Bourbaki

Exercise 15, § 3

Commutative Algebra, II

Bourbaki marks this one as harder than the rest.

Let A be a (commutative) ring, M an A-module.

a) In order that a submodule M′M' of M be pure (Chapter I, § 2, Exercise 24), it is necessary and sufficient that, for every maximal ideal mm of A, Mm′M'_m be a pure sub-AmA_m-module of MmM_m (use Theorem 1 of No. 3).

b) Suppose that A is a local ring with maximal ideal mm, and M a free a-module of finite type. In order that a submodule M′M' of finite type of M be pure, it is necessary and sufficient that it be a direct factor of M. (Using Corollary 1 of Proposition 5 of No. 2, reduce to proving that, if M′⊂mM,M′M' \subset mM, M' can only be a pure submodule of M if M′=0M' = 0.)

Solution

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Machine translation of the English, by gpt-5-6. Not checked by a person.