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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.

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up vote 2 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).

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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.

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