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Possible Duplicate:
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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marked as duplicate by Arturo Magidin, Asaf Karagila, Nate Eldredge, t.b., Jonas Meyer Jan 28 '12 at 16:02

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    $\begingroup$ Dear Adrian: please see math.stackexchange.com/questions/75866/…. $\endgroup$ – Pierre-Yves Gaillard Jan 27 '12 at 10:40
  • $\begingroup$ 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... $\endgroup$ – Adrian Manea Jan 27 '12 at 10:47
  • $\begingroup$ Dear Adrian: You're welcome! No apology is needed. By the way, I upvoted your outstanding question. $\endgroup$ – Pierre-Yves Gaillard Jan 27 '12 at 10:55
  • $\begingroup$ Thank you, sir! But why is it outstanding, if I may ask? :) $\endgroup$ – Adrian Manea Jan 27 '12 at 10:58
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    $\begingroup$ ... I see that as a kind of Egg of Columbus phenomenon. $\endgroup$ – Pierre-Yves Gaillard Jan 27 '12 at 11:17

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