Abstract Nonsense

Crushing one theorem at a time

Adjoint Isomorphism for Tensor and Hom (Pt. II)


Point of Post: This is a continuation of this post.

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From all of this we can conclude that \eta is a well-defined map (yay!). Let’s now verify that it is a group isomorphism. To see that \eta is a homomorphism we merely note that

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

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and since m,n were arbitrary the desired equality holds. Lastly, let’s check that \eta is a bijection. Injectivity is simple enough, if \eta(f)=\eta(g) then we see that f and g agree on all simple tensors in M\otimes_R N and so (because the simple tensors are a generating set) must be equal on all of M\otimes_R N. For surjectivity, let F:M\to\text{Hom}_S(N,L) be an arbitrary (A,R)-map. Define the map f:M\times N\to L given by f(m,n)=F(m)(n). We claim that f is an R-biadditive (A,S)-bimap. Indeed:

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\begin{aligned}f(amr,ns) &= F(amr)(ns)\\ &=F(amr)(n)s\\ &(aF(m)r)(n)s\\ &=aF(m))(rn)s\\ &=af(m,rn)s\end{aligned}

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Thus, by the universal properties of the tensor product f lifts to an (A,S)-map \widetilde{f}:M\otimes_R N\to L such that \widetilde{f}(m\otimes n)=F(m)(n). Note then that \eta(\widetilde{f})(m)(n)=\widetilde{f}(m\otimes n)=F(m)(n) for all m,n and so \eta(\widetilde{f})=F. Thus, surjectivity, and thus bijectivity, and thus isomorphism follow.

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Let us fix L,N and show that in M \eta_{(M,N,L)} is a natural transformation. Indeed, suppose that f:M\to M' is an (A,R)-map then we wish to show the commutativity of the following diagram

\text{ }

\begin{matrix}\text{Hom}_{(A,S)}(M'\otimes_R N,L) & \overset{\eta_{(M',N,L)}}{\longrightarrow} & \text{Hom}_{(A,R)}(M',\text{Hom}_S(N,L))\\ ^{(f\otimes1_N)^\ast}\big\downarrow & & \big\downarrow ^{f^\ast}\\ \text{Hom}_{(A,S)}(M\otimes_R N,L) & \underset{\eta_{(M,N,L)}}{\longrightarrow} & \text{Hom}_{(A,R)}(M,\text{Hom}_S(M,L))\end{matrix}

\text{ }

but this is easy:

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f^\ast((\eta_{(M',N,L)})(g))(m)(n)=(\eta_{(M',N,L)}(g)(f(m)))(n)=g(f(m)\otimes n)

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and

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 \begin{aligned}(\eta_{(M,N,L)}(f\otimes 1_N)^\ast(g))(m)(n) &=(\eta_{(M,N,L)}(g\circ(f\otimes 1_N))(m)(n)\\ &=(g\circ (f\otimes 1_N))(m\otimes n)\\ &=g(f(m)\otimes n)\end{aligned}

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The other naturalities follow similarly.

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The case for \nu is basically the same as the case for \eta, and so we admit it. \blacksquare

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Ok, so getting a little more down to earth, the above basically says that if we take a commutative ring R and and take three R-modules (thought of, in the usual way, as (R,R)-bimodules) and noting that \eta and \nu are both R-maps in this context, that’s pretty easy to see

\text{ }

\eta(rf)(m)(n)=(rf)(m\otimes n)=rf(m\otimes n)=r\eta(f)(m)(n)

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for example, we have the following corollary (which is the case we’ll most often use):

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Theorem: Let R be a commutative unital ring and M,N,L be unital R-modules. Then, the above maps \nu and \eta give R-isomorphisms

\text{ }

\begin{aligned}&\text{Hom}_R(M\otimes_R N,L)\cong\text{Hom}_R(M,\text{Hom}_R(N,L))\\ &\text{Hom}_R(M\otimes_R N,L)\cong\text{Hom}_R(N,\text{Hom}_R(M,L))\end{aligned}

\text{ }

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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 10, 2012 - Posted by | Uncategorized | , , , , , , , , ,

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