Abstract Nonsense

Crushing one theorem at a time

The Construction of the Tensor Product of Modules (Pt. IV)

Point of Post: This is a continuation of this post.

\text{ }

Thus, we see that despite there being, given a right R-module M and a left R-module N a lot of categories,multiplications, etc. etc. floating around, we see that we must only keep the (relatively) simple construction of M\otimes_R N and the obvious (A,B)-multiplication in mind to create any type of tensor product.

\text{ }

Let’s look at some particular ways in which bimodules naturally arise, and talk about how their tensor products might work.

\text{ }

Let R be a commutative ring, then any left R-module M with multiplication rm, can be given the R-bimodule structure defined by mr\overset{\text{def.}}{=}rm–the only axiom that needs to be checked is that r(ms)=(rm)s for all r,s\in R and m\in M. But, this follows from commutativity of the R-multiplication: r(ms)=r(sm)=(rs)m=(sr)m=s(rm)=(rm)s. Of course, this is what allows us to make no distinction between left and right R-modules. Thus, given two R-modules M and N we see then that the R-biadditive R-bimaps f:M\times N\to L (where L is just any R-module, thought of as an R-bimodule in the usual way) is nothing more than the classic notion of a bilinear i.e. linear in each variable–f(rm+m',n)=rf(m,n)+f(m',n) and f(m,rn+m')=rf(m,n)+f(m,n'). We then note that we have made M\otimes_R N to be just the usual abelian group with an R-bimodule structure (i.e. just an R-module structure, I want to grill this in) defined to be such that r(m\otimes n)=(rm)\otimes n=m\otimes (rn). Moreover, we see that with this R-module structure on M\otimes_R N one has that M\otimes_R N is universal with respect to bilinear maps out of M\times N.

\text{ }

Generalizing, any left or right R-module M can be given the structure of a (Z(R),Z(R))-bimodule (where, of course, Z(R) is the center of R) defined, as usual, by mr\overset{\text{def.}}{=}rm–once again we note that this is practically equivalent to considering M as a Z(R)-module. We see then that given a left R-module M and a left R-module N, both thought of as (Z(R),Z(R))-bimodule (i.e. a Z(R)-module) , an R-biadditive (Z(R),Z(R))-bimap f:M\times N\to L where L is some (Z(R),Z(R))-bimodule (i.e. a Z(R)-module) is nothing more than an R-biadditive map which is Z(R)-bilinear.

\text{ }

Let M and N be a right and left R-module respectively, then we can consider them as (\mathbb{Z},R)-modules and (R,\mathbb{Z})-modules respectively (in the obvious ways). We see that this line of thinking, noting that (\mathbb{Z},\mathbb{Z})-modules are just abelian groups, is related to our first construction of the tensor product.

\text{ }

If R is a unital ring and \mathfrak{a} a two-sided ideal of R and S,S' two unital subrings of R. Then \mathfrak{a} is an (S,S')-bimodule defined by just left and right multiplication–the compatibility condition between the left and right multiplication is merely just associativity.

\text{ }

Remark: I want to make a point about the comments made above. One may get from my above comments that the ONLY (R,R)-bimodule structures are the ones that came from R-modules. Of course, this is not at all true. My attempt to conflate the notions of (R,R)-bimodules and R-module structures (for commutative rings) is merely that in our future encounters we shall rarely be presented, on a silver platter one may say, bimodules. More commonly we shall just be presented, for example, a right R-module M and a left R-module N where R is commutative.  In this case, we’d like to think of M and N as just being (R,R)-bimodules in the natural way, so that if we consider M\otimes_R N it will be as an R-module, instead of just an abelian group. More generally, given any ring R we shall always consider (for the sake of tensor products) any right R-module M and left R-module N as (Z(R),Z(R))-bimodules so that we can consider M\otimes_R N as a (Z(R),Z(R))-bimodule, or more realistically just a Z(R)-module.

\text{ }

\text{ }


[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. Module Theory. Clarendon, 1990. Print.

[4] Lang, Serge. Algebra. Reading, MA: Addison-Wesley Pub., 1965. Print.

[5] Grillet, Pierre A. Abstract Algebra. New York: Springer, 2007. Print.


January 3, 2012 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , , , ,


  1. […] theorem will show. Moreover, if is commutative and and we give every module in sight the usual -bimodule structure then these all collapse to the same definition we have mentioned before: […]

    Pingback by Adjoint Isomorphism for Tensor and Hom « Abstract Nonsense | January 10, 2012 | Reply

  2. […] and . We can then consider the set-valued functor given by on objects (where, as we have previously discussed, is the set of bilinear maps ) and given an arrow we define where (as usual) . So, what does a […]

    Pingback by Universal Arrows and Universal Elements (Pt. II) « Abstract Nonsense | January 12, 2012 | Reply

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: