8
$\begingroup$

Let $K$ be a field and let $A$ be a $K$-algebra which is finite dimensional as $K$-vector space. Then the nice structure theorem for artinian rings says that we can write $A$ as the direct product of local $K$-algebras. Is there also a structure theorem for finitely generated modules over such $A$?

If this is too vague, my naive hope would be that, similar to the case of principal ideal domains, we can write every finitely generated $A$-module $M$ as $$\oplus_{k=1}^n I_k/J_k$$ where $J_k\subseteq I_k$ are ideals of $A$ and $I_k/J_k$ should denote the ideal in $A/J_k$ given by all $x+J_k$ for $x\in I_k$.

$\endgroup$
3
  • 1
    $\begingroup$ There is no such structure theorem for Artinian rings unless you're assuming that $A$ is commutative. And no, there's no such classification, even for $K[x,y]/(x^3,y^2)$, which already has wild representation type. $\endgroup$ Nov 29, 2023 at 12:49
  • 1
    $\begingroup$ The title is a bit confusing to me. You're interested in finitely generated modules over (commutative) artinian $K$-algebras, not specifically in artinian modules over some given algebras (although f.g. modules over an artinian ring is artinian). (For commutative rings of positive Krull dimension there are infinitely generated artinian modules and a good theory for artinian modules in general, described by Matlis duality. Your title suggest you're asking about this, but the question is something else.) $\endgroup$
    – YCor
    Nov 29, 2023 at 14:06
  • $\begingroup$ I am sorry if this was confusing. I thought that these things were kind of equivalent because you can view the finitely generated module over $A$ also as a $K[x_1,\ldots,x_n]$ where $x_1,\ldots,x_n$ are mapped to generators of $A$. $\endgroup$
    – Hans
    Dec 8, 2023 at 8:48

2 Answers 2

6
$\begingroup$

No, in general local finite dimensional commutative $K$-algebras are wild and their modules can not be classified. For example when $m$ is the maximal ideal and $m/m^2$ has $K$-dimension at least 3, then the algebra is wild and there is no classification of indecomposable $A$-modules. On the other hand, one can classify when local algebras are tame in which case one can classify the modules, I think this was done by Ringel in an article about local tame algebras.

See also the answer of Mariano Suárez-Álvarez in What are tame and wild hereditary algebras? for more about tame/wild algebras.

$\endgroup$
3
$\begingroup$

We shouldn't expect a general classification theorem for modules. Take $A=K[x,y]/(x^2,y^2)$. If $K$ is characteristic 2 this is the group algebra for the Klein 4 group.

There are infinitely many indecomposable modules (no nontrivial direct summands) of arbitrarily large dimension over $K$, the classification is ad-hoc, and that there is such a classification makes it tame representation type. Contrast with other local artinian rings which can be wild representation type, (for example adjoin another nilpotent to $A$ above) meaning nobody can/has ever/should be expected to reasonably classify all the finitely generated indecomposable modules. The way an algebra is proven to be wild is to relate the problem of classifying modules to the classification of $k\langle x, y\rangle $ modules, i.e. pairs of matrices up to simultaneous equivalence.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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