Decomposition of rings gives decomposition of modules Let $R_1$, $R_2$ be rings containing unity (not necessarily commutative), and let $R=R_1\times R_2$. How to show that 
a) Every left $R$-module $M$ decomposes as $M\cong M_1\oplus M_2$, where each $M_i$ is an $R_i$-module, which is also viewed as an $R$-module via the natural projection map from $R$ to $R_i$.
b) If $e$ is in $R$ and $e$ is central idempotent, then $R\cong R_1 \times R_2$ where $R_1$ = $Re$ and $R_2=R(1-e)$.
Central element means, it commutes with all the elements of $R$.
I have noted that $Re$ and $R(1-e)$ are rings with unit $e$ and $1-e$ respectively.
 A: a) The  $R$-module  $M$ is the internal direct sum  $M=M_1\oplus M_2$ (equality, not only isomorphism!) of its $R$-submodules $$M_1=(1,0)M\; \text {and}\; M_2=(0,1)M$$  The $R$-module $M_1$ becomes  an $R_1$-module through the recipe $$r_1\cdot (1,0)m=(r_1,0)m$$ and similarly for $M_2$.
By "restricting" these new scalars" $R_1$ for $M_1$  to  $R$ through the first projection $pr_1:R=R_1\times R_2\to R_1$ you get the original structure of  of $M_1$ as an $R$-module, more precisely as an $R$-submodule  of $M$.
["Restriction of scalars" is a correct but  weird terminology, since $R$ is "bigger" than $R_1$.]
This construction is completely trivial but nevertheless rather subtle.  
b) This is straightforward: the required isomorphism is $$R\stackrel {\cong} {\to} R_1\times R_2: r\mapsto (re,r(1-e))$$with inverse $$R_1\times R_2\stackrel {\cong} {\to}R:(re,s(1-e))\mapsto  re+s(1-e)$$
A: Hint for 1: Look at $M_i=MR_i$.
Hint for 2: Verify that $R_1+R_2=R$ and that $R_1\cap R_2=0$. ( Where are you stuck? ) Everything to verify this should flow from the facts that $(1-e) + e=1$ and $e(1-e)=0$.
