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I've been refreshing my linear algebra, and this is a question of curiosity I have.

Let $U:=U_n(F)$ be the algebra of upper triangular $n\times n$ matrices over a field $F$. Is there a classification of all simple $U$-modules (up to isomorphism of course)? I've been researching around, but didn't find any relevant results on first look. I'd also be happy for a reference showing such classification if one exists. Thank you.

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    @Changwei Since $U$ can be thought of as a ring, the phrase "simple $U$-module" refers to a simple module over the ring $U$. See http://en.wikipedia.org/wiki/Simple_module.2012-04-21

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The Jacobson radical $J(R)$ of a ring $R$ consists of all elements which act by zero in all simple left $R$-modules. It follows that the simple modules of $R$ are naturally identified with the simple modules of $R/J(R)$.

The Jacobson radical has an important alternate characterization as consisting of all elements $r$ such that $1 - xr$ is invertible for all $x$. Using this characterization, I claim that the Jacobson radical of $R = U_n(F)$ consists precisely of the strictly upper triangular matrices. The quotient $R/J(R)$ is isomorphic to $F^n$, where the isomorphism sends an upper triangular matrix to its diagonal, and the simple modules of $F^n$ are identified with the $n$ copies of $F$.

$U_n(F)$ happens to be a quiver algebra of a finite acyclic quiver, and a similar statement is true in this generality: the simple modules of such an algebra are naturally in bijection with the vertices of the corresponding quiver.

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    @QiaochuYuan Nice answer! Sorry I have a question. How can we find the projective covers of these simple modules?2018-05-15
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The answer you're not hoping to hear probably is: for any ring with 1, the (right) simple modules are characterized as quotients R/M where M is a maximal right ideal. So, if you can determine all the maximal right ideals, you have all the simple right modules, and similarly for the left side. I'm sure you can immediately find several, but I'll have to apologize I can't remember if there's a slick way to point out all the simple modules at once.

It may also be helpful to know that the radical is the set of all strictly upper triangular matrices (since all the maximal left/right ideals will have to contain these). I think selecting any diagonal entry and looking at the subset of matrices zero on that diagonal entry is a maximal ideal. I thought this might be a complete set, but then I read that not all simple modules embed into an upper triangular matrix ring, so there must be a few more hiding in there.

I know that's pretty lackluster, so I'll tell you everything else I know about this ring. It's an Artinian serial ring (this means it's a direct sum of right ideals, each of whom has a finite linearly ordered set of submodules). It's also hereditary, and is a ring of finite representation type.

I look forward to seeing a more confident and complete solution for this problem :)