Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This might be a little bit spesific but here it goes. While reading a paper (Brauer-Manin pairing...) by Yamazaki, I encountered this definition.

Let $V$ be a variety. $y$ be a one dimensional point on $V$, i.e. $dim\overline{ \{ y \} } = 1$. Then let $C(y)$ be its closure in $V$ and $\tilde{C}(y)$ be normalization and $\bar C(y)$ be smooth completion. Let $C_\infty : = \bar C(y) - \tilde{C}(y)$. Then he definied the following group:

$UK^M_{r+1}:= ker [K_r^M(k(y)) \to \bigoplus_{x \in C_\infty}(K_{r-1}^M(k(x)) \oplus K_{r}^M(k(x))]$.

First component is the tame symbol at $x$. He defined the second component as $a \to \partial_x(a \cup \pi_x)$ where $\partial_x$ is the tame symbol at $x$ and $\pi_x$ is a uniformizer at $x$.

He says this group does not depend on the choice of the uniformizer, I couldn't see why it doesn't.

Is there an easy way to tell this group does not depend on the choice of the uniformizer?

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.