Koszul Euler Characteristics and Hilbert–Samuel Multiplicity

If $I=R$ or $M=0$, every quotient in the statement is zero, giving the polynomial zero. More generally put $N=M\mathrm{/}IM$. If $N=0$, then $M=IM$ implies $M=I^{a}M$ by repeated multiplication for every $a\ge 1$, so again all quotients are zero. Uniqueness in these cases follows from the polynomial argument at the end. Henceforth $I\mathrm{\ne }R$ and $c={\mathrm{\ell }}_R(N)>0$.

For a simple $R$-module $S$ and $0\mathrm{\ne }s\in S$, $Rs=S$; the map $R\to S$ taking $a$ to $as$ is onto and its kernel is maximal, since a nontrivial intermediate ideal would give a nontrivial submodule. Thus the kernel is $\mathfrak{m}$ and $S\cong k$. A composition series $0=N_0\subset \cdots \subset N_c=N$ therefore satisfies $\mathfrak{m}{N}_j\subseteq N_{j-1}$. Iteration gives ${\mathfrak{m}}^{c}N=0$.

Set $A=R\mathrm{/}{\mathfrak{m}}^c$, a Noetherian ring. Each ${\mathfrak{m}}^j\mathrm{/}{\mathfrak{m}}^{j+1}$ for $0\le j<c$ is generated by finitely many elements and killed by $\mathfrak{m}$, hence is a finite-dimensional $k$-space. Indeed a finite spanning list can be reduced, by deleting any member in the span of the others, to a basis; the flag of initial basis spans has simple factors $k$. Its $R$-length is its dimension. Applying length additivity to the finite $\mathfrak{m}$-adic filtration gives ${\mathrm{\ell }}_R(A)<\mathrm{\infty }$.

Choose generators $f_1,\dots ,f_s$ of $I$. For each $j\ge 0$, multiplication by the finitely many monomials $f^{\alpha }$ with $\mathrm{\mid }\alpha \mathrm{\mid }=j$ gives a surjection from copies of $N$ onto $G_j=I^{j}M\mathrm{/}{I}^{j+1}M$: coefficients modulo $IM$ suffice because $f^{\alpha }IM\subseteq I^{j+1}M$. In particular every $G_j$ has finite length and is killed by ${\mathfrak{m}}^c$. The actions $X_i\bar{m}=\overline{f_{i}m}$ commute, and define $G={⨁}_{j\ge 0}{G}_j$ as a finite graded module over $A[X_1,\dots ,X_s]$, generated by a finite generating list of $N$ in degree zero.

For any finite graded $A[X_1,\dots ,X_s]$-module $V$, replacing a finite generating list by all its homogeneous components gives homogeneous generators. Their degrees have a lower bound. Each $V_j$ is a quotient of finitely many copies of $A$, indexed by monomials of the required degree times these generators. Thus $H_V(t)={\sum }_j{\mathrm{\ell }}_R(V_j)t^j$ is a well-defined formal Laurent series. When $s=0$, only the finitely many generator degrees can occur, so $H_V$ is a Laurent polynomial.

Suppose $s>0$ and rationality with denominator $(1-t{)}^{s-1}$ has been proved for every finite graded module over $A[X_1,\dots ,X_{s-1}]$. Put $V(-1{)}_j=V_{j-1}$, $U=\ker (X_s:V(-1)\to V)$ and $W=V\mathrm{/}{X}_{s}V$. Iterating polynomial extension from the Noetherian ring $A$ makes $A[X_1,\dots ,X_s]$ Noetherian. Consequently $U$ is finite as a submodule of finite $V(-1)$; $W$ is finite as a quotient. Both are killed by $X_s$, so their same finite generators generate them over the ring with $s-1$ variables.

In degree $j$, the exact sequence $0\to U_j\to V_{j-1}\stackrel{X_s}{\to }{V}_j\to W_j\to 0$, split at its image, yields ${\mathrm{\ell }}_R(V_j)-{\mathrm{\ell }}_R(V_{j-1})={\mathrm{\ell }}_R(W_j)-{\mathrm{\ell }}_R(U_j)$. Hence $(1-t)H_V=H_W-H_U$. The induction hypothesis gives a Laurent polynomial numerator divided by $(1-t{)}^s$. Together with the base case this proves that form for all $s$.

The filtration of $M\mathrm{/}{I}^{n+1}M$ has successive factors $G_0,\dots ,G_n$, so its length is $h(n)={\sum }_{j=0}^n{\mathrm{\ell }}_R(G_j)$. Its generating series is $H_G(t)\mathrm{/}(1-t)=Q(t)\mathrm{/}(1-t{)}^{s+1}$ for a Laurent polynomial $Q(t)={\sum }_a{q}_a{t}^a$. The geometric series and repeated convolution give the coefficient $\left(\genfrac{}{}{0px}{}{n-a+s}{s}\right)$ of $t^n$ in $t^a\mathrm{/}(1-t{)}^{s+1}$ when $n\ge a$. One can count the convolution terms as nonnegative $(s+1)$-tuples with sum $n-a$, separated by $s$ dividers. Thus for $n$ beyond the finite set of exponents of $Q$, $h(n)={\sum }_a{q}_a\left(\genfrac{}{}{0px}{}{n-a+s}{s}\right)$, a rational polynomial in $n$. For $s=0$ the binomial is $1$, as required for $I=0$.

If two rational polynomials eventually equal $h(n)$, their difference vanishes at infinitely many distinct integers. Division by $T-a$ at a root lowers the degree by one, so a nonzero polynomial of degree $d$ has at most $d$ distinct roots. The difference is therefore zero. This also proves the uniqueness left open in the zero cases.

For $N$ of finite length $c$, each simple factor is $k=R\mathrm{/}\mathfrak{m}$: a nonzero vector generates the factor, whose annihilator is maximal and therefore is $\mathfrak{m}$. Thus a composition series shows ${\mathfrak{m}}^{c}N=0$. If $c=0$, then $N=0$. If $\mathfrak{p}\mathrm{\ne }\mathfrak{m}$, choose $a\in \mathfrak{m}\setminus \mathfrak{p}$; $a^c$ is invertible in $R_{\mathfrak{p}}$ and kills $N_{\mathfrak{p}}$, so $N_{\mathfrak{p}}=0$. This proves the forward direction of the finite-length criterion.

Conversely suppose $N\mathrm{\ne }0$ is finite with support contained in $\{\mathfrak{m}\}$. Its annihilator is proper and hence is contained in the unique maximal ideal $\mathfrak{m}$ (maximal-ideal existence is available under AC). The support formula then says the set of primes containing $\mathrm{Ann}(N)$ is exactly $\{\mathfrak{m}\}$. Radical intersection gives $\sqrt{\mathrm{Ann}(N)}=\mathfrak{m}$. This is the separating-prime use of AC.

Exact localization identifies $(M\mathrm{/}IM{)}_{\mathfrak{p}}$ with $M_{\mathfrak{p}}\mathrm{/}{I}_{\mathfrak{p}}{M}_{\mathfrak{p}}$. If $I\subseteq ̸\mathfrak{p}$, one of its elements becomes a unit, so this quotient is zero. If $I\subseteq \mathfrak{p}$, then $R_{\mathfrak{p}}$ is local with maximal ideal $\mathfrak{p}{R}_{\mathfrak{p}}$: a fraction whose numerator is outside $\mathfrak{p}$ is invertible, while the fractions with numerator in $\mathfrak{p}$ form a proper ideal with field quotient. Thus $I_{\mathfrak{p}}$ lies in the Jacobson radical. The module $M_{\mathfrak{p}}$ is finite, generated by localized generators. Nakayama says the quotient is zero only if $M_{\mathfrak{p}}=0$; the reverse implication is immediate. This proves both inclusions of the support identity. This application of the published Nakayama interface uses AC.

Choose generators $a_1,\dots ,a_v$ of $\mathfrak{m}$, and for each choose $b_j\ge 1$ with $a_j^{b_j}N=0$. Put $c=1+{\sum }_j(b_j-1)$. A monomial of degree $c$ must have an exponent at least $b_j$, since otherwise its degree is at most $c-1$. These monomials generate ${\mathfrak{m}}^c$, giving ${\mathfrak{m}}^{c}N=0$. If $v=0$, then $\mathfrak{m}=0$ and $c=1$ works as well.

Koszul terms are finite direct sums of $M$. Their kernels, images and homology are finite by Noetherianity. If $M_{\mathfrak{p}}=0$, the localized complex is zero, and hence its homology is zero. Also $I$ kills every homology module, so at a prime outside $V(I)$ an invertible annihilator forces the localized homology to vanish. We have proved $\mathrm{Supp}(H_i)\subseteq \mathrm{Supp}(M)\cap V(I)=\mathrm{Supp}(M\mathrm{/}IM)$.

Each ${\mathfrak{m}}^{j}N\mathrm{/}{\mathfrak{m}}^{j+1}N$ for $0\le j<c$ is finite and killed by $\mathfrak{m}$. It is therefore a finite-dimensional $k$-space: delete dependent elements from a finite spanning list to get a basis; its basis flag has one simple $k$ factor for each vector. It has finite $R$-length. Repeated additivity along the finite filtration proves that $N$ has finite length. For $N=0$ the support is empty and the length is zero. Together with the forward direction this proves the iff assertion.

The hypothesis and the forward criterion put this last support inside $\{\mathfrak{m}\}$. The reverse criterion applies to each finite $H_i$, proving its finite length; boundedness then defines the Euler sum. If $I=R$, the annihilation assertion makes every $H_i=0$. If $r=0$, the only homology is $M$, whose finite length is exactly the hypothesis. If $M=0$, every term vanishes.

Set $Z_i=\ker (d_i)$ and $B_i=\mathrm{im}(d_{i+1})$. There are short exact sequences $0\to Z_i\to D_i\to B_{i-1}\to 0$ and $0\to B_i\to Z_i\to H_i(D)\to 0$, with maps induced by the differential and quotient. Since $D_i$ has finite length, all these modules do. Additivity gives ${\mathrm{\ell }}_R(D_i)={\mathrm{\ell }}_R(H_i(D))+{\mathrm{\ell }}_R(B_i)+{\mathrm{\ell }}_R(B_{i-1})$.

For any finite exact sequence $0\to E_0\to \cdots \to E_t\to 0$ of finite-length modules, set $J_j$ to be the image in $E_j$, so $J_0=J_{t+1}=0$ and $0\to J_j\to E_j\to J_{j+1}\to 0$ is exact. Hence ${\mathrm{\ell }}_R(E_j)={\mathrm{\ell }}_R(J_j)+{\mathrm{\ell }}_R(J_{j+1})$, and summation with alternating signs cancels every image length, yielding ${\sum }_j(-1{)}^j{\mathrm{\ell }}_R(E_j)=0$.

Choose integers $a\le b$ with $D_i=0$ outside $[a,b]$. Then $B_{a-1}=B_b=0$. In the alternating sum of the preceding equality, ${\mathrm{\ell }}_R(B_j)$ has coefficient $(-1{)}^j+(-1{)}^{j+1}=0$. Only ${\sum }_{i=a}^b(-1{)}^i{\mathrm{\ell }}_R(H_i(D))$ remains. This equals $\chi (D)$, including the zero complex and a complex with only one nonzero term.

The long exact sequence for $0\to A\to B\to C\to 0$ has successive blocks $H_i(A),H_i(B),H_i(C),H_{i-1}(A)$. Boundedness permits cutting it between zero endpoints, and all its terms have finite length by hypothesis. The alternating signs on each block can be taken as $(-1{)}^i,-(-1{)}^i,(-1{)}^i$: the next block starts with $(-1{)}^{i-1}$, the opposite of the previous block's last sign. The exact-sequence cancellation therefore gives $\chi (A)-\chi (B)+\chi (C)=0$.

The shift differential has the same kernels and images as $d_D$ in the corresponding degrees, so $H_i(D[1])=H_{i-1}(D)$. Reindexing the finite Euler sum gives $\chi (D[1])={\sum }_j(-1{)}^{j+1}{\mathrm{\ell }}_R(H_j(D))=-\chi (D)$. These establish all three assertions.

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.

If $M=0$, both sides are zero. If $I=R$, the polynomial is zero and the Koszul complex is acyclic. If $r=0$, then $I=0$ and $M$ has finite length; the complex is $M[0]$, its Euler characteristic is ${\mathrm{\ell }}_R(M)$ and $P$ is that constant, so $e_0={\mathrm{\ell }}_R(M)$. The remaining argument concerns $r\ge 1$ and a proper ideal.

Put $c={\mathrm{\ell }}_R(M\mathrm{/}IM)$. For $j\ge 0$, degree-$j$ monomials in the $r$ generators give a surjection $(M\mathrm{/}IM{)}^{a_j}\to I^{j}M\mathrm{/}{I}^{j+1}M$, where $a_j$ is the number of tuples $({\alpha }_1,\dots ,{\alpha }_r)$ of nonnegative integers summing to $j$. Encode a tuple by $j$ marks with $r-1$ separators, including adjacent separators for zero entries. This is a bijection with choices of separator positions, giving $a_j=\left(\genfrac{}{}{0px}{}{j+r-1}{r-1}\right)$. Consequently ${\mathrm{\ell }}_R(I^{j}M\mathrm{/}{I}^{j+1}M)\le c\left(\genfrac{}{}{0px}{}{j+r-1}{r-1}\right)$.

Choose $p>r$ deep enough for the shifted-adic comparison and for $P(p-i-1)={\mathrm{\ell }}_R(M\mathrm{/}{I}^{p-i}M)$ for every $0\le i\le r$. This is possible since only finitely many inequalities are required. The term in cochain degree $-i$ of $K\mathrm{/}{F}^{p}K$ is $(M\mathrm{/}{I}^{p-i}M{)}^{\left(\genfrac{}{}{0px}{}{r}{i}\right)}$. Homology comparison and term cancellation therefore give $\chi (K)={\sum }_{i=0}^r(-1{)}^i\left(\genfrac{}{}{0px}{}{r}{i}\right)P(p-i-1)\mathrm{.}$ The AC hypothesis supplies that in the finite-length and tail lemmas.

Define $\mathrm{\Delta }Q(t)=Q(t)-Q(t-1)$. For $k=0$ the identity ${\mathrm{\Delta }}^{k}Q(t)={\sum }_{i=0}^k(-1{)}^i\left(\genfrac{}{}{0px}{}{k}{i}\right)Q(t-i)$ is the single term $Q(t)$. If it holds at $k$, subtract its value at $t-1$ from its value at $t$. The coefficient of $Q(t-i)$ becomes $(-1{)}^i(\left(\genfrac{}{}{0px}{}{k}{i}\right)+\left(\genfrac{}{}{0px}{}{k}{i-1}\right))=(-1{)}^i\left(\genfrac{}{}{0px}{}{k+1}{i}\right)$, with out-of-range binomials zero; the two endpoint coefficients are $1$ and $(-1{)}^{k+1}$. This proves the identity for every $k$ by induction.

Summing along the $I$-adic filtration gives $0\le {\mathrm{\ell }}_R(M\mathrm{/}{I}^{n+1}M)\le c{\sum }_{j=0}^n\left(\genfrac{}{}{0px}{}{j+r-1}{r-1}\right)=c\left(\genfrac{}{}{0px}{}{n+r}{r}\right)$. For the last identity, tuples in $r$ variables of total at most $n$ correspond bijectively to tuples in $r+1$ variables of total exactly $n$, by adjoining the slack $n-j$; the same separator count applies. This holds for every $n\ge 0$.

The eventual polynomial is nonnegative at all sufficiently large integers. If it is nonzero, its leading coefficient is positive: division by its highest power of $n$ makes the lower terms tend to zero, so the sign is eventually the leading sign. If its degree exceeded $r$, that same division would make the upper bound from the preceding step tend to zero while the polynomial tends to a positive leading coefficient. This is impossible. Hence $P=0$ or $\mathrm{deg}(P)\le r$.

For $d\ge 1$, the binomial expansion gives $t^d-(t-1{)}^d=d{t}^{d-1}$ plus terms of degree at most $d-2$; a constant has difference zero. By linearity, $r$ differences annihilate every monomial of degree less than $r$ and take $t^r$ to $r!$. The degree bound thus gives ${\mathrm{\Delta }}^{r}P=r![T^r]P(T)$, including the zero polynomial. At $t=p-1$ the preceding finite-difference identity is precisely the Euler sum, so it equals $e_r(I,M)$. Together with the initial cases this proves the theorem.

The modules $C$ and $T$ are finite, the latter as a submodule of $M$. Direct quotienting gives $C\mathrm{/}JC=M\mathrm{/}(x,J)M$, of finite length by hypothesis. Further, $xT=0$ and $T\subseteq M$, so $\mathrm{Supp}(T)\subseteq \mathrm{Supp}(M)\cap V(x)$: the first inclusion follows by exact localization of the injection and the second by the annihilator formula.

Let $B=[M\stackrel{x}{\to }M]$ in degrees $1,0$. It contains the subcomplex $T[1]$, with $T$ only in degree one. Its quotient is $Q=[M\mathrm{/}T\stackrel{\bar{x}}{\to }M]$, where $\bar{x}(m+T)=xm$. This is well-defined and injective: $xm=0$ holds exactly for $m\in T$. The map $Q\to C[0]$, zero in degree one and quotient in degree zero, is onto with kernel $D=[M\mathrm{/}T\stackrel{\bar{x}}{\to }xM]$. This differential is an isomorphism, since every element of $xM$ is $xm$ and its kernel is zero. Thus $D$ is acyclic.

For a prime $\mathfrak{p}\mathrm{\ne }\mathfrak{m}$ containing $(x,J)$, the finite-length hypothesis makes $M_{\mathfrak{p}}\mathrm{/}(x,J{)}_{\mathfrak{p}}{M}_{\mathfrak{p}}=0$. Here the local ring $R_{\mathfrak{p}}$ has maximal ideal $\mathfrak{p}{R}_{\mathfrak{p}}$ containing $(x,J{)}_{\mathfrak{p}}$ and $M_{\mathfrak{p}}$ is finite. Nakayama under AC gives $M_{\mathfrak{p}}=0$, hence $T_{\mathfrak{p}}=0$. If $\mathfrak{p}$ does not contain $x$, $T_{\mathfrak{p}}=0$ because $x$ is an invertible annihilator; if it does not contain $J$, then $(T\mathrm{/}JT{)}_{\mathfrak{p}}=0$. These cases cover every $\mathfrak{p}\mathrm{\ne }\mathfrak{m}$, so $T\mathrm{/}JT$ has closed-point support and finite length. Consequently all three Koszul complexes in the statement have finite-length homology.

Put $P=K(\mathbf{y};R)$. Define the total tensor differential on $B_i\otimes P_j$ by $d_B\otimes 1+(-1{)}^{i}1\otimes d_P$. Identifying $m\otimes (e_x\wedge w)$ with the corresponding ordered wedge places the $x$ term first. Deleting that first factor gives $xm\otimes w$, and deleting a $y$ factor has the extra sign $(-1{)}^i$ from passing the $i$ $x$-factors. Thus this total complex is $K(x,\mathbf{y};M)$, as in the concatenation interface (the coefficient $M$ may be moved between tensor factors via $m\otimes (a\otimes w)\mapsto a\otimes (w\otimes m)$).

Every $P_j$ is finite free. Tensoring either $0\to T[1]\to B\to Q\to 0$ or $0\to D\to Q\to C[0]\to 0$ with $P_j$ gives a finite direct sum of that exact sequence. Taking finite sums in each total degree therefore preserves exactness, yielding short exact total complexes. No flatness of $T$ or $M\mathrm{/}T$ is needed.

To prove $D\otimes P$ acyclic, filter it by columns $j\le k$ for $k=-1,0,\dots ,s$. The differential in $D$ preserves $j$, and that in $P$ lowers it, so these are subcomplexes. The initial subcomplex is zero. The quotient at stage $k$ is $D\otimes P_k$ shifted in total degree by $k$, with only the $D$ differential. It is a finite direct sum of the two-term isomorphism $D$, and hence acyclic. The LES at each of the finitely many stages shows that the total complex is acyclic. Applying the LES to the second tensor exact sequence gives $H_i(Q\otimes P)\cong H_i(K(\mathbf{y};C))$.

The first tensor exact sequence has left complex $T[1]\otimes P=K(\mathbf{y};T)[1]$: in its terms the total differential on $P$ is $-d_P$, exactly the shift convention. The middle complex is $K(x,\mathbf{y};M)$ and the right has the homology just computed. All these homologies have finite length by the earlier support calculation. Euler additivity and the shift sign give precisely $\chi (K(x,\mathbf{y};M))=\chi (K(\mathbf{y};C))-\chi (K(\mathbf{y};T))$.