Tame symbol

Remarks

The tame symbol is a Steinberg symbol: it is multiplicative in each variable, and $(a, 1 - a)_v = 1$ for $a \ne 0, 1$. So it induces a homomorphism $\partial_v \colon K_2(F) \to k^\times$. If $a$ and $b$ are both units of $\mathcal{O}$, then $(a, b)_v = 1$. Some authors use the inverse of this element; the two conventions differ by a sign in additive notation.

Example: $F = \Q$, $v = v_p$ the $p$-adic valuation, and $k = \F_p$. For an integer $u$ prime to $p$, $v_p(p) = 1$ and $v_p(u) = 0$, so $(p, u)_p = u^{-1} \bmod p$.

Non-example: the map $(a, b) \mapsto a^{v(b)} b^{-v(a)} \bmod \mathfrak{m}$, without the sign, is multiplicative in each variable but is not a Steinberg symbol in general. For $F = \Q$, an odd prime $p$ and $a = 1/p$, one has $1 - a = (p - 1)/p$ and $v_p(a) = v_p(1 - a) = -1$. The map without the sign gives $a^{-1}(1 - a) = p - 1 \equiv -1 \pmod p$, which is not $1$; the sign $(-1)^{1}$ corrects it to $1$.