Theorem 1.19 of Representation Theory of Finite Groups: Algebra and Arithmetic is the Krull-Schmidt theorem, which I screenshotted and uploaded it here, I don't have any problem with this theorem and the proof is clear to me.
But I have some questions about the following remark, which follows the theorem,
I know why $R$ is free, but I wonder why $M$ is not free? some information is given but I am not convinced and I cannot see the details.
Again I know why $R$ is indecomposable as a left $R$- module, but I wonder why $M$ is also indecomposable as left $R$- module?
And finally the final isomorphism is not clear to me, can anyone please give me some details?
Thanks in advance for any help!
The hint they give isn't aimed at showing that $M$ isn't free, rather that it isn't free of rank one. The point is that if $m \in M$ then the submodule $Rm$ cannot be all of $M$: there is some $x \in [0,1]$ with $m(x)=0$ by the intermediate value theorem, thus every element of $Rm$ vanishes at $x$, and it follows $Rm \neq M$ as $M$ contains functions not vanishing at $x$.
It is true that $M$ is not free. We've already shown it is not free of rank one, and we show it is not free of any higher rank by proving any two elements are not $R$-linearly independent. Let $m,n \in M$ be nonzero, let $0\neq g \in R$ such that $g(0)=g(1)=0$ and $gm,gn \neq 0$, and note $gm,gn \in R$. Then $(gm)n = (gn)m$ which shows $m,n$ are not $R$-linearly independent.
I think a similar argument will show that $M$ is indecomposable by showing any two nonzero submodules have a nonzero element in common.
To understand the definition of the isomorphism $\varphi$, identify $R\oplus R$ with the continuous functions $a: [0,1]\to \mathbb{R}^2$ such that $a(0)=a(1)$. The pair $(r,s) \in R\oplus R$ corresponds to the function $x \mapsto (r(x),s(x))$ under this identification. Then if $f,g \in M$, $\varphi(f,g)$ is the function $[0,1] \to \mathbb{R}^2$ sending $x$ to the point of $\mathbb{R}^2$ obtained by rotating $(f(x),g(x))$ by $\pi x$ about the origin. This is continuous as a composition of continuous functions, and it satisfies the endpoint condition: $0$ gets sent to $(f(0),g(0))$, and $1$ to $(f(1),g(1))= (-f(0),-g(0))$ rotated by $\pi$, which is $(f(0),g(0))$.