# modules over decomposable rings

Let $R_{1},R_{2}$ be two, nontrivial, unital rings and $R=R_{1}\oplus R_{2}$. If $M_{1}$ is a $R_{1}$-module and $M_{2}$ is a $R_{2}$-module, then obviously $M_{1}\oplus M_{2}$ is a $R$-module via $(r,s)\cdot(m_{1},m_{2})=(r\cdot m_{1},s\cdot m_{2})$. We will show that every $R$-module can be obtain in this way.

If $M$ is a $R$-module, then there exist submodules $M_{1},M_{2}\subseteq M$ such that $M=M_{1}\oplus M_{2}$ and for any $r\in R_{1}$, $s\in R_{2}$, $m_{1}\in M_{1}$ and $m_{2}\in M_{2}$ we have

 $(r,s)\cdot m_{1}=(r,0)\cdot m_{1}\ \ \ \ (r,s)\cdot m_{2}=(0,s)\cdot m_{2},$

i.e. ring action on $M_{1}$ (respectively $M_{2}$) does not depend on $R_{2}$ (respectively $R_{1}$).

Proof. Let $e=(1,0)\in R$ and $f=(0,1)\in R$. Of course both $e,f$ are idempotents  and $(1,1)=e+f$. Moreover $ef=fe=0$ and $e,f$ are central, i.e. $e,f\in\{c\in R\ \big{|}\ \forall_{x\in R}\ cx=xc\}$. We will use $e,f$ to construct submodules $M_{1},M_{2}$. More precisely, let $M_{1}=eM$ and $M_{2}=fM$. Because $e,f$ are central, then it is clear that both $M_{1}$ and $M_{2}$ are submodules. We will show that $M_{1}+M_{2}=M$. Indeed, let $m\in M$. Then we have

 $m=(1,1)\cdot m=(e+f)\cdot m=e\cdot m+f\cdot m.$

Thus $M_{1}+M_{2}=M$. Furthermore, assume that $m\in M_{1}\cap M_{2}$. Then there exist $m_{1},m_{2}\in M$ such that

 $e\cdot m_{1}=m=f\cdot m_{2}$

and therefore

 $e\cdot m_{1}-f\cdot m_{2}=0.$

Now, after multiplying both sides by $e$ we obtain that

 $0=(ee)\cdot m_{1}-(ef)\cdot m_{2}=e\cdot m_{1}-0\cdot m_{2}=e\cdot m_{1}=m,$

thus $M_{1}\cap M_{2}=0$. This shows that $M=M_{1}\oplus M_{2}$. To finish the proof, we need to show that the ring action on $M_{1}$ does not depend on $R_{2}$ (the other case is analogous). But this is clear, since for any $(r,s)\in R$ and $m\in M$ we have

 $(r,s)\cdot(e\cdot m)=\big{(}(r,s)(1,0)\big{)}\cdot m=(r,0)\cdot m=\big{(}(r,0)% (1,0)\big{)}\cdot m=(r,0)\cdot(e\cdot m).$
Title modules over decomposable rings ModulesOverDecomposableRings 2013-03-22 18:50:05 2013-03-22 18:50:05 joking (16130) joking (16130) 4 joking (16130) Theorem msc 20-00 msc 16-00 msc 13-00