Theorem 2

Let A be a noetherian ring, q an ideal of A, M a finitely generated A-module such that M/qM is nonzero and of finite length, and F a q-good filtration on M.

a) There exist an integer d $\geq 0$, and an element R of $\mathbf{Z}[T, T^{-1}]$, uniquely determined, such that $R(1) > 0$ and $H_{M,F} = (1 - T)^{-d}R$.

b) The integers d and R(1) are independent of the q-good filtration F chosen.

a) Let us consider the graded ring gr(A) such that $\operatorname{gr}_n(A) = q^n/q^{n+1}$ and the graded gr(A)-module gr(M) such that $\operatorname{gr}_n(M) = F_n/F_{n+1}$. Since one has $F_{n_1} = M$ and $qF_n = F_{n+1}$ for $n \geq n_0$, gr(M) is generated by $\bigoplus_{n_1 \leq n \leq n_0} \operatorname{gr}_n(M)$, hence is finitely generated. Moreover, if $(x_1, ..., x_r)$ is a finite system of generators of the A-module q, gr(A) is generated by $\operatorname{gr}_0(A)$ and the classes of the $x_i$ modulo $q^2$, hence is isomorphic to a quotient graded ring of $H = (A/q)[X_1, ..., X_r]$. By Theorem 1 of No. 2, one has

One has $H_{M,F} \neq 0$ and there therefore exist $d \in \mathbf{N}$ and $R \in \mathbf{Z}[T, T^{-1}]$, uniquely determined, such that $R(1) > 0$ and $H_{M,F} = (1 - T)^{-d}.R$ (Lemma 2 of No. 1).

b) Let $F'$ be another q-good filtration and let us write analogously

By Proposition 3, b), there exists an integer $m$ such that $(1 - T)^{-d' - 1} R' \geq T^m (1 - T)^{-d - 1} R$. By Lemma 2, b) of No. 1, this implies $d' \geq d$ and, if $d' = d$, $R'(1) \geq R(1)$. Exchanging the roles of $F$ and $F'$, one obtains $d = d'$ and $R(1) = R'(1)$.