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

Let $V$ be a variety and $y$ 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 defined the following group:

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

The first component is the tame symbol at $x$ and the second component is defined 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 that this group doesn't depend on the choice of the uniformizer, but I couldn't see why. Is there an easy way to tell that this group doesn't depend on the choice of the uniformizer?

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It follows from a derivation-like property of the tame symbol (which can be found e.g. in Remark 4.6. of Bass-Tate "The Milnor ring of a global field"). To fix notation, let $C$ be a smooth affine curve, $\overline{C}$ the completion and $C_\infty=\overline{C}\setminus C$ the boundary divisor. For a point $x\in C_\infty$, any uniformizer is of the form $u\pi_x$ for $\pi_x$ a specific choice of uniformizer and $u\in k(C)^\times$ with $v_x(u)=0$. For a symbol $a\in K^M_r(k(C))$ and $u\pi_x\in k(C)^\times$ we get $$ \partial_x(a\cup u\pi_x)=\lambda(a)\partial_x(u\pi_x)-\partial_x(a)\rho(u\pi_x)\in K^M_r(k(x)). $$ The map $\lambda:K^M_r(k(C))\to K^M_r(k(x))$ is induced from $k(C)^\times\to k(x)^\times:u\pi^n\mapsto u$, and $\rho$ is induced from mapping $k(C)^\times\to k(x)^\times:u\pi^n\mapsto (-1)^nu$.

An element $a\in UK^M_{r+1}$ will have $\partial_x(a)=0$, so the second term of $\partial_x(a\cup u\pi_x)$ is $0$ independent of the choice of uniformizer. In the first term, $\partial_x(u\pi_x)=1\in K_0(k(x))$ independently of the generator. We get that $\partial_x(a\cup u\pi_x)=\lambda(a)$ only depends on $a$, and hence is independent of the choice of uniformizer whenever $\partial_x(a)=0$. (This, however, requires to use both the tame symbol and the second map, the second map as well as its kernel may still depend on the choice.)

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