Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi,

I have two commutative finite-dimensional $k$-Algebras $A$ and $B$ ($k$ is a field).

I wonder, whether there is a way to get the finite-dimensional indecomposable modules of $A \otimes_k B$,
if I know the finite-dimensional indecomposable modules of $A$ and $B$.

For example, is there a way to do this, if $A=k[x]/(x^n)$ and $B=k[y]/(y^n)$?

Thank you very much.

share|improve this question

2 Answers 2

up vote 1 down vote accepted

No, in general there isn't. In your particular case by accident for $n=1$ this is possible. But even for $n=2$, there are only $2$ indecomposable $A$-modules up to isomorphism. But Kronecker already clasified the $A\otimes_k B$-modules and there are infinitely many. For $n>2$ there are still only finitely many indecomposable $A$-modules (namely $n$) up to isomorphism, but the situation is in some sense worse. The indecomposable $A\otimes_k B$-modules are not even classifiable (one says $A\otimes B$ is wild).

share|improve this answer

There are a lot of indecomposable modules of $R=k[x]/x^n \otimes k[y]/y^n = k[x,y]/(x^n,y^n)$. If $k$ is infinite then there are infinitely many: $R/(x-ay)$ is a distinct indecomposable module for each $a\in k$. There is also $R$ mod any complicated but irreducible polynomial. I don't think you'll get a full classification that is nice in any way.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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