Bourbaki

Exercise 2, § 3

Algebra, VII, p. 59

Bourbaki marks this one as harder than the rest.

Let M be a module over a principal ideal domain A ; suppose M is a direct sum of cyclic submodules. Show that every submodule N of M is a direct sum of cyclic submodules (if T is the torsion submodule of N, show first of all that N/T is a free module, using Th. 1, and deduce that T has a complement in N ; then show that T is a direct sum of cyclic submodules, using Ex. 12, c) of VII, p. 57).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.