Commutative Algebra, X, p. 154
Bourbaki marks this one as harder than the rest.
Let A be a noetherian local ring, M a finitely generated a-module, x=(x1,…,xn) a sequence of elements of mA, generating an ideal x; suppose that the a-module M/xM is of finite length (which implies n⩾dim(M)). We denote by gr(A) and gr(M) the associated graded modules of A and M for the x-adic topology, ξi (1⩽i⩽n) the image of xi in x/x2, ξ the sequence (ξ1,…,ξn). a) For p∈Z, put FpKi(x,M)=Ki(x,xp−iM) (with the convention xs=A for s⩽0). Prove that FpK∙(x,M) is a subcomplex of K∙(x,M), and that the complex ⨁pFpK∙(x,M)/Fp+1K∙(x,M) is identified with K∙gr(A)(ξ,gr(M)). b) For every complex of A-modules C such that H(C) is of finite length, put χ(C)=∑i(−1)ilg(Hi(C)). Prove that there exists an integer p0 such that the complex
FpK∙(x,M)/Fp+1K∙(x,M) has zero homology for p⩾p0; deduce the equalities
χ(K∙gr(A)(ξ,gr(M)))=χ(K∙(x,M)/FpK∙(x,M))=∑(−1)i(in)lg(M/xp−iM)
for p⩾p0.
c) Prove that the expression above is zero if n>dim(M), and equal to cx(M) if n=dim(M).
d) Let h⩾0, and p⩾p0; prove that the canonical injection of FpK∙(x,M) into Fp+hK∙(x,M) is a homology equivalence. Deduce that Hi(FpK∙(x,M)) is discrete for the x-adic topology. Prove moreover that the filtration of this module defined by the images of the mappings Hi(Fp+hK∙(x,M))→Hi(FpK∙(x,M)) is x-good, and deduce that FpK∙(x,M) has zero homology. Conclude that χ(K∙(x,M)) is zero if n>dim(M), and equal to ex(M) if n=dim(M).