Galois symbols and Milnor K-group I was reading the paper Swan conductors for characters of degree one in the imperfect residue field case by Kato. Is it easy to prove the property that the symbol {...} has the property if any two elements sum up to 1 or 0, then the symbol is 0? Here, $R$ is a ring (scheme) is over $\mathbf{Q}$, or a smooth ring over a field characteristic p > 0.

(The case $p,n$ not coprime. The result is also "correct", but we need to redefine the part of $\mathbf Z/p^s\mathbf Z$. It is not the sheaf $\pmb\mu_{p^s}$; it involves a more complicated object.)
 A: This is pretty classical. Firstly, note that it suffices to prove the result for $r=2$, since $\{a_1,\ldots,a_r\} = \{a_1,a_2\} \smile \{a_3,\ldots,a_r\}$ and $\{\cdot,\ldots,\cdot\}$ is skew-symmetric. For $r = 2$, we only need to show that $\{a,1-a\} = 0$ for all $a \in R^\times$ with $1-a \in R^\times$. Indeed, note that $1-a^{-1} = \tfrac{a-1}{a} \in R^\times$, so we also get $\{a^{-1},1-a^{-1}\} = 0$. Bilinearity of $\{\cdot,\cdot\}$ gives $\{a,1-a^{-1}\} = 0$, and then the relation $(1-a^{-1}) \cdot (-a) = 1-a$ gives
$$\{a,-a\} = \{a,1-a\} - \{a,1-a^{-1}\} = 0.$$
This then gives the required result for $r=2$.
To show that $\{a,1-a\} = 0$, first assume that $n$ is invertible in $R$. Then $S = R[x]/(x^n-a)$ is a finite étale extension of $R$, and $x \in S^\times$ maps to $a$ under $(-)^n \colon S^\times \to S^\times$. Now the key observation is that $1-a$ is a norm for $f \colon \operatorname{Spec} S \to \operatorname{Spec} R$:
$$N_{S/R}\big(1-x\big) = \prod_{i=1}^n \big(1-\zeta_n^ix\big) = 1 - x^n = 1-a.$$
Thus the projection formula for the pullback $f^* \colon H^*(R,\pmb \mu_n) \to H^*(S,\pmb \mu_n)$ and the trace map $\operatorname{tr}_f \colon H^*(S,\pmb \mu_n) \to H^*(S,\pmb \mu_n)$ shows
$$\{a,1-a\} = \{a\} \smile \operatorname{tr}_f\{1-x\} = \operatorname{tr}_f\big(f^*\{a\} \smile \{1-x\}\big) = 0$$
since $f^*\{a\} = 0 \in H^1(S,\pmb \mu_n)$. (Presumably this can also be translated into a Čech-theoretic argument, but I found it notationally a bit heavy.)
It remains to discuss the case $n = p^s$ for $p = \operatorname{char} R$. Let me sketch the result by sticking to $s=1$; the general case should be similar (but I am not really an expert, so don't take my word for it!). For $n = p$ we're just looking at the subsheaf $\Omega^1_{R,\log}\subseteq\Omega^1_R$ of differentials that are locally $R$-linear combinations of terms of the form $\tfrac{\mathrm df}{f}$. At any rate, the point is supposed to be that
$$\mathrm d\log(a) \wedge \mathrm d\log(1-a) = \frac{\mathrm d a}{a} \wedge \frac{\mathrm d a}{a-1} = 0$$
since both are multiples of $\mathrm da$.
