Exercise 15, § 4

a) Let M be a module of finite length and rank n over a principal ideal domain A, and let Aα_i (1 ≤ i ≤ n) be its invariant factors in decreasing order. Let (β_i) (1 ≤ i ≤ n) be a sequence of elements of A such that β_i divides a, for 1 ≤ i ≤ n and a, divides β_{i−1}, for 1 ≤ i ≤ n − 1. Let (M_i)_{≤ i ≤ n} be a sequence of submodules of M such that M is the direct sum of the M_i and each M_i is isomorphic to A/Aα_i; let a_i be a generator of M_i for 1 ≤ i ≤ n. Put

Show that the quotient of M by the cyclic submodule Ab is isomorphic to the direct sum of the n modules A/Aβ_i (1 ≤ i ≤ n). b) Let M and N be two modules of finite length over A; let Aα_i (1 ≤ i ≤ n) and Aβ_k (1 ≤ k ≤ p) be the invariant factors of M and N respectively, arranged in decreasing order; suppose that p ≤ n, that β_k divides α_{k+n−p} for 1 ≤ k ≤ p, and that a_i divides β_{k+(p+q−n)} for 1 ≤ k ≤ n − q, where q is an integer such that

Show that there exists a submodule P of M, isomorphic to N, such that M/P has rank ≤ q (decompose each of the two modules M and N as a direct sum of q modules, in such a way as to reduce the problem to the case q = 1; then use a) and Ex. 13, g)). c) For a module N over A to be isomorphic to a direct factor of a module M of finite length, it is necessary and sufficient that every elementary divisor of N be an elementary divisor of M, and that its multiplicity in N be at most equal to its multiplicity in M. Hence obtain an example of a module M of finite length and a submodule N of M such that the rank of M is equal to the sum of the ranks of N and M/N, and such that N does not admit a complement in M (use a)).