rings whose every module is free
Recall that if is a (nontrivial) ring and is a -module, then (nonempty) subset is called linearly independent if for any and any the equality
implies that . If is a linearly independent subset of generators of , then is called a basis of . Of course not every module has a basis (it even doesn’t have to have linearly independent subsets). -module is called free, if it has basis. In particular if is a field, then it is well known that every -module is free. What about the converse?
Proof. ,,” First assume that is a divison ring. Then obviously has only two (left) ideals, namely and (because every nontrivial ideal contains invertible element and thus it contains , so it contains every element of ). Let be a -module and such that . Then we have homomorphism of -modules such that . Note that (because ) and thus (because is a left ideal). It is clear that this implies that is linearly independent subset of . Now let
Therefore we proved that . Note that is a poset (where ,,” denotes the inclusion) in which every chain is bounded. Thus we may apply Zorn’s lemma. Let be a maximal element in . We will show that is a basis (i.e. generates ). Assume that is such that . Then is linearly dependent (because is maximal) and thus there exist and such that and
Since , then is invertible in (because is a divison ring) and therefore
,,” Assume now that every left -module is free. In particular every left -module is projective, thus is semisimple and therefore is Noetherian. This implies that has invariant basis number. Let be a nontrivial left ideal. Thus is a -module, so it is free and since all modules are projective (because they are free), then is direct summand of . If is proper, then we have a decomposition of a -module
but rank of is and rank of is at least . Contradiction, because has invariant basis number. Thus the only left ideals in are and . Now let . Then , so there exists such that
so is right invertible. Thus 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 is a divison ring if and only if every right -module is free.
|Title||rings whose every module is free|
|Date of creation||2013-03-22 18:50:20|
|Last modified on||2013-03-22 18:50:20|
|Last modified by||joking (16130)|