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I am trying to solve part (c) of the following Representation Theory question:

Let $D$ be a division ring and let $n$ be a positive integer. For $ 1 \leq l \leq n $ let $$C_l= \{A = (a_{ij}) \in {M_n}{(D)} : a_{ij} = 0 \text{ for } j \neq l \} .$$

(a) Show that each $C_l$ is a simple submodule of ${M_n}{(D)}$.

(b) Deduce that ${M_n}{(D)}$ is a semisimple ring.

(c) Also deduce that all simple modules of ${M_n}{(D)}$ are isomorphic to $D^n$.

Now, I have already managed to show (a) and (b), but I am struggling with (c) and specifically with defining a homomorphism. Any help with (c) would be a great benefit, thanks.

The Lost Unicorn
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  • Are you familiar with the Artin-Wedderburn theorem? – anomaly Feb 05 '16 at 16:05
  • Please make use of the search feature and the related questions that pop up while you ask a question. http://math.stackexchange.com/a/874720/29335 and http://math.stackexchange.com/q/1060896/29335 will also be helpful (they are related to the dupe.) – rschwieb Feb 05 '16 at 16:06
  • Strategy in a nutshell: every simple right module embeds as a minimal right ideal, and then the product of two nonisomorphic minimal right ideals has to be zero. This cannot occur in a simple ring, since if $B\neq {0}$, $AB={0}$ implies that $A={0}$. – rschwieb Feb 05 '16 at 16:08
  • Obviously any of the $C_l$ are isomorphic to $D^n$. – rschwieb Feb 05 '16 at 16:15
  • @anomaly Yes we have covered Artin-Wedderburn in lectures. – The Lost Unicorn Feb 06 '16 at 13:56
  • @rschwieb thanks again for commenting. I have showed that $D^n$ and $C_l$ are isomorphic, but not all simple modules of $M_n {(D)}$ are necessarily of the form $C_l$. I also don't see how this is a duplicate, as (c) specifically asks to deduce from (a), (b) that that all simple modules of $M_n {(D)}$ are isomorphic to $D^n$, so this confuses me a little bit. Thank you for your help. :) – The Lost Unicorn Feb 06 '16 at 14:06
  • The solutions I linked to do prove c) from b), and you're right that not all minimal ideals are in the collection of $C_l$, but that is irrelevant. I just meant to point out how one minimal right ideal is isomorphic to $D^n$ and then the solutions demonstrate there can't be other types. – rschwieb Feb 06 '16 at 14:39
  • Check also this topic – Ender Wiggins Mar 24 '20 at 13:48

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