shifted adic koszul filtration euler comparison

A differential term deletes $e_j$ and multiplies its coefficient by $f_j$, with sign $(-1{)}^{j-1}$ in the ordered wedge. Thus $d(I^{\max (0,p+n)}{K}^n)\subseteq I^{\max (0,p+n+1)}{K}^{n+1}$: if $p+n\ge 0$ multiplication raises the power by one, and if $p+n<0$ the target required power is zero. This proves the subcomplex assertion for every $p$.

If $M=0$, all complexes vanish. If $r=0$, $I=0$, $K=M[0]$ and $F^{p}K=0$ for $p\ge 1$; the finite-length hypothesis is exactly that on $M$. If $I=R$, there are $a_j$ with ${\sum }_j{a}_j{f}_j=1$. The map $h={\sum }_j{a}_j{h}_j$ satisfies $dh+hd=\mathrm{id}$, so every cycle $z$ is the boundary $d(hz)$. Every tail equals $K$ and the quotient is zero. This proves all conclusions in these cases. Henceforth $r\ge 1$ and $I\subseteq \mathfrak{m}$.

Let $S={\mathrm{gr}}_I(R)$ and $G={\mathrm{gr}}_I(M)$, extending $G_j=0$ for $j<0$. In degree $p$ the graded complex $F^{p}K\mathrm{/}{F}^{p+1}K$ has cochain term $G_{p+n}^{\left(\genfrac{}{}{0px}{}{r}{-n}\right)}$. For $p+n<0$ this is zero since the two filtration terms coincide. A representative $m\in I^{p+n}M$ in a wedge summand maps to the sum of $(-1{)}^{j-1}{f}_{j}m$ in the deleted wedge summands, modulo $I^{p+n+2}M$. This is exactly multiplication by $f_j^{\ast }\in S_1$. Therefore the direct sum over $p$ is the Koszul complex on ${\mathbf{f}}^{\ast }$ with coefficients $G$, giving wedge degree $i$ weight $i$, so the total internal degree $p$ is preserved.

The map $(R\mathrm{/}I)[X_1,\dots ,X_r]\to S$ taking $X_j$ to $f_j^{\ast }$ is onto: every element of $I^a\mathrm{/}{I}^{a+1}$ is a sum of degree-$a$ monomials in these initial forms. Thus $S$ is Noetherian by quotient preservation and iterated Hilbert basis. A finite generating list of $M$ gives generators of $G$ in degree zero, by expressing elements of $I^{a}M$ as monomials times those generators. All terms, cycles and homologies of the graded Koszul complex are therefore finite over $S$.

Each such graded homology is killed by all $f_j^{\ast }$, hence by $S_{+}$, since the $f_j^{\ast }$ generate the positive-degree ideal. Replacing its finite generating list by its finitely many homogeneous components gives homogeneous generators; kernels and images are graded because the differential preserves internal degree. Only their degrees can occur: positive-degree scalars act as zero and degree-zero scalars preserve degree. There are only $r+1$ homology modules. Choose $b$ above all their generator degrees (take $b=0$ if all are zero). Then ${\mathrm{gr}}_F^{p}K$ is acyclic for every $p\ge b$.

In $0\to F^{p+1}K\to F^{p}K\to {\mathrm{gr}}_F^{p}K\to 0$, the last complex is acyclic for $p\ge b$. The LES therefore makes $H^n(F^{p+1}K)\to H^n(F^{p}K)$ an isomorphism. Finite composition gives the same for $H^n(F^{q}K)\to H^n(F^{p}K)$ whenever $q\ge p\ge b$.

Fix $p\ge \max (b,r)$ and put $E^n=F^p{K}^n$, $Z^n=\ker (d:E^n\to E^{n+1})$. These and $L^n=H^n(F^{p}K)$ are finite $R$-modules. For $q\ge p$ all the exponents are nonnegative and $F^q{K}^n=I^{q-p}{E}^n$. Artin–Rees gives $c_n\ge 0$ such that, when $q-p\ge c_n+1$, $Z^n\cap F^q{K}^n=I^{q-p-c_n}(Z^n\cap I^{c_n}{E}^n)\subseteq I{Z}^n$. Choose one $q$ satisfying this for the finitely many $n$ between $-r$ and $0$.

A cycle from $F^q{K}^n$ represents, in $L^n$, the class of an element of $Z^n\cap F^q{K}^n\subseteq I{Z}^n$. Its class is in $I{L}^n$, since the quotient map $Z^n\to L^n$ is linear. Surjectivity in the preceding LES comparison gives $L^n=I{L}^n$. The ring $R$ is local, so $J(R)=\mathfrak{m}$; $I\subseteq \mathfrak{m}$ and $L^n$ is finite. Nakayama, under AC, gives $L^n=0$. Since $p$ was arbitrary above the bound, every such tail is acyclic.

The LES of $0\to F^{p}K\to K\to K\mathrm{/}{F}^{p}K\to 0$ now gives the claimed homology isomorphisms. For any $a\ge 1$, each factor $I^{j}M\mathrm{/}{I}^{j+1}M$ of $M\mathrm{/}{I}^{a}M$ is a quotient of finitely many copies of $M\mathrm{/}IM$, via degree-$j$ monomials in the $r$ generators. All factors, and hence $M\mathrm{/}{I}^{a}M$, have finite length. The case $a=0$ gives zero. Each quotient term is a finite direct sum of such modules. Its Euler characteristic is therefore computable by term lengths, and equals that of $K$ by the homology isomorphisms and finite-length original homology.