Question about tensor definition I am a newbie in tensor. In the Hungerford Grad. Algebra book, tensor product of modules is defined as follows.
Let $A$ and $B$ be, respectively, right and left $R$-modules. Let $F$ be the free abelian group on $A \times B$. Let $K$ be the subgroup of $F$ generated by elements of form: (with $a,a'\in A,b,b'\in B,r\in R$)


*

*$(a+a',b)-(a,b)-(a',b)$;

*$(a,b+b')-(a,b)-(a,b')$;

*$(ar,b)-(a,rb)$
And so on.
In the end, I think I understand that tensor product of $A$ and $B$ over $R$ are linear combinations of cosets of the above elements (please correct me if I'm wrong).
My question is very elementary.
For number 1. above, is it equal to $(0,-b)$ or not (as I am just summing componentwise)? The question is also for number 2. Why must they be written in the above form instead if they are the same?
Thanks in advance.
 A: In the free module on $A\times B$ that we will denote $F_R(A\times B)$, you have only formal linear combinations so $(a+a',b)-(a,b)-(a',b) \neq (0,-b)$ (else you would be summing component wise, which is not what you do here : you are not looking at the group $A\times B$ but at the free module generated by $A\times B$). 
Let $G$ be the group generated by $\{(a+a',b)-(a,b)-(a',b),(a,b+b')-(a,b)-(a,b') , (a,rb)-(ra,b) \text{ such that } (a,b)\in A\times B ,\, r \in R\}$.
Then $A\otimes B$ is $ F_R(A\times B)$ quotiented out by $G$ .
The idea of tensor product is to construct a space $A\otimes B$ such that for every $R$-module $P$ we have a "natural" (I dont use this term randomly) bijection between $Bil_R(A\times B, P)$ and $Hom_R (A\otimes B, P)$. This is indeed the case with our construction, since we did exactly what was needed to have this bijection (check it if you want).
A: It is better to start with the goal: you are looking for an $R$-module $V\otimes W$ and a bilinear map $f: V\times W \to V\otimes W$ that has the property that for any bilinear map $g$ out of $V\times W$ into an $R$-module $X$, there is a unique $\textit{linear}$ map $u:V\otimes W\to X$ such that the following diagram commutes. That is, $g$ factors through $f$ uniquely:

Of course, for this to make sense, you actually have to prove that there $\textit{is}$ such an object and an associated map that has the above property. 
The construction itself is done by starting with the free group and the word "free" tells the story: the free group consists formal sums; there are no relations. By taking the quotient, you are identifying elements within the free group, so the object you get is not "free" any more. But the quotient and the associated map it induces prove the existence of an object that satisfies the universal property above. I imagine the next thing Hungerford does give is the map $f$ and prove that the quotient satisfies the UMP. After you do this, the quotient itself is never really used again.
