Let $S$ be a (commutative) ring with identity, and let $M$, $N$ be $S$-modules. (I guess if $S$ isn't commutative, I want $M$ to be a right $S$-module an $N$ a left $S$-module.) In the definition of the tensor product $M \otimes_S N$ in terms of generators and relations, one starts with the free abelian group $F(M \times N)$, where $M \times N$ is the set on which $F(M \times N)$ is defined, and then one defines a certain subgroup $H$. Namely, $H$ is the subgroup of $F(M \times N)$ generated by all elements of the forms:
- $A_{m,m',n} := (m+m',n) - (m,n) - (m',n)$, where $m,m' \in M$ and $n\in N$,
- $B_{m,n,n'} := (m, n+n') - (m,n)-(m,n')$, where $m\in M$ and $n,n'\in N$, and
- $C_{m,s,n} := (ms,n) - (m,sn)$, where $m\in M$, $s\in S$, and $n \in N$.
Then $M \otimes_S N = F(M \times N) / H$, we establish the universal property (and in the commutative case, the $S$-module structure), and typically we forget all about the groups we just defined. But my question is this:
$H$ is a free abelian group, since it is a subgroup of a free abelian group. But is it generated freely by the $A$'s, the $B$'s, and the $C$'s? That is, is the set of all elements of these three types (labelled as above) linearly independent over $\mathbb Z$? And if not, is some subset of them linearly independent over $\mathbb Z$?
$\newcommand{\Z}{\mathbb{Z}}$Consider the case of finite abelian groups. Let $M = \Z/2\Z$, $N = \Z/n\Z$. Then the rank of $F(M \times N)$ is $2n$, so any subgroup has rank at most $2n$. But already there are $2\binom{n}{2}$ elements of the form $B_{m,n,n'}$, which is $> 2n$ if $n > 3$, in which case the $B_{m,n,n'}$ cannot be linearly independent.