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I'm trying to prove that if $M$ is a simple module over $M_n(D)$ where $D$ is a division algebra, then $M\cong D^n$. I know that if $M$ is a simple module over $R$ then it is isomorphic to $Rv=\{rv|r\in R\}$ for all $v\in M$. I'm trying to find a $M_n(D)$ module isomorphism $f:M_n(D)v\rightarrow D^n$. I've been told to try the map that maps $Xv\mapsto Xe_1$, where $e_1$ is the column vector with 1st entry 1 and the rest 0. I can prove that this is a homomorphism but I'm stuck on the injectivity. Any help would be well appreciated!

Lammey
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    There is a very quick way to prove this (which may not be very satisfactory to you). $M_{n}(D)$ and $D$ are Morita equivalent and the functor $F$ that establishes this equivalence takes a $D$-vector space $V$ to the $M_{n}(D)$-module $V^{n}$ (where $M_{n}(D)$ acts by matrix multiplication on column vectors.) Thus, since the only simple left-module over $D$ is $D$ itself, the only simple $M_{n}(D)$ module is $D^{n}$. – Siddharth Venkatesh May 06 '14 at 00:59

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The kernel of the map is itself a submodule of M. It must be trivial because M is simple.

Jake Koenig
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  • The domain of the map, $M$, is by definitino cyclic, but cyclic modules and simple modules are not the same thing. – Jo Mo Jun 25 '21 at 08:21