Exercise 6, § 5
Commutative Algebra, VI, 456
Bourbaki marks this one as harder than the rest.
(a) Let A be a valuation ring, M an A-module and M' a submodule of M. Show that for M' to be a pure submodule of M (Chapter I, § 2, Exercise 24), it is necessary and sufficient that, for all , .
(b) Let M' be a pure submodule of M such that, if M' is given the topology for which the (where ) form a fundamental system of neighbourhoods of 0, M' is linearly compact. Show that for every element there exists such that has the following property: for all such that , . (For each of the such that , consider the subset S, of M' consisting of the such that .)
(c) Suppose that A is a linearly compact valuation ring; let K be the field of fractions of A. Show that, if M is a torsion-free A-module such that admits a countable basis, M is the direct sum of a countable family of A-modules of rank 1. (Consider M as the union of an increasing sequence of pure submodules such that is of rank i and for each i apply (b) to and its submodule .
(d) With the same hypotheses on A and M as in (c), let N be a pure submodule of M of finite rank. Show that N is a direct factor of M (observe that every A-module which is the direct sum of a finite number of A-modules of rank 1 is linearly compact and use (b)).
(e) With the same hypotheses on M and M' as in (b), show that for all there exists such that the annihilator of is equal to the annihilator of the element . (Let a be the annihilator of ; for all , let T, be the subset of M' consisting of they' such that ; show that the intersection of the T, is non-empty.)
(f) Suppose that A is a valuation ring such that for every ideal the A-module A/a is linearly compact (with the discrete topology). Show that every finitely generated torsion A-module M is the direct sum of a finite number of monogenous A-modules. (If is a system of generators of M, argue by induction on n, considering one of the whose annihilator is the least and noting that the submodule of M which it generates is pure.)
Solution
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