The Stacks project

Lemma 42.5.2. The tame symbol ( 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 ( The other cases are handled in exactly the same manner. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 42.5: Tame symbols

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0EAS. Beware of the difference between the letter 'O' and the digit '0'.