Lemma 42.5.2. The tame symbol (42.5.0.1) satisfies (4), (5), (6), (7) and hence gives a map $\partial _ A : Q(A)^* \times Q(A)^* \to \kappa (\mathfrak m)^*$ satisfying (1), (2), (3).
Proof. Let us prove (4). Let $a_1, a_2, a_3 \in A$ be nonzerodivisors. Choose $A \subset B$ as in Lemma 42.4.4 for $a_1, a_2, a_3$. Then the equality
\[ \partial _ A(a_1a_2, a_3) = \partial _ A(a_1, a_3) \partial _ A(a_2, a_3) \]
follows from the equality
\[ (-1)^{(e_{1, j} + e_{2, j})e_{3, j}} (u_{1, j}u_{2, j})^{e_{3, j}}u_{3, j}^{-e_{1, j} - e_{2, j}} = (-1)^{e_{1, j}e_{3, j}} u_{1, j}^{e_{3, j}}u_{3, j}^{-e_{1, j}} (-1)^{e_{2, j}e_{3, j}} u_{2, j}^{e_{3, j}}u_{3, j}^{-e_{2, j}} \]
in $B_ j$. Properties (5) and (6) are equally immediate.
Let us prove (7). Let $a_1, a_2, a_1 - a_2 \in A$ be nonzerodivisors and set $a_3 = a_1 - a_2$. Choose $A \subset B$ as in Lemma 42.4.4 for $a_1, a_2, a_3$. Then it suffices to show
\[ (-1)^{e_{1, j}e_{2, j} + e_{1, j}e_{3, j} + e_{2, j}e_{3, j} + e_{2, j}} u_{1, j}^{e_{2, j} - e_{3, j}} u_{2, j}^{e_{3, j} - e_{1, j}} u_{3, j}^{e_{1, j} - e_{2, j}} \bmod \mathfrak m_ j = 1 \]
This is clear if $e_{1, j} = e_{2, j} = e_{3, j}$. Say $e_{1, j} > e_{2, j}$. Then we see that $e_{3, j} = e_{2, j}$ because $a_3 = a_1 - a_2$ and we see that $u_{3, j}$ has the same residue class as $-u_{2, j}$. Hence the formula is true – the signs work out as well and this verification is the reason for the choice of signs in (42.5.0.1). The other cases are handled in exactly the same manner. $\square$
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