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Let $R$ be a ring (with identity, but not necessarily commutative). If given the presumption that for any left free $R$-module $M$, every submodule of $M$ is free, what can we say about $R$?

Apparently any left principal ideal domain fulfills the above requirement and it is the only possibility for $R$ commutative. But I have no idea about the case where $R$ is noncommutative. Does anyone have some thought of this problem?

Censi LI
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  • Related http://math.stackexchange.com/questions/1002778/is-every-submodule-of-a-projective-module-projective/1002817#1002817 – Hanno Apr 24 '16 at 09:04
  • @Hanno Thanks for pointing out the relation. But I think through substituting "projective" by "free", the constraint put by me is much more restrictive. – Censi LI Apr 24 '16 at 09:22
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    Not a characterization, but a name: if you add the extra condition of having IBN (every free module has unique dimension) then this what you describe is called a left fir (fir stands for "free ideal ring"). There is a lot of literature about these (at least two books by Paul Cohn). – Jeremy Rickard Apr 24 '16 at 09:44
  • @CensiLI Your question differs from the one referenced in that you additionally require projectivity and freeness to be equivalent, which is somehow an orthogonal question. – Hanno Apr 24 '16 at 12:03
  • @JeremyRickard Thanks a lot. It seems that I accidentally run into a complicate problem. – Censi LI Apr 24 '16 at 16:07

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