Consider $R=M_{n}(k)$ which $k$ is a field. We know $M=M_{n \times r}(k)$ is an $R$-Module.
When $M$ is Cyclic?
When $M$ is Free?
When $M$ is Simple?
My ideas:
For the first one, I think if $n=r$, then we can generate $M$ by identity element.
For the second one, If $M$ is Free, then it should be the direct sum of copies of $R$. In this case, $r$ should be equal to $n$. Otherwise, we have missed elements in some columns in our module.
For the third one, $M$ must be equal to $\{0 \}$, otherwise, there are many submodules, pick any matrices with the last zero columns.
But I am feeling that I have missed important things, which I know that I didn't use $k$ as a field. So, it will be great if you help me with that.
(1) True, $M$ is cyclic if $r=n$. What about if $r<n$? What about if $r>n$?
(2) If $M\cong R^m$ as an $R$-module then they have the same dimension as $k$-vector spaces. What does this require about $r$ with respect to $n$? What about the converse?
(3) True, but you're missing one important case: what if there is only one column, i.e. $r=1$? Can you show this is a simple $R$-module? Indeed, then $M\cong k^n$ as an $R$-module. Moreover,
$$ M\cong \underbrace{(k^n)\oplus\cdots\oplus(k^n)}_r $$
as $R$-modules for general $r$. Do you see how?