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If M is a maximal ideal of Z(R), the center of a ring with identity R, and R_M, the localization of R at M, is a commutative field, what can we say about R?

My guess is that we can's say that much because even for a commutative ring R, the only thing I can say is that for every x in M there exists some y in R\M such that xy = 0.

Let me put a condition on R: suppose that R is von Neumann regular, that is for every r in R there exists some s in R such that r = rsr. If R_M is a commutative field for some maximal ideal M of Z(R), what can we say about R?

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    $\begingroup$ When $R$ is commutative, we can also say that $\mathfrak{m}$ is a minimal prime. Also, if the localization of $R$ at every maximal ideal is a field, it does not follow that $R$ is itself a field. Consider $\mathbf{Q}\times \mathbf{Q}$. $\endgroup$
    – Harry Gindi
    Commented Oct 18, 2010 at 6:31
  • $\begingroup$ Thank you. You are right. Actually, just a few minutes ago I found out that Kaplansky has already characterized those "commutative" rings R in which R_M is a field for "all" maximal ideals: they are exactly commutative von Neumann regular rings! This is a surprising coincidence for me because, as I mentioned in my question, I'm also interested in (non-commutative) von Neumann regualr rings. Anyway, I guess my question is either too hard to answer or basically there isn't that much to say! $\endgroup$
    – iravan
    Commented Oct 18, 2010 at 8:56

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