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We know this is true for commutative ring, but if $S\subset R$ is a left and right Ore set, and $S^{-1}R$ its localization by this Ore set, is this always a flat $R$-module?

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2 Answers 2

up vote 8 down vote accepted

This is Proposition 2.1.16 in McConnel+Robson's book on Noncommutative Noetherian Rings.

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Easy enough, thanks! –  BBischof Sep 19 '10 at 2:12
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The proof was easy, thanks for the tip, here is the link for others who may be looking and don't own the book: books.google.com/… –  BBischof Sep 19 '10 at 2:16

As I mentioned in a prior post here there is a wealth of information on noncommutative localizations in Ranicki, A.(ed). Noncommutative localization in algebra and topology. ICMS 2002. In particular, there you will find an interesting paper on this very topic by Beachy: "On flatness and the Ore condition". Below is general reference information for flatness in the commutative case.

There is a very nice treatment of flatness in Bourbaki's "Commutative Algebra" - which begins with an excellent chapter on flat modules before turning to localizations in Chapter 2 (see Theorem 2.41. p. 68 for the result you seek). Also perhaps of interest is the following motivational remark from the introduction

The study of the passage from a ring $\rm A$ to a local ring $\rm A$, or to a completion $\rm \hat A$ brings to light a feature common to these two operations, the property of flatness of the $\rm A$-modules $\rm A$, and $\rm \hat A$, which allows amongst other things the use of tensor products of such $\rm A$-modules with arbitrary $\rm A$-modules somewhat similar to that of tensor products of vector spaces, that is, without all the precautions surrounding their use in the general case. The properties associated with this notion, which are also applicable to modules over non-commutative rings, are the object of study in Chapter I.

See also Atiyah and Macdonald, Corollary 3.6 and Proposition 3.10 pp. 40-41.

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Note that Bryan is asking specifically about localization in the noncommutative setting. –  user126 Sep 26 '10 at 9:14
    
@97832123: Yes, I realized that later. Somehow I missed it at first glance - perhaps due to the absence of "noncommutative" in the title combined with too small a phone LCD. Since a couple readers found the answer useful I decided against deleting it. –  Bill Dubuque Sep 26 '10 at 20:56
    
I agree that you should not delete it. Mostly because somebody trying to understand the commutative version might stumble across this and find it useful. I hope that you understand if I don't vote it up though. –  BBischof Sep 27 '10 at 15:18
    
@BBischof: My answer is now updated for the noncommutative case. –  Bill Dubuque Sep 27 '10 at 15:36

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