TITLE: Brauer group of a field of rational numbers
QUESTION [10 upvotes]: Can we say anything about Brauer group of $\mathbb{Q}$? And how can we construct it?

REPLY [5 votes]: The Brauer group of $\mathbb Q$ (and more generally, of any finite extension of $\mathbb Q$) is described by class field theory.
The answer is that it is isomorphic to the direct sum of $\mathbb Z/2 \mathbb Z$
and  a countable number of copies of $\mathbb Q/\mathbb Z$.
More canonically, consider the direct sum
$$\dfrac{1}{2}\mathbb Z/\mathbb Z \oplus \bigoplus_p \mathbb Q/\mathbb Z,$$
where $\dfrac{1}{2}\mathbb Z/\mathbb Z$ denotes the (unique) cyclic
subgroup of order $2$ in $\mathbb Q/\mathbb Z$, and the direct sum is indexed by primes.
There is a natural map from this direct sum to $\mathbb Q/\mathbb Z$, given
by summing components, and the Brauer group of $\mathbb Q$ is canonically identified with its kernel.
The elements of order $2$ correspond to quaternion algebras over $\mathbb Q$,
and these are constructed explicitly in Serre's book A course in arithmetic.
Off the top of my head I'm not sure about computing them explicitly more generally.  
[Note: I wrote this answer several hours ago, but just got around to submitting it now.  Thus it likely duplicates the other answers.]