Tensor Products II - Tensor products of bimodules

Motivation

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 MβŠ—SN 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].

The construction of MβŠ—SN

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 U1(M)Γ—U2(N), where U1 and U2 are the appropriate forgetful functors to Set. We then form the free Z-module on this set, obtaining the abelian group F(U1(M)Γ—U2(N)) whose elements consist of all formal finite sums of the form βˆ‘iki(mi,ni) where ki∈Z, mi∈M and ni∈N. 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

The resulting quotient group is then denoted MβŠ—RN. If we write mβŠ—n for the coset represented by the pair (m,n) in this quotient group (and extend that notation linearly), then every element in MβŠ—RN can be represented (non-uniquely!) by a finite sum of the form βˆ‘imiβŠ—ni for some mi∈M and ni∈N.[2]

How about the (R,T)-bimodule structure? You could probably guess it: for simple tensors we define rβ‹…(mβŠ—n)=rmβŠ—n and (mβŠ—n)β‹…t=mβŠ—nt for r∈R and t∈T, 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 MβŠ—SN to have these properties:

  1. (m1+m2)βŠ—n=m1βŠ—n+m2βŠ—n for every m1,m2∈M and n∈N
  2. mβŠ—(n1+n2)=mβŠ—n1+mβŠ—n2 for every m∈M and n1,n2∈N
  3. msβŠ—n=mβŠ—sn for every m∈M, n∈N, and s∈S
  4. r(mβŠ—n)=rmβŠ—n for every m∈M, n∈N, and r∈R
  5. (mβŠ—n)t=mβŠ—nt for every m∈M, n∈N, and t∈T

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 (Z,Z)-bimodule structures, and so we can form their tensor product AβŠ—ZB. This is another (Z,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,Z)-bimodule structure on M, so we can form the tensor product SβŠ—RM. The result is an (S,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 SβŠ—RM) 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,R)-bimodule structure on M and the canonical (R,Z)-bimodule structure on N, we can form the tensor product MβŠ—RN. The result is an (R,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 MβŠ—RN, 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. Of course, category theory would chastise us for our sloppiness, as properly employing the appropriate forgetful functors would clear up any such confusion.

Tensor products of rings

Suppose R and S are rings (with unity). We can then also view R and S as (Z,Z)-bimodules, by forgetting their internal multiplicative operations and remembering only their additive structures.[4] We can then form the (Z,Z)-bimodule RβŠ—ZS. We can give this abelian group the structure of a ring by defining the product on simple tensors by

(r1βŠ—s1)β‹…(r2βŠ—s2)=r1r2βŠ—s1s2

and then extending linearly to all of RβŠ—ZS. One can verify that this makes RβŠ—ZS into a ring (with unity given by 1RβŠ—1S).

This tensor product construction on rings is usually simply denoted RβŠ—S.


Suggested next note

Tensor Products III - Balanced Maps and a Universal Property of the Tensor Product


  1. In fact, our previous construction will reduce to a special case of this more general construction β†©οΈŽ

  2. Note that the integer coefficients in the free group can now be "absorbed" into either the M- or N-component of the tensors, via the relations we imposed above. For example, 2(mβŠ—n)=(mβŠ—n)+(mβŠ—n)=(m+m)βŠ—n=(2m)βŠ—n, and similarly 2(mβŠ—n)=mβŠ—(2n). β†©οΈŽ

  3. Of course, these actions should really be defined on the free abelian group (pre-quotient) and shown to descend to an action on the quotient, but let's not worry about checking those details for now. β†©οΈŽ

  4. So really we're considering the abelian groups U(R) and U(S), where U is the forgetful functor from rings to abelian groups. β†©οΈŽ