## Using Partial Exactness to Compute Things (Pt. I)

**Point of Post: **In this post we show how one can use partial exactness to actually compute, explicitly, the isomorphism type of certain tensor products.

*Motivation*

Up until this point I have committed a sort of embarrassing crime–I have yet to explicitly compute a tensor product! This is not good, because I strongly believe that if one doesn’t stop and compute some things, one can get a little lost in the abstraction. So, this post is devoted primarily to discussing a way in which we can use partial exactness to explicitly compute the isomorphism type of some actual examples. Roughly the idea is that if we can express a module in terms of an exact sequence of the form then we know that is the cokernel of the initial, right exactness tells us then that is exact and so will be the cokernel of the first map. After showing some examples where this is useful we shall describe a general technique for computing tensor products by using “free presentations” which ultimately amount to expressing a module in terms of generators and relations.

*Some Computations*

Let’s get warmed up by using partial exactness (as described above) to prove the following theorem. Oh, although I mentioned it already, in case one has forgotten, if we have an -map we define the *cokernel* of , denoted , to be . With this in mind we get on proving:

**Theorem: ***Let be a unital ring and and ideals of . Then,*

**Proof: **We have the following exact sequence

tensoring with and using right exactness gives

Now we have that but if we consider the isomorphism we see then that

So, what is ? We claim that where, as usual,

To prove that it suffices to show that the image of simple tensors are contained, but this follows since

Conversely, we see that if is in , then this is the image of and so the opposite inclusion follows. Thus, we have that

To finish we prove that the right hand side of the above is isomorphic to . To prove this note that since is contained in the kernel of the canonical map it descends to an epimorphism . To finish we prove that from where the rest will follow from the first isomorphism theorem. Indeed, suppose that is in then we see that

conversely, suppose that is in , then which tells us that so that for and so that . 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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