Definition of Tensor Product in Hungerford’s Algebra

Question

Let $A$ be a right module and $B$ a left module over a ring $R$. Let $F$ be the free abelian group on the set $A\times B$. Let $K$ be the subgroup of $F$ generated by all elements of the following forms (for all $a,a'\in A$; $b,b'\in B$; $r \in R$):

(i) $(a+a’,b)-(a,b)-(a’,b)$;

(ii) $(a,b+b’)-(a,b)-(a,b’)$;

(iii) $(ar,b)-(a,rb)$.

The quotient group $F/K$ is called the tensor product of $A$ and $B$; it is denoted $A \otimes_R B$ (or simply $A\otimes B$ if $R = \Bbb{Z}$). The coset $(a,b) + K$ of the element $(a,b)$ in $F$ is denoted $a\otimes b$; the coset of $(0,0)$ is denoted $0$.

Exercise 2 section 4.2 states “If $R$ is an arbitrary ring and $X$ is any set, then there exists a free $R$-module on $X$”. So existence of free $\Bbb{Z}$-module (abelian group) $F$ on set $A\times B$ follows from exercise 2 section 4.2. Let $F$ with injection $\iota :A\times B\to F$ be free $\Bbb{Z}$-module. Since $\iota$ is injective, $A\times B$ is identified as subset of $F$. So (i) $(a+a’,b)-(a,b)-(a’,b)$ means $\iota(a+a’,b)-\iota(a,b)-\iota(a’,b)$. Similarly for (ii) and (iii).

Questions: (1) Is it okay to use $\Bbb{Z}$-module in place of abelian group and vice versa? (2) Let $Y$ be generators of $K$. I think, in general we have $Y\nsubseteq A\times B$. (3) Just to clear, in last sentence of definition, Author consider only coset of $(a,b)\in A\times B\subseteq F$, not coset of arbitrary elements of $F$.

Answer 1

1. Yes, $\mathbb{Z}$-modules are exactly the same thing as abelian groups.
2. Yes, but I don't see any problem in that since $K$ is just the submodule of $F$ generated by the elemnents you listed.
3. Yes, but since a generic element of $F$ is written as linear combination of elements of the form $(a,b)$, a generic element of $F/K$ will be written as linear combination of elements of the form $a\otimes b$.