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 be a linearly compact module such that there exists a family of simple submodules whose sum is dense in . Now prove that: 1° is strictly linearly compact (apply a) to the quotient of by an open submodule) ; 2° is isomorphic to the (topological) product of a family of discrete simple modules. (Consider the sets of maximal open submodules of such that for every finite sequence of distinct elements of , is a direct sum of simple submodules. Now prove that there exists a maximal set (for the relation of inclusion), and that the intersection of all the submodules belonging to such a set is reduced to 0 ; one will use the fact that is intersection of maximal open submodules, and that for every maximal open submodule of , there is at least one simple submodule of not contained in . Conclude using the fact that is strictly linearly compact.)
(d) Deduce from c) that every linearly compact vector space over a field is strictly linearly compact and isomorphic to a product .
Solution
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