I'm trying to reach the solution of 1 problem from the book "Modules and Rings: A translation of Moduln und Ringe" of F. Kasch. The problem is:
"Let $_RM\ne 0$ (left $R$-module) such that $M\cong M\oplus M$ and be $S=End_R(M)$. Prove that $_SS$ are free (as S-module) and has a basis of lenght $n$ for every natural number $n$".
I've seen in other post that the hypotesis implies $S\cong S\oplus S$ and from there i tried:
"As $1_M$ (the identity map) is such that for every $f\in S:f(m)=(1_M)f(m)$, $B_1=\{1_M\}$ is a basis of lenght 1. Then if we take $S\cong S^2$ so we have that $B_2=\{(1_M,0),(0,1_M)\}$ is again a basis for $S$ of lenght 2 and so on". But sincerely I'm not convinced that this is the correct way.
Dummit and Foote in "Abstract Algebra" give a similar excersice using that $M=\bigoplus\limits_{i\in \Bbb{N}}{\Bbb{Z}}$ and given $S=End_R(M)$ then they prove that $S$ are free, $S\cong S\oplus S$ and for every $n\in\Bbb{N}:S\cong S^n$. In this case that is finite I'm really stuck. Please help!.
If $M, N$ are $R$-modules such that $M \cong N$, it follows that for any $R$-module $P$, $M \oplus P \cong N \oplus P$.
From that, we get that $S \cong S \oplus S \implies S \oplus S \cong S \oplus S \oplus S$. Therefore, $S \cong S \oplus S \oplus S$.
With that in mind, an inductive argument shows that $S \cong S^n$, for all $n \in \mathbb{N}$.
Now, let $\Phi : S^n \rightarrow S$ be an isomorphism. Given a basis for $S^n$, such as $\{(1_M, 0,...,0), (0,1_M,...,0), ..., (0,0,...,1_M)\}$, we can get an $n$-element basis for $S$ by considering its image under $\Phi$.