If $M$ is a $R$-module with $R=R_1\times \cdots \times R_n$ then $M\simeq M_1\times \cdots \times M_n$ where $M_i$ is a $R_i$-module? Let $R_1, \ldots, R_n$ be rings and consider the ring $R:=R_1\times \cdots\times R_n$. How can I show every $R$-module $M$ is isomorphic to a product $M_1\times \cdots\times M_n$ where each $M_i$ is a $R_i$-module?
Obs: By module I always mean left module.

It's easier with just two rings and then you can do induction.
So consider $R=R_1\times R_2$ and the two idempotents $e_1=(1,0)$, $e_2=(0,1)$. If $M$ is an $R$-module, you can consider $M_1=e_1M$ and $M_2=e_2M$.
For all $r\in R$, $e_1r=re_1$, and similarly for $e_2$, so $M_1$ and $M_2$ are $R$-submodules of $M$ and $M=M_1\oplus M_2$ is clear:
$$
x=1x=(e_1+e_2)x=e_1x+e_2x\in M_1+M_2
$$
Note that $M_1=\{x\in M:e_1x=x\}$ and similarly for $M_2$. Thus if $x\in M_1\cap M_2$, then $x=e_1x=e_1e_2x=0$.
That $M_1$ and $M_2$ are in a natural way modules over $R_1$ and $R_2$ is straightforward.
