K2 of a field

Example

Let $F$ be a field. By Matsumoto's theorem, $K_2(F)$ is generated by symbols $\set{a, b}$ with $a, b \in F^\times$, subject to bimultiplicativity and $\set{a, 1 - a} = 1$.

Finite fields: $K_2(\F_q) = 0$ for every finite field $\F_q$. Check: the group $\F_q^\times$ is cyclic with a generator $g$, so $K_2(\F_q)$ is generated by $\set{g, g}$. From $\set{g, -g} = 1$ one gets $\set{g, g} = \set{g, -1}$, and $\set{g, -1}^2 = \set{g, 1} = 1$, so $\set{g, g}^2 = 1$. If $q$ is even, $\set{g, g}^{q - 1} = \set{g^{q-1}, g} = 1$ with $q - 1$ odd, so $\set{g, g} = 1$. If $q$ is odd, a counting argument gives $a \in \F_q$ such that $a$ and $1 - a$ are both non-squares, that is, odd powers of $g$. Then $1 = \set{a, 1 - a} = \set{g, g}^{m}$ with $m$ odd, so again $\set{g, g} = 1$.

The real numbers: $\set{-1, -1} \ne 1$ in $K_2(\R)$. Check: the real Hilbert symbol is a Steinberg symbol, so it induces a homomorphism $K_2(\R) \to \set{\pm 1}$, and it sends $\set{-1, -1}$ to $-1$.