Questions related to a map tensored with $\mathbb R$ Context 
I'm studying Audin's book - Torus Action on Symplectic manifolds, chapter VII where she builds Toric varieties from fans. 
It is defined on page 231 (2 ed) a linear map 
$$\pi : \mathbb Z ^N \to \mathbb Z^n \tag{*}$$
and also by the same letter $\pi$ the maps $\pi \otimes \mathbb Q$, $\pi \otimes \mathbb R$, $\pi \otimes \mathbb C$. Moreover if $\mathbb K \subset \mathbb Z $ is the kernel of $\pi$ as in (*) then one gets $K \subset \mathbb T^N$ which is the kernel of 
$$\mathbb R^N / \mathbb Z^N \to \mathbb R^n / \mathbb Z^n$$
Remark: Assume the fan $\Sigma$ is complete so that $\pi$ is surjective. 

Questions:
1) Since I'm not familiar with algebraic notation how does one define a map tensored with a field, say, $\mathbb R$ as in $\pi \otimes \mathbb R$? 
2) How is it that from the kernel $\mathbb K$ one can deduce $K$? For instance, say $\pi : \mathbb Z^2 \to \mathbb Z$ takes the canonical basis in $(e_1, e_2)$ to the basis vectors of the 1-skeleton $\Sigma^{(1)} = \{x_1 = e_1, x_2 = -e_1\}$, that is, 
  $$\pi = \begin{pmatrix}1 & -1 \end{pmatrix} : \mathbb Z^2 \to \mathbb Z$$ 
  whose kernel $\mathbb K$ is the span of the vector $(1,1)$, i.e., $\mathbb K \cong \mathbb Z$. How would one get that $K = \mathbb C^*$?

 A: I believe the set $\mathbb{T}^{N}$ is not $\mathbb{R}^{N}/\mathbb{Z}^{N},$ as you have indicated. Instead, $\mathbb{T}=\mathbb{C}^{\ast},$ and $\mathbb{T}^{N}$ is the algebraic $N$-torus $(\mathbb{C}^{\ast})^{N}.$ Up to isomorphism, $\mathbb{T}^{N}=\mathbb{Z}^{N}\otimes_{\mathbb{Z}}\mathbb{C}^{\ast}$ (in response to your comment: yes, this is a tensor product of $\mathbb{Z}$-modules; since $\mathbb{C}^{\ast}$ is an abelian group, it is a $\mathbb{Z}$-module, the module structure being given by $n\cdot t=t^{n}$).
So now we have a map $\pi\colon\mathbb{Z}^{N}\to\mathbb{Z}^{n}$ with kernel $\mathbb{K}\subseteq\mathbb{Z}^{N},$ and the same letter $\pi$ is being used to denote an induced map $$\pi\colon \mathbb{T}^{N}=\mathbb{Z}^{N}\otimes_{\mathbb{Z}}\mathbb{C}^{\ast}\to\mathbb{T}^{n}=\mathbb{Z}^{n}\otimes_{\mathbb{Z}}\mathbb{C}^{\ast}$$ with kernel $K\subseteq\mathbb{T}^{N}.$ Your question is: what is this induced map, and how does $K$ relate to $\mathbb{K}?$
The induced map. For clarity, I will use $\pi_{\mathbb{C}^{\ast}}$ to denote the induced map. The map $\pi_{\mathbb{C}^{\ast}}$ is the unique $\mathbb{Z}$-linear map satisfying, for all $u\in\mathbb{Z}^{n}$ and $t\in\mathbb{C}^{\ast},$
$$\pi_{\mathbb{C}^{\ast}}(u\otimes t)=\pi(u)\otimes t.$$
That is, for all $m>0,$ $r_{i}\in\mathbb{Z},$ $u_{i}\in\mathbb{Z}^{n}$ and $t_{i}\in\mathbb{C}^{\ast},$ we have
$$\pi_{\mathbb{C}^{\ast}}\left(\sum_{i=1}^{m}a_{i}(u_{i}\otimes t_{i})\right)=\sum_{i=1}^{m}a_{i}(\pi(u_{i})\otimes t_{i}).$$
Frequently such a map is also written $\pi\otimes 1$ or $\pi\otimes \operatorname{id}.$
The kernel $K.$ It remains to see how $K\subseteq \mathbb{T}^{N}$ relates to $\mathbb{K}\subseteq\mathbb{Z}^{N}.$ The answer is that since $\mathbb{C}^{\ast}$ is a torsion-free abelian group, it is a torsion-free $\mathbb{Z}$-module, and torsion-free $\mathbb{Z}$-modules are necessarily flat; that is, the functor $-\otimes_{\mathbb{Z}}\mathbb{C}^{\ast}$ that takes a $\mathbb{Z}$-module $M$ to $M\otimes_{\mathbb{Z}}\mathbb{C}^{\ast}$ and takes a $\mathbb{Z}$-linear map $\pi$ to $\pi\otimes\operatorname{id}$ as above, is an exact functor. In particular, the exact sequence
$$0\to \mathbb{K}\to \mathbb{Z}^{N}\to \mathbb{Z}^{n}\to 0$$
is taken to an exact sequence
$$1\to \mathbb{K}\otimes_{\mathbb{Z}}\mathbb{C}^{\ast}\to \mathbb{Z}^{N}\otimes_{\mathbb{Z}}\mathbb{C}^{\ast}\to \mathbb{Z}^{n}\otimes_{\mathbb{Z}}\mathbb{C}^{\ast}\to 1,$$
and it follows that $\mathbb{K}\otimes_{\mathbb{Z}}\mathbb{C}^{\ast} \cong K,$ by definition of $K.$
Unfortunately I have not been able to find a super-simple proof or reference for the fact that $\mathbb{C}^{\ast}$ is flat, but see this mathoverflow question for a relevant discussion with references.
Your example. In your example, since $\mathbb{K}\cong\mathbb{Z},$ it follows that $K\cong\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{C}^{\ast}\cong\mathbb{C}^{\ast}.$ More explicitly: every element of $\mathbb{K}$ is of the form $(u,u)$ for some $u\in\mathbb{Z},$ so a general element of $\mathbb{K}\otimes_{\mathbb{Z}}\mathbb{C}^{\ast}$ looks like
$$\sum_{i=1}^{m}a_{i}((u_{i},u_{i})\otimes t_{i}) = \sum_{i=1}^{m}a_{i}u_{i}((1,1)\otimes t_{i}) = (1,1)\otimes\prod_{i=1}^{m}t_{i}^{a_{i}u_{i}}.$$
The map $\sum_{i=1}^{m}a_{i}((u_{i},u_{i})\otimes t_{i})\mapsto \prod_{i=1}^{m}t_{i}^{a_{i}u_{i}}$ is the isomorphism $\mathbb{K}\otimes_{\mathbb{Z}}\mathbb{C}^{\ast}\to\mathbb{C}^{\ast}.$
