Bourbaki

Exercise 20, § 2

Commutative Algebra, III

Bourbaki marks this one as harder than the rest.

(a) Now prove that a discretely linearly compact module cannot be a direct sum of an infinity of submodules not reduced to 0.

(b) Give an example of a minimal linear topology (exerc. 17) which is not linearly compact (consider an infinite direct sum of pairwise non-isomorphic simple modules).

(c) Let EE be a linearly compact module such that there exists a family of simple submodules whose sum is dense in EE. Now prove that: 1° EE is strictly linearly compact (apply a) to the quotient of EE by an open submodule) ; 2° EE is isomorphic to the (topological) product of a family of discrete simple modules. (Consider the sets D\mathfrak{D} of maximal open submodules of EE such that for every finite sequence (Gk)1≤k≤n(G_k)_{1 \leq k \leq n} of nn distinct elements of D\mathfrak{D}, E/(⋂k=1nGk)E / (\bigcap_{k=1}^n G_k) is a direct sum of nn simple submodules. Now prove that there exists a maximal set D\mathfrak{D} (for the relation of inclusion), and that the intersection of all the submodules GG belonging to such a set is reduced to 0 ; one will use the fact that {0}\{0\} is intersection of maximal open submodules, and that for every maximal open submodule GG of EE, there is at least one simple submodule of EE not contained in GG. Conclude using the fact that EE is strictly linearly compact.)

(d) Deduce from c) that every linearly compact vector space over a field KK is strictly linearly compact and isomorphic to a product Ks1K_s^1.

Solution

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