A note on free products of linear groups

For a field K, let K denote its algebraic closure. Assume that IK: KI = oo. Then for any linear groups G, H C GL (K) their free product G * H can be embedded into GLN(K(t)). Here N is an integer depending on K only and t stands for an indeterminate. If K is a field, let K denote its algebraic closure. We are going to prove THEOREM. Let G, H C GLn (K) be subgroups such that all entries of matrices g E G, h E H lie in a subfield Ko C K with IKo: Kol = ox. Then the free product G * H can be embedded into GLN(K(t)). Here t is an indeterminate and N is an integer depending on Ko and n only. We need two lemmata. LEMMA A. Let L be a field satisfying IL :LI = ox. For any integer m there exists a E L such that degL a > m. PROOF. If L is separable over L it follows from the Theorem of Primitive Element [1, Chapter VII, ?6, Theorem 14, p. 185]. If not, i.e. if L is not perfect, say a E L\LP, then the elements al/p, al/P,2. ... have degrees p, p2, ... over L. D LEMMA B. LetL be afield satisfying IL: LI = ox. If S C GLn(L) is a subgroup with S n L * ln = {ln,} then there exists an embedding S C GLn(L) such that (i) IL: LI n over the previous field L C L(all) C L(all, a12) C . C L(all,... ,a) = L Before we conclude the proof let us make a REMARK. If p E L[xij] is a nonzero polynomial and degp 91,2n+1 #: 0, (iii) h EH\111 =:>h2n+l,l 780. These are precisely the assumptions of Shalen's Theorem [2, Proposition 1.3]. It follows that the map p: G * H -* GL2n+1(KO[t, t']) defined by 0] [1 01t-2 G9gt, . .9. . ~~~, H9h-h, ? ~~t2n+ 1 Ot-(2n+l has the following property: If a = p(gi)p(hi) * * . p(gr)p(hr) and all gi, hi :A 1, then Trace(a) = a. t2nr+ lower (possibly negative) powers of t; a + 0. Clearly, p is an embedding of G U H. Further, each element w of G * H\(G U H) is conjugate to one in the form g1hl * *. grhr with gi E G\{1}, hi E H\{1}. Therefore, Tracep(w) $8 Tracep(1), hence p(w) $& 1. Thus p embeds G * H into GL2n+1(Ko(t)). To finish the proof, notice that Ko(t) = Ko(t)(aij) and so Therefore, IKo(t): Ko(t) = d < oo. G * H C GL(2n+l)d(K0(t)) C GL(2n+1)d(K(t)). D This content downloaded from 157.55.39.238 on Tue, 22 Nov 2016 04:07:23 UTC All use subject to http://about.jstor.org/terms