Suppose $R$, $S$, and $T$ are rings (with unity), $M$ is an $(R,S)$-bimodule, and $N$ is an $(S,T)$-bimodule. If we trace through our construction of the tensor product in the special case of extending scalars, we see that we can create all of the various constructions in a more general setting, ultimately forming a new module. In this new general setting, that module will be denoted $M{\beta \x8a\x97}_{S}N$ and called the tensor product of $M$ and $N$ over $S$. It will be an $(R,T)$-bimodule and will satisfy a universal property similar to that of our previous construction.^{[1]}. We describe the general construction here, but the vast majority of examples we consider will be less general.

The construction of $M{\beta \x8a\x97}_{S}N$

As a set, we first consider the Cartesian product of the elements of the underlying sets of $M$ and $N$, i.e., we consider the set $U(M)\Gamma \x97U(N)$. We then form the free $\mathbf{Z}$-module on this set, obtaining the abelian group $F(U(M)\Gamma \x97U(N))$ whose elements consist of all formal finite sums of the form $\underset{i}{\beta \x88\x91}{k}_{i}({m}_{i},{n}_{i})$ where ${k}_{i}\beta \x88\x88\mathbf{Z}$, ${m}_{i}\beta \x88\x88M$ and ${n}_{i}\beta \x88\x88N$. We then prepare to "recover" the lost additive structures of $M$ and $N$ by letting $H$ denote the subgroup of this abelian group generated by all elements of the form

$({m}_{1}+{m}_{2},n)\beta \x88\x92({m}_{1},n)\beta \x88\x92({m}_{2},n)$ with ${m}_{1},{m}_{2}\beta \x88\x88M$ and $n\beta \x88\x88N$

$(m,{n}_{1}+{n}_{2})\beta \x88\x92(m,{n}_{1})\beta \x88\x92(m,{n}_{2})$ with $m\beta \x88\x88M$ and ${n}_{1},{n}_{2}\beta \x88\x88N$
As well as all elements of the form

$(mr,n)\beta \x88\x92(m,rn)$ for $r\beta \x88\x88R$, $m\beta \x88\x88M$, and $n\beta \x88\x88N$
This last set of elements is to recover the right $R$-action on $M$ and left $R$-action on $N$.

The resulting quotient group is then denoted $M{\beta \x8a\x97}_{R}N$. If we write $m\beta \x8a\x97n$ for the coset represented by the pair $(m,n)$ in this quotient group (and extend that notation linearly), then every element in $M{\beta \x8a\x97}_{R}N$ can be represented (non-uniquely!) by a finite sum of the form $\underset{i}{\beta \x88\x91}{m}_{i}\beta \x8a\x97{n}_{i}$ for some ${m}_{i}\beta \x88\x88M$ and ${n}_{i}\beta \x88\x88N$.^{[2]}

How about the $(R,T)$-bimodule structure? You could probably guess it, in that for simple tensors we define $r\beta \x8b\x85(m\beta \x8a\x97n)=rm\beta \x8a\x97n$ and $(m\beta \x8a\x97n)\beta \x8b\x85t=m\beta \x8a\x97nt$ for $r\beta \x88\x88R$ and $t\beta \x88\x88T$, and then extend to all tensors linearly.^{[3]}

Basic properties of tensors

It's worth listing for easy reference the basic properties of simple tensors, as the construction was tailored especially for $M{\beta \x8a\x97}_{S}N$ to have these properties:

$({m}_{1}+{m}_{2})\beta \x8a\x97n={m}_{1}\beta \x8a\x97n+{m}_{2}\beta \x8a\x97n$ for every ${m}_{1},{m}_{2}\beta \x88\x88M$ and $n\beta \x88\x88N$

$m\beta \x8a\x97({n}_{1}+{n}_{2})=m\beta \x8a\x97{n}_{1}+m\beta \x8a\x97{n}_{2}$ for every $m\beta \x88\x88M$ and ${n}_{1},{n}_{2}\beta \x88\x88N$

$ms\beta \x8a\x97n=m\beta \x8a\x97sn$ for every $m\beta \x88\x88M$, $n\beta \x88\x88N$, and $s\beta \x88\x88S$

$r(m\beta \x8a\x97n)=rm\beta \x8a\x97n$ for every $m\beta \x88\x88M$, $n\beta \x88\x88N$, and $r\beta \x88\x88R$

$(m\beta \x8a\x97n)t=m\beta \x8a\x97nt$ for every $m\beta \x88\x88M$, $n\beta \x88\x88N$, and $t\beta \x88\x88T$

Special cases

Tensor products of abelian groups

Suppose $A$ and $B$ are abelian groups. We have seen that $A$ and $B$ then also have unique $(\mathbf{Z},\mathbf{Z})$-bimodule structures, and so we can form their tensor product $A{\beta \x8a\x97}_{\mathbf{Z}}B$. This is another $(\mathbf{Z},\mathbf{Z})$-bimodule, i.e., abelian group.

Extension of scalars

Suppose $M$ is an $R$-module and $R$ is a subring of $S$. We have seen that we have a natural $(S,R)$-bimodule structure on $S$ and $(R,\mathbf{Z})$-bimodule structure on $M$, so we can form the tensor product $S{\beta \x8a\x97}_{R}M$. The result is an $(S,\mathbf{Z})$-bimodule, i.e., a left $S$-module. Comparing the construction above with our previous construction, we can see that this the same $S$-module we constructed (and also denoted $S{\beta \x8a\x97}_{R}M$) previously.

Tensor product of left $R$-modules over a commutative ring

Suppose $R$ is a commutative ring and $M$ and $N$ are left $R$-modules. By taking the standard $R$-module structure (i.e., $(R,R)$-bimodule structure) on $M$ and the canonical $(R,\mathbf{Z})$-bimodule structure on $N$, we can form the tensor product $M{\beta \x8a\x97}_{R}N$. The result is an $(R,\mathbf{Z})$-bimodule, i.e., a left $R$-module.

We can also consider the standard $(R,R)$-bimodule structures on both $M$ and $N$ and form the tensor product $M{\beta \x8a\x97}_{R}N$, which is now an $(R,R)$-bimodule. It is a bit unfortunate that this notation is identical to the one above, and is the first situation in which one should be careful to distinguish the type of modules being considered for the tensor product construction.

Tensor products of rings

Suppose $R$ and $S$ are rings (with unity). We can then also view $R$ and $S$ as $(\mathbf{Z},\mathbf{Z})$-bimodules, by forgetting their internal multiplicative operations and remembering only their additive structures.^{[4]} We can then form the $(\mathbf{Z},\mathbf{Z})$-bimodule $R{\beta \x8a\x97}_{\mathbf{Z}}S$. We can give this abelian group the structure of a ring by defining the product on simple tensors by

and then extending linearly to all of $R{\beta \x8a\x97}_{\mathbf{Z}}S$. One can verify that this makes $R{\beta \x8a\x97}_{\mathbf{Z}}S$ into a ring (with unity given by ${1}_{R}\beta \x8a\x97{1}_{S}$).

This tensor product construction on rings is usually simply denoted $R\beta \x8a\x97S$.