Proposition 12

Let $A$ be a commutative Noetherian ring, $m$ an ideal of $A$, $S$ the multiplicative subset $1 + m$ of $A$ and $E$ a finitely generated $A$-module. Under these conditions:

(i) $S^{-1}A$ is a Zariski ring with the ($S^{-1}m$)-adic topology.

(ii) The canonical mapping $f : E \to S^{-1}E$ is continuous if $E$ is given the m-adic topology and $S^{-1}E$ the $(S^{-1}m)$-adic topology and $\hat{f} : E \to (S^{-1}E)^{\wedge}$ is an isomorphism.

Every element of $1 + (S^{-1}m)$ is of the form

where $m \in m$ and $m' \in m$; it is therefore invertible in $S^{-1}A$, which proves that $S^{-1}m$ is contained in the Jacobson radical of $S^{-1}A$; as $S^{-1}A$ is Noetherian (Chapter II, § 2, no. 4, Corollary 2 to Proposition 10), $S^{-1}A$ is a Zariski ring with the $(S^{-1}m)$-adic topology, which proves (i). Let us show (ii). For all $n > 0$, for clearly first of all $f(m^nE) \subset S^{-1}m^nE$; conversely, let $x$ be an element of $\bar{f}^{-1}(S^{-1}m^nE)$; then there exist elements $m', m''$ of $m$ and $x'' \in m^nE$ such that $(1 + m')((1 + m'')x - x'') = 0$, whence $(1 - m)x = x'$, where and $x' = (1 + m')x'' \in m^nE$; we conclude that This proves that $f$ is a strict morphism. Moreover, the kernel of $f$, which is the set of $x \in E$ for which there exists some $s \in S$ such that $sx = 0$, is identical with the kernel of the canonical mapping $j : E \to \hat{E}$ (no. 2, Proposition 5). Then there is a topological isomorphism $f_0 : j(E) \to f(E)$ such that $f = f_0 \circ j$; as $\hat{f}$ is a topological isomorphism, the problem reduces to verifying that $f(E)$ is dense in $S^{-1}E$. Now every element of $S^{-1}E$ may be written as $x/(1 - m)$, where $m \in m$, and it is immediately verified that which completes the proof.