Let $D$ be a division algebra. I'm trying to show that all simple $M_n(D)$ modules are isomorphic to $D^n$. I know that $M_n(D)=D^n\oplus D_n \oplus \dots \oplus D^n$ (n terms).
I have that if $S$ is a simple, then $S\cong M_n(D)/M$ for some maximal ideal $M$. I think the logic goes that $M$ has to be $D^n\oplus D_n \oplus \dots \oplus D^n$ (with $n-1$ terms), and for some reason $M_n(D)/M\cong D^n$.
I'm not sure if this is correct, but if it is, I still don't understand why a maximal ideal of a matrix ring has to be $D^n\oplus D^n \oplus \dots \oplus D^n$ (with $n-1$ terms), or why the quotient would be $D^n$. Thanks for any replies!
Lemma 1: Show that $D^n$ is a simple right $M_n(D)$ module where you just multiply row vectors from $D^n$ on the right by matrices.
Lemma 2: If $R$ has a simple right ideal $S$, the sum of all right ideals isomorphic to $S$ is atwo-sided ideal of $R$. If two such ideals arise from two collections of nonisomorphic simple right ideals, then the two ideals have intersection zero. Consequently, a simple ring has exactly one isotype of simple right module.
Corollary: all simple right $R$ modules of a simple ring are mutually isomorphic, and so in particular for a matrix ring over a division ring, all simple right modules are isomorphic to $D^n$ with matrix multiplication on the right.