rings whose every module is free

Recall that if R is a (nontrivial) ring and M is a R-module, then (nonempty) subset SM is called linearly independentMathworldPlanetmath if for any m1,,mnM and any r1,,rnR the equality


implies that r1==rn=0. If SM is a linearly independent subset of generatorsPlanetmathPlanetmathPlanetmath of M, then S is called a basis of M. Of course not every module has a basis (it even doesn’t have to have linearly independent subsets). R-module is called free, if it has basis. In particular if R is a field, then it is well known that every R-module is free. What about the converseMathworldPlanetmath?

PropositionPlanetmathPlanetmath. Let R be a unital ring. Then R is a division ring if and only if every left R-module is free.

Proof. ,,” First assume that R is a divison ring. Then obviously R has only two (left) ideals, namely 0 and R (because every nontrivial ideal contains invertible element and thus it contains 1, so it contains every element of R). Let M be a R-module and mM such that m0. Then we have homomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of R-modules f:RM such that f(r)=rm. Note that ker(f)R (because f(1)0) and thus ker(f)=0 (because ker(f) is a left idealMathworldPlanetmathPlanetmath). It is clear that this implies that {m} is linearly independent subset of M. Now let

Λ={PM|P is linearly independent}.

Therefore we proved that Λ. Note that (Λ,) is a poset (where ,,” denotes the inclusion) in which every chain is bounded. Thus we may apply Zorn’s lemma. Let P0Λ be a maximal element in Λ. We will show that P0 is a basis (i.e. P0 generates M). Assume that mM is such that mP0. Then P0{m} is linearly dependent (because P0 is maximal) and thus there exist m1,,mnM and λ,λ1,,λnR such that λ0 and


Since λ0, then λ is invertiblePlanetmathPlanetmathPlanetmathPlanetmath in R (because R is a divison ring) and therefore


Thus P0 generates M, so every R-module is free. This completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath this implicationMathworldPlanetmath.

,,” Assume now that every left R-module is free. In particular every left R-module is projective, thus R is semisimplePlanetmathPlanetmathPlanetmathPlanetmath and therefore R is NoetherianPlanetmathPlanetmathPlanetmath. This implies that R has invariant basis number. Let IR be a nontrivial left ideal. Thus I is a R-module, so it is free and since all modules are projective (because they are free), then I is direct summandMathworldPlanetmath of R. If I is proper, then we have a decomposition of a R-module


but rank of R is 1 and rank of II is at least 2. ContradictionMathworldPlanetmathPlanetmath, because R has invariant basis number. Thus the only left ideals in R are 0 and R. Now let xR. Then Rx=R, so there exists βR such that


Thus every element is left invertible. But then every element is invertible. Indeed, if βx=1 then there exist αR such that αβ=1 and thus


so x is right invertible. Thus R is a divison ring.

Remark. Note that this proof can be dualized to the case of right modules and thus we obtained that a unital ring R is a divison ring if and only if every right R-module is free.

Title rings whose every module is free
Canonical name RingsWhoseEveryModuleIsFree
Date of creation 2013-03-22 18:50:20
Last modified on 2013-03-22 18:50:20
Owner joking (16130)
Last modified by joking (16130)
Numerical id 10
Author joking (16130)
Entry type Example
Classification msc 16D40