2
$\begingroup$

We consider finitely generated modules over an Artin algebra. Let $X$ be an indecomposable module such that the radical $\text{rad} \,X$ is a submodule of the socle $\text{soc}\,X$. What can we say about $X$? In particular, can we bound the length of $X$?

$\endgroup$

1 Answer 1

1
$\begingroup$

No. Let $(R,\mathfrak m)$ be commutative local Artin ring, then the radical of $M$ is $\mathfrak mM$ and your condition is equivalent to $\mathfrak m^2M=0$. One can not bound the length of such indecomposable modules in general. In particular, if you take $R=k[x,y]/(x,y)^2$ then it has $\mathfrak m^2=0$, so any module works, and there are indecomposable modules of arbitrary length over $R$, as it does not have finite representation type (plus Brauer-Thrall).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .