Proposition 3

Let $A$ be a Noetherian ring and $m$ an ideal of $A$ such that $A$ is Hausdorff and complete with respect to the m-adic topology. Let $E$ be a filtered $A$-module over the filtered ring $A$, the filtration $(E_n)$ of $A$ being such that $E_0 = E$ and $E$ is Hausdorff with respect to the topology defined by $(E_n)$. Then the following conditions are equivalent: (a) $E$ is a finitely generated $A$-module and $(E_n)$ is an m-good filtration. (b) $\mathrm{gr}(E)$ is a finitely generated $\mathrm{gr}(A)$-module. (c) For all $n \geq 0$, $\mathrm{gr}_n(E)$ is a finitely generated $A$-module and there exists $n_0$ such that for $n \geq n_0$ the canonical homomorphism

is surjective.

It follows immediately from the definitions that (a) implies (c). The fact that (b) implies (c) is a consequence of § 1, no. 3, Lemma 1; conversely, if (c) holds, clearly $\mathrm{gr}(E)$ is generated as a $\mathrm{gr}(A)$-module by the sum of the $\mathrm{gr}_p(E)$ for $p \leq n_0$ and hence by hypothesis admits a finite system of generators. It remains to prove that (c) implies (a); as the $\mathrm{gr}_n(E)$ are finitely generated and $E_n = E$, clearly first, by induction on $n$, $E/E_n$ is a finitely generated $A$-module for all $n$; it will therefore be sufficient to prove that, for $n > n_0$, $E_n$ is a finitely generated $A$-module and that $mE_n = E_{n+1}$. Now, consider the $A$-module $E_{n+1}$ with the exhaustive and separated filtration $(E_{n+k})(k \geq 1)$; $mE_n \subset E_{n+1}$; hypothesis (c) implies that the image of $mE_n$ in $\mathrm{gr}_{n+1}(E) = E_{n+1}/E_{n+2}$ is equal to $\mathrm{gr}_{n+1}(E)$ and generates the graded $\mathrm{gr}(A)$-module $\mathrm{gr}(E_{n+1})$. As $\mathrm{gr}_{n+1}(E)$ is by hypothesis a finitely generated $A$-module, it follows from § 2, no. 9, Proposition 12 that $mE_n = E_{n+1}$ and that $E_{n+1}$ is a finitely generated $A$-module.