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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?

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  • $\begingroup$ I wonder if this is always true (ring = limit) for a Noetherian ring, or a Noetherian ring where the closed points are dense in the spectrum... I'm pretty sure for analytic reasons that it's true for the coordinate ring of an affine variety over $\mathbb{C}$. $\endgroup$
    – Andrew Critch
    Dec 2, 2009 at 18:26
  • $\begingroup$ @Andrew: Can you elaborate on the analytic reasons? $\endgroup$
    – Dinakar Muthiah
    Dec 3, 2009 at 0:35
  • $\begingroup$ My mistake, localizing at a minimal prime gives us a field when the ring is reduced. =\. $\endgroup$
    – Harry Gindi
    Dec 3, 2009 at 1:41

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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.

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

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  • $\begingroup$ Good point, and it also explains how to make sure my example is not a product of fields. If k is finite, then any product of copies of k will be finite or uncountable, and the ring I used will be countable. $\endgroup$
    – Mark Hovey
    Dec 2, 2009 at 15:40
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    $\begingroup$ Right. (this space unintentionally not left blank) $\endgroup$
    – Pete L. Clark
    Dec 2, 2009 at 15:50

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