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.


2 Answers 2


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 $$R_1\times R_2\stackrel {\cong} {\to}R:(re,s(1-e))\mapsto re+s(1-e)$$

  • $\begingroup$ I wrote this answer because I wanted to make the point that everything above is literally true: there is no abuse of language or identification or isomorphism in a). For some reason, this matters for me. $\endgroup$ Feb 21, 2013 at 20:05

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 $e(1-e)=0$.

  • $\begingroup$ I got part b, after realising that direct sum and product is the same thing for finitely many elements. But for part a) let 1=r_1 +r_2. Then m=r_1*m + r_2*m, which is the decomposition. But how to show that M_1 and M_2 are disjoint. $\endgroup$
    – user12345
    Feb 21, 2013 at 19:24
  • $\begingroup$ Use $R_1 \cdot R_2 = 0$ in $R$. $\endgroup$ Feb 21, 2013 at 19:37

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