1
$\begingroup$

I was asked to show the following: Let $A$ be a finite dimensional algebra. If $S$ is a finite dimensional semisimple $A$-module, $M$ is also a finite dimensional $A$-module and $nM\cong nS$ for some $n\in \mathbb{N}$ then $M\cong S$.

The hint suggests me to use Jordan-Holder series. I don't why it leads to $M\cong S$. I can only do the following: Let $S=\oplus S_i$. Since $nM\cong nS$, we know that the $S_i$ are all the composition factors of $M$. And I argue randomly like following: If $M\not\cong \oplus S_i$, then we have $[M]\ne0\in \mathrm{Ext}(\oplus S_{i-1},S_i)$, so $[nM]\ne0\in\mathrm{Ext}(n(\oplus S_{i-1}),nS_i)$ which leads to a contradiction. Is it true? Also I'm not allowed to use $\mathrm{Ext}$. How to prove that under this condition?

$\endgroup$
2
  • 1
    $\begingroup$ A problem with the Ext argument is that $[nM]\neq 0$ just tells you that a specific short exact sequence does not split. it doesn't tell you that $nM$ can't be isomorphic to a direct sum of the $nS_i$ in a different way. See for instance math.stackexchange.com/questions/135444/… $\endgroup$
    – Eric Wofsey
    Commented May 11, 2022 at 14:24
  • 1
    $\begingroup$ note that it is possible to prove this without any semisimplicity assumptions by using Krull-Schmidt. $\endgroup$
    – Lukas Heger
    Commented May 11, 2022 at 20:56

1 Answer 1

Reset to default
0
$\begingroup$

Just observe that since $S$ is semisimple, so is $nS\cong nM$, and thus $M$ is semisimple since a submodule of a semisimple module is semisimple. So, any Jordan-Holder series for $M$ automatically splits.

$\endgroup$

You must log in to answer this question.

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