Exercise 8, § 2

Let $A$ be a principal ideal domain, let $\pi$ be an irreducible element of $A$ and let $M$ be a $\pi$-primary module. Then an element $x$ of $M$ is said to have height $n$ if $x \in \pi^n M$ and $x \notin \pi^{n+1} M$. If $x \in \pi^n M$ for all $n \geq 1$ then $x$ is said to have infinite height.

u) Show that $M$ is divisible if and only if all its elements have infinite height (Ex. 3).

b) Let $x$ be an element of $M$ of height $n$, such that $\pi x = 0$; show that if $x = \pi^n y$ then the cyclic submodule $Ay$ of $M$ is a direct factor of $M$ (notice that $Ay$ is a pure submodule, and apply Ex. 7, d)).

c) Show that a nondivisible $n$-primary module $M$ always contains at least one cyclic nonzero direct factor (if $x \in M$ has finite height, and there exist integers $m$ such that $\pi^m x \neq 0$ has infinite height, then lets be the least such integer ; write $\pi^s x = \pi y$, where $y$ has height strictly greater than $\pi^{s-1} x$, and apply b) to $y - \pi^{s-1} x$). Deduce that any indecomposable $n$-primary module is either cyclic or isomorphic to a module $U_\tau$ (Ex. 3).

d) Show that if every pure submodule of the $n$-primary module $M$ is a direct factor in $M$, then $M$ is a direct sum of a divisible module $P$ and a $\pi$-primary module $Q$ with nonzero annihilator. (Reduce to the case where $M$ contains no nonzero divisible submodule, and argue by contradiction. Use b) to show that there would exist an infinite sequence $(x_i)$ of elements of height 1 in $M$ such that : 1) the annihilator of $x_k$ is $A \pi^{n_k}$, where the sequence $(n_i)$ of integers is strictly increasing ; and 2) the sum of the $Ax_k$ is direct, and the sum of any finite number of these submodules is a direct factor in $M$. Then show that the submodule $N$ of $M$ generated by the elements $x_i - \pi^{n_k-1} n_k x_{k-1}$, is pure, but not a direct factor of $M$, arguing as in Ex. 7, e)).