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Let $k$ be a field, $A$ a $k$-algebra of finite length and $M$ an $A$-module of finite length. When does it happen, that all endomorphisms of $M$ are either an isomorphism or zero? It is easy to see that if $M$ is simple, then it happens and if it happens, then $M$ must be indecomposable.

kevkev1695
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  • You're looking for a condition on the indecomposable module $M$ for such an $A$ so that $End(M_A)$ is a division ring, right? Can't say I've seen the solution, but maybe it is possible for a finite dimensional algebra.. – rschwieb Nov 25 '20 at 13:49
  • Yes that it correct. It came up in the context of torsion pairs. – kevkev1695 Nov 25 '20 at 14:30
  • Crossposted. kevkev, are you aware it is in general bad form to crosspost without special circumstances? – rschwieb Dec 01 '20 at 15:35
  • Should I close the original post before "crossposting"? Or what is the norm? – kevkev1695 Dec 01 '20 at 20:42
  • Usually it's permissible to crosspost if you've had a question open a long time without reasonable answers (say about two weeks to a month.) If and when you crosspost, leave a note that you're doing so on both questions linking them together. That way whoever reads can choose which site to answer on, and not duplicate work. You can delete if there are no answers, but definitely don't delete if someone already put work into writing an answer for you. – rschwieb Dec 01 '20 at 22:16

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Modules whose endomorphism algebra is a division ring are called bricks. As you mentioned, simple modules are bricks, but the converse is far from true. They are related to torsion pairs and $\tau$-tilting theory. While I will not attempt to outline the theory here, I can perhaps point to this paper by Demonet, Iyama and Jasso and this other paper by Asai on semibricks.

Pierre-Guy Plamondon
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