Here's an exercise from my book, which only gives a brief solution which leaves me very confused.
Let $R$ be a simple Artinian ring, say $R=K_r$. Show that there is only one simple right $R$-module up to isomorphism, $S$ say, and that every finitely generated right $R$-module $M$ is a direct sum of copies of $S$. If $M\cong S^k$ say, show that $k$ is uniquely determined by $M$. What is the condition on $k$ for $M$ to be free?
$K_r$ denotes the $r\times r$ matrices over the field $K$ (I imagine).
Each row is a simple right ideal and all are isomorphic. $S^k$ is free iff $r\mid k$.
$R$ being a simple Artinian ring means it has no proper submodules, and it has an ascending chain of ideals which eventually breaks off. An ideal is also a $R-$submodule, so if we take the minimal ideal in a chain of the Artinian ring then it can be viewed as a simple $R$-module. A finitely generated module has a generating set (kind of like a basis, but without the condition of linear independence).
Is my answer to the first question on the right track? How do I answer the other three questions?
Why should the $r\mid k$ imply that $S^k$ has a basis?