K2 and the Brauer group

Remarks

By Matsumoto's theorem, it is enough to check that $(a, b) \mapsto [A_\zeta(a, b)]$ is a Steinberg symbol. Bimultiplicativity holds in the Brauer group: $A_\zeta(a, b) \otimes A_\zeta(a', b)$ is Brauer equivalent to $A_\zeta(aa', b)$, and similarly in $b$. For the relation $\set{a, 1 - a}$, consider the main case where $X^n - a$ is irreducible and let $\alpha^n = a$. Then $A_\zeta(a, b)$ is the cyclic algebra of $F(\alpha)/F$ with $u^n = b$, and $1 - a = \prod_{i=0}^{n-1}(1 - \zeta^i \alpha)$ is a norm from $F(\alpha)$, so $A_\zeta(a, 1 - a)$ is split.

Example: $F = \R$, $n = 2$, $\zeta = -1$. The algebra $A_{-1}(-1, -1)$ is Hamilton's quaternions $\mathbb{H}$, a division algebra, so $R_{2, -1}(\set{-1, -1}) \ne 0$ in $\operatorname{Br}(\R) \cong \Z/2\Z$.

Going deeper

For a student

The algebras $A_\zeta(a, b)$ are called symbol algebras or power norm residue algebras; for $n = 2$ they are the quaternion algebras $(a, b)_F$. The Merkurjev–Suslin theorem says that the induced map $K_2(F)/n \to \operatorname{Br}(F)[n]$ is an isomorphism. Over a local field the Brauer class of $A_\zeta(a, b)$ is determined by the norm residue symbol of $a$ and $b$.