modules over decomposable rings

Let R1,R2 be two, nontrivial, unital rings and R=R1R2. If M1 is a R1-module and M2 is a R2-module, then obviously M1M2 is a R-module via (r,s)(m1,m2)=(rm1,sm2). We will show that every R-module can be obtain in this way.

Proposition. If M is a R-module, then there exist submodules M1,M2M such that M=M1M2 and for any rR1, sR2, m1M1 and m2M2 we have

(r,s)m1=(r,0)m1  (r,s)m2=(0,s)m2,

i.e. ring action on M1 (respectively M2) does not depend on R2 (respectively R1).

Proof. Let e=(1,0)R and f=(0,1)R. Of course both e,f are idempotentsPlanetmathPlanetmath and (1,1)=e+f. Moreover ef=fe=0 and e,f are central, i.e. e,f{cR|xRcx=xc}. We will use e,f to construct submodules M1,M2. More precisely, let M1=eM and M2=fM. Because e,f are central, then it is clear that both M1 and M2 are submodules. We will show that M1+M2=M. Indeed, let mM. Then we have


Thus M1+M2=M. Furthermore, assume that mM1M2. Then there exist m1,m2M such that


and therefore


Now, after multiplying both sides by e we obtain that


thus M1M2=0. This shows that M=M1M2. To finish the proof, we need to show that the ring action on M1 does not depend on R2 (the other case is analogous). But this is clear, since for any (r,s)R and mM we have


This completesPlanetmathPlanetmathPlanetmath the proof.

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