Bourbaki

Exercise 6, § 3

Tag 073H, permanent. Cite it as [Bourbaki, Tag 073H].

Algebra, VII, p. 60

Bourbaki marks this one as harder than the rest.

Let AA be a principal ideal domain with a unique maximal ideal AτA \tau, and complete with respect to the topology T\mathscr{T} defined in Ex. 5. a) Show that any torsion-free AA-module EE of finite rank, containing no nonzero divisible submodules, is free (consider a submodule MM of EE generated by a maximal pseudofree family; if M≠EM \neq E show, with the help of Ex. 4, b) and 5, b), that EE contains a nonzero divisible submodule). Generalise this result to the case where EE has countable rank (proceed by induction). b) In the free module E=A(N)E = A^{(\mathbf{N})} (the direct sum of countably infinitely many copies of AA), let (an)n=1∞(a_n)_{n=1}^\infty be the canonical basis, and let en=an−πane_n = a_n - \pi a_n for each integer n≥1n \geq 1. Show that the submodule MM of EE generated by the ene_n is pure, and that the ene_n form a maximal pseudofree family, but that E/ME/M is a divisible module, and hence MM does not admit a complement in EE.

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Transcribed from the printed volume. Not translated.