This question is really just a curiosity, but I'm really interested in the answer.

Given a field $K$, we can form the set$^*$ $Br(K)$ consisting of equivalence classes of finite-dimensional central simple $K$-algebras which split over some Galois extension of $K$, modulo "are Morita-equivalent" (I hope I have that right, it's been a while). This set is actually a group, in a natural way: the tensor product over $K$ is well-defined on the equivalence classes, and has identity (the equivalence class of $K$ as an algebra over itself) and inverses (given by $R\mapsto R^{op}$). Actually $Br(K)$ turns out to be a second cohomology group, in a natural and useful way, but I don't really have a good understanding of that part.

My main question is, what groups are the Brauer group of some field? I know a couple trivial bits of the answer to this: $Br(K)$ is always abelian, and of cardinality at most $\aleph_0\times\vert K\vert$; and can be trivial (e.g., if $K$ is algebraically closed), or $\mathbb{Z}/2\mathbb{Z}$ ($Br(\mathbb{R})$, by Frobenius' Theorem on division algebras over $\mathbb{R}$). I also recall that $Br(K)$ is torsion, although I can't seem to find a reference for that right now. Is there a known list of properties which are necessary and sufficient for a group to be $\cong Br(K)$ for some $K$?

Now, there are many possible variations/elaborations of this question, which may not have deep significance but seem kind of interesting. For example, leaving the context of fields for a moment, there is an analogous notion of Brauer group for groups: is there a group which is its own Brauer group? Going back to the context of fields, given a field $K$, what groups are $Br(H)$ for some $K\le H\le \overline{K}$? My second question is just: is there a good resource for this type of question, that is, for constructing Brauer groups to specification? I imagine the opposite direction (finding Brauer groups of fields we already care about) is much more useful, but I'm personally interested in this direction.

$^*$ Size issues don't arise here: since we specify "finite-dimensional," there are at most $\aleph_0\times\vert K\vert$ many such algebras up to isomorphism; using Scott's trick then lets us represent these equivalence classes in perfectly fine way.


1 Answer 1


There is a paper, that may partially answer your question.

The paper is entitled "Every finite abelian group is the Brauer group of a ring" by T. J. Ford.

You can find it here.

  • $\begingroup$ This is exactly the sort of thing I was interested in! $\endgroup$ Apr 16, 2014 at 6:24

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