When is a commutative ring the limit of its local rings? - MathOverflow

When is a commutative ring the limit of its local rings?

2009-12-02T05:32:02Z

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?

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?

Answer by Mark Hovey for When is a commutative ring the limit of its local rings?

2009-12-02T14:56:22Z

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.

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.

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.

Answer by Pete L. Clark for When is a commutative ring the limit of its local rings?

2009-12-02T15:24:30Z

This is a minor variation of MH's response:

A ring R is Boolean if x^2 = x for all x in R. (This implies R is commutative.)

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.

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.

For more details on Boolean rings, see e.g. Section 4.5 of

http://math.uga.edu/~pete/integral.pdf