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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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Could you explain what is a "simple U-module"? Sorry never heard of it before.. – Kerry Apr 21 '12 at 20:38
@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 – Jim Belk Apr 21 '12 at 21:37
up vote 6 down vote accepted

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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This is the path to an easier answer I knew I was forgetting :) Now we can see that $U/rad(U)$ has exactly $n$ maximal ideals, and that the ideals I had in mind below exhaust the possibilities. It should also be clarified that they are all nonisomorphic, despite looking like "copies of F". – rschwieb Apr 23 '12 at 18:55
Thank you Qiaochu. So the simple modules of $U$ can be identified with the simple modules of $F^n$, which are in turn identified with the $n$ copies of $F$. But then how could they be described, informally even, in $U$ itself? As in, "the simple modules of $U$ are those that look like this: _____"? – Danielle Intal Apr 25 '12 at 2:31
@Danielle: the $k^{th}$ simple module is one-dimensional, with each upper-triangular matrix acting by multiplication by its $k^{th}$ diagonal entry. (This can be deduced from what I said above about the quotient map $R \to R/J(R)$ sending an upper-triangular matrix to its diagonal.) – Qiaochu Yuan Apr 25 '12 at 2:33
This is clear to me now, thank you. – Danielle Intal Apr 25 '12 at 2:43

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 :)

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Thank you rschwieb. – Danielle Intal Apr 25 '12 at 2:32

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