Brauer group of a field of rational numbers

Can we say anything about Brauer group of $\mathbb{Q}$? And how can we construct it?

Solution 1:

$\DeclareMathOperator{\br}{Br}$ This can be done using the Brauer-Hasse-Noether theorem. One statement of this theorem is that for $k$ a global field, there is a canonical exact sequence $$ 0 \to \br(k) \to \bigoplus_v \br(k_v) \to \mathbb Q/\mathbb Z \to 0 . $$ where $v$ ranges over all places (finite and infinite) of $k$. It is known (see for example Serre's Local Fields, chapter 13) that $\br(k_v)\simeq \mathbb Q/\mathbb Z$. So for $k=\mathbb Q$, our sequence is $$ 0 \to \br(\mathbb Q) \to \mathbb Z/2\times \bigoplus_p \mathbb Q/\mathbb Z \to \mathbb Q/\mathbb Z\to 0. $$ So we have $$ \br(\mathbb Q) = \left\{(a,x):a\in \{0,1/2\}\text{ , }x\in \bigoplus_p \mathbb Q/\mathbb Z\text{ and }a+\sum x_p=0\right\}. $$ We have similar descriptions of $\br(k)$ for any global field $k$.

Solution 2:

The Brauer group of $\mathbb{Q}$ can be identified with a subgroup of $\bigoplus_vBr(\mathbb{Q}_v)$, the direct sum taken over all places of $\mathbb{Q}$ (including the Archimedean one). The map from $Br(\mathbb{Q})$ into this direct sum is given by tensoring any central simple algebra representing a class of the Brauer group with all the completions, to get central simple algebras over the completions. This tensor product will be a split CSA for almost all $v$, so you do land in the direct sum, rather than in the direct product.

Each of the local Brauer groups at the finite places is isomorphic to $\mathbb{Q}/\mathbb{Z}$, while $Br(\mathbb{R})\cong\frac12\mathbb{Z}/\mathbb{Z}$, the isomorphism given by the Hasse invariant. From the direct sum, you have a map to $\mathbb{Q}/\mathbb{Z}$ that simply adds the Hasse invariants, and the Brauer group of $\mathbb{Q}$ in the direct sum is precisely the kernel of this map.

A source for all of this would be, for example, the book on algebraic number theory edited by Cassels and Froehlich, specifically the chapters on local and global class field theory.

Solution 3:

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.]