On free subsemigroups of skew fields

I will prove in this note that the multiplicative group of a skew field with uncountable center either contains a free nonabelian subsemigroup or is commutative. Let T be a skew field with center Z and let us suppose that Z is an uncountable set. Then the following statement is true. THEOREM. IfT ^ Z then the multiplicative group ofT contains a free subsemigroup with two generators. The proof of the theorem is broken into the following three lemmas. LEMMA 1. Suppose T has no such subsemigroup. Then for any pair of elements x,y from T there exists a semigroup relation A(a, b) = B(a,b) such that this relation is valid for every pair x + c, y + dx, where c, d € Z, and that dega A = degas, degaA = deg6s. PROOF. Let us fix a pair x,y and consider all pairs x + c, y + dx. Each pair fails to span a free semigroup (with two generators) so it satisfies some relation which of course depends on c and d. Now we have only a countable number of possible relations. This means that if we fix, for example, c = en then there exists a relation which is satisfied by an infinite number of pairs x + en, y + dx. Let us represent this relation as a polynomial in d with coefficients depending on x.y and t'oThis polynomial has infinitely many roots so all of its coefficients must be zeros (we can use Vandermond's determinant with entries {d^}). So this relation is satisfied by the pair x + cq, y + dx for every d in Z. Thus for every c there exists a relation P(c) of this type which is satisfied by every pair x + c, y + dx (where c is fixed and d can be taken arbitrarily). Let us call such relations good for x,y and c. Among the countably many good relations we can find a relation P which is satisfied by x + c, y + dx for infinitely many c (and any d). Then the same reasoning shows that this relation P is satisfied by x + c, y + dx for every c, d (because for every fixed d this relation is satisfied for infinitely many c and thus for every c). Let us call such relations better for x,y. Now for each pair ex, fy, where e, / € Z, there exists a better relation for ex, fy and again we can see that there exists a better relation which is satisfied by every pair ex, fy. By factoring out e and / we can see that this relation has the same degrees in a and in b on its left and right sides. Received by the editors May 13, 1983 and, in revised form, September 22, 1983. 1980 Mathematics Subject Classification. Primary 16A39, 16A70, 16A25; Secondary 20M05, 20M99. lrThe author is supported by NSF Grant No. MCS-8201115. ©1984 American Mathematical Society 0002-9939/84 $1.00 + $. 25 per page 189 License or copyright restrictions may apply to redistribution; see http://www.ams. org/journal-terms-of-use 190 L. MAKAR-LIMANOV LEMMA 2. If the relation A(a, b) = B(a, b) is nontrivial and satisfies the conclusion of Lemma 1 for a pair x, y € T then there exists some n = n(x, y) such that (adx)ny — 0, where adxy = [x, y] = xy — yx. PROOF. We can write A(a,b + da) as V■ Aj(a,b)dJ~3, where degb Aj = j. Let us show first that Ai(a,b) uniquely defines A(a,b) if degbA ^ 0. Suppose that A(a,b) = a1^1 ■■■alkbjk. Then Ai{a,b) = J2meMambaI+J~m~^ where 7 = 1C *s » J = YlJsThe set M can be represented as the union of the longest possible arithmetic progressions with difference equal to one. The first of these progressions starts with ? i and contains j\ terms, the second one starts with i\ + j\ + i2 and contains j2 terms, the mth progression starts with ii + ji + ■ ■ • + jm-i + im and contains jm terms. This information, together with the known £Z(is +js), uniquely defines all ? 's and j's. Now we may assume that both dega A and deg6 A are nonzero because otherwise the relation is trivial. So A\ — B\ is a nontrivial relation. We can rewrite this relation as the equivalent relation i+J (*) J2 ksasbaI+J-s = 0. a=0 (Here all the ks are 0 or ±1 but we are going to use only that ks G Z.) We can rewrite relation (*) as i+j-i J2 fc>56ia/+J-1-s + k'I+JaI+Jb = 0,