Milnor K-theory

Remarks

Equivalently, $K_*^M(K) = \bigoplus_{n \ge 0} K_n^M(K)$ is the quotient of the tensor algebra of $K^\times$ over $\Z$ by the two-sided ideal generated by the elements $a \otimes (1 - a)$ with $a \ne 0, 1$. The relation $\set{a, 1 - a} = 0$ is the Steinberg relation. Consequences: $\set{a, -a} = 0$ and $\set{a, b} = -\set{b, a}$ for all $a, b \in K^\times$. Example: in $K_2^M(\Q)$, $\set{2, -1} = 0$ since $2 + (-1) = 1$. Non-example: the element $2 \otimes (-1)$ of $\Q^\times \otimes_\Z \Q^\times$ is not zero. Indeed $\Q^\times$ is the product of $\set{\pm 1}$ and the free abelian group on the primes. So $2 \otimes (-1)$ generates a direct summand $\langle 2 \rangle \otimes \set{\pm 1} \cong \Z/2\Z$, where $\langle 2 \rangle$ is the cyclic group generated by $2$. Thus the tensor product alone is not $K_2^M(\Q)$. For a finite field $\F_q$, $K_n^M(\F_q) = 0$ for $n \ge 2$. By a theorem of Matsumoto, $K_2^M(K)$ equals Quillen's $K_2(K)$.