$a\otimes(-a) = a \otimes (1-a) + a^{-1}\otimes(1-a^{-1})$??? J. Browkin writes in his article "K-Theory, Cyclotomic Equations, and Clausen's Function" (which appears on Chapter 11 of Lewin's book "Structural properties of polylogarithms") that for any field $F$ and $a\in F^\times$, $a\neq 1$,
$$a\otimes(-a) = a \otimes (1-a) + a^{-1}\otimes(1-a^{-1}).$$
I have two questions:


*

*Is this tensor over $\mathbb Z$?

*How is this equality true?

 A: This is a 'guess' - I'm afraid that I don't know anything about $K$-theory, for instance - but I did get the expected equality.
View $F^\times$ as a (multiplicative) abelian group, and hence a $\mathbb Z$ module: e.g., $n\cdot(a)= (a^n)$, etc... The tensor product is indeed over $\mathbb Z$, but above, in your question (and in what follows), the group operation on $F^\times\otimes F^\times$ is written using addition, while the group structure on $F^\times$ is written using multiplication. So, for instance,
$$ a\otimes 1 = 0, $$ 
and $$(a\otimes b) - (a \otimes c) = a\otimes (b/c),\tag{*}$$ 
and
$$ a\otimes b^{-1}= a^{-1}\otimes b= - (a\otimes b)\tag{**}$$
all hold true.
Therefore, as a special case of $(*)$:
$$ a\otimes (-a) - a\otimes (1-a) = a \otimes (-a /( 1-a)). $$
But, of course, $$-a/ (1-a)= 1/ (1-a^{-1}) = (1-a^{-1})^{-1}.$$
Therefore, from $(**)$, one gets 
$$ a \otimes (-a / (1-a))= a\otimes (1-a^{-1})^{-1} = a^{-1}\otimes (1-a^{-1}),$$ 
from which follows the desired equality.
