Lemma 4

Let $F$ be a $q$-good filtration on $M$. If $M/qM$ is of finite length, $M/F_{n+1}$ and $F_n/F_{n+1}$ are of finite length for every $n \in \mathbf{Z}$. With the notation of (7), we have $\operatorname{long}(M/F_{n+1}) \leq \operatorname{long}(M/q^{n-n_1+1}M)$ and it is enough to prove that $q^nM/q^{n+1}M$ is of finite length for every $n$. We are thus reduced to the case of the $q$-adic filtration. Let $(x_1, ..., x_r)$ be a finite generating system of the $A$-module $q$, and let $I$ be the finite set of monomials of total degree $n$ in $r$ variables

X_1, ..., X_r. The homomorphism from (M/qM)^l into q^nM/q^{n+1}M which maps the family (u_m)_{m\in I} to the element $\sum m(x_1, ..., x_r)\ u_m$ is surjective. Since M/qM is of finite length, q^nM/q^{n+1}M is so also.

Henceforth suppose M/qM of finite length. Let F be a q-good filtration on M. There exists $n_1 \in \mathbf{Z}$ such that $F_{n_1} = M$, hence $F_n = M$ for $n \leq n_1$; we therefore define an element $H_{M,F}$ of $\mathbf{Z}((T))$ by putting

(8)