1550075568-C-Algebras_and_Finite-Dimensional_Approximations__Brown_

6.4. Factorization and property (T) 233

is continuous with respect to the minimal norm for all hyperbolic groups

r. However, if one passes to the Calkin algebra (instead of stopping in

lBl(£^2 (r))), then the map is min-continuous for all hyperbolic groups (Corol-

lary 5.3.20). This remark is unlikely to help resolve the question of residual

finiteness for hyperbolic groups - it's just a cute observation.

As promised, we now give the proof of Theorem 6.4.12.^13 We need two

simple lemmas.

Lemma 6.4.17. Let R be a unique factorization domain with infinitely

many distinct irreducibles. Then R's quotient field can't be finitely generated

as a ring.

In particular, K(x1, x2, ... ) - the field of rational functions in (possibly

infinitely many) indeterminants x1, x2, ... over a field K - is not finitely

generated. Neither are the rational numbers Q.

Proof. Every element in the quotient field F of R has an essentially unique

representation as a fraction

P1P2 · · · Ps

qlq2 ... qt

for some irreducibles p1, ... ,ps and qi, ... , qt. Let {r1, r2, ... , rn} C F be

a finite set. One easily checks that if an irreducible q E R does not appear

in the denominator of any of the ri's, then ~ will not belong to the ring

generated by ri, ... , rn· Hence F can't be finitely generated.

Euclid proved that Z has infinitely many primes (= irreducibles); the

same proof shows the polynomial ring K[x1, x2, ... ] does too. D

Lemma 6.4.18. Assume K is a finite field extension of another field F and

K is finitely generated as a ring. Then so is F.

Proof. By induction, we may assume that K = (F, x) is a simple extension

• i.e., generated by F and one element x t/:. F. Since [K: F] < oo, x must
  be algebraic over F. Let P(X) = Xd + fd-1Xd-l + · · · + fo, fi E F, be its
  minimal polynomial -hence {1, x, x^2 , ... , xd-l} is a basis for K as a vector
  space over F ([87, Theorem V.1.6]). Let kl, ... , km be a finite generating
  set of K (as a ring). Find coefficients gi,j E F such that ki = I:,J:_~ gi,jXj
  and let R be the subring of F generated by the fi's and gi/s. We will show
  R=F.

13our proof is borrowed from lecture notes for a geometric group theory course (Math 257)

taught by Stallings at Berkeley in the fall of 2000. (Notes are available at Stallings's website.)

He, in turn, credits the argument to Shalen.