Exercise 23, § 2

Let $A$ be a commutative ring endowed with a linear topology $\mathscr{C}$.

a) Now prove that the ideals $a$ of $A$, closed for $\mathscr{C}$ and such that $A/a$ is artinian, form a fundamental system of neighbourhoods of $0$ for a linear topology $\mathscr{C}^{(c)}$ on $A$, less fine than $\mathscr{C}$ (use Alg., chap. VIII, § 2, exerc. 12); one has $(\mathscr{C}^{(c)})^{(c)} = \mathscr{C}^{(c)}$. Let $B$ be the separated completion of the ring $A$ for the topology $\mathscr{C}^{(c)}$; now prove that $B$ is strictly linearly compact; one says that $B$ is the strictly linearly compact ring associated with $A$.

b) Let $\mathscr{C}_\omega(A)$ be the discrete topology on $A$, $\mathscr{C}_c(A)$ the topology $(\mathscr{C}_\omega(A))^{(c)}$. For the topology $\mathscr{C}^{(c)}$ on $A$ to be separated, it is necessary and sufficient that $\mathscr{C}$ be separated and that, for every ideal $b$ of $A$, open for $\mathscr{C}$, the topology $\mathscr{C}_c(A/b)$ be separated. If $A$ is a non-discrete valuation ring of rank 1 (chap. VI), $\mathscr{C}_c(A)$ is not separated.

c) Suppose that $A$ is linearly compact for $\mathscr{C}$; then $A$ is complete (but in general non-separated) for $\mathscr{C}^{(c)}$. For $\mathscr{C}^{(c)}$ to be then separated, it is necessary and sufficient that there exist on $A$ a linear topology, strictly linearly compact, and less fine than $\mathscr{C}$; $\mathscr{C}^{(c)}$ is then the unique topology having these properties.

d) Let $E$ be an $A$-module linearly topologized and strictly linearly compact. Now prove that when $A$ is endowed with the topology $\mathscr{C}_c(A)$, $E$ is a topological $A$-module (note that if $F$ is an $A$-module artinian, then, for every $x \in F$, the annihilator $a$ of $x$ is such that $A/a$ is artinian). Deduce that, if $B$ is the separated completion of $A$ for the topology $\mathscr{C}_c(A)$, $E$ can be considered as a topological $B$-module; the ring $B$ being isomorphic to a product of local rings $B_\lambda$ strictly linearly compact (Exercise 21 d)), now prove that $E$ is isomorphic to a product of strictly linearly compact submodules $E_\lambda$, where $E_\lambda$ is annihilated by the $B_\mu$ of index $\mu \neq \lambda$ and can consequently be considered as a topological $B_\lambda$-module (if $e_\lambda$ is the unit element of $B_\lambda$, take $E_\lambda = e_\lambda . E$).