Remark 42.27.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $k \in \mathbf{Z}$. We claim that there is a complex

Here we use notation and conventions introduced in Remark 42.19.2 and in addition

$K_2^ M(\kappa (x))$ is the degree $2$ part of the Milnor K-theory of the residue field $\kappa (x)$ of the point $x \in X$ (see Remark 42.6.4) which is the quotient of $\kappa (x)^* \otimes _\mathbf {Z} \kappa (x)^*$ by the subgroup generated by elements of the form $\lambda \otimes (1 - \lambda )$ for $\lambda \in \kappa (x) \setminus \{ 0, 1\} $, and

the first differential $\partial $ is defined as follows: given an element $\xi = \sum _ x \alpha _ x$ in the first term we set

\[ \partial (\xi ) = \sum \nolimits _{x \leadsto x',\ \delta (x') = k + 1} \partial _{\mathcal{O}_{W_ x, x'}}(\alpha _ x) \]where $\partial _{\mathcal{O}_{W_ x, x'}} : K_2^ M(\kappa (x)) \to K_1^ M(\kappa (x))$ is the tame symbol constructed in Section 42.5.

We claim that we get a complex, i.e., that $\partial \circ \partial = 0$. To see this it suffices to take an element $\xi $ as above and a point $x'' \in X$ with $\delta (x'') = k$ and check that the coefficient of $x''$ in the element $\partial (\partial (\xi ))$ is zero. Because $\xi = \sum \alpha _ x$ is a locally finite sum, we may in fact assume by additivity that $\xi = \alpha _ x$ for some $x \in X$ with $\delta (x) = k + 2$ and $\alpha _ x \in K_2^ M(\kappa (x))$. By linearity again we may assume that $\alpha _ x = f \otimes g$ for some $f, g \in \kappa (x)^*$. Denote $W \subset X$ the integral closed subscheme with generic point $x$. If $x'' \not\in W$, then it is immediately clear that the coefficient of $x$ in $\partial (\partial (\xi ))$ is zero. If $x'' \in W$, then we see that the coefficient of $x''$ in $\partial (\partial (x))$ is equal to

The key algebraic Lemma 42.6.3 says exactly that this is zero.

## Comments (0)

There are also: