koszul euler characteristic first element reduction

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 $\chi (K(x,\mathbf{y};M))=\chi (K(\mathbf{y};C))-\chi (K(\mathbf{y};T))$.

If $s=0$, then $P=R[0]$, and the calculation reads $\chi ([M\stackrel{x}{\to }M])={\mathrm{\ell }}_R(C)-{\mathrm{\ell }}_R(T)$; their lengths are finite by the support argument. If $M=0$, every complex is zero. If $(x,J)=R$, the same support cases prove the required finiteness and the tensor argument still applies; no step required this ideal to be proper. If $x$ is a unit, then $C=T=0$ and $B$ is an isomorphism complex. If $x=0$, then $C=T=M$ and the formula gives zero by cancellation. Thus all asserted cases are included.