example of a projective module which is not free

Let R1 and R2 be two nontrivial, unital rings and let R=R1R2. Furthermore let πi:RRi be a projectionPlanetmathPlanetmath for i=1,2. Note that in this case both R1 and R2 are (left) modules over R via


where on the right side we have the multiplication in a ring Ri.

PropositionPlanetmathPlanetmath. Both R1 and R2 are projective R-modules, but neither R1 nor R2 is free.

Proof. Obviously R1R2 is isomorphic (as a R-modules) with R thus both R1 and R2 are projective as a direct summands of a free moduleMathworldPlanetmathPlanetmath.

Assume now that R1 is free, i.e. there exists ={ei}iIR1 which is a basis. Take any i0I. Both R1 and R2 are nontrivial and thus 10 in both R1 and R2. Therefore (1,0)(1,1) in R, but


This situation is impossible in free modules (linear combinationMathworldPlanetmath is uniquely determined by scalars). ContradictionMathworldPlanetmathPlanetmath. Analogously we prove that R2 is not free.

Title example of a projective moduleMathworldPlanetmath which is not free
Canonical name ExampleOfAProjectiveModuleWhichIsNotFree
Date of creation 2013-03-22 18:49:55
Last modified on 2013-03-22 18:49:55
Owner joking (16130)
Last modified by joking (16130)
Numerical id 6
Author joking (16130)
Entry type Example
Classification msc 16D40