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).



Normalization of a Noetherian local domain and line bundles on the punctured spectrum
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](http://stacks.math.columbia.edu/tag/05E5)). 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!


