I have been reading a book of module theory. So it says that if $M$ is a module over a non-commutative ring,then it may have two bases with different number of elements. To prove this, it says that let $V$ be a vector space over a field $K$ with countable infinite basis $\{e_n \mid n\geq 1\}$.
So we will have a bijection between the sets $\{e_n | n \in \Bbb N\}$ and $\{(e_n,0)|n \in \Bbb N\} \cup \{(0,e_n)|n \in \Bbb N\}$. The latter will form a basis for $V \oplus V$.
So this bijection can be extended to a $K$-module isomorphism say $$f: V \rightarrow V \oplus V$$ This isomorphism gives a $K$-module isomorphism say $$f': \operatorname{Hom}_K(V,V) \rightarrow \operatorname{Hom}_K(V,V \oplus V)$$(Note that here $\operatorname{Hom}_K(V,V)$ denotes the set of all $K$-homomorphisms from $V$ to $V$). This isomorphism $f'$ is defined as $f'(g)=gf$ for all $g \in \operatorname{Hom}_K(V,V)$ (Verify that this indeed a $K$-module isomorphism)
Also we have a $K$-module isomorphism say $$h: \operatorname{Hom}_K(V \oplus V) \rightarrow \operatorname{Hom}_K(V,V) \oplus \operatorname{Hom}_K(V,V)$$ which is given by $h(s)=(sp_1,sp_2)$ for all $s \in \operatorname{Hom}_K(V \oplus V)$ and here $p_1$ and $p_2$ are projection maps from $V \oplus V$ to $V$ on first and second components respectively.(Why is this an isomorphism?)
Now the composition $i=f'h$ is a $K$-module isomorphism $\operatorname{Hom}_K(V,V) \rightarrow \operatorname{Hom}_K(V,V) \oplus \operatorname{Hom}_K(V,V)$. Now we denote $\operatorname{Hom}_K(V,V)$ as $R$ and we have that $R$ is isomorphic to $R \oplus R$ (preseves addition and scalar multiplication). We can show that $R$ is a ring with the identity linear transformation as its unity.
Main problem of mine is this: Now it says that clearly $R$ is not commutative(Why?). Hence $R$ and $R \oplus R$ can be considered as modules of $R$(Why?) In the next step,we prove that $i$ is $R$-linear(Why?). Hence $R$ and $R \oplus R$ are isomorphic as $R$-modules. Now since $R$ is free with basis $\{1\}$ , $R \oplus R$ has basis $i(1)$ (How?) ie. It has singleton basis. Also $\{(1,0),(0,1)\}$ is an $R$-basis for $R \oplus R$.
So we conclude that $R \oplus R$ has bases with distinct number of elements. Please explain the proof and also what's the use of non-commutativity here?