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 $\text{End}(M)$ is a division ring? Notice if $M$ is simple, then it happens and if it happens, then $M$ but be indecomposable. So this property is something inbetween simple and indecomposable, and it came up in the context of torsion pairs.

  • $\begingroup$ When $A$ is an arbitrary commutative (associative unital) ring, it's equivalent to being simple. Indeed, in this case, every length 1 submodule of $M$ is isomorphic to a quotient of $M$. $\endgroup$ – YCor Nov 26 '20 at 11:58
  • $\begingroup$ If $A$ is the algebra of upper triangular $2\times 2$ matrices and $M=k^2$, then $M$ has this property (although it's not simple). This is because the centralizer of $A$ in $M_2(k)$ is reduced to scalars, and this is precisely the endomorphism $k$-algebra of the $A$-module $M$. $\endgroup$ – YCor Nov 26 '20 at 12:01
  • $\begingroup$ Crosspost $\endgroup$ – rschwieb Dec 1 '20 at 15:36

Such modules are called bricks for finite dimensional algebras and there are in general very many of them.

Having a division ring as the endomorphism ring is equivalent to the condition that every non-zero endomorphism morphism is invertible.

For hereditary and tilted algebras they are quite interesting, see https://link.springer.com/article/10.1007/BF03323325 where the name brick might also appeared first.

A classification for given classes of such algebras is an interesting problem, see for example https://arxiv.org/pdf/1712.08311.pdf .

Theorem 1.4 in https://arxiv.org/pdf/1503.00285.pdf gives a characterisation when an algebra has only finitely many bricks.


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