# Abstract Nonsense

## 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.