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?

Question

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.

Answer 1

Hint: Let $M_i = e_i M$, where $e_i = (0,\dotsc,1,\dotsc,0) \in R$ with $1$ in the $i$th entry. Use that $e_i$ are pairwise orthogonal idempotents with $\sum_i e_i = 1$ to show $M = \oplus_i M_i$.

Answer 2

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.