Definition of a $\Bbb{Z}$-bilinear map? I am looking at a definition of a tensor space and came across the expression of a $\Bbb{Z}$-bilinear map in the following context:

Let $M$ and $N$ be R-module and $G$ and abelian group (i.e.
  $\Bbb{Z}$-module). A $\Bbb{Z}$-bilinear map, $\beta:M\times N\rightarrow G$ is called $R$-balanced
  if... (paraphrased from Wisbauer, 1988; pg90)

What is the meaning of a $\Bbb{Z}$-bilinear map in this context? (p.s. I am asking about what the qualifier $\Bbb{Z}$ indicates - I know what a Bilinear map is) 
 A: Here's the way I like to write it (for a commutative ring $R$):
Suppose $f:M\times N\to G$, $M,N,T$ all $R$ modules over a commutative ring $R$.
$f$ is called a bi-additive mapping if
$$f(m+m',n)=f(m,n)+f(m',n)$$
and
$$f(m,n+n')=f(m,n)+f(m,n')$$
$f$ is called an $R$-bilinear mapping if 
$$f(mr,n)=rf(m,n)$$
and
$$f(m,nr)=rf(m,n)$$
for all $r\in R$.
$f$ is called an $R$-balanced mapping if
$$f(mr,n)=f(m,rn)$$
for all $r\in R$.
It is easy to see that when $R=\mathbb Z$, a bi-additive map is already $\mathbb Z$-bilinear and $\mathbb Z$-balanced.
Now, the last condition is the interesting one. An $R$-bilinear map might be $R$ balanced, or it might not be.  That is the difference between $M\otimes_\mathbb ZN$ and $M\otimes_R N$.
You can generalize this way out for noncommutative rings and say that if $_SM_R$ and $_RN_T$ are bimodules for rings $S,R,T$, then you can talk about bi-additive, left-$S$-linear right-$T$-linear $R$-balanced maps and the natural bimodule structure of $_S(M\otimes_R N)_T$.
Explicitly:
$$f(sm,n)=sf(m,n)$$
and
$$f(m,nt)=f(m,n)t$$
and 
$$f(mr,n)=f(m,rn)$$
for all $(r,s,t)\in R\times S\times T$.
A: A map $f:M\times N\to A$ (for an Abelian group $A$) is
$\Bbb Z$-bilinear if
$$f(m+m',n)=f(m,n)+f(m',n)$$
and
$$f(m,n+n')=f(m,n)+f(m,n')$$
for all $m$, $m'\in M$ and all $n$, $n'\in N$.
In other words, for fixed $m$, $n\mapsto f(m,n)$ is a $\Bbb Z$-linear
map from $N$ to $A$, and ditto for the other argument.
