Example wanted

Can somebody give me an explicit example of a Noetherian local domain A which does not have resolution of singularities? In particular, I would like an example where the completion A* of A defines an isolated singularity, i.e., where Spec(A*) – {m*} is a regular scheme?

A related question (I think) is this: Does there exist a Noetherian local ring A such that there exists a proper morphism Y —> Spec(A*) whose source Y is an algebraic space which is an isomorphism over the punctured spectrum such that Y is not isomorphic to the base change of a proper morphism X —> Spec(A) whose source is an algebraic space? [Edit 20 October 2014: This cannot happen; follow link below.]

I wanted to make an example for the second question by taking a Noetherian local ring A which does not have a resolution whose completion A* is a domain and defines an isolated singularity. Then a resolution of singularities Y of A* would presumably not be the base change of an X, because if so then X would presumably be a resolution for A. [Edit 20 October 2014: This cannot happen; follow link below.]

Anyway, I searched for examples of this sort on the web but failed to find a relevant example. Can you help? Thanks!

[Edit 23 July 2014: The discussion is continued in this blog post.]

6 thoughts on “Example wanted

  1. I know that there are examples where $R$ is factorial, yet the completion $R’$ is not factorial. Maybe that could be the origin of an example: let $X$ be the blowing up of a Weil divisor in $\text{Spec}(R’)$ that is not Cartier. The Weil divisor does not descend to $R$, or else it would be Cartier.

    • This won’t work (I think) because there are such examples where the ring is excellent and I think for excellent rings every Y as in the second question is the base change of an X over A.

      • I understand what you say, but I think it may be complicated. After all, if $R$ is factorial, yet $R’$ is not factorial, that means that there is an ideal sheaf $I$ of a Weil divisor on $R’$ that does not descend to $R$, at least not as an ideal sheaf. If descent can fail for ideal sheaves, why should descent hold for algebraic spaces? To make one more point, for the blowing up $Y$ of $R’$ along $I$, there is an invertible sheaf $\mathcal{O}(-D)$ on $Y$ whose pushforward to $R’$ equals $I$, the ideal sheaf of the total transform of $I$. Either descent fails for $Y$ or descent fails for the invertible sheaf $\mathcal{O}(-D)$ on $Y$. Although not a proof, my intuition suggests that descent for algebraic spaces is more likely to fail than descent for invertible sheaves.

      • Hmm, now I take back what I said in my last reply. I guess you should be able to use “Algebraization of Formal Moduli, II” to descend $Y$ to $R$, at least in the excellent case. First you dominate $Y$ by something that descends (certainly possible in the examples I was proposing), i.e., $\nu: Z\otimes_R R’ \to Y$, and then you use “Algebraization II” to descend the contraction $\nu$.

        • Hi Jason! I hadn’t realized that this should be so, but now that you point it out, yes, the excellent case should/does follow from Artin’s results! Thank you for pointing this out. I have now written up the arguments showing this type of descent in the excellent case (not using Artin’s paper but using a little bit of material on coequalizers in algebraic spaces). Thanks again.

          • Happy to point that out — sorry I doubted in my first reply! I did try to find a non-excellent (honest) counterexample, but so far no luck.

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