Let $A$ be a Noetherian local domain ($2$-dimensional if needed) such that its punctured spectrum $U$ is regular, and let $A'$ be the normalization of $A$.

1) Is it possible for $A'$ to have infinitely many maximal ideals? (I know that this is not possible if, say, $A$ is universally Japanese, but I am interested in the general case.)

2) If $A$ is excellent, so that $A'$ has only finitely many maximal ideals $\mathfrak{m}_1, \dotsc, \mathfrak{m}_n$, is the restriction map $\mathrm{Pic}(U) \rightarrow \prod_{i = 1}^n \mathrm{Pic}(U_i)$ injective, where $U_i$ denotes the punctured spectrum of the localization of $A'$ at its $i$-th maximal ideal $\mathfrak{m}_i$?

Partial answers or comments on these questions are very welcome.

EDIT: Thanks to Mr Klutz's insight, the remaining question is 2).


2 Answers 2


Let $M$ be a finite $A$ module which induces an invertible module $\mathcal{L}$ on $U$. Then $\mathcal{L}$ is trivial if and only if you can find $A$-module maps $M \to M'$ and $A \to M'$ with $M'$ finite and kernel and cokernel annihilated by a power of the maximal ideal of $A$. (This works for any Noetherian local ring.)

Then you can easily see that the existence of $M \to M'$ and $A \to M'$ over the henselization $A^h$ descends to $A$ (essentially because $A \to A^h$ is flat and because the categories of finite length modules over $A$ and $A^h$ are the same -- although this should be avoided there is a big hammer you can use here if you like, which goes under the name formal glueing). Hence Mr Klutz is right that question 2 can be reduced to the henselian case.

This is not the optimal answer in fact I think you could have answered this entire question yourself in a more elementary way without asking on mathoverflow. Certainly there are many papers talking about the relationship between local Picard groups, henselization, completion, etc, etc. Cheers!


Instead of assuming $A$ is a local domain with $U$ regular, let's just assume that the punctured spectrum is $U$ is normal and not assume that $A$ is a domain. Instead of the integral closure of $A$ in its fraction we field we let $A'$ be the integral closure of $A$ in $\Gamma(U, \mathcal{O}_U)$. The key observation that should make things work is that the inverse image $U'$ of $U$ in $\text{Spec}(A')$ maps isomorphically to $U$.

For example, to answer the first question we may assume that $A$ is henselian. (Details of reduction omitted.) In this case, if the closed fibre $F$ of $\text{Spec}(A') \to \text{Spec}(A)$ is disconnected, then $\text{Spec}(A')$ is disconnected. This is one of a long list of characterizations of henselian local rings. In fact $\pi_0(F) = \pi_0(\text{Spec}(A'))$. Since every irreducible component of $\text{Spec}(A')$ meets $U'$ we conclude that $\pi_0(F) = \pi_0(\text{Spec}(A')) = \pi_0(U') = \pi_0(U)$. Since $U$ is a Noetherian scheme we see that $\pi_0(U)$ is finite.

Still in the henselian case we see that the second question is true because $U$ decomposes into the corresponding pieces. Then to prove it in general we can try to show that the local Picard group (Pic of punctured spec) of a Noetherian local ring maps injectively into the local Picard group of its henselization. I think this is true (look at cohomology...)

  • $\begingroup$ Thank you very much! In your answer to the first question, the omitted details are just the fact that the formation of integral closure commutes with etale base change (+ a limit argument), right? $\endgroup$ Oct 17, 2015 at 20:47
  • $\begingroup$ I can't see how to reduce the second question to the Henselian case though, i.e., how to argue your second to last sentence... $\endgroup$ Oct 17, 2015 at 21:57

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