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Background/Motivation: The facts about the Brauer groups I will be using are mainly in Chapter IV of Milne's book on Etale cohomology (unfortunately it was not in his online note).

Let $R$ be a Noetherian normal domain and $K$ its quotient field. Then there is a natural map $f: Br(R) \to Br(K)$. In case $R$ is regular, a well known result by Auslander-Goldman says that $f$ is injective. The natural question is when can we drop the regularity condition? But anything in the $Cl(R)/Pic(R)$ will be in the kernel of $f$, so we need that quotient to be $0$ to make it interesting.

Also, it is known that even assuming said quotient to be $0$, one has example like $R=\mathbb R[x,y,z]_{(x,y,z)}/(x^2+y^2+z^2)$ which is UFD, but the kernel contains (I think) the quaternion algebra over $R$. The trouble in this case is that $\mathbb R$ is not algebraically closed. In fact, if $R$ is local, then $ker(f)$ injects into $Cl(R^{sh})/Cl(R)$, here $R^{sh}$ is the strict henselization. Most of the examples with non-trivial kernel I know seem a little ad-hoc, so:

Question: Are there more general methods to generate examples of $R$ such that $f$ is not injective?

I would love to see answers with more geometric/arithmetic flavors. I am also very interested in the positive and mixed charateristics case. Thanks in advance.

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@Hailong: Wonderful question! +1 For those readers looking for an online source, Auslander and Goldman's classic paper "The Brauer Group of a Commutative Ring" is available online (it gives an example which is essentially the same as that of Hailong): jstor.org/stable/pdfplus/1993378.pdf –  Ben Linowitz Feb 11 '10 at 2:29
    
@Ben: Thanks for the comment and the reference, I forgot to put it in the question. –  Hailong Dao Feb 11 '10 at 5:21

1 Answer 1

up vote 6 down vote accepted

As R is normal all localizations at height one primes are DVR whence regular and so the question is really one of describing the kernel of the morphism

Br(R) ---> beta(R) = \cap_P Br(R_P) (intersection over all ht 1 primes)

beta(R) is called the 'reflexive Brauer group' (see for example ancient papers by Orzech and Hoobler, in a Brauer-group proceedings of a conference in Antwerp, in the Springer LNM 917 begin 80ties)

anyway, I've once given a cohomological description of beta(R) in this paper and also one for the exact sequence related to your question

0-->Pic(RT)--->Cl(R)-->BCl(R)--->Br(R)--->beta(R)

here BCl(R) is the 'Brauer-class group'. It consists of reflexive R-modules M such that End(M) is a projectie R-module, made into a group under the modified tensor product (see paper mentioned before).

as you will see in that paper, these cohomological descriptions have everything to do with the smooth locus of R. For example, beta(R) is the Brauer group of the regular open piece and BCl(R) can 'in principle' be calculated from cohomology with support on the singular locus.

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As for my question, it was exactly this group BCl(R) I want to understand. I still could not find a way to build ease example so that BCl(R)/Cl(R) is not zero. Would you care to explain more? Thanks. –  Hailong Dao Feb 11 '10 at 6:48

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