Abstract Nonsense

Crushing one theorem at a time

Functorial Properties of the Tensor Product (Pt. I)


Point of Post: In this post we discuss some of the more “functorial” properties of the tensor product–namely that the tensor product defines a bifunctor, additive in each entry between certain categories.

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Motivation

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Now that we have defined tensor products we’d like to discuss some of the more category theoretic aspects of the construction. The tensor product functor shall serve as one of the prime examples of an additive functor.

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The Functorial Properties of the Tensor Product

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We begin by discussing the tensor product of two maps. Namely, suppose that M_1,M_2 are (A,R)-bimodules and N_1,N_2 are (R,B)-bimodules. Furthermore, suppose that we have an (A,R)-map f:M_1\to N_1 and an (R,B)-map g:M_2\to N_2. We then want to define a map M_1\otimes_R M_2\to N_1\otimes_R N_2 which is an (A,B)-map such that m_1\otimes m_2\otimes f(m_1)\otimes f(m_2). To prove the existence of such a map it suffices to prove that there exists an R-biadditive (A,B)-bimap j:M_1\times M_2\to N_1\otimes_R N_2 such that j(m_1\otimes m_1)=f(m_1)\otimes g(m_2).  To do this we merely define j(m_1,m_2)=f(m_1)\otimes g(m_2). Let’s now show that j is R-biadditive as well as an (A,B)-bimap. To see that j is R-biadditive we note that

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\begin{aligned}&f(m_1+m_1')\otimes g(m_2)=(f(m_1)+f(m_1'))\otimes g(m_2)=f(m_1)\otimes g(m_2)+f(m_1')\otimes g(m_2)\\ &f(m_1)\otimes g(m_2+m_2')=f(m_1)\otimes (g(m_2)+g(m_2'))=f(m_1)\otimes g(m_2)+f(m_1)\otimes g(m_2')\\ & f(m_1r)\otimes g(m_2)=f(m_1)r\otimes g(m_2)=f(m_1)\otimes rg(m_2)=f(m_1)\otimes g(rm_2)\end{aligned}

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to see that j is a (A,B)-bimap we merely note that

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f(am_1)\otimes g(m_2b)=(af(m_1))\otimes (g(m_2)b)=a(f(m_1)\otimes g(m_2))b

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Thus, j is an R-biadditive (A,B)-bimap and so ascends to an (A,B)-map \widetilde{j}:M_1\otimes_R M_2\to N_1\otimes_R N_2 just that \widetilde{j}(m_1\otimes m_2)=f(m_1)\otimes g(m_2)–moreover \widetilde{j} is unique to this property. We call this mapping \widetilde{j} the tensor product of the maps f and g and denote it by f\otimes g.

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Now that we have figured out how to “tensor” two maps we can describe the tensor product as a functor:

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Theorem: Let A,R,B be unital rings. Then, the tensor product \otimes is a covariant bifunctor (A,R)\text{-}\mathbf{Mod}\times (R,B)\text{-}\mathbf{Mod}\to (A,B)\text{-}\mathbf{Mod}, which is additive in each variable.

Proof: The functor (just to be clear) takes (M,N) to M\otimes_R N and takes \left(M_1\xrightarrow{f}_1M_2,N_1\xrightarrow{g}N_2\right) to M_1\otimes_R N_1\xrightarrow{f\otimes g}M_2\otimes_R N_2. To prove that this is, in fact, a functor it suffices to prove that 1_{M}\otimes 1_{N}=1_{M\otimes_R N} and (f_2\otimes g_2)\circ (f_1\otimes g_1)=(f_2\circ f_1)\otimes (g_2\circ g_1). Now, to prove these identities we note that since everything in sight is an (A,B)-map it suffices to check these equalities on simple tensors. To do this we merely note that

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(1_M\otimes 1_N)(m\otimes n)=1_M(m)\otimes 1_N(n)=m\otimes n=1_{M\otimes_R N}(m\otimes n)

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and

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\begin{aligned}(f_2\otimes g_2)((f_1\otimes g_1)(m\otimes n)) &=(f_2\otimes g_2)(f_1(m)\otimes g_1(n))\\ &=f_2(f_1(m))\otimes g_2(g_1(n))\\ &=((f_2\circ f_1)\otimes (g_2\circ g_1))(m\otimes n)\end{aligned}

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By additive in each variable we mean that the functors M\otimes_R \bullet:(R,B)\text{-}\mathbf{Mod}\to (A,B)\text{-}\mathbf{Mod} and \bullet\otimes_R N:(A,R)\text{-}\mathbf{Mod}\to (A,B)\text{-}\mathbf{Mod} defined, as per usual with bifunctors, by (M\otimes_R\bullet)(N)=M\otimes_R N and (M\otimes_R \bullet)(g)=1_M\otimes g (and similarly for \bullet\otimes_R N) are additive functors. We prove this for \bullet\otimes_R N since the other case is similar. So, we want to prove that \bullet\otimes_R N induces a group homomorphism \text{Hom}_{(A,R)}(M_1,M_2)\to\text{Hom}_{(A,B)}(M_1\otimes_R N,M_2\otimes_R N). Of course, this is equivalent to showing that if f_1,f_2:M_1\to M_2 are (A,R)-maps then (f_1+f_2)\otimes 1_N= f_1\otimes 1_N+f_2\otimes 1_N. Once again, since both of these are (A,B)-maps and so it suffices to check this on simple tensors. This is easy though since

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\begin{aligned}((f_1+f_2)\otimes 1_N)(m\otimes n) &=((f_1+f_2)(m))\otimes 1_N(n)\\ &=(f_1(m)+f_2(m))\otimes n\\ &=f_1(m)\otimes n+f_2(m)\otimes n\\ &=(f_1\otimes 1_N)(m\otimes n)+(f_2\otimes 1_N)(m\otimes n)\end{aligned}

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so that (f_1+f_2)\otimes 1_N=f_1\otimes 1_N+f_2\otimes 1_N. \blacksquare

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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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January 4, 2012 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , , , ,

4 Comments »

  1. […] Functorial Properties of the Tensor Product (Pt. II) Point of Post: This is a continuation of this post. […]

    Pingback by Functorial Properties of the Tensor Product (Pt. II) « Abstract Nonsense | January 4, 2012 | Reply

  2. […] 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 […]

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  3. […] consider this as a functor . Suppose we then had some -bimodule . Then, post-composing with the tensor functor  gives us a functor which, upon inspection, looks like . What we’d now like to prove is […]

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  4. […] functors, two of which are the covariant and contravariant Hom functor. The last, of course, is the tensor functor . The tensor functor shows up everywhere in mathematics and is, dare I say it, “more […]

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