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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?
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  • $\begingroup$ 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
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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 $k$.)

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    $\begingroup$ 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

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