Exercise 7, § 2
Algebra, VII, p. 55
Let M be a module over a principal ideal domain A; a submodule N of M is said to be pure if, for all , we have . a) The submodule N of M is pure if and only if, for all and every element whose annihilator is , there exists an element whose annihilator is . b) Show that the torsion submodule of M is pure. A family of elements of M is pseudofree (Ex. 4) if and only if the sum of the submodules is direct and is a pure submodule of M. Show that if P and Q are submodules of M such that and are pure then P and Q are pure. If Q is a pure submodule of M and a submodule of M, then P is pure in M if and only if is pure in . c) Show that every submodule N of M which is a direct factor of M is pure. The same is true for the union of a filtered family of direct factors of M. Conversely, if N is a pure submodule of M, and is a direct sum of cyclic submodules, show that N is a direct factor of M (use a)). Give examples of direct factors P and Q of M such that (resp. ) is not a pure submodule (for the first example, take , and for the second, take ). d) Let N be a pure submodule of M, whose annihilator is nonzero. Iff is the natural homomorphism from M onto , show that f restricts to an isomorphism from N onto , and that is a pure submodule of (use Bezout's identity). Deduce (with the help of c) and Ex. 4, c)) that is a direct factor in , and that is a direct factor in . In particular, if the torsion submodule of a module has nonzero annihilator (which is the case when it is finitely generated), then it is a direct factor in (cf. VII, p. 20, Cor. 1 to Th. 2).
e) Let be a prime number; let denote the direct sum of the cyclic -modules (), and let denote the element of whose coordinates are all zero except for the -th, which is equal to the class of 1 mod . Let be the submodule of the -module generated by the elements . Show that the torsion submodule of is not a direct factor in (notice that every element of is divisible by for all , and that no element of has this property except for multiples of ).
Solution
No solution has been written yet.