ModulesOverDecomposableRings 2009-02-20 05:44:39 2009-02-20 05:44:39 Theorem module % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here 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. \textbf{Proposition.} 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 $R_1$). \textit{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 $$(r,s)\cdot (e\cdot m)=\big( (r,s)(1,0)\big) \cdot m=(r,0)\cdot m=\big( (r,0)(1,0)\big)\cdot m=(r,0)\cdot(e\cdot m).$$ This completes the proof. $\square$