## Basic Properties of the Tensor Product (Pt. I)

**Point of Post: **In this post we prove some of the basic properties of tensor products: commutativity (in some cases), associativity, identity (i.e. that acts as an identity for tensor product), etc.

*Motivation*

This is just the standard post when we define a new operation, the checking that the basic properties of the operation hold: associativity, commutativity, etc.

*The Basics*

**Theorem (Identity): ***Let be a ring and a left -module. Then, if we consider as an -bimodule in the usual way so that is considered as a left -module. Then, . Similarly, if is a right -module then as right -modules. Moreover, these isomorphisms are natural.*

**Proof: **We do the case when is a left -module because the other case is done similarly. So, define by . Note then that

so that is an -biadditve -map. Thus, by the universal characterization of tensor products lifts to a map such that . We now claim that is an isomorphism. Indeed, to see that is surjective we merely note that for any one has that . Now, to see that is an injection we merely note that if then

and so is trivial. Thus, is an isomorphism as desired.

Now, what we now want to prove is that the tensor functor is naturally equivalent to the identity functor on . To prove this we let (where is the above isomorphism) and merely have to check that if and are two left -modules and is an map then the following diagram commutes

but since this is equivalent to checking two maps out of a tensor product are equal, we must only check this commutativity is true on simple tensors. But, and from where the conclusion follows.

**Theorem(Commutativity): ***Let be a -bimodule and a -bimodule where is commutative. Then is a -bimodule and a -bimodule in a natural way. Then,*

*as -bimodules such that . Moreover, this isomorphism is natural.*

**Proof: **Define by . To see that this is an -biadditive -bimap we note that

and so is lifted to a map such that . Now, we could repeat the same procedure to produce a map such that . It’s easy then to see that and are mutual inverses and so is an isomorphism.

Now we wish to show that the two functors sending and are naturally isomorphic. This amounts to showing that if is the set of isomorphisms and if is a morphism in the product category then the follwoing diagram commutes

But, as always, since we are dealing with maps out of tensor products it suffices to check this is true on simple tensors. But, this is easy since

and

The conclusion follows.

**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.

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