When is a commutative ring the limit of its local rings? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T07:18:49Z http://mathoverflow.net/feeds/question/7558 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7558/when-is-a-commutative-ring-the-limit-of-its-local-rings When is a commutative ring the limit of its local rings? Dinakar Muthiah 2009-12-02T05:32:02Z 2009-12-02T15:24:30Z <p>Let $A$ be a commutative ring. Then we get local rings $A_p$ by localizing at each prime ideal $p$. Moreover, we get $A_p \rightarrow A_q$ when $p$ contains $q$. So we get a big diagram indexed by the inclusion poset of prime ideals. When is $A$ the limit of this diagram?</p> <p>When $A$ is a local ring or an integral domain it's true. I don't see any reason why it should be true for arbitrary rings. What's going on here?</p> http://mathoverflow.net/questions/7558/when-is-a-commutative-ring-the-limit-of-its-local-rings/7580#7580 Answer by Mark Hovey for When is a commutative ring the limit of its local rings? Mark Hovey 2009-12-02T14:56:22Z 2009-12-02T14:56:22Z <p>The map from A to the inverse limit of all its localizations is always injective. This boils down to the fact that the global sections of the structure sheaf O on Spec A are just A. The map from A to the global sections of O just takes A to the section which is the image of a in A_p on each stalk. So it is the map from A to the product of all its localizations, and this is therefore injective, so the map to the inverse limit will also be. </p> <p>But it does not always have to be surjective. Indeed, we can just take a commutative von Neumann regular ring that is not a product of fields. The reason this will work is that every prime ideal in a commutative VNR ring is maximal, and every localization is a field. So the inverse limit of the localizations will be a product of fields. </p> <p>Here is a commutative VNR that is not a product of fields; take the subring of a countably infinite product of copies of a fixed field k consisting of sequences that are eventually constant. I got this from Lam, Lectures on Modules and Rings, Example 7.54p. 263. It is easy to see this is VNR, and surely it is not a product of fields. </p> http://mathoverflow.net/questions/7558/when-is-a-commutative-ring-the-limit-of-its-local-rings/7582#7582 Answer by Pete L. Clark for When is a commutative ring the limit of its local rings? Pete L. Clark 2009-12-02T15:24:30Z 2009-12-02T15:24:30Z <p>This is a minor variation of MH's response:</p> <p>A ring R is Boolean if x^2 = x for all x in R. (This implies R is commutative.)</p> <p>In a Boolean ring R, every prime ideal is maximal, Moreover, the only local Boolean ring is Z/2Z. Therefore, R' := inverse limit_{p \in Spec R} R_p = (Z/2Z)^{# Spec R}. In particular, R' is either finite or uncountably infinite.</p> <p>But there are certainly countably infinite Boolean rings (a fancy justification for this is the Lowenheim-Skolem theorem in model theory): take an uncountable Boolean ring, and consider the subring generated by a countably infinite set of generators. </p> <p>For more details on Boolean rings, see e.g. Section 4.5 of </p> <p><a href="http://math.uga.edu/~pete/integral.pdf" rel="nofollow">http://math.uga.edu/~pete/integral.pdf</a></p>