H Ž J . g L a, b , then each g J g Žker wa, b < A .w z , zy1 x. The claim finishes the proof of Theorem A by letting g J s Ý I ; J g I X 0tŽ I . g A : Aw z , zy1 x. We now turn to the proof of the claim. Suppose that the claim is not true. In a counterexample l, we may clearly suppose that if each g J Ž . r Žwhere l F m are occurring in l is written in the form g J s Ý m rsl c r, J z Ž . . Ž . integers and c r, J g A , with c r, J f ker wa, b < A if cŽ r, J . / 0. Evidently, this implies that g J f Žker wa, b < A .w z , zy1 x for all J. In addition, choose the counterexample l g L a, b with the properties Ža. t is minimal and Žb. the total number of cŽ r, I . with l Ž I . s t is minimal. Fix J 0 with l Ž J 0 . s t. By multiplying l by a suitable power of z , if necessary, we may suppose that cŽ0, J 0 . f ker wa, b < A . Then cŽ0, J 0 . s l y cŽ0, J 0 . zls g L a, b and s

c Ž 0, J 0 . l y c Ž 0, J 0 . zls t

s

Ý Ý Ž c Ž 0, J0 .

s

g J H Ž J . y c Ž 0, J 0 . z H Ž J .

s

.

is0 l Ž J .si

s

Ý Ž c Ž 0, J0 .

s

g J y c Ž 0, J 0 . g Js . H Ž J .

l Ž J .st

q

Ý l Ž L .sty1

ž c Ž 0, J . 0

s

g L y c Ž 0, J 0 . g Ls q

tŽ J . Ý c Ž 0, J0 . g Js YyN XyN X

J sL

ty2

=H Ž L . q

Ý Ý

gJ H Ž J . ,

is0 l Ž J .si

where g J g Aw z , zy1 x. Write c s cŽ0, J 0 . w . We now have to consider several possibilities. Suppose first that  cŽ r, J .: l Ž J . s t 4 s  cŽ0, J 0 .4 . Then g J 0 s cŽ0, J 0 ., and therefore the above becomes s

c Ž 0, J 0 . l y c Ž 0, J 0 . zls s

ty1

Ý Ý

fJ H Ž J . .

is1 l Ž J .si

By the minimality of t, each f J g Žker w .w z , zy1 x. In particular, for L s J 0X , we have s

tŽ J 0 . c Ž 0, J 0 . g J 0X y c Ž 0, J 0 . g Js0X q c Ž 0, J 0 . g Js0 YyN XyN g Ž ker w . z , zy1 .

/

305

FREE GROUP ALGEBRAS

Thus, s

s

s

tŽ J 0 . c Ž 0, J 0 . c Ž 0, J 0X . y c Ž 0, J 0 . c Ž 0, J 0X . q c Ž 0, J 0 . c Ž 0, J 0 . YyN XyN

g ker w . Set d s cŽ0, J 0X . w . Then c u d y cd u q cc u Ž batŽ J 0 . . u s 0. Multiplying by yN Ž cc u .y1 , we obtain cy1 d y Ž cy1 d . u s yŽ batŽ J 0 . . u , contradicting the hypotheses. Thus the number of cŽ r, J . with l Ž J . s t is at least two. In this case, cŽ0, J 0 . s y cŽ0, J 0 . zls has one less cŽ r, J . with l Ž J . s t than the original l, so yN

s

c Ž 0, J 0 . g J y c Ž 0, J 0 . g Js g Ž ker w . z , zy1 for all J with l Ž J . s t and s

c Ž 0, J 0 . g L y c Ž 0, J 0 . g Ls q

tŽ J . g Ž ker w . Ý c Ž 0, J0 . g Js YyN XyN X

z , zy1

J sL

for all L with l Ž L. s t y 1. We claim that there exists J / J 0 with l Ž J . s t. For if not, then there must exist j / 0 such that cŽ j, J 0 . f ker w . However, s

s

c Ž 0, J 0 . c Ž j, J 0 . y c Ž 0, J 0 . c Ž j, J 0 . g ker w , and s

c Ž 0, J 0 . c Ž j, J 0X . y c Ž 0, J 0 . c Ž j, J 0X . s

tŽ J . q c Ž 0, J 0 . c Ž j, J 0 . YyN XyN g Ž ker w . .

Let d s cŽ j, J 0 . w and e s cŽ j, J 0X . w . Then, as before, we have c u d y cd u yN s 0, whence cy1 d g k. Also, c u e y ce u q cd u Ž batŽ J 0 . . u s 0. Multiplying both sides of the above equality by Ž cc u .y1 , we obtain y1

žc

e y Ž cy1 e .

u

y1

/ qc

d Ž batŽ J 0 . .

uyN

s 0.

The first term belongs to K d , and the second summand is a nonzero element of bkw ax. This is a contradiction. Thus, there exists J1 / J 0 and some cŽ j0 , J1 . f ker w . Therefore, s

s

c Ž 0, J 0 . c Ž j0 , J . y c Ž 0, J 0 . c Ž j0 , J . g ker w

FIGUEIREDO, GONC ¸ALVES, AND SHIRVANI

306

for all J with l Ž J . s t, and s

c Ž 0, J 0 . c Ž j0 , L . y c Ž 0, J 0 . c Ž j0 , L . q

s

tŽ J . g ker w Ý c Ž 0, J0 . c Ž j0 , J . YyN XyN X

J sL

for all L with l Ž L. s t y 1. Take L s J1X , and set d J s cŽ j0 , J . w and e L s cŽ j0 , L. w . Then, as before, c u d J y cd uJ s 0 implies that cy1d J g k for all J with l Ž J . s t. We also have the equation c u e L y ce Lu q

Ý X

J sL

cd uJ Ž batŽ J 0 . .

uyN

s 0,

which on multiplication by Ž cc u .y1 yields y1

žc

e L y Ž cy1 e L .

u

/q

yN

u Ý Ž cy1d uJ . Ž batŽ J . . 0

X

s 0.

J sL

Now, all the t Ž J . in the above formula are distinct, and at least one of the coefficients of a tŽ J . is nonzero. Moreover, the first summand belongs to d Ž K ., and the second summand is a nonzero element of bkw ax, which is impossible. The theorem is, therefore, established for positive N. When N F 0, use the K-isomorphism c : K Ž x; u . ª K Ž y; uy1 . defined by x c s yy1 . Then  a, bx N Ž1 y x .y1 4c s  a, ybyyN q1 Ž1 y y .y1 4 , and the latter set is k-free by the first part of the proof, since yN q 1 ) 0, a has infinite orbit under uy1 , and if d X is defined on K by c d X s c y cuy1 , then K d X l bkw ax s  04 . Before we prove Theorem B, we turn to a generalization of a result of Lorenz. In w6, Lemma 2x, Lorenz proves that for every infinite-order k-automorphism u of the rational function field k Ž t ., there is a generator a of k Ž t . satisfying the property stated in Theorem A in the form K d l akw ax s  04 . We refer to such an a as a distinguished generator of k Ž t .. COROLLARY 2.1. Let k be a field, u be a k-automorphism of k Ž t . of infinite order, and a be a distinguished generator of k Ž t . as abo¨ e, and choose any g g k w t x with g Ž0. s 0. Then for any integer N,  a, g Ž a. X N Ž1 y X .y1 4 is a k-free set in D s k Ž t .Ž X; u .. Moreo¨ er, k Ž t .Ž X; u . contains a free group algebra. Proof. Only the last statement requires proof. Put c s aX Ž1 y X .y1 . Then  a, c4 is a k-free set in D. The elements ac and c are also k-free and have valuation 1 relative to `

¨

žÝ / jGn

X j a j s n, where a n / 0

FREE GROUP ALGEBRAS

307

Žwhere we regard D as a subset of the ring of skew Laurent series.. The existence of the free group algebra now follows from Lichtman’s proposition. COROLLARY 2.2. Let G be a torsion-free soluble-by-finite group, and let D be the ring of fractions of the group algebra kG. Then D contains a k-free group algebra if and only if G is not abelian-by-finite. Proof. It is proved in w2x that kG has no zero divisors, and it is then well known that kG is an Ore domain, so the existence of D is clear. Also, D is finite dimensional over its centre if G is abelian-by-finite, so assume otherwise. By a theorem of Lorenz w5x, if G contains no nilpotent normal subgroup of finite index, then D contains the group algebra of a free group. We are left with the case where G contains a normal nilpotent nonabelian subgroup N of finite index. Then kN contains the skew polynomial ring k Ž t .w x; u x, where t u s l t and l is not a root of unity. It follows from Corollary 2.1 that D contains the group algebra of a free group, as required. Proof of Theorem B. To begin with, if u / 1, then by w12, Theorem 1x, R s k Ž t .w x; u , c x , k Ž t .w y; u , 0x, where 0 denotes the zero derivation. The latter ring is finite dimensional over its centre if u has finite order, and it contains a free group algebra if u has infinite order by Corollary 2.1 above. This proves Ži. and part of Žiii.. As for Žii., if ­ denotes formal differentiation on k Ž t ., then it is known that every k-derivation of k Ž t . is of the form g ­ for some g g k Ž t .. It follows that for any two k-derivations c 1 and c 2 of k Ž t ., the rings k Ž t .w x 1; 1, c 1 x and k Ž t .w x 2 ; 1, c 2 x are k-isox morphic wuse the k Ž t .-map x 1 ¬ g 1 gy1 2 x 2 . Therefore, it suffices to prove Žii. for any one derivation. Choose the k-derivation D with tD s t. Then in k Ž t .w x; 1, D x we have tx s xt q t s Ž x q 1. t, so conjugation by t defines the automorphism a : x ¬ x q 1 of the rational function field k Ž x .. Thus, k Ž t .Ž x; 1, D . s k Ž x .Ž t; a .. The latter is clearly finite dimensional over its centre if char k ) 0, and contains a free group algebra otherwise, again by Corollary 2.1. ŽAlternatively, since t and xty1 generate the first Weyl algebra over k, one may use w9x for this part.. This proves Žii. and the remaining part of Žiii..

3. FREE GROUP ALGEBRAS IN DIVISION RINGS GENERATED BY POLYCYCLIC-BY-FINITE GROUPS If G is abelian-by-finite, then D is finite dimensional over its centre, so suppose that G is not abelian-by-finite. For the proof of Theorem C we require the following well-known result of Bergman Žcf. w11x, 9.3.9..

308

FIGUEIREDO, GONC ¸ALVES, AND SHIRVANI

THEOREM ŽBergman.. Let A be a finitely generated torsion-free abelian group, and let U be a group of automorphisms of A. Assume that U and all its subgroups of finite index act rationally irreducibly on A. If I / A is a U-in¨ ariant ideal of kA, then either I s 0 or dim k kArI - `. Proof of Theorem C. Our argument is along the lines of Lorenz w5, Theorem 2.3x. Suppose first that G is not nilpotent-by-finite and that all its nilpotent subgroups are abelian-by-finite, for otherwise D contains a skew polynomial ring of the form appearing in the proof of Theorem B. Then, G contains a subgroup H s A² z :, with A free abelian of rank at least 2, and z and all of its powers acting rationally irreducibly on A. Now, the inclusion H ; DU induces an embedding kH ¨ D. For if the kernel I of the induced k-algebra map kH ª D is nonzero, then I is completely prime and I l kA s 0. ŽThis can be seen by choosing a nonzero element a s Ý qis0 a i z i g I with a i g kA and minimal q, and using the usual argument to find a nonzero element with smaller q on the assumption that q ) 0.. Thus, by the above theorem of Bergman, the field F s kArŽ I l kA. is a finite extension of k, which implies that some power of z acts trivially on F. This contradicts the way H was chosen. Hence, kH embeds into D. However, by Corollary 2.2, the field of fractions of kH contains a free group algebra, as required. Remark. It follows from Theorem C that DU always contains a free group w3x. When G is abelian-by-finite, D is finite dimensional over k and the claim is a consequence of the Tits alternative for linear groups. Otherwise, Theorem C is complementary to the main result of w5x. Note, however, that the Tits alternative is not used in our proof when D is finite dimensional over k.

4. RINGS OF FRACTIONS OF UNIVERSAL ENVELOPING ALGEBRAS The proof of Theorem D once again depends on the result of Lichtman, to go from suitable free k-sets to free group algebras. In order to produce the required free generators, we follow the construction of Makar-Limanov and Malcolmson, but use a different method to verify the freeness of the proposed generators. The following technical result is true in even greater generality, but already subsumes many of the freeness results in the literature. THEOREM 4.1. Let F be a field, u and w commuting automorphisms of F, E the fixed field of w , a a gi¨ en element of F, and suppose that, for each multi-index I / B and integer r g w0, l Ž I . y 1x, we are gi¨ en an element

309

FREE GROUP ALGEBRAS

dŽ a, I, r . g F. Write dŽ a, I . s dŽ a, I, l Ž I . y 1. and let B s  04 if u s 1, and B s Z otherwise. Assume that the following hold: Ža. E is contained in the fixed field of u . Žb. dŽ a, I, r . s dŽ a, I X , r . u for all I with l Ž I . G 2 and r g w0, l Ž I . y 2x. j Žc. For a fixed I X , the elements dŽ a, I . u , j g B, are linearly independent o¨ er E. Žd. For a fixed I X , if V Ž I X . is the E-subspace of F spanned by all the j dŽ a, I . u for j g B and d is defined on F by x d s x w y x for all x g F, then V Ž I X . l Fd s  04 . Define M Ž a, I . as follows: M Ž a, B. s 1 and for I / B, M Ž a, I . s

l Ž I .y1

Ł

d Ž a, I, r . Ž 1 y t .

y1

g FŽ t; w. .

rs0

Then, for different I, the elements M Ž a, I . are linearly independent o¨ er F. Proof. Write M Ž a, I . s Ý`ns 0 t n f nŽ a, I . g F ŽŽ t; w .. and consider the corresponding generating function GŽ a, I . s Ý`ns 0 f nŽ a, I . z n g F ŽŽ z .., where z is central over F. Clearly, it is enough to show that the GŽ a, I . are linearly independent over F. Let I be such that l Ž I . G 2. Then assumption Žb. implies that M Ž a, I . s M Ž a, I X . u dŽ a, I .Ž1 y t .y1 s M Ž a, I X . u dŽ a, I .w1 q t Ž1 y t .y1 x s M Ž a, I X . u dŽ a, I . q tM Ž a, I X . uw dŽ a, I . w Ž1 y t .y1 s M Ž a, I X . u dŽ a, I . q tM Ž a, I . w. On equating the powers of t, we obtain u

f n Ž a, I . s f n Ž a, I X . d Ž a, I . q f ny1 Ž a, I .

w

for all n G 0,

where fy1 s 0. Therefore w

u

G Ž a, I . s G Ž a, I X . d Ž a, I . q z G Ž a, I . .

Ž 5.

So far, this has only been proven for l Ž I . G 1. However, it may be verified directly that Ž5. also holds when l Ž I . s 1, since GŽ a, B. s 1. j We prove that GŽ a, B. and the GŽ a, I . u , for all l Ž I . G 2 and j g B, are linearly independent over F. If not, there exist nontrivial relations of the form

Ý Ý l j Ž I . G Ž a, I . u l Ž I .sm jgB

j

s

Ý

Ý l j Ž J . G Ž a, J . u

j

ql

Ž 6.

1Fl Ž J .-m jgB

with l g F w z x and all l j Ž I ., l j Ž J . g F. ŽThere is clearly nothing lost by allowing l to belong to F w z x rather than F.. Among all such relations, choose one with the minimal m, and among these, one with the least total number of l j Ž I ., with l Ž I . s m occurring on the left. Since F is a field, we

FIGUEIREDO, GONC ¸ALVES, AND SHIRVANI

310

may suppose that for some j0 g B and I0 with l Ž I0 . s m, we have l j 0Ž I0 . s 1. Apply w to Ž6., multiply by z , substitute from Ž5. into the left-hand side of the resulting equation, and rearrange. This yields

Ý Ý lj Ž I . l Ž I .sm

w

G Ž a, I . u

j

j

s

Ý Ý

w

l j Ž I . G Ž a, I X . u

jq 1

d Ž a, I . u

j

j l Ž I .sm

q

w

w

Ý l j Ž J . Ž G Ž a, J . z .

Ý 1Fl Ž J .-m

uj

q lw z .

Ž 7.

j

Suppose first that m s 1. Then the second double sum on the right-hand side of Ž7. is vacuous. Also, each GŽ a, I X . s GŽ a, B. s 1, so Ž7. reduces to

Ý Ý lj Ž I . I

w

w Ý Ý l j Ž I . d Ž a, I . u

G Ž a, I . u s j

j

I

j

q lwz .

Ž 8.

j

Subtraction of Ž6. from Ž8. yields

Ý Ý ž lj Ž I . I

w

y l j Ž I . G Ž a, I . u s j

/

j

w Ý Ý l j Ž I . d Ž a, I . u

I

j

q lwz y l .

j

Ž 9. Equation Ž9. has the same form as Ž6., but fewer terms on the left, and so must be the zero relation. It follows that

l j Ž I . g E for all j g B and l Ž I . s m

Ž 10 .

and

lwz y l q

w Ý Ý l j Ž I . d Ž a, I . u

I

j

s 0.

Ž 11 .

j

Since the sum in Ž11. is in F, the equation is impossible unless l is the j zero polynomial. Thus we obtain Ý I, j l j Ž I . dŽ a, I . u s 0, which by assumption Žc. implies that all l j Ž I . s 0, a contradiction. Thus m G 2. In this case, the second summation on the right-hand side of Ž7. may also be expanded using Ž5.. Do this and subtract Ž6. from it to obtain an equation of the same form as Ž6., but with fewer terms on the j left, so the coefficient of every G u must be the zero. Equation Ž10. then holds. Next, for each J with l Ž J . s m y 1 and each j g B, the coefficient j of GŽ a, J . u on the right-hand side of the new equation is easily seen to be w

lj Ž J . y lj Ž J . q

Ý X

I sJ

l jy1 Ž I . d Ž a, I . u

jy 1

s0

Ž 12 .

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FREE GROUP ALGEBRAS

if u / 1 and w

l0 Ž J . y l0 Ž J . q

Ý X

I sJ

l0 Ž I . d Ž a, I . s 0

Ž 13 .

if u s 1. Applying uyŽ jy1. to whichever equation we have and using the assumptions Ža. and Žd., we find that Ý I XsJ l jy1Ž I . dŽ a, I . s 0 Žrespectively, Ý I XsJ l 0 Ž I . dŽ a, I . s 0. if u / 1 Žresp. u s 1.. The E-linear independence of the dŽ a, I . now implies that the coefficients are all zero, which is the final contradiction. The result follows. COROLLARY 4.1. With the assumptions of the pre¨ ious theorem, if a is transcendental o¨ er E, then the elements GŽ a, I X . ayt Ž I . are linearly independent o¨ er E. Proof. Recall the equivalence relation ; defined by I ; J if and only if I X s J X , with I a complete set of representatives of the equivalence classes. Suppose we are given elements a Ž I . g E such that

Ý a Ž I . G Ž a, I X . ayt Ž I . s 0. I

Then 0 s Ý I a Ž I .GŽ a, I X . ayt Ž I . s Ý I g I ŽÝ J ; I a Ž I . ayt Ž I . .GŽ a, I X .. The elements in the inner summation belong to F. By the previous theorem, each Ý J ; I a Ž I . ayt Ž I . s 0 and so all a Ž I . s 0 by the transcendence of a over E. Proof of Theorem D. There is nothing to prove in those cases where the universal enveloping algebra of L contains a copy of the first Weyl algebra Že.g., when L is soluble.. In the remaining cases, we modify the proof of Malcolmson and Makar-Limanov w10x in order to be able to apply Lichtman’s argument. We follow the notation of w10, pp. 319 et seq.x. We have an element a transcendental over the field K. Let us choose some f Ž a. g K Ž a. _ K and see whether  f Ž a., f Ž a. by1 4 is a K-free set. If not, repeat the proof from the introduction of the m I on page 319 Žthere denoted m i .. The final conclusion is that, for certain I s Ž i 0 , . . . , i k ., the monomials f Ž a y Ž k y 1. b .

yi 0

Ž1 y t .

y1

??? f Ž a y b .

yi ky 1

Ž1 y t .

y1

f Ž a.

yi k

Ž 14 .

are linearly dependent over K. Note that i 0 , . . . , i ky1 G 1, and that t acts on K Ž a. as the K-automorphism w : a ¬ a y g . Recall also that b , g g K with g / 0. Define d Ž x, I, r . s f Ž x y Ž l Ž I . y r y 1 . b .

yi r

Ž 15 .

312

FIGUEIREDO, GONC ¸ALVES, AND SHIRVANI

for almost all x g K Ž a.. The conditions of Theorem 4.1 are satisfied: here F s K Ž a. and the automorphisms are defined by a w s a y g and au s a y b . Conditions Ža. and Žb. are trivial to verify. As for Žc., we have j dŽ a, I . s f Ž a.yi ky 1 and f Ž a. u s f Ž a y j b ., so, having fixed I X , we require the K-linear independence of rational functions of the form f Ž a y j b .yn , for j g B and n G 0. If 1rf is not a polynomial, then the usual argument with poles Žin case u / 1. and the transcendence of the powers of f over K imply the result. Finally, for condition Žd. of the theorem, we need to choose f in such a way that the only solution to xw y x s

Ý

a m fym ,

mG1

for x g K Ž a. and a m g K, is a m s 0. ŽNote that the summation excludes m s 0, since i ky 1 G 1.. The function f may be chosen in many ways, e.g., f Ž a. s Ž1 q a n .ra for all n G 0. With this choice of f, note that Ž14. has the form of an M Ž a, I . f Ž a.yt Ž I . as defined in Theorem 4.1, so the set  f, fby1 4 is K-free by Corollary 4.1, which contradicts the assumption of linear dependence. In other words, the set  f Ž a., f Ž a. by1 4 is K-free. Now use the degree function on U Ž L., with a large enough n in f Ž a. s Ž1 q a n .ra, and Lichtman’s result, to conclude the proof of Theorem D. EXAMPLE. Malcolmson and Makar-Limanov choose f Ž a. s ay1 in w10x, so the final part of their argument may be deduced from ours. Similarly in w8x, one may deduce the main result by choosing dŽ t, I, r . s Ž1 y t .yi r , tu s t, t w s l t, and applying Theorem 4.1, while the result of w7x may be obtained by choosing dŽ t, I, r . s tyi r , tu s t, and t w s t q 1.

REFERENCES 1. J. Z. Gonc¸alves and M. Shirvani, On free group algebras in division rings with uncountable center, Proc. Amer. Math. Soc., 124 Ž1996., 685]687. 2. P. H. Kropholler, P. A. Linnell, and J. A. Moody, Applications of a new K-theoretic theorem to solvable group rings, Proc. Amer. Math. Soc. 104 Ž1988., 675]684. 3. A. I. Lichtman, On normal subgroups of multiplicative groups of skew fields generated by polycyclic-by-finite groups, J. Algebra 78 Ž1982., 548]577. 4. A. I. Lichtman, On matrix rings and linear groups over fields of fractions of group rings and enveloping algebras, II, J. Algebra 90 Ž1984., 516]527. 5. M. Lorenz, ‘‘Methods in Ring Theory,’’ pp. 265]280, Reidel, Dordrecht, 1984. 6. M. Lorenz, On free subalgebras of certain division algebras, Proc. Amer. Math. Soc. 98 Ž1986., 401]405. 7. L. Makar-Limanov, The skew field of fractions of the Weyl algebra contains a free non-commutative algebra, Comm. Algebra 17 Ž1983., 2003]2006.

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8. L. Makar-Limanov, On group rings of nilpotent groups, Israel J. Math. 48 Ž1984., 245]248. 9. L. Makar-Limanov, On subalgebras of the first Weyl skew field, Comm. Algebra 19 Ž1991., 1971]1982. 10. L. Makar-Limanov and P. Malcolmson, Free subalgebras of enveloping algebras, Proc. Amer. Math. Soc. 111 Ž1991., 315]322. 11. D. Passman, ‘‘The Algebraic Structure of Group Rings,’’ Wiley, New York, 1977. 12. K. C. Smith, Algebraically closed noncommutative polynomial rings, Comm. Algebra 5 Ž1977., 331]346.