Example 85.4.1. Let $R$ be a linearly topologized ring and let $M$ be a linearly topologized $R$-module. Let $I_\lambda$ be a fundamental system of open ideals in $R$ and let $M_\mu$ be a fundamental system of open submodules of $M$. The continuity of $+ : M \times M \to M$ is automatic and the continuity of $R \times M \to M$ signifies

$\forall f, x, \mu \ \exists \lambda , \nu ,\ (f + I_\lambda )(x + M_\nu ) \subset fx + M_\mu$

Since $fM_\nu + I_\lambda M_\nu \subset M_\mu$ if $M_\nu \subset M_\mu$ we see that the condition is equivalent to

$\forall x, \mu \ \exists \lambda \ I_\lambda x \subset M_\mu$

However, it need not be the case that given $\mu$ there is a $\lambda$ such that $I_\lambda M \subset M_\mu$. For example, consider $R = k[[t]]$ with the $t$-adic topology and $M = \bigoplus _{n \in \mathbf{N}} R$ with fundamental system of open submodules given by

$M_ m = \bigoplus \nolimits _{n \in \mathbf{N}} t^{nm}R$

Since every $x \in M$ has finitely many nonzero coordinates we see that, given $m$ and $x$ there exists a $k$ such that $t^ k x \in M_ m$. Thus $M$ is a linearly topologized $R$-module, but it isn't true that given $m$ there is a $k$ such that $t^ kM \subset M_ m$. On the other hand, if $R \to S$ is a continuous map of linearly topologized rings, then the corresponding statement does hold, i.e., for every open ideal $J \subset S$ there exists an open ideal $I \subset R$ such that $IS \subset J$ (as the reader can easily deduce from continuity of the map $R \to S$).

Comment #5982 by Dario Weißmann on

typo: A-module should be R-module

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