Let $R = M_n(k)$, where $k$ is a field. Then any $R$-module that is finite dimensional over $K$ is a direct sum of isomorphic copies of $V$, where $V = k^n$.

I was able to show that $R$ has a unique simple module $V$, but I'm stuck on the rest of the exercise.

Any help is appreciated.

  • $\begingroup$ Uh, what's the exercise? $\endgroup$ – user98602 Dec 14 '15 at 3:02
  • $\begingroup$ I have edited for clarity. $\endgroup$ – 3-in-441 Dec 14 '15 at 3:25

The full matrix ring over a field is a semisimple ring, so all of its modules are direct sums of simple submodules.

Since there is exactly only one isotype of simple module, the submodules in those decompositions must be copies of the one simple module. It does not even have anything to do with finite generation.

  • $\begingroup$ That's what I thought, but how does one show that the matrix ring is semi-simple? $\endgroup$ – 3-in-441 Dec 14 '15 at 4:01
  • $\begingroup$ The Artin-Wedderburn theorem characterizes semisimple rings as finite products of matrix rings over division rings. A single matrix ring over a field is just a very special case. $\endgroup$ – rschwieb Dec 14 '15 at 4:08

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