Bourbaki

Exercise 20, § 2

Commutative Algebra, III, 239

Bourbaki marks this one as harder than the rest.

(a) Show that a discrete linearly compact module cannot be the direct sum of an infinity of submodules which are not reduced to 0.

(b) Give an example of a minimal linear topology (Exercise 17) which is not linearly compact (consider an infinite direct sum of simple modules no two of which are isomorphic).

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

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

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.