Bourbaki

Exercise 22, § 3

Commutative Algebra, III, 252

Bourbaki marks this one as harder than the rest.

(*) Let AA be a ring, commutative or not; all the AA-modules to be considered are left AA-modules. Let PP be a property such that: (α) if f:M→Nf : M \to N is an injective AA-module homomorphism and NN has property PP, then MM has property PP; (β) the direct sum of two AA-modules with property PP has property PP.

(a) Let MM be an AA-module. Show that the submodules M′M' of MM such that M/M′M/M' has property PP form a fundamental system of neighbourhoods of 0 for a linear topology TP(M)\mathscr{T}_P(M) on MM. Show that TP(As)\mathscr{T}_P(A_s) is compatible with the ring structure on AA and that MM with the topology TP(M)\mathscr{T}_P(M) is a topological module over AA with TP(As)\mathscr{T}_P(A_s). Every AA-module homomorphism f:M→Nf : M \to N is continuous with the topologies TP(M)\mathscr{T}_P(M) and TP(N)\mathscr{T}_P(N).

(b) Suppose that the following condition holds: (γ) if NN is a submodule of MM with property PP and, for every submodule L≠0L \neq 0 of MM, N∩L≠0N \cap L \neq 0, then MM has property PP. Show that with these conditions, if FF is a submodule of an AA-module EE, the topology TP(F)\mathscr{T}_P(F) is induced by TP(E)\mathscr{T}_P(E). (Let F′F' be a submodule of FF such that F/F′F/F' has property PP; consider a maximal element GG among the submodules of EE such that G∩F=F′G \cap F = F' and show that E/GE/G has property PP.)

Solution

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Transcribed from the printed volume. Not translated.