# 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.

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

$$(r,s)\cdot {m}_{1}=(r,0)\cdot {m}_{1}\mathit{\hspace{1em}\hspace{1em}}(r,s)\cdot {m}_{2}=(0,s)\cdot {m}_{2},$$ |

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. $e,f\in \{c\in R|{\forall}_{x\in R}cx=xc\}$. 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

$$(r,s)\cdot (e\cdot m)=\left((r,s)(1,0)\right)\cdot m=(r,0)\cdot m=\left((r,0)(1,0)\right)\cdot m=(r,0)\cdot (e\cdot m).$$ |

This completes^{} the proof. $\mathrm{\square}$

Title | modules over decomposable rings |
---|---|

Canonical name | ModulesOverDecomposableRings |

Date of creation | 2013-03-22 18:50:05 |

Last modified on | 2013-03-22 18:50:05 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 20-00 |

Classification | msc 16-00 |

Classification | msc 13-00 |