Math, Ann. 179 295 (1969)

On a Paper of Ribenboim GEORGE MAXWELL

In a recent paper [1], P. Ribenboim proved by an inductive argument that if E/K is an infinite Galois extension of fields and f and g are distinct K-auto- morphisms of E, then there exists a subset Q c E such that if S is the symmetric difference o f f ( Q ) and g(Q), then [K(S):K] = oo. The point was that distinct automorphisms have "very distinct" values on some subset of E. In this paper, we demonstrate that this fact is an easy corollary of a much more general phe- nomenon.

Let E and F be sets and f , 9 : E--,F two functions from E to F. Consider W = {Q c E [ f(Q) n 9(Q) = 0}, ordered by inclusion. Then W # 0 since 0 ~ W and the order is inductive: for let (Qi)i~t be a completely ordered subset of W and put Q = U Qi. If xef (Q)~o(Q) then x ef(Qi ) and x Eg(Qj) for some i, j ~ I ;

iEI letting Qk be the greater of Q i and Qj we see that k e I and x ~ f(Qk)nO(Qk)! Thus f(Q)ng(Q)= 0, proving our assertion. Take a maximal element Q0 of w. If x ~ E~Qo, then f(Qou{X})~g(Qou{x})4: o, so that either f (x)~ o(Qo) or 9(x) ~ f(Qo) or f(x) = 9(x). Putting E r ' o = {x ~ E If(x) = 9(x)} we have therefore the decomposition

E= Ef . , uQou f - l g(Qo)ug-l f(Qo). (*) Secondly, we remark that if E is a group and E' a proper subgroup, then

every element in E can be written as a product of at most two elements of Ek E'. For take an x E E; if x ~ E~E', we are finished; if x ~ E', choose y ~ E~E' and write x = ( x y ) y -1. Then y - l ~ E~E' and also x y ~E~E' for otherwise y = x - 1 (x y) ~ E' ! Afort iori , E\ E' generates E.

Prolmsition. Let K be a rin9, E a K-module which is not of fnite type, and f and g distinct K-automorphisms of E. Then there exists a subset Qo c E such that f (Qo)ng(Qo) = 0 and the submodule (Go) 9enerated by Qo is not of finite type.

Proof. Apply (.) and note that Ey, g is a proper additive subgroup of E so that Qowf- lg(Qo)wg-af - l (Qo) generates E as an abelian group. Since f - l g and g - i f are also K-automorphisms of E, if (Go) were of finite type so would be ( f - x g(Qo)) and (O- 1 f(Qo)) and thus E itself! Q.E.D.

To obtain Ribenboim's result, let K be a field and E an infinite Galois extension of K; note that the symmetric difference S of f(Qo) and g(Qo) is simply their union so that certainly [K(S) : K] = oo.

References 1. Ribenboim, P. : A property of automorphisms of infinite Galois extensions. Math. Ann. i66,

54---55 (1966). Prof, George Maxwell Queen's University Department of Mathematics Kingston, Ontario, Canada

(Received January 17, 1968)