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  1. Let R be the ring of $n\times $n upper triangular matrices over a field Q ,and let $\wp$ be the class of simple left R-modules.Prove that: $(1) Tr_{R}(\wp)=\{[[a_{ij}]]\in R\mid a_{ij}=0 (i\geq2)\}.\\$ $(2) Rej_{R}(\wp)=\{[[a_{ij}]]\in R\mid a_{kk}=0 (k=1,...,n)\}.$

The definition :$Tr_{M}(\wp)=\Sigma\{Im h \mid h :U \longrightarrow M for some U\in\wp\},\\ Rej_{M}(\wp)=\bigcap\{Ker h|h:M \longrightarrow U for some U\in\wp\}.$

Now,I state my opinion.First,we should found all the simple left R-modules.$\\$ Clearly,we can found a class simple modules of R,that is$\\$ \begin{pmatrix} Q & 0 & ... & 0 \\ 0 & 0 & ... & 0\\ ... & ... & ... &...\\ 0 & 0 & ... & 0 \end{pmatrix}$\quad$ , \begin{pmatrix} 0 & Q & ... & 0 \\ 0 & 0 & ... & 0\\ ... & ... & ... &...\\ 0 & 0 & ... & 0 \end{pmatrix}$\quad$ ... \begin{pmatrix} 0 & 0 & ... & Q \\ 0 & 0 & ... & 0\\ ... & ... & ... &...\\ 0 & 0 & ... & 0 \end{pmatrix}$\\$ But these simple modules is not all. For example ,for a $3\times$3 upper triangular matrices,besides them, $\\$we also can find other simple modules,$\\$ \begin{pmatrix} 0 & Q & 0\\ 0 & Q & 0\\ 0 & 0 & 0 \end{pmatrix} is a left R-module,we denote it by $M_{1}\\$ \begin{pmatrix} 0 & Q & 0 \\ 0 & 0 & 0\\ 0 &0 & 0 \end{pmatrix} is a maximal submodule of $M_{1}$,we denote it by $M_{2}$,$\\$ clearly $M_{1}/M_{2}$is also a simple module. So ,I can't found all simple left R-modules. I want some one can help me solve the problem,or give me some ideas.

guojm
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  • I suspect you are trying to find all minimal ideals of $R$, rather than all simple $R$-modules. For instance, in the first bunch of examples you give, those are different left $R$-ideals, but they are isomorphic as left $R$-modules. – Angina Seng Apr 16 '17 at 08:30
  • They are also the simple modules,do you think so ? – guojm Apr 16 '17 at 08:58
  • @guojm You can easily see what the simple modules look like, and an explanation why the upper triangular ring only contains one isoclass of $n$ different isoclasses of simple modules here All that has been answered before, but I guess we leave this question open to deal with the original question at the top. – rschwieb Apr 16 '17 at 20:16

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