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?
  • $\begingroup$ I guess one could introduce the idea of resolutions and show how hereditary algebras have dimension at most $1$ (this is fairly trivial from the definition). If they know Maschke's theorem, one can be a bit more refined and show that in the modular case when the trivial module has infinite projective dimension, so there one has a wide example of non-hereditary algebras which one can "prove" are indeed non-hereditary. $\endgroup$ – Pedro Tamaroff Dec 21 '15 at 2:23
  • $\begingroup$ The example mentioned here is an answer to the linked question math.stackexchange.com/questions/2153476/… . Have you been able to find a simpler example since the time this question was asked? $\endgroup$ – user3281410 Feb 24 '17 at 11:04

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 $k$.)

  • 1
    $\begingroup$ Dear Jeremy: This was the first thing that occurred to me also, but at the time I didn't realize why its modules would have that property. Then I remembered that artinian rings whose ideals are linearly ordered have this property. That fact is somewhat nontrivial and warrants mention, don't you think? Regards $\endgroup$ – rschwieb Dec 22 '15 at 2:44
  • $\begingroup$ Actually I think this follows from the fact that $A = k[X]/(X^2)$ is a quotient of the PID $k[X]$. Any free module over $A$ is a module over $k[X]$, and the structure theorem for PIDs plus some basic algebra is enough to prove the result. The same should work for any quotient of $k[X]$. $\endgroup$ – Pablo Zadunaisky Dec 24 '15 at 18:11
  • $\begingroup$ @PabloZadunaisky I see, there may be an elementary way in the special case of quotients of PIDs, but it still seems worth elaborating about. $\endgroup$ – rschwieb Dec 28 '15 at 4:13
  • $\begingroup$ @PabloZadunaisky The structure theorem for PIDs will only help for finitely generated modules. And even for finitely generated modules, I think the classification of $k[x]/(x^2)$-modules is more elementary (but probably less well-known) than the classification of $k[x]$-modules. $\endgroup$ – Jeremy Rickard Jan 6 '16 at 22:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.