Corollary

Let $A$ be a noetherian ring, $q$ an ideal of $A$, $M$ a finitely generated $A$-module such that $M/qM$ is of finite length, and $F$ a q-good filtration on $M$. a) In order that one have $d_q(M) \leq 0$, it is necessary and sufficient that the sequence $(q^n M)$ be stationary, or again that the sequence $(F_n)$ be stationary. One then has, for every sufficiently large $n$,

b) Suppose that one has $d_q(M) > 0$. One then has

$$\text{(13)} \quad \operatorname{long}_A(F_n/F_{n+1}) = e_q(M) n^{d_q(M) - 1}/(d_q(M) - 1)! + \rho_n n^{d_q(M) - 2},$$

$$\text{(14)} \quad \operatorname{long}_A(M/F_{n+1}) = e_q(M) n^{d_q(M)}/d_q(M)! + \sigma_n n^{d_q(M) - 1},$$

where $\rho_n$ and $\sigma_n$ tend to a limit as $n$ increases indefinitely. This results at once from Theorem 2 and formula (2) of No. 1.