This redirect does not require a rating on Wikipedia's content assessment scale. It is of interest to the following WikiProjects:

Mathematics Low‑priority This redirect is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.MathematicsWikipedia:WikiProject MathematicsTemplate:WikiProject Mathematicsmathematics Low This redirect has been rated as Low-priority on the project's priority scale.

I am not sure whether R-module (as in AtiyahEisenbud) or S^-1R-module (as in Atiyah) should be used in the definition of the localization of modules. --Kompik (talk) 12:41, 27 May 2008 (UTC)[reply]

The current article is fairly ambiguous about this. Both structures are extremely important, and there is a unique way to go from one to the other, so the current ambiguous approach might be best. JackSchmidt (talk) 13:38, 27 May 2008 (UTC)[reply]

What are necessary and sufficient conditions on M such that the homomorphism from M to S^-1M is one-one? Druiffic (talk) 16:12, 8 December 2008 (UTC)Druiffic[reply]

I'm not sure why this is showing up in code mode. :? --Druiffic (talk) 00:39, 9 December 2008 (UTC)Druiffic[reply]

(there was an initial space making it appear funnily)

If R is commutative, then it is necessary and sufficient that if s in S, m in M, and sm=0, then m=0. I think the same is true if S is an Ore set in general. This is often phrased as "S⁻¹ kills the S-torsion in M". The proofs could probably be done directly or by replacing R by its image in the endomorphism ring of the additive group of M. JackSchmidt (talk) 07:04, 20 December 2008 (UTC)[reply]

Perhaps this article could be combined with the article on the localization on the ring, after all, the two constructions are very related. LkNsngth (talk) 05:56, 4 February 2009 (UTC)[reply]

The localisation of a module already uses the construction of localisation of a ring. To the best of my knowledge, it can not be defined standalone.--Mathmensch (talk) 13:55, 8 June 2016 (UTC) Or otherwise, we don't have a generalisation, as it's done in the article as I just saw.--Mathmensch (talk) 13:56, 8 June 2016 (UTC)[reply]