1550075568-C-Algebras_and_Finite-Dimensional_Approximations__Brown_

234 6. Amenable Traces

Given z E F, we can write z as an integer polynomial in the k/s. Hence,

substituting in our decompositions of the k/s,

n

z = Lrnxn

s=O

for some r/s coming from R. If n < d, then linear independence of the
elements {1, x, x^2 , ... , xd-l} implies z = ro. If n :2: d, we can apply Euclid's
algorithm and divide by the minimal polynomial P. Since the coefficients of
P belong to R, we can rewrite z as a polynomial of degree < d with coeffi-
cients in R, and this completes the proof. (If T = QP + T' are polynomials
and all coefficients of T, Q and P come from R - as they do in this case,
since Q will be a polynomial in k1, ... , km -then the coefficients of T' also
come from R.) D

Now we can complete the proof of Theorem 6.4.12.

Proof of Theorem 6.4.12. Let K be a finitely generated field and P be
its prime subfield. We will show [K : P] < oo and Pis a finite field; evidently
this implies the theorem.

Let z1, z2, · · · be a (possibly finite, maybe even empty) transcendence
basis for K over P. The field generated by P and {z1, z2, · · ·} is isomor-
phic to the field of rational functions P(x1,x2, ... ) ([87, Theorem VI.1.2]).
Every element of K is algebraic over P(z1, z2, ... ) (by maximality of a tran-
scendence basis); in particular, every element in a generating set of K is
algebraic over P(z1, z2, ... ), which implies [K : P(z1, z 2 , ... )] < oo. Thus
Lemma 6.4.18 implies P(z1, z2, ... ) is finitely generated. Hence, by Lemma
6.4.17, the transcendence basis must be empty -that is, [K : P] < oo. Since
P is either the rational numbers or the field of p elements for some prime
number p, similar reasoning shows P is a finite field. D

Exercises

Exercise 6.4.1. Show that a discrete group I' has the factorization property
if and only if every finitely generated subgroup does. (Hint: You will need
the fact that if I'o c I' is a subgroup, then C(I'o) c C(I'), by Proposition
2.5.8.)

Exercise 6.4.2. A group is called maximally almost periodic if it has a sepa-
rating family of finite-dimensional unitary representations (this is equivalent
to being a subgroup of a compact group). Show that every maximally almost
periodic group has the factorization property.