Brauer group of rational numbers

Question

Reading about the calculation of the Brauer group of rational numbers, the calculations of the group are extremely lengthy and technical. First of all, it will be very helpful to me if someone can explicitly give me a class which corresponds to a nontrivial member of the Brauer group of rational numbers. Secondly, I appreciate it if someone can sketch the calculation if we assume we know the Brauer group of every local field is $\mathbb{Q}/\mathbb{Z}$.

Answer 1

What reference are you reading?

For a field $K$, every finite-dimensional central simple $K$-algebra $A$ is isomorphic to ${\rm M}_n(D)$ where $n$ is a positive integer and $D$ is a division ring with center $K$ where $\dim_K(D)$ is finite. The number $n$ is unique and $D$ is unique up to $K$-algeba isomorphism. The equivalence classes defining ${\rm Br}(K)$ just involve declaring ${\rm M}_n(D) \sim D$. That is, we identify two $A$'s if they have isomorphic $D$'s. Thus representatives for ${\rm Br}(K)$ are the finite-dimensional $K$-central division algebras, up to isomorphism.

The group law on ${\rm Br}(K)$ uses the tensor product of representatives: $[A][A'] = [A \otimes_K A']$ (isomorphic central simple $K$-algebras have isomorphic tensor products over $K$). Since we can represent each element of ${\rm Br}(K)$ by a unique finite-dimensional $K$-central division algebra up to isomorphism, in terms of $K$-central division algebras you could say the group law in ${\rm Br}(K)$ is $[D][D'] = [D'']$ where $D \otimes_K D' \cong {\rm M}_n(D'')$ for some $n$ and $D''$. (Note: a tensor product of two $K$-central division rings need not be a division ring, e.g., ${\mathbf H} \otimes_{\mathbf R} {\mathbf H} \cong {\rm M}_{4}(\mathbf R)$, so in ${\rm Br}(\mathbf R) = \{[\mathbf R],[\mathbf H]\} =\{1,[\mathbf H]\}$ we have $[\mathbf H]^2 = 1$.)

To ask for a nontrivial element of ${\rm Br}(K)$ is to ask for a finite-dimensional $K$-central division algebra other than $K$ itself. The simplest examples besides $K$ are the quaternion algebras over $K$, which are defined to be the $4$-dimensional central simple $K$-algebras. Every quaternion algebra over $K$ that's not isomorphic to ${\rm M}_2(K)$ will be a division ring. (For some $K$ there are no quaternion algebras over $K$ except for ${\rm M}_2(K)$, such as when $K$ is algebraically closed or finite, in fact in these cases ${\rm Br}(K)$ is trivial: all finite-dimensional $K$-central simple algebras are just the rings ${\rm M}_n(K)$ up to isomorphism.) A concrete account of quaternion algebras, which avoids working in characteristic $2$ (even though the general theory can handle this case if you develop it in a more abstract way) is here. In ${\rm Br}(K)$, the quaternion algebras (including the trivial or "split" case ${\rm M}_2(K)$) are representatives of the $2$-torsion subgroup: every element of ${\rm Br}(K)$ with order dividing $2$ is represented by a unique quaternion algebra over $K$, so the tensor product operation is a group law on quaternion algebras over $K$.

It's very helpful to be able to understand quaternion algebras well before trying to deal with the full Brauer group of a field, just as quadratic fields are a good testing ground to understand theorems about general number fields.

The construction called cyclic algebras gives you an $n^2$-dimensional $K$-central simple algebra from a cyclic Galois extension $L/K$ of degree $n$. (The Hamilton quaternions come in this way from the quadratic extension $\mathbf C/\mathbf R$: $\mathbf H = \mathbf C + \mathbf Cj$ where $j^2 = -1$ and $jz = \overline{z}j$ for all $z \in \mathbf C$.) Number fields and local fields have lots of cyclic Galois extensions (think of the unramified extensions of $\mathbf Q_p$ or suitable subfields of cyclotomic extensions of $\mathbf Q$), so you can get lots of central simple algebras of number fields and local fields this way. An account of cyclic algebras is in Section 14 of Chapter 5 of Lam's A First Course in Noncommutative Rings, and in that section there is an example of a $9$-dimensional division ring over $\mathbf Q$ built from a cyclic cubic extension of $\mathbf Q$ (the cubic subfield of the $7$th cyclotomic field). Also see Definition 7.10 and Theorem 7.13 here. There is a more general way of building $n^2$-dimensional central simple algebras from degree-$n$ Galois extensions (not necessarily having a cyclic Galois group) called a crossed product algebra. It is not always clear when these constructions are nontrivial, meaning they are not isomorphic to ${\rm M}_n(K)$

The general concepts I have mentioned here can be applied to an arbitrary base field $K$: $\mathbf Q$, $\mathbf R$, $\mathbf Q_p$, finite fields, and so on. The fact that ${\rm Br}(\mathbf Q_p) \cong \mathbf Q/\mathbf Z$ amounts to showing every $n^2$-dimensional division ring over $\mathbf Q_p$ can be built as a cyclic algebra using the degree-$n$ unramified extension of $\mathbf Q_p$ (which is a cyclic extension). See Theorem 7.14 here. Figuring out when two such constructions can be isomorphic, in terms of the parameters that show up in the construction, turns out to be the same as asking what the elements of order $n$ look like in $\mathbf Q/\mathbf Z$. A detailed proof of the calculation of ${\rm Br}(K)$ when $K$ is a local field is in Jacobson's Basic Algebra II: see the sections on division rings and the Brauer group near the end of Chapter $9$.