TITLE: Is Kaplansky's theorem for hereditary rings a characterization?
QUESTION [10 upvotes]: This question came up during a first course on rings and modules I TA'd at.
Kaplansky's Theorem for hereditary rings states that

If $A$ is a hereditary ring, and $F$ is a free left $A$-module, then every submodule $M \subset F$ is isomorphic to a direct sum $\bigoplus_{i \in I} J_i$, where every $J_i$ is a left ideal of $A$.

See for example Lam's Lectures on modules and rings, (2.24). Recently a student asked me for an example of a submodule of a free module that was not a direct sum of ideals, and the best I could come up with was the following: Let $A = \mathbb Z_4[X]/(X^2)$, and let $M = \langle (2,X) \rangle \subset A^2$; then $M$ is not isomorphic to a direct sum of ideals. My proof is long and tedious, and besides $A$ is very very far from being hereditary, since it has infinite global dimension. Hence the question:

Can we find a simpler example of a submodule of a free module that is not isomorphic to a direct sum of ideals (say, over $\mathbb Z[X]$)?

In fact, I was wondering if there are examples for all non-hereditary rings. 

is the converse of Kaplansky's theorem true? if a ring $A$ is such that all submodules of a free module are isomorphic to a direct sum of ideals, does it follow that $A$ is hereditary?

REPLY [4 votes]: To answer the second question, no, it is not a characterization.
For example, let $k$ be a field and let $A=k[x]/(x^2)$. Then $A$ is not hereditary, but every $A$-module is a direct sum of copies of $A$ and of $k=A/(x)\cong Ax$, both of which are ideals.
(As mentioned by rschwieb in comments, my claimed classification of $A$-modules follows from more general results. But there's a fairly simple direct proof. Let $M$ be an $A$-module. Choose a basis $\{n_i\}$ of $Mx$ together with a choice of elements $\{m_i\}$ such that $n_i=m_ix$. Now extend $\{n_i\}$ to a basis $\{n_i\}\cup\{k_j\}$ of the kernel of multiplication by $x$. Then $\{m_i\}\cup\{n_i\}\cup\{k_j\}$ is a basis of $M$, for each $i$ the elements $m_i$ and $n_i$ span a submodule isomorphic to $A$, and for each $j$ the element $k_j$ spans a submodule isomorphic to $k$.)