Abstract Nonsense

Crushing one theorem at a time

Basic Properties of Tensor Products (Pt. II)


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

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Theorem(Associativity): Let M,N,L be (A,R),(R,S) and (S,B) bimodules respectively. Then, there is an isomorphism

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M\otimes_R(N\otimes_S L)\cong (M\otimes_R N)\otimes_S L

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of (A,B)-bimodules such that m\otimes (n\otimes \ell)\mapsto (m\otimes n)\otimes\ell

Proof: Fix \ell\in L and define

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\displaystyle f_\ell:M\times N\to M\otimes_R(N\otimes_S L):(m,n)\mapsto m\otimes(n\otimes\ell)

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We claim that f_\ell is an R-biadditive A-map. Indeed,

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\displaystyle \begin{aligned}&f_\ell(mr,n)=mr\otimes(n\otimes\ell)=m\otimes r(n\otimes\ell)=m\otimes(rn\otimes\ell)=f_\ell(m,rn)\\ &f_\ell(m+m',n)=(m+m')\otimes(n\otimes\ell)=m\otimes(n\otimes_\ell)+m'\otimes(n\otimes\ell)=f_\ell(m,n)+f_\ell(m',n)\\&f_\ell(m,n+n')=m\otimes((n+n')\otimes\ell)=m\otimes(n\otimes\ell+n'\otimes\ell)=m\otimes(n\otimes\ell)+m\otimes(n'\otimes\ell)=f_\ell(m,n)+f_\ell(m,n')\\& f_\ell(am,n)=am\otimes(n\otimes\ell)=a(m\otimes(n\otimes\ell))=af_\ell(m,n)\end{aligned}

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Thus, by the universal properties of tensor products we are afforded an A-map \widetilde{f}_\ell:M\otimes_R\to M\otimes_R(N\otimes_S L) with the property that \widetilde{f}_\ell(m\otimes n)=m\otimes(n\otimes\ell). We now define a map

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g:(M\otimes_R N)\times L\to M\otimes_R(N\otimes_S L):(x,\ell)\mapsto \widetilde{f}_\ell(x)

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We claim that g is S-biadditive and B-linear. Indeed,

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g(x+x',\ell)=\widetilde{f}_\ell(x+x')=\widetilde{f}_\ell(x)+\widetilde{f}_\ell(x')=g(x,\ell)+g(x',\ell)

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To prove that \widetilde{f}_{\ell}(xs)=g(xs,\ell)=g(x,s\ell)=\widetilde{f}_{s\ell}(x) we merely prove that for each fixed s one has that the maps x\to \widetilde{f}_{\ell}(xs) and x\mapsto \widetilde{f}_{s\ell}(x) are equal. Since both of these maps are R-biadditive, it suffices to check that they agree on simple tensors. To do this we note that \widetilde{f}_\ell((m\otimes n)s)=\widetilde{f}_\ell(m\otimes (ns))=m\otimes(ns\otimes\ell)=m\otimes(n\otimes s\ell) and \widetilde{f}_{s\ell}(m\otimes n)=m\otimes(n\otimes s\ell). To prove that \widetilde{f}_{\ell+\ell'}(x)=g(x,\ell+\ell')=g(x,\ell)+g(x,\ell')=\widetilde{f}_\ell(x)+\widetilde{f}_{\ell'}(x) we once again, show that \widetilde{f}_{\ell+\ell'} and \widetilde{f}_\ell+\widetilde{f}_{\ell'} agree on simple tensors, but this is trivial. Lastly, we want to show that \widetilde{f}_{\ell b}(x)=g(x,\ell b)=g(x,\ell)g=\widetilde{f}_\ell(x)b but this merely requires, once again, checking on simple tensors, which is simple enough.

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From all of this we see that g lifts to an (A,B)-map \widetilde{g}:(M\otimes_R N)\otimes_S L\to M\otimes_R (N\otimes_S L) such that (m\otimes n)\otimes\ell=m\otimes(n\otimes\ell). Now, it’s clear that we could have repeated this same procedure, in the opposite direction, to produce a map

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q:M\otimes_R(N\otimes_S L)\to (M\otimes_R N)\otimes_S L

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with the property that m\otimes(n\otimes\ell)=(m\otimes n)\otimes\ell). It’s easy to see then that \widetilde{g} and q are inverses, and so \widetilde{g} is an isomorphism. \blacksquare

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Of course, by induction the above theorem allows us to freely associate any collection of tensor products and unambiguously denote a tensor product without parenthesising. Of course, the theoretical interpretation of what the tensor product of more than two modules means can be interpreted two-by-two, in the sense that M\otimes_R (N\otimes_S L) M\otimes_R (N\otimes_L) can be seen as the module for which all R-biadditive maps M\times N\otimes_S L factor through. But, as we can see we can “further factor” this to see that we are really looking at modules which are universal for maps out of M\times N\times L which are additive in each variable (with the others fixed) and which the appropriate scalars switch between adjacent entries. To be more explicit, let M_1,\cdots,M_n be (A,R_1),(R_1,R_2),\cdots,(R_n,B) modules respectively. Then, we call an n-additive (A,B)-bimap to be a map f:M_1\times\cdots\times M_n\to L such that f is additive in each variable, and such that

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f(x_1,\cdots,x_{j-1}r,x_j,\cdots,x_n)=f(x_1,\cdots,x_{j-1},rx_j,\cdots,x_n)

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for all j and r\in R_j. We see then M_1\otimes\cdots\otimes M_n, which in literal constructive terms is any associate of M_1\otimes(M_2\otimes(\cdots(\otimes M_{n-1}\otimes M_n))))), is universal with respect to n-additive (A,B)-bimaps out of M_1\times\cdots\times M_n.

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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 5, 2012 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , , , , , , , , ,

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