## Tensor Product of Algebras

**Point of Post: **In this post we discuss the tensor product of algebras, and discuss their universal characterization in .

*Motivation*

In this post we will discuss how we can take the tensor product of algebras, and get an algebra. The important thing to note is that while the construction of the tensor product of algebras shall not really differ from our previous construction of the tensor product of modules, the “place” that tensor product occupies in shall be entirely different than the place it occupies in . This will be clear from the universal characterization of the tensor product in .

*Tensor Product of Algebras*

Suppose that have a commutative unital ring and two -algebras and , we can then define their usual tensor product where, since is commutative, this naturally has the structure of an -module, such that . The question is whether or not we can define a ring structure on which is compatible with this -module structure, so that we get an -algebra. This is what the following theorem says:

**Theorem: ***Let and be -algebras, then , with the usual -module structure, is an -algebra in such a way that .*

The idea behind the theorem is straightforward, and just busy work so I leave it’s actual proof to anyone interested/motivated enough.

For us, the most interesting thing about the tensor product of modules is a certain universal mapping problem that it solves. Before we can say exactly what we want to, let’s first note that there are natural maps and given by and respectively. It’s easy to check that these maps, which we shall denote and respectively, are -algebra maps. So with these defined, we can state the theorem that makes the tensor product of algebras so fundamentally interesting:

**Theorem: ***Let and be commutative -algebras. Then, given any -algebra and -algebra maps and there exists a unique -algebra map such that and .*

**Proof: **We define by . We note that evidently is -bilinear and so, by definition, lifts to a map such that . We claim then that is actually an -algebra map. Since is already an -map it suffices to check that it is a ring map. To do this we merely note that

Note next that and and so also satisfies the appropriate diagram. To see that is unique we merely note that the image of and generated as an -algebra, and so clearly any two -algebra maps agreeing on the union of their images must be equal.

There is a classic, somewhat related, theorem that is made simple once one thinks about the tensor product of two algebras in the above way. Indeed, suppose that we have some field and three field extensions with . Suppose that the are finite extensions with . We can then form the “compositum” of the two fields and which is nothing more than the subfield of generated by . It is easy to see that since if and are bases for and respectively then , but it is less obvious when this inequality is actually equality. From basic field theory we know that and so in some easy cases, such as when we have equality, but this is a far cry from a characterization of when equality holds. Somewhat surprisingly we have the following characterization in terms of the tensor product of algebras:

**Theorem: ***Let and be as above. Then, is a field if and only if .*

**Proof: **Note that is obviously a commutative -algebra, and since we have the obvious inclusions (which are evidently -algebra maps) we are granted by the previous theorem an -algebra map with and for each and and thus, in particular, . From this we may conclude that is actually a surjective -algebra map (since it hits the basis as described above). So, assume first that is a field. Then, must be an injection (since it’s a ring map emanating from a field!) and since is an -map this implies that (since we know ) , but since the reverse inequality holds (by above discussion) we may conclude that as desired. Conversely, suppose that . Since is a surjective -map and the domain and codomain spaces are of the same finite dimension, we may conclude from basic linear algebra that must actually be a bijection. Thus, is a bijective ring map, and so as rings, but since is a field this implies that is as well.

While billed as a neat way to tell when the compositum of two extension fields has degree equal to the product of its factors, the above is actually neat in the sense that it also allows us to tell when such a tensor product of fields (over a field) is still a field and, in fact, gives us a nice description of it. For example, since and are finite extensions of of coprime degree the above tells not only that is a field, but that it’s (in fact) isomorphic as -algebras to !

**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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