Bourbaki

Exercise 18, § 2

Commutative Algebra, III

Bourbaki marks this one as harder than the rest.

In an AA-module EE, a submodule M≠EM \neq E is said to be sheltered if there exists a smallest element in the set of submodules ≠0\neq 0 of E/ME/M.

(a) Now prove that every submodule N≠EN \neq E of EE is an intersection of sheltered submodules (for every x∉Nx \notin N, consider a maximal element in the set of submodules of EE containing NN and not containing xx).

(b) Let C\mathscr{C} be a separated linear topology on EE, and let C∗\mathscr{C}^* be the linear topology having as fundamental system of neighbourhoods of 00 the filter basis generated by the submodules of EE open for C\mathscr{C}, and sheltered. Now prove that C∗\mathscr{C}^* is separated and that every submodule closed for C\mathscr{C} is also closed for C∗\mathscr{C}^* (remark that every submodule closed for C\mathscr{C} is an intersection of submodules open for C\mathscr{C}). Deduce that if EE is complete for C∗\mathscr{C}^*, it is complete for C\mathscr{C} (Top. gén., Chap. III, 3rd ed., § 3, No. 5, Prop. 9).

(c) Suppose that C\mathscr{C} is linearly compact. Now prove that C∗\mathscr{C}^* is linearly compact and is the least fine of the separated linear topologies less fine than C\mathscr{C}; in particular C∗\mathscr{C}^* is a minimal linear topology (Exercise 17). (Let B\mathscr{B} be a filter basis formed of submodules closed for C\mathscr{C}, having 00 for intersection; prove that for every submodule UU open for C\mathscr{C} and sheltered, there exists M∈BM \in \mathscr{B} such that M⊂UM \subset U, using Exercise 16 d)).

(d) Let FF be a separated linearly topologized AA-module, and let C1\mathscr{C}_1 be its topology; prove that if u:E→Fu : E \to F is a linear mapping continuous for the topologies C\mathscr{C} and C1\mathscr{C}_1, uu is also continuous for the topologies C∗\mathscr{C}^* and C1∗\mathscr{C}_1^*.

Solution

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Machine translation of the English, by gpt-5-6. Not checked by a person.