Tensor product of two left $R$-modules $A$ and $B$ In the case $R$ is not a commutative ring.
I don't know what happens if I define the tensor product of two left $R$-modules $A$ and $B$ as follows:
Step 1: Let $F(A,B,R)$ be the free $R$-modules with basis $A\times B$
step 2: Consider the submodule $N$ of $F(A,B,R)$ which is generated by elements with the following forms:
$(a+a',b)-(a,b)-(a',b)$
$(a,b+b')-(a,b)-(a,b')$
$(ra,b)-r(a,b)$
$(a,rb)-r(a,b)$
for $a\in A$, $b\in B$ and $r\in R$.
Then tensor product of the two modules $A$ and $B$ is $F/N$.
 A: Nobody stops you from making such a definition. However, what you get is not really useful. Let's see why.
Let $a\mathbin{\tau}b$ denote the element $(a,b)+N\in F/N$. Suppose $r,s\in R$; then, by definition,
$$
rs(a\mathbin{\tau}b)=r(sa\mathbin{\tau}b)=(rsa)\mathbin{\tau}b
$$
but also
$$
r(sa\mathbin{\tau}b)=(sa)\mathbin{\tau}(rb)=
s(a\mathbin{\tau}(rb))=
s(r(a\mathbin{\tau}b))=sr(a\mathbin{\tau}b)
$$
Thus
$$
rs(a\mathbin{\tau}b)=sr(a\mathbin{\tau}b)
$$
Hence $F/N$ is a module with the following property:

for each $x\in F/N$ and $r,s\in R$, $rsx=srx$.

When the ring is not commutative, this limits very much a possible usefulness of the concept.
For instance, suppose $A=B=R$. Then it is easily seen that $F/N$ is generated by $1\mathbin{\tau}1$, so there is a surjective module homomorphism $f\colon R\to F/N$ with $f(1)=1\mathbin{\tau}1$. The kernel is a left ideal $I$ such that $rs-sr\in I$, for every $r,s\in R$.
Now take the easiest noncommutative ring, the quaternions $\mathbb{H}$. The left ideal above is not $\{0\}$, so it is the whole ring and $F/N$ is the zero module. Similarly for the ring of $2\times2$ matrices over a field.
