Proposition 14

Let $(A_i, h_{ij})$ be an inverse system of discrete commutative rings, whose indexing set is $\mathbf{N}$ and let $(M_i, u_{ij})$ be an inverse system of modules over the inverse system of rings $(A_i, h_{ij})$. Let $n_j$ denote the kernel of $h_{0j}: A \to A$, and set $A = \lim A_i$, $M = \lim M_i$. Suppose that

(a) for all $i \in \mathbf{N}$, $h_{ii}$ is the identity mapping on A, and, for $i \leq j$, $h_{ij}$ and $u_{ij}$ are surjective; (b) for $i \leq j$, the kernels of $h_{ij}$ and $u_{ij}$ are $n_j^{i+1}$ and $n_j^{i+1} M_j$ respectively.

Then: (i) $A$ is a complete Hausdorff topological ring, $M$ is a complete Hausdorff topological A-module and the canonical mappings $h_i : A \to A_i$, $u_i : M \to M_i$ are surjective. (ii) If $M_0$ is a finitely generated A,-module, $M$ is a finitely generated A-module;

to be precise, every finite subset S of M such that $u_0(S)$ generates M, is a system of generators of M.

The assertions in (i) follow from General Topology, Chapter II, § 3, no. 5, Corollary to Proposition 10 and Corollary 1 to Theorem 1.

For all $i \in \mathbf{N}$, let us write $m_{i+1} = \mathrm{Ker}(h_i), N_{i+1} = \mathrm{Ker}(u_i)$; then

and $N_{i+1} = \lim_{\leftarrow} n_{i+k}^{t+1} M_{i+k}$; as $h_{i+k}$ and $u_{i+k}$ are surjective,

Let us show that $m_i N_j \subset N_{i+j}$ for $i \geq 1$ and $j \geq 1$, which amounts to proving that $u_{i+j-1}(m_i N_j) = 0$; now

is equal to $n_{i+j-1}^{t'}(n_{i+j-1}^{t'} M_{i+j-1}) = 0$, as, for all $k \geq 0$, $n_k^{k+1}$, which is the kernel of $h_{kk}$, is equal to 0. We see similarly that $m_i m_j \subset m_{i+j}$. If for $i \leq 0$ we set $m_i = A$ and $N_i = M$, $(m_i)_{i \in \mathbf{Z}}$ is a filtration of A and $(N_i)_{i \in \mathbf{Z}}$ a filtration of M compatible with the filtration on A; the topologies on A and M are obviously those defined by these filtrations. This being so, let $a$ be an ideal of A such that $h_1(a) = n_1$ and $M'$ the submodule of M generated by S; we are going to prove that

Let us write $a_i = h_i(a), M'_i = u_i(M')$; it is sufficient to show that

This is true if $i = 0$, for $N_0 = M$ and $M'_0 = M$, by hypothesis. If $i \geq 1$, then $u_i(N_i) = n_i^i M_i$ by (16). As $h_{1i}$ is surjective and $h_{0i} = h_{01} \circ h_{1i}$, $h_{1i}$ maps the kernel $n_i$ of $h_i$, onto the kernel $n_1$ of $h_1$, and $n_i = h_{1i}(n_1)$; then

and, as the kernel of $h_{1i}$ is $n_i^2$, $n_i \subset a_i + n_i^2$ and $a_i \subset n_i$, whence $n_i = a_i + n_i^2$. Moreover $u_{0i}(M'_i) = u_0(M') = M, = u_{0i}(M_i)$ and, as $\mathrm{Ker}(u_{0i}) = n_i M_i$, $M_i = M'_i + n_i M_i$; whence

Now, $a_i^k n_i^{i+1-k} \subset n_i^{i+1} = 0$ for $0 \leq k \leq i$; then it certainly follows that $u_i(N_i) = n_i M_i = a_i^i M'_i$, which proves (17).

Moreover $m_1 = h_1(n_1)$, whence $a \subset m_1$ and therefore $a^i \subset m_1^i \subset m_i$, whence

$N_i \subset m_i M' + N_{i+1}$; on the other hand obviously $m_i M \subset N_i$ and hence $N_i = m_i M' + N_{i+1}$ for all $i \geq 0$; then it follows from Corollary 3 to Proposition 12 of no. 9 that $M' = M$, which completes the proof.