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  "body": "\n## Definition\n\nLet $K$ be a [[field|0QA8CHH8MG]] and let $K^\\times$ be its [[group of units|05THZKR9AN]], an [[abelian group|071DDFVG4J]].\nLet $n \\ge 2$.\nLet $T_n$ be the [[tensor product|F0HQXQ0HCD]] over $\\Z$ of $n$ copies of $K^\\times$.\nLet $R_n \\subseteq T_n$ be the subgroup generated by all $a_1 \\otimes \\cdots \\otimes a_n$ with $a_i + a_j = 1$ for some $i \u003c j$.\nThe *$n$-th Milnor K-group* of $K$ is the quotient\n$$K_n^M(K) = T_n / R_n.$$\nAlso $K_0^M(K) = \\Z$ and $K_1^M(K) = K^\\times$.\nThe class of $a_1 \\otimes \\cdots \\otimes a_n$ is the *symbol* $\\set{a_1, \\ldots, a_n}$, and the group law of $K_n^M(K)$ is written additively.\n\n## Remarks\n\nEquivalently, $K_*^M(K) = \\bigoplus_{n \\ge 0} K_n^M(K)$ 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 $a \\ne 0, 1$.\nThe relation $\\set{a, 1 - a} = 0$ is the Steinberg relation.\nConsequences: $\\set{a, -a} = 0$ and $\\set{a, b} = -\\set{b, a}$ for all $a, b \\in K^\\times$.\nExample: in $K_2^M(\\Q)$, $\\set{2, -1} = 0$ since $2 + (-1) = 1$.\nNon-example: the element $2 \\otimes (-1)$ of $\\Q^\\times \\otimes_\\Z \\Q^\\times$ is not zero.\nIndeed $\\Q^\\times$ is the product of $\\set{\\pm 1}$ and the free abelian group on the primes.\nSo $2 \\otimes (-1)$ 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 $2$.\nThus the tensor product alone is not $K_2^M(\\Q)$.\nFor a finite field $\\F_q$, $K_n^M(\\F_q) = 0$ for $n \\ge 2$.\nBy a theorem of Matsumoto, $K_2^M(K)$ equals Quillen's $K_2(K)$.\n",
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