Theorem 1

Let $A$ be a commutative ring, $m$ an ideal of $A$, $E$ an $A$-module and $(E_n)$ a filtration of the additive group $E$ consisting offinitely generated sub-$A$-modules. Suppose that $m E_n \subset E_{n+1}$ for all $n$. Let $A'$ be the graded subalgebra $\sum_{n \geq 0} m^n X^n$ of $A[X]$ and $E'$ the graded $A'$-module $\sum_{n \geq 0} E_n \otimes_A A X^n$. The twofollowing conditions are equivalent: (a) The filtration $(E_n)$ is m-good. (b) $E'$ is a finitely generated $A'$-module.

Suppose that $m E_{n-1} = E_n$ for $n > n_0 \geq 0$. For $i \leq n_0$, let $(e_i)_{i \leq n_0}$ be a finite system of generators of the $A$-module $E_i$. As the $A$-module $E_n \otimes_A A X^n$ is generated by the elements $e_{n,j} \otimes X^n$ for $0 \leq n \leq n_0$ and is equal to

for $n > n_0$, the $A'$-module $E'$ is generated by the elements $e_{n,j} \otimes X^n$ for $0 \leq n \leq n_0$ and $1 \leq j \leq r_n$; then it is certainly finitely generated.

Conversely, if $E'$ is a finitely generated $A'$-module, it is generated by a finite family of elements of the form $e_k \otimes X^{n(k)}$, where $e_k \in E_{n(k)}$. Let $n_0$ be the greatest of the integers $n(k)$. Then for $n \geq n_0$ and $f \in E_n$,

where $t_k \in A'$; replacing if need be $t$, by its homogeneous component of degree $n - n(k)$, we may assume that $t_k = a_k X^{n-n(k)}$, where $a_k \in m^{n-n(k)}$.

As the unique element $X^n$ forms a basis of the A-module AX", the equation $f \otimes X^n = \left( \sum_k a_k e_k \right) \otimes X^n$ implies $f = \sum_k a_k e_k$. Then $E_n \subset m^{n-n_0} E_{n_0}$; since the opposite inclusion is obvious,

whence $E_n = m E_{n-1}$ for $n > n_0$.