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The Stacks project

Example 87.4.1. 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).


Comments (2)

Comment #5982 by Dario Weißmann on

typo: A-module should be R-module

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  • 2 comment(s) on Section 87.4: Topological rings and modules

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