This question is just a curiosity, but I'm really interested in the answer. It was originally posted on math.stackexchange (https://math.stackexchange.com/questions/368897/inverse-problem-for-brauer-groups), but hasn't received any responses despite some upvotes, so I'm posting it here.

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 $Br(K)$ is always torsion. Within those constraints, I only know of one specific nontrivial Brauer group: $Br(\mathbb{R})=\mathbb{Z}/2\mathbb{Z}$ by Frobenius' Theorem on division algebras over $\mathbb{R}$. (I've seen the Brauer group of $\mathbb{Q}$ described by a short exact sequence, but I wasn't able to get an explicit description from that; is it known?) EDIT: As Emerton points out in a comment below, the Brauer group of $\mathbb{Q}$ (and much more) is known: it is $Br(\mathbb{Q})=\mathbb{Z}/2\mathbb{Z}\oplus\bigoplus_1^\infty\mathbb{Q}/\mathbb{Z}.$

My main question is: is there a known list of properties which are necessary and sufficient for a group to be $\cong Br(K)$ for some $K$?

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, and we can ask (although I'm not sure *why* we would ask): is there a group which is its own Brauer group? My second question is just: is there a good resource for this type of question, that is, for constructing Brauer groups of various objects 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.

(As an aside, I'm not sure whether the "group theory" tag is appropriate here; if it is not, feel free to delete it, or let me know and I will delete it.)

$^*$ As a very minor aside, note that 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.

equalto the direct sum you write in the edit in response to Emerton's comment. It injects into the direct sum, and the image is equal to the kernel of the "sum of the components" from this direct sum onto Q/Z. This is a way of expressing the fact that a central simple algebra over Q gives rise to a family of central simple algebras over Q_p (for every prime p, including p=\infty), but you don't get all such local families from a global object, $\endgroup$ – Chandan Singh Dalawat May 16 '13 at 3:05abstractlyisomorphic to this direct sum (and that might suffice for an algebraist), but the point is that B(Q_p) isnaturallyisomorphic to Q/Z and B(R) to Z/2Z, and that there arenaturalmaps from B(Q) to B(Q_p) and to B(R) which embed B(Q) into the direct sum; this natural embedding isnotan isomorphism. $\endgroup$ – Chandan Singh Dalawat May 16 '13 at 4:09placesof Q, not by the natural numbers. $\endgroup$ – Chandan Singh Dalawat May 16 '13 at 4:133more comments