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Every $R$-module is free $\implies$ $R$ is a division ring

Prove that if a (generally noncommutattive) ring $R$, any $R$-module is free then $R$ is a field.

The commutative case is fairly easy, but I don't know how to deal with the noncommutative one. What could be the tools, or in what context is this problem solved most clearly?

Although I am somehow a novice in noncommutative ring theory, the problem looks very interesting and I am willing to study something new even for this problem only.

Thank you.

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Adrian Manea
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    Dear Adrian: please see http://math.stackexchange.com/questions/75866/every-r-module-is-free-implies-r-is-a-division-ring. – Pierre-Yves Gaillard Jan 27 '12 at 10:40
  • Great! Thank you! It is even better that I expected, since it fits my knowledge of Wedderburn-Artin theory. Sorry if my question seems a repost... – Adrian Manea Jan 27 '12 at 10:47
  • Dear Adrian: You're welcome! No apology is needed. By the way, I upvoted your outstanding question. – Pierre-Yves Gaillard Jan 27 '12 at 10:55
  • Thank you, sir! But why is it outstanding, if I may ask? :) – Adrian Manea Jan 27 '12 at 10:58
  • Dear Adrian: I had nothing deep in mind. I just told myself this. The question was asked before on MathOveflow - see here - where it got 16 votes. If you look at the first answers given either on MSE or on MO, you see that they were very complicated, and suddenly a simple argument emerged. I find interesting the fact that there are problems which have a simple solution, but, for some reason, it takes a lot of time to find this solution... – Pierre-Yves Gaillard Jan 27 '12 at 11:17
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    ... I see that as a kind of Egg of Columbus phenomenon. – Pierre-Yves Gaillard Jan 27 '12 at 11:17

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