2
$\begingroup$

I'm reading the following paper: http://math0.bnu.edu.cn/~huwei/paper/Holm-Hu-1.pdf

On page 795 and 796 there are the definitions (in a diagrammatical way) of some $A_n$ modules, whereupon $A_n:=k[x,y]/\langle x^2,xy^{n+1},y^{n+2}\rangle$.

All the $A_n$ modules defined there should be indecomposable.

Question: Is there an easy way to prove this?

Of course, one could show that the endomorphism ring is local, but that seems to end in a lot of extensive computations.

Can we use anyhow that $A_n$ is local and commutative, or the fact that $x$ and $y$ are nilpotent?

Thanks for the help.

$\endgroup$
1
  • $\begingroup$ I think the best approach is staring at the module and convincing yourself it can't be written non-trivially as a direct sum. $\endgroup$ Dec 16, 2012 at 20:59

1 Answer 1

3
$\begingroup$

Hello.

I believe that the endmorphism ring may not be hard to compute in this case.

Let $R = k[x,y]$ where $k$ is a field, and let $I = (x^2, xy^n, y^{n-1})$. Then $End_R(R/I) = Hom_R(R/I,R/I) \cong Hom_{R/I} (R/I,R/I) \cong R/I$. But $R/I$ is a local ring since $\sqrt{I} = (x,y)$ which is a maximal ideal in $R$.

You may also want to take a look at Section 2 of http://arxiv.org/abs/1212.2912 by Yi Zhang. It discusses indecomposibility of a module in terms of a presentation matrix of the module.

$\endgroup$

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.