Exercise 23, § 2
Algebra, II, 394
Bourbaki marks this one as harder than the rest.
Let M be an A-module, the direct sum of a family (finite or otherwise) of submodules such that for all the non-invertible elements of form a two-sided ideal of this ring (which implies that is indecomposable).
(a) Let be two endomorphisms of such that . Show that for every finite sequence of distinct indices in there exists a family of submodules of such that for every one of the endomorphisms restricted to is an isomorphism of this submodule onto and that is the direct sum of the and the for distinct from . (Reduce it to the case ; if is the canonical projection of onto corresponding to the decomposition as a direct sum of and , note that either or , restricted to , is an automorphism of .)
(b) Let be an idempotent endomorphism of . Show that there exists at least one index such that the restriction of to is an isomorphism of onto ; moreover is a direct factor of . (If , note that there exists a finite sequence of indices such that the intersection of and the direct sum of the is and use (a).)
(c) Deduce from (b) that every indecomposable direct factor of is isomorphic to one of the .
(d) Let be a family of indecomposable submodules of of which is the direct sum; every is therefore isomorphic to an and vice versa, by virtue of (c). Let be the set of classes of indecomposable submodules of (under the relation of isomorphism) such that an (or an ) belongs to one of these classes. For all , let (resp. ) be the set of (resp. ) such that (resp. ). Show that, for all , . (In the notation of (a), let be the set such that the restriction of to is an isomorphism of onto ; show that is finite and that the cover when runs through . Deduce that when is infinite. When is finite, show with the aid of (a) that is the direct sum of the for and a subfamily of of cardinal equal to that of .)
(e) Deduce from (d) and Exercises 22 and 21 (d) that if and are two reduced irreducible decompositions of in a module , and are equipotent.
Solution
No solution has been written yet.