Exercise 15, § 1
Tag 05MI, permanent. Cite it as [Bourbaki, Tag 05MI].
Algebra, II, 383
Bourbaki marks this one as harder than the rest.
Let c be an infinite cardinal.
(a) Let I be a set and R a relation between elements of I such that for all the set of for which holds has cardinal . Show that there exists a well-ordered set L and an increasing family of subsets of I such that: (1) there exists a such that ; (2) I is the union of the for ; (3) if is such that the set of has a greatest element , ; (4) if is the least upper bound of the set of , is the union of the for ; (5) if and holds, then . (Note first that for all the set of for which there exists an integer n and a sequence of elements of I such that , and holds for , has cardinal . Then construct the by transfinite induction.)
(b) Let E be an A-module the direct sum of a family of submodules such that for all (cf. Exercise 6). Let f be an endomorphism of E. Show that there exists a well-ordered set L and an increasing family of subsets of I satisfying properties (1), (2), (3) and (4) of (a) and such
LINEAR ALGEBRA
moreover that if we write , then . (Apply (a) to the relation R: "there exists such that the component of in is ".) For all such that the set of has a greatest element , write . Show that E is the direct sum of the (apply Exercise 14).
(c) With the hypotheses and notation of (b), suppose further that is a projector and write . Let ; show that if is such that the set of has a greatest element , is a direct factor of (cf. Exercise 12), that a supplementary submodule of in is isomorphic to a direct factor of and that . Show that is the direct sum of the (apply Exercise 14).
(d) Deduce from (c) that if an A-module E is the direct sum of a family of submodules such that , then every direct factor P of E is also the direct sum of a family of submodules such that .
Solution
No solution has been written yet.