I was reading about semisimple modules, and there's a fact relating to simple modules that I don't recall.
Suppose you have a ring $R=M_n(D)$, by which I mean the $n\times n$ ring of matrices with entries in a division ring $D$. Then actually, $R\cong M^n$ for some simple module $M$, and this $M$ is unique up to isomorphism.
I'm hoping to understand this before proceeding, but I don't see this isomorphism, nor why such a module is unique up to isomorphism. Can someone please explain? A kind thanks all.
Best Answer
Let $D$ be a division ring, $n\geq1$ and let $R=M_n(D)$.
Let $M=D^n$, the set of column vectors with $n$ coordinates in $D$. It is a left $D$-module in the obvious way, and the usual matrix multiplication gives us a map $R\times M\to M$ which turns $M$ into an $R$-module. It is in fact a simple $R$-module, as you can check easily using linear algebra.
There is moreover an isomorphism of left $R$-modules $R\to M^n$, mapping each element $a\in R$ into the column vector $(a_1,\dots,a_n)\in M^n$ whose elements are the columns of $a$ in order.
Suppose now that $N$ is another simple $R$-module such that $R\cong N^k$ for some $k$. Then $M^n\cong N^k$ and the Krull–Schmidt theorem (which applies, because both $M^n$ and $N^k$ are of both noetherian and artinian) tells us that $n=k$ and $M\cong N$.
All this should be explained in pretty much any book dealing with the subject!