MHB Tensor Products of Modules and Free Abelian Groups based on Cartesian Product

[Math Amateur]

I am reading Donald S. Passmore's book "A Course in Ring Theory" ...

I am currently focussed on Chapter 9 Tensor Products ... ...

I need help in order to get a full understanding of the free abelian group involved in the construction of the tensor product ... ...

The text by Passmore relevant to the free abelian group involved in the construction of the tensor product is as follows:View attachment 5591My question is about the nature of the free abelian group S whose elements are finite sums of the form ...$$\sum_{ (a,b) \in A \times B } z_{ a,b } (a, b) \text{ with } z_{ a,b } \in \mathbb{Z}
$$
So $$3(a_3, b_5) + 4(a_7, b_5)$$ and $$1(a_1, b_1)$$ are members of $$S$$ ...But how does this group 'work' ... and how do we do arithmetic (if we can?) with the elements ... and how do we end up with elements like $$(a_1 + a_2, b)$$ that play a role in forming the quotient for the tensor product ... ... Hope someone can help ...

(Apologies if I have asked a similar question before ... suspect the answer may be that we can only form formal sums and can do no arithmetic with the elements ... ... )

Peter

*** NOTE ***

To allow MHB members to be aware of the context of my question I am providing Passmore's introduction to his chapter on tensor products which includes the text given above ... ... as follows:https://www.physicsforums.com/attachments/5592
View attachment 5593

 

[Deveno]

Yes, formal sums. We can only "combine" these sums if the $(a,b)$ part of the $z_{a,b}(a,b)$ terms is identical:

$3(a_1,b_3) + 2(a_1,b_3) = 5(a_1,b_3)$.

A summand like $(a_1+a_2,b)$ arises just as a "general term" in $\mathcal{S}(A\times B)$, the addition between $a_1$ and $a_2$ takes place in $A$, not in $\mathcal{S}(A \times B)$.

After we take the quotient, we have more "options" for combining terms.

Balanced maps are introduced (instead of $R$-bilinear maps) because we are "combining" left and right modules. When $R$ is a commutative ring (such as a field, for example) we can ignore such niceties.

 

[Math Amateur]

Yes, formal sums. We can only "combine" these sums if the $(a,b)$ part of the $z_{a,b}(a,b)$ terms is identical:

$3(a_1,b_3) + 2(a_1,b_3) = 5(a_1,b_3)$.

A summand like $(a_1+a_2,b)$ arises just as a "general term" in $\mathcal{S}(A\times B)$, the addition between $a_1$ and $a_2$ takes place in $A$, not in $\mathcal{S}(A \times B)$.

After we take the quotient, we have more "options" for combining terms.

Balanced maps are introduced (instead of $R$-bilinear maps) because we are "combining" left and right modules. When $R$ is a commutative ring (such as a field, for example) we can ignore such niceties.

Thank you, Deveno ...

Most helpful and also reassuring ...

Peter