The Stacks project

Lemma 42.68.37. Let $A$ be a discrete valuation ring with fraction field $K$. For nonzero $x, y \in K$ we have

\[ d_ A(x, y) = (-1)^{\text{ord}_ A(x)\text{ord}_ A(y)} \frac{x^{\text{ord}_ A(y)}}{y^{\text{ord}_ A(x)}} \bmod \mathfrak m_ A, \]

in other words the symbol is equal to the usual tame symbol.

Proof. By multiplicativity it suffices to prove this when $x, y \in A$. Let $t \in A$ be a uniformizer. Write $x = t^ bu$ and $y = t^ bv$ for some $a, b \geq 0$ and $u, v \in A^*$. Set $l = a + b$. Then $t^{l - 1}, \ldots , t^ b$ is an admissible sequence in $(x)/(xy)$ and $t^{l - 1}, \ldots , t^ a$ is an admissible sequence in $(y)/(xy)$. Hence by Remark 42.68.14 we see that $d_ A(x, y)$ is characterized by the equation

\[ [t^{l - 1}, \ldots , t^ b, v^{-1}t^{b - 1}, \ldots , v^{-1}] = (-1)^{ab} d_ A(x, y) [t^{l - 1}, \ldots , t^ a, u^{-1}t^{a - 1}, \ldots , u^{-1}]. \]

Hence by the admissible relations for the symbols $[x_1, \ldots , x_ l]$ we see that

\[ d_ A(x, y) = (-1)^{ab} u^ a/v^ b \bmod \mathfrak m_ A \]

as desired. $\square$


Comments (0)


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Your email address will not be published.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 02Q6. Beware of the difference between the letter 'O' and the digit '0'.