Exercise 18, § 2
Commutative Algebra, III
Bourbaki marks this one as harder than the rest.
In an -module , a submodule is said to be sheltered if there exists a smallest element in the set of submodules of .
(a) Now prove that every submodule of is an intersection of sheltered submodules (for every , consider a maximal element in the set of submodules of containing and not containing ).
(b) Let be a separated linear topology on , and let be the linear topology having as fundamental system of neighbourhoods of the filter basis generated by the submodules of open for , and sheltered. Now prove that is separated and that every submodule closed for is also closed for (remark that every submodule closed for is an intersection of submodules open for ). Deduce that if is complete for , it is complete for (Top. gén., Chap. III, 3rd ed., § 3, No. 5, Prop. 9).
(c) Suppose that is linearly compact. Now prove that is linearly compact and is the least fine of the separated linear topologies less fine than ; in particular is a minimal linear topology (Exercise 17). (Let be a filter basis formed of submodules closed for , having for intersection; prove that for every submodule open for and sheltered, there exists such that , using Exercise 16 d)).
(d) Let be a separated linearly topologized -module, and let be its topology; prove that if is a linear mapping continuous for the topologies and , is also continuous for the topologies and .
Solution
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