Corollary 1

With the notation and hypotheses of Proposition 14 suppose further that $M_0$ is a finitely generated $A$-module and that the ideal $n_1$ of $A$, is finitely generated. Let $m_1$ be the kernel of $h_1$; the topologies on $A$ and $M$ are then the $m_1$-adic topologies on this ring and this module respectively; to be precise, for all $i \geq 0$, the kernels of $h_i$ and $u_i$ are $m_1^{i+1}$ and $m_1^{i+1} M$ respectively; further $m_1 / m_1^2$ is a finitely generated $A$-module.

We preserve the notation of the proof of Proposition 14; the hypotheses here allow us to assume that the ideal $a$ is finitely generated. Let $i \geq 0$ be any integer; for all $j \geq 0$, by (17), $N_{i+j} = a^j(a^i M) + N_{i+j+1} \subset m_j(a^i M) + N_{i+j+1}$; conversely, $m_j(a^i M) \subset m_j m_i M \subset m_{i+j} M \subset N_{i+j}$, whence

As $a$ and $M$ are finitely generated $A$-modules, so is $a^i M$. Applying Corollary 3 to Proposition 12 of no. 9 to the module $N_i$ with the filtration $(N_{ij})_{j \in \mathbf{Z}}$ defined by $N_{ij} = N_i$, if $j < 0$, $N_{ij} = N_{i+j}$ if $j \geq 0$, we obtain $N_i = a^i M$, whence $N_i \subset m_1^i M$. But also $m_1^i M \subset m_i M \subset N_i$, whence $N_i = m_1^i M$. Applying this to the case where $M = A$, $u_{ij} = h_{ij}$, we obtain $m_i = m_1^i$. Moreover, $m_1 = a + m_1^2$ by (17), which proves the last assertion of the corollary.