Bourbaki

Exercise 16, § 2

Commutative Algebra, III

Bourbaki marks this one as harder than the rest.

a) Let E be a separated linearly topologized A-module. Now prove that the following properties are equivalent α) E is linearly compact (exerc. 15). β) For every continuous linear mapping u from E into a separated linearly topologized A-module F, u(E) is a closed submodule of F. γ) For every linear topology (separated or not) on E, less fine than the given topology, E is complete. δ) E is complete, and there is a fundamental system (U_λ) of open neighbourhoods of 0 in E, formed by submodules, such that the E/U_λ are discrete linearly compact A-modules (cf. § 3, exerc. 5). (To see that γ implies δ), consider an open submodule F of E, and a filter basis B on E/F formed by affine linear varieties. For every V ∈ B, let M_V be the inverse image in E of the direction submodule of V in E/F; consider on E the linear topology whose M_V form a fundamental system of neighbourhoods of 0.)

b) Let E be a separated linearly topologized A-module, M a linearly compact submodule of E. Now prove that for every closed submodule F of E, M + F is closed in E (consider the image of M in E/F).

c) Let E be a linearly compact A-module, F a separated linearly topologized A-module. Now prove that for every closed submodule M of E × F, the projection of M on F is closed. Converse (cf. Top. gén., chap. I, 3e éd., § 10, n° 2).

d) Let E be a linearly compact A-module, u a continuous linear mapping from E into a separated linearly topologized A-module F. Now prove that for every filter basis B on E, formed by affine linear varieties, the image by u of the set of adherent points to B is the set of adherent points to u(B)u(\mathscr{B}). In particular, for every closed submodule MM of EE, one has ⋂N∈B(M+N)=M+⋂N∈BN\bigcap_{N \in \mathscr{B}} (M + N) = M + \bigcap_{N \in \mathscr{B}} N.

Solution

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.