Tensor product is o? 
$K$ generated by $(ar,rb)$ for every $a\in A$, $b\in B$, $r\in R$.
$F$ generated by $r(a,b)$ for every $a\in A$, $b\in B$, $r\in R$.
$(a,rb)=(ar,b)=r(a,b)$,then $K=F$, $F/K=0$   
Apparently I misread the definition.I must be doing something wrong. 
Thanks in advance
 A: No, the tensor product isn't zero. Be more careful about the definition. The way tensor product is constructed is to take the free group generated by elements of $A \times B$ and what we really want is to have identities such as $(a+a',b)=(a,b)+(a',b)$ so we take the subgroup generated by those three above types of elements and then taking quotient would kill all those elements of $F$ where $(a+a',b)=(a,b)+(a',b)$ doesn't hold. So in fact, in $F/K$ we have $(a+a',b)=(a,b)+(a',b), (a,b+b')=(a,b)+(a,b')$ and $(ar,b)=(a,rb)$ which is exactly what we hoped for.
A: You're being misled by the notation. The module $F$ is not $A\times B$, but the free $R$-module with the set $A\times B$ as basis. So $F$ is the set of “formal expressions“ of the form
$$
r_1(a_1,b_1)+r_2(a_2,b_2)+\dots+r_k(a_k,b_k)
$$
where $k$ is any natural number, $a_i\in A$, $b_i\in B$, $r_i\in R$ (for $i=1,2,\dots,k$). If $k=0$ the formal sum is empty and it is the zero element of $F$.
More precisely, the elements of $F$ are functions
$$
f\colon A\times B\to R
$$
such that $f(a,b)=0$ for all but a finite number of elements in the domain. The sum of two functions is defined pointwise. The element $(a,b)$ can be identified with the function which is $1$ on $(a,b)$ and $0$ elsewhere, making it possible to use the “simplified notation” as formal sums above.
In particular
$$
(a+a',b)\ne(a,b)+(a',b)
$$
for all $a,a'\in A$ and $b\in B$.
The tensor product can be zero, but in general it isn't. However, proving facts about the tensor product is never done with this representation, which is just a way to show it exists.
