Abstract Nonsense

Crushing one theorem at a time

Adjoint Isomorphism for Tensor and Hom (Pt. I)

Point of Post: In this post we show that the tensor functor and Hom functor form an adjoint pair.

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In this post we prove one of the fundamental theorems of module theory–that the Hom and tensor functors form an adjoint pair. I do not want to spend a great deal of time explaining what this precisely means, because I will eventually discuss adjoint functors in their full generality, and anything I say here will be nothing more than a specific, less enlightened case of what I say there. That said, saying nothing at all would be a crime. In a very practical sense  the adjointness of Hom and tensor says that there is a one-to-one correspondence of maps of the form M\otimes N\to L and maps of the form M\to \text{Hom}(N,L). The vague reasoning for this should be clear enough, namely a mapping M\otimes N\to L can be thought of as a mapping M\times N\to L (via the universal characterization) and the rest is the common techinique of “currying” where a two-variable map can be thought of as a mapping to a set of maps. Namely, if we have a map f(m,n) fixing one of the variables, say m_0\in M, gives us a map N\to L:n\mapsto f(m_0,n). Thus, we can think that a function M\times N\to L is nothing more than a paramateried set of functions \{g_m:N\to L\}_{m\in M} or a mapping M\to\text{Hom}(N,L). In fact, as an equality of sets this is eqiuvalent to saying that \#(Z^{(X\times Y)})=\#((Z^X)^Y)) or, with cardinal numbers, that \kappa^{(\lambda\mu)}=(\kappa^\lambda)^\mu. The amazing thing is that we aren’t going to just get a bijection of some typese of Hom sets, but an isomorphism. This connection between tensor and Hom will often allow us to make arguments about one in terms of the other, often allowing us to take difficult problems involving, say, tensor and make them into easy looking problems involving Hom.

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We begin by noting how the advent of bimodules allows us to endow Hom sets with more than an abelian group structure. In particular:

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If M be an (R,S)-bimodule and N a left R-module. Then, \text{Hom}_R(M,N) is a left S-module with (sf)(a)=f(as).

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If M is an (R,S)-bimodule and N a right S-module then \text{Hom}_S(M,N) is a right R-module with (fr)(m)=f(rm).

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If M is a right R-module and N an (S,R)-bimodule then \text{Hom}_R(M,N) is a left S-module with (sf)(m)=sf(m).

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If M is a left S-module and N an (S,R)-bimodule then \text{Hom}_S(M,N) is a right R-module with (fr)(m)=f(m)r.

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These are all easily verified. While these may seem awfully contrived (we just multiplied where we could, on the side that made associativity work out) they do have meaning, as the following theorem will show. Moreover, if R is commutative and R=S and we give every module in sight the usual (R,R)-bimodule structure then these all collapse to the same definition we have mentioned before: (rf)(m)=rf(m)=f(rm)=f(mr)=f(m)r.

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Theorem(Adjointness for Hom and Tensor): Let M be an (A,R)-bimodule, N an (R,S)-bimodule, and L an (A,S)-bimodule. Then, there exists group isomorphisms

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\begin{aligned}&\eta_{(M,N,L)}=\eta:\text{Hom}_{(A,S)}(M\otimes_R N,L)\overset{\approx}{\longrightarrow}\text{Hom}_{(A,R)}(M,\text{Hom}_S(N,L))\\ &\nu_{(M,N,L)}=\nu:\text{Hom}_{(A,S)}(M\otimes_R N,L)\overset{\approx}{\longrightarrow} \text{Hom}_{(R,S)}(N,\text{Hom}_A(M,L))\end{aligned}

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such that fixing any two M,N,L and letting the third range over the appropriate set of modules makes the corresponding set of isomorphisms \eta_{(M,N,L)} and \nu_{(M,N,L)} into natural equivalences between the associated functors.

Proof: We define, given an (A,S)-map f:M\otimes_R N\to L a map \eta(f):M\to\text{Hom}_S(N,L) by (\eta(f)(m))(n)=f(m\otimes n). For completeness, let’s check that \eta(f)(m) really is an S-map N\to L. To see that \eta(f)(m) is a group homomorphism we merely note that

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\begin{aligned}\eta(f)(m)(n+n') &=f(m\otimes (n+n'))\\ &=f(m\otimes n+m\otimes n')\\ &=f(m\otimes n)+f(m\otimes n')\\ &=\eta(f)(m)(n)+\eta(f)(m)(n')\end{aligned}

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checking that \eta(f)(m) is a right S-map then comes via

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\eta(f)(m)(ns)=f(m\otimes ns)=f(m\otimes n)s=\eta(f)(m)(n)s

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Thus, we see that \eta(f)(m) really is a right S-map N\to L. Let’s now verify that the correspondence m\mapsto \eta(f)(m) really is an (A,R)-map. To do this we must merely note that

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\begin{aligned}\eta(f)(am+m'r)(n) &=f((am+m'r)\otimes n)\\ &=f(am\otimes n+m'r\otimes n)\\ &=af(m\otimes n)+f(m'\otimes rn)\\ &=a\eta(f)(m)(n)+\eta(f)(m')(rn)\\ &=(a\eta(f)(m))(n)+(\eta(f)(m')r)(n)\end{aligned}

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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.


January 10, 2012 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , , , , ,


  1. […] Adjoint Isomorphism for Tensor and Hom (Pt. II) Point of Post: This is a continuation of this post. […]

    Pingback by Adjoint Isomorphism for Tensor and Hom (Pt. II) « Abstract Nonsense | January 10, 2012 | Reply

  2. […] different directed systems). The idea will go as follows. We’ll basically show using the adjointness of Hom and tensor and previous results on the commuting of Hom and direct limits that the covariant Hom functors […]

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  3. […] We have proven that the tensor functor is ‘right exact’, and moreover we have seen that this partial exactness can be a great boon in our quest to compute things. But, there are other things besides tensor products for which we would like to compute. In particular, we like to compute the homomorphism group/module if not for practical reasons then for purely philosophical ones. Thus, in this post we shall prove that the Hom functor itself has a certain kind of ‘partial exactness’, one which is dual to that of the tensor product: ‘left exactness’. In other words, the Hom functor (in either variance [i.e. the covariant or contravariant]) takes exact sequences of the form to exact sequences of the same form. There should be, at least some, lack of surprise at this fact considering the somewhat dualistic nature that the Hom and tensor functors share, via their adjointness. […]

    Pingback by The Hom Functor is Left Exact « Abstract Nonsense | January 26, 2012 | Reply

  4. […] just a particular case of a more general theorem we have previously discussed the tensor functor and where is some commutative ring and is some -module (thought of in the […]

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  5. […] must show that is an exact functor. But, by adjointness of Hom and tensor we have that but this is clearly just isomorphic to (since is naturally isomorphic to the […]

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