Equivalence of definitions of the Milnor $K$-groups

Question

In Kurihara's paper: "The exponential homomorphisms for the Milnor $K$-groups and an explicit reciprocity law" he difines, in the first page, the $q$-th Milnor K-group for the ring $R$ as

$(R^{\times}\otimes\cdots \otimes R^{\times})/J$

where $R^{\times}$ is the unit group of $R$ and $J$ is the subgroup generated by the elements of the form $a_1\otimes\cdots \otimes a_q$ such that $a_i+a_j=0$ or $1$ for some $i\neq j$.

In all the definitions I've seen about the Milnor $K$-group they take $J$ to be the subgroup of the elements of the form $a_1\otimes\cdots \otimes a_q$ such that $a_i+a_j=1$ for some $i\neq j$ (i.e, the Steinberg relation).

Are both definitions equivalent? how can you prove this?

Answer 1

Write $(a,b)$ for the clas of $a\otimes b$ in $K_2(R)$. It is known that the Steinberg relation $(x,1-x)=1$ in $K_2(R)$ implies that $(x,-x)=1$ for every $x\in R\setminus\{0,1\}$. Indeed, since $(1-x)/(1-1/x)=-x$, one has $$ (x,-x)=(x,1-x)(x,1-1/x)^{-1}=(x,1-1/x)^{-1}=(1/x,1-1/x)=1 $$ for every $x\in F\setminus\{0,1\}$. NB. This implies that the symbol $(a,b)$ is skew-symmetric: $$ (a,b)(b,a)=(a,b)(a,-a)(b,a)(b,-b)= (a,-ab)(b,-ab)=(ab,-ab)=1. $$

NB. All of this is well explained in introductory texts in K-theory, such as Weibel's book (to which I borrowed this computation).