## Download Analytic K-Homology by Nigel Higson PDF

00 nl·····n d ~ PROOF. Proof by induction on n > 0, we have integer d = 1. M) = l(M/x nM) - l«O:x n )). 12)(i),we have n > m. Thus d. Suppose (O:x n ) = (O:x m) M M M for all n > m. M) = l(M/~nM) + C/n, where C is for i~dependent of n.

Proof by induction on d. Suppose d=1. M) = Q,R(M/xlM) - Q,R«O:x l )). Therefore, we M get Q,R«O:x l )) = 0, that is, (O:x l ) = O. M M Now suppose that d = s + 1 and the result holds for d = s. 31 Let nl, ••• ,n d be arbitrary positive integers. 12)(i) we have nl nd nl nd eR(x l "",xd )~~(M/(xl "",x d )M)~nl·····nd~(M/(xl"",xd)M nl nd = nl .... ·n d eR(xl'""xdIM)=eR(xl "",xd 1M). nl nd (O:x l ) C(x l , ••• ,xd)M and hence integers M for arbitrary positive nl, ••• ,n d • Then we get that (O:X l ) C n (x~, ••• ,X~)MCn qnM = 0 by Krull's Intersection M n>O n>O - - Now, e R/ Xl (x2"",xdIM/xIM)=eR(xl"",xdIM)=~(M/(xl"",xd)M) =~(M/xl M/(x 2, ••.