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

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

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

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

Let be a commutative ring, then any left -module with multiplication , can be given the -bimodule structure defined by –the only axiom that needs to be checked is that for all and . But, this follows from commutativity of the -multiplication: . Of course, this is what allows us to make no distinction between left and right -modules. Thus, given two -modules and we see then that the -biadditive -bimaps (where is just any -module, thought of as an -bimodule in the usual way) is nothing more than the classic notion of a *bilinear *i.e. linear in each variable– and . We then note that we have made to be just the usual abelian group with an -bimodule structure (i.e. just an -module structure, I want to grill this in) defined to be such that . Moreover, we see that with this -module structure on one has that is universal with respect to bilinear maps out of .

Generalizing, any left or right -module can be given the structure of a -bimodule (where, of course, is the center of ) defined, as usual, by –once again we note that this is practically equivalent to considering as a -module. We see then that given a left -module and a left -module , both thought of as -bimodule (i.e. a -module) , an -biadditive -bimap where is some -bimodule (i.e. a -module) is nothing more than an -biadditive map which is -bilinear.

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

If is a unital ring and a two-sided ideal of and two unital subrings of . Then is an -bimodule defined by just left and right multiplication–the compatibility condition between the left and right multiplication is merely just associativity.

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

**References:**

[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.

[…] 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: […]

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[…] 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 […]

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