If $R$ is a simple Artinian ring, then when is a finitely generated module free?

Question

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).

Book's solution:

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?

Answer 1

Is my answer to the first question on the right track? (Show that there is only one simple right R−module up to isomorphism)

You made a comment about a chain and a minimal ideal, but this isn't going anywhere. The question has been asked a few times before, and you can find an explanation here: https://math.stackexchange.com/a/1011301/29335

How do I answer the other three questions?

My advice would be to go read a bit about semisimple rings in a book like Isaac's Algebra: a graduate course or Jacobson's Basic Algebra II or Lam's First course in noncommutative rings. All of these questions are special cases of the theory of semisimple rings.

Every finitely generated right $R$−module $M$ is a direct sum of copies of $S$

Every (not just the finitely generated ones) $R$ module has to be a direct sum of copies of $S$. But the finitely generated ones are the ones that look like $S^k$ for a natural number $k$. I can't think of a better way to prove this than to learn that $R$ is a semisimple ring, and that the modules of semisimple rings are direct sums of simple submodules. If there is only one simple module up to isomorphism (as is the case for a simple semisimple ring) then naturally any module is a direct sum of copies of that module ($S$, in our case.) You'll be able to find this in any of the references I gave before. It's not hard, but it doesn't fit into a post either.

show that $k$ is uniquely determined by $M$

If $M\cong S^k\cong S^j$, then the endomorphism rings $End(S^k_R)\cong End(S^j_R)$ as rings. But also you can say that $End(S^n_R)\cong M_n(End(S_R))$. Now $End(S_R)$ is a division ring by Schur's lemma, and unless $k=j$, the two matrix rings have different dimension over that division ring. (There are many details to be filled in here: take your time.)

What is the condition on $k$ for $M$ to be free? (Why should the $r∣k$ imply that $S^k$ has a basis?)

Thinking in terms of bases is not nearly as big of a win as thinking of free modules as "sums of copies of $R$." Now, $R$ is certainly a finitely generated $R$ module, so $R\cong S^k$ for some $k$. Every finitely generated free module is a finite direct sum of copies of $R$. So... how many copies of $S$ does such a module have?