tensor product writing of $\mathbb{C} / (2\pi i)^n \mathbb{Q}$ Can someone explain to me why $\mathbb{C}/(2\pi i)^n \mathbb{Q}$ is isomorphic to the tensor product $(2\pi i)^{n-1} \mathbb{C}^{\times} \otimes_{\mathbb{Z}} \mathbb{Q}$? I don't understand well the meaning of this tensor product. What I figured out is that an element $z \otimes \lambda$ somehow corresponds to $z^\lambda$. At least when $n=1$ I guess one has to use the fact that $exp(2\pi i \cdot \mathbb{Q})$ are roots of unity...
 A: In the Abelian group $\Bbb C^\times$, the operation is the multiplication, probably it gets clearer if we write $\def\pl{\overset\bullet+} \pl$ for this operation, $\def\egy{\overset\bullet0} \egy$ for $1\ $ and $\ n\def\pt{\mathop\bullet}\pt z$ for $z\pl z\pl \dots \pl z$ if $n\in\Bbb N$.
Of course, we will get $n\pt z=z^n$ for all $n\in\Bbb Z$, so if $a,b$ are integers, $b>0$, then 
 $b\pt w=a\pt z$ means $w^b=z^a$, i.e. it is indeed something like $w=z^{\frac ab}$, but notice that generally it has $b$ solutions for $w$ if $z$ is given.
So, in this tensor product we have $z\otimes n=z\otimes (n\cdot 1)=(n\pt z)\otimes  1=z^n\otimes 1$. 
Let $U:=\{e^{(2\pi im)/n}\mid n,m\in\Bbb N\}$ be the subgroup of the roots of unity (taken for all $n$(!!)) in $\Bbb C^\times$. 
Verify that we must have $u\otimes 1=0$ for all $u\in U$ in $\Bbb C^\times\otimes\Bbb Q$.
Then, we can define a map $\varphi:\Bbb C^\times\otimes\Bbb Q\to\Bbb C^\times/U$, and the kernel of $\psi_0:\Bbb C^\times\to\Bbb C^\times\otimes\Bbb Q,\ z\mapsto(z\otimes 1)$ will be just $U$, yielding the inverse of $\varphi$.
