koszul homology finite length for an ideal of definition

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.