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Additional resources for Algebra, K-Theory, Groups, and Education: On the Occasion of Hyman Bass's 65th Birthday
Will work to 1 — 1 — 1 I prove that (g,m,y,r) has property (n^). CLAIM 2. m^ ^ m . X^ and let i > r. ^ 1=0 I - If a . = m^,i a . = a , = a, then since i. > i > r, there exists an integer m,i. n,k. l,s. I — — ^ ^•'i I ^i t . such that : (i) ^ ^ (ii) t. Ii = a^ . + a for some a ^ I m,j. s. -I I I t . , and I. — I *^1 (ill) b- ^ I T for 0 < X < t. -I — I. I •'i PATHOLOGY IN R[[X]] 26 Taking T. = t. and using the fact that a . = a . , we see that satisfies the properties (i), (ii) and (iii) of the definition and so (g,m^,y,r) has property (n).
That a . = a , = a . = a, • m,i n,k. v,X. l,s. ¿ ^ k ^ ^ у}. X^, then we wish to Thus, suppose that i ^ r^ and By assumption, there exists an ' ^ * integer t. such that b^ = a^ . + a for some a ^ I ^ and such that ^ i t. -l I I 2 2 2 t. < yk. < k.. Since X. > k. + I and t. , it follows that X. > t.. 1 — 1 — 1 1 — 1 1 — 1" 1 1 Thus, for 0 < i < t . < X . , w e have that a . is "to the left" of — — I I v,j a^ = v,X^ I,S^ . Therefore, a . € I -. Note that here we have 2 used the precise form of the recursive relation to say that X^ — Consequently, i f h = z"?
M,i ^I We first prove several properties which are formal consequences of the definition. CLAIM I. n Suppose that n and n^ are positive integers with If (g,m,y,r) has property (n), then (g,m,p,r) has property (np . Proof, Properties (i) and (iii) of the definition are indepen dent of the choice of n. Thus, we have only to verify that (ii) holds. Suppose that i > r and that a . = a , = a . m,i n,k^ ^l"^i Then k. < v. and 1 - 1 hence t. < pk. < yv.. Therefore, the same choice of t. will work to 1 — 1 — 1 I prove that (g,m,y,r) has property (n^).