Abstract Nonsense

Crushing one theorem at a time

Tensor Product of Algebras

Point of Post: In this post we discuss the tensor product of algebras, and discuss their universal characterization in R\text{-}\mathbf{Alg}.

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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 R\text{-}\mathbf{Alg} shall be entirely different than the place it occupies in R\text{-}\mathbf{Mod}. This will be clear from the universal characterization of the tensor product in R\text{-}\mathbf{Alg}.

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Tensor Product of Algebras

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

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Theorem: Let A and B be R-algebras, then A\otimes_R B, with the usual R-module structure, is an R-algebra in such a way that (a\otimes b)(a'\otimes b')=(aa)\otimes (bb').

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

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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 A\to A\otimes_R B and B\otimes_R B given by a\mapsto a\otimes1 and b\mapsto 1\otimes b respectively. It’s easy to check that these maps, which we shall denote \iota_A and \iota_B respectively, are R-algebra maps. So with these defined, we can state the theorem that makes the tensor product of algebras so fundamentally interesting:

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Theorem: Let A and B be commutative R-algebras. Then, given any R-algebra C and R-algebra maps f:A\to C and g:B\to C there exists a unique R-algebra map j:A\otimes_R B\to C such that j\circ\iota_A=f and j\circ\iota_B=g.

Proof: We define t:A\times B\to C by t(a,b)=f(a)g(b). We note that evidently t is R-bilinear and so, by definition, lifts to a map j:A\otimes_R B\to C such that j(a\otimes b)=f(a)g(b). We claim then that j is actually an R-algebra map. Since j is already an R-map it suffices to check that it is a ring map. To do this we merely note that

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\begin{aligned}j\left(\left(\sum_i a_i\otimes b_i\right)\left(\sum_j c_j\otimes d_j\right)\right) &=j\left(\sum_{i,j}(a_i\otimes b_i)(c_j\otimes d_j)\right)\\ &=\sum_{i,j}j((a_ic_j)\otimes (b_id_j))\\ &=\sum_{i,j}f(a_i c_j)g(b_id_j)\\ &= \sum_{i,j}f(a_i)f(c_j)g(b_i)g(d_j)\\ &= \left(\sum_i f(a_i)g(b_i)\right)\left(\sum_j f(c_j)b(d_j)\right)\\ &= j\left(\sum_i a_i\otimes b_i\right)j\left(\sum_j c_j\otimes d_j\right)\end{aligned}

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Note next that j(a\otimes 1)=f(a)g(1)=f(a) and j(1\otimes b)=f(1)g(b)=g(b) and so j also satisfies the appropriate diagram. To see that j is unique we merely note that the image of \iota_A and \iota_B generated A\otimes_R B as an R-algebra, and so clearly any two R-algebra maps agreeing on the union of their images must be equal. \blacksquare

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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 F and three field extensions K,k_1,k_2/F with k_i\subseteq K. Suppose that the k_i are finite extensions with [k_i:F]=n_i. We can then form the “compositum” k_1k_2 of the two fields k_1 and k_2 which is nothing more than the subfield of K generated by k_1\cup k_2. It is easy to see that [k_1k_2:F]\leqslant nm since if \{e_1,\cdots,e_n\} and \{f_1,\cdots,f_m\} are bases for k_1 and k_2 respectively then k_1k_2=\text{span}_F\{e_if_j:i\in[n]\text{ and }j\in[m]\}, but it is less obvious when this inequality is actually equality. From basic field theory we know that n,m\mid [k_1k_2:F] and so in some easy cases, such as when (n,m)=1 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:

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Theorem: Let F,k_1,k_2 and K be as above. Then, k_1\otimes_F k_2 is a field if and only if [k_1k_2:F]=nm.

Proof: Note that k_1k_2 is obviously a commutative F-algebra, and since we have the obvious inclusions k_i\hookrightarrow k_1k_2 (which are evidently F-algebra maps) we are granted by the previous theorem an F-algebra map j:k_1\otimes_F k_2\to k_1k_2 with j(a\otimes 1)=a and j(1\otimes b)=b for each a\in k_1 and b\in k_2 and thus, in particular, j(a\otimes b)=j(a\otimes 1)j(1\otimes b)=ab.  From this we may conclude that j is actually a surjective F-algebra map (since it hits the basis \{e_if_j\} as described above). So, assume first that k_1\otimes_F k_2 is a field. Then, j must be an injection (since it’s a ring map emanating from a field!) and since j is an F-map this implies that [k_1k_2:F]\geqslant nm (since we know \dim_F k_1\otimes_F k_2=nm) , but since the reverse inequality holds (by above discussion) we may conclude that [k_1k_2:F]=nm as desired. Conversely, suppose that [k_1k_2:F]=nm. Since j is a surjective F-map and the domain and codomain spaces are of the same finite dimension, we may conclude from basic linear algebra that j must actually be a bijection. Thus, j is a bijective ring map, and so k_1k_2\cong k_1\otimes_F k_2 as rings, but since k_1k_2 is a field this implies that k_1\otimes_F k_2 is as well. \blacksquare

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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 \mathbb{Q}(\sqrt{2}) and \mathbb{Q}(\sqrt[3]{2}) are finite extensions of \mathbb{Q} of coprime degree the above tells not only that \mathbb{Q}(\sqrt{2})\otimes_\mathbb{Q}\mathbb{Q}(\sqrt[3]{2}) is a field, but that it’s (in fact) isomorphic as \mathbb{Q}-algebras to \mathbb{Q}(\sqrt{2},\sqrt[3]{2})!

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[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.


February 7, 2012 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , , , , ,


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