Suppose the sequence x is regular for M. The element of ExtAn(M/(x)M,M) associated with the exact sequence (23) is (−1)n(n+1)/2θx. For i=0,1,…,n, define an A-linear mapping
ai:Ki(x,M)→Mi=M/(x1M+⋯+xi−1M) as follows: if
m∈Ki(x,M),
ai(m) is the class in
Mi of the element
m(1,2,…,i) of
M. It is clear that
a0 is the identity mapping of
M and that
p∘an=q. Moreover
ai+1∘∂i(m) is the image in
Mi+1 of the element
k=1∑i+1(−1)k+1xkm(1,2,…,k−1,k+1,…,i+1). Since
xk annihilates
Mi+1 for
k=1,…,i,
ai+1∘∂i(m) is also the image of
(−1)ixi+1m(1,2,…,i), hence
ai+1∘∂i=(−1)ixi+1∘ai. By X, p. 120, cor. 1 and 2, the element of
ExtAn(M/(x)M,M) associated with (23) is equal to
∏i=1n(−1)i⋅θx, whence the assertion.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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