Abstract Nonsense

The Construction of the Tensor Product of Modules (Pt. II)

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

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Of course, also unapparent from the above definition is to whether or not tensor products of two given modules always exist since, as is common knowledge, not all categories have initial objects. We thus have now the burden of proving existence of tensor products. To this end, given a right $R$-module $M$ and a left $R$-module $N$ let $M\otimes_R N$ be the group $\mathbb{Z}[M\times N]/A$ where $\mathbb{Z}[M\times N]$ denotes the free abelian group (i.e. free $\mathbb{Z}$-module) on the set of $M\times N$ and $A$ the subgroup of $\mathbb{Z}[M\times N]$ generated by the set of all elements of the form

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\begin{aligned}&\mathbf{(1)}\quad(m+m', n)-(m, n)-(m',n)\\ &\mathbf{(2)}\quad (m,n+n')-(m,n)-(m,n')\\ &\mathbf{(3)}\quad (mr,n)-(m,rn)\end{aligned}

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as $m,m'$, $n,n'$, and $r$ ranges over $M$, $N$, and $R$ respectively. We define then a map $\otimes:M\times N\to M\otimes_R N$ by taking $(m,n)$ to its coset $(m,n)+A$ in $\mathbb{Z}[M\times N]/A$–in other words $\otimes$ is the composition of the universal injection $M\times N\hookrightarrow \mathbb{Z}[M\times N]$ and the canonical projection $\pi:\mathbb{Z}[M\times N]\to \mathbb{Z}[M\times N]/A$. Thus, we see that, roughly, $M\otimes_R N$ is the ‘largest’ possible abelian group which has a biadditive map into it from $M\times N$. This is solidified in the following theorem:

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Theorem: Let $M$ be a right $R$-module and $N$ a left $R$-module. Then, $M\otimes_R N$ is a tensor product of $M$ by $N$.

Proof: Let us first prove that $\otimes:M\times N\to M\otimes_R N$ really is a biadditve map. To check this we must check that the image of the elements of the form $\mathbf{(1)},\mathbf{(2)}$ and $\mathbf{(3)}$  under $\otimes$ are zero, but since $\otimes=\pi\circ\iota$ this is equivalent to showing that the elements of the described form are in $A$, which is true by definition.

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Suppose now that we are given another biadditive map $f:M\times N\to G$ where $G$ is some abelian group. By the definition of free modules we get from this map a unique group homomorphism $g:\mathbb{Z}[M\times N]\to G$ such that $g\circ\iota=f$ where, as per usual, $\iota$ denotes the universal injection $M\times N\hookrightarrow \mathbb{Z}[M\times N]$. What we now claim is that $g$ descends to a map $\widetilde{f}:M\otimes_R N\to G$ given by $\widetilde{f}(x+A)=g(x)$. To prove that this map is well-defined we must show that $A\subseteq\ker g$. To do this it suffices to show that $g$ annihilates each of the elements of type $\mathbf{(1)},\mathbf{(2)}$ and $\mathbf{(3)}$, but since $g\circ\iota=f$ we see that the image of these elements under $g$ is

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\begin{aligned}&f(m+m',n)-f(m,n)-f(m',n)\\ &f(m,n+n')-f(m,n)-f(m,n')\\ &f(mr,n)-f(m,rn)\end{aligned}

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which are all zero since $f$ is$R$-biadditive. Thus, we have that $\widetilde{f}:M\otimes_R N\to G$ is really an $R$-map. Moreover, we see that

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$\widetilde{f}(\otimes(m,n))=\widetilde{f}(m\otimes n)=\widetilde{f}((m,n)+A)=g(m,n)$

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for all $(m,n)$ so that $\widetilde{f}\circ \otimes=g$. Moreover, the uniqueness of such a map is clear since the image of $\otimes$ generates $M\otimes_R N$. $\blacksquare$

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Since tensor products are unique up to isomorphism there is no harm in identifying all tensor products of $M$ by $N$ with $M\otimes_R N$, the same way in which we identify direct and inverse limits with their particular, concrete constructions.

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From this particular construction, and the uniqueness statement of tensor products we get the following theorem:

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Theorem: Let $f:M\times N\to G$ be a tensor product of $M$ by $N$. Then, $\text{im }f$ generates $G$.

Proof: We know that there exists an isomorphism $\phi:M\otimes_R N\to G$ such that $\phi\circ\otimes=f$. But, since the image of $\otimes$ generates $M\otimes_R N$ and $\phi$ is an isomorphism we see that $\text{im }\phi\circ\otimes=\text{im }f$ generates $G$. $\blacksquare$

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Elements of the image of $\otimes:M\times N\to M\otimes_R N$ are called simple tensors. The important thing to note is that while the set of all simple tensors generate $M\otimes_R N$ they do not, in general, constitute all of $M\otimes_R N$. For example, we shall eventually show that if $\{e_1,e_2\}$ is the standard basis of $\mathbb{R}^2$ then $e_1\otimes e_2+e_2\otimes e_1$ is not a simple tensor. For this (among others) reason, it is important to note that we cannot just define a map on simple tensors and hope for it to be well-defined. To define a map $m\otimes n\mapsto f(m\otimes n)$ on simple tensors alone we have to use the universal property of tensor maps and show that we have an $R$-biadditive map out of $M\times N$ such that $(m,n)\mapsto f(m,n)$ and then lift it to our desired map.

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Bimodules and Tensor Products

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Perhaps the most confusing thing about tensor products is what ‘objects’ they are. Namely, from the above construction the answer is that a tensor product is an abelian group. Of course, doing module theory we’d like to have extra structure on the tensor product, namely consider it as a module over some ring other than $\mathbb{Z}$. For example, it seems desirable to find when we can consider the tensor product $M\otimes_R N$ as being a left $R$-module. We might try to take a naive approach and, taking a “hint” from direct products, try to define  $R$-multiplication on $M\otimes_R N$ in the “obvious way”. In other words, do “some” kind of multiplication such that $r(m\otimes n)=(mr)\otimes n$. The obvious problem with this is in the associativity axioms, namely we need that

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$m(rs)\otimes n=(rs)(m\otimes n)=r(s(m\otimes n))=r(ms\otimes n)=m(sr)\otimes n$

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and so we see that, in general, defining this $R$-multiplication will only make sense if $R$ is commutative.

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The answer to these questions about tensor prodcuts with extra module structure  is answered by introducing the notion of bimodules. Roughly, a bimodule is an abelian group with the module structure from two rings, one acting on the left and one acting on the right. The necessity for considering this can be conceptualized by thinking that  if $M$ and $N$ are just right and left $R$-modules respectively, then $M\otimes_R S$ “eats” the right $R$-modules structure of $M$ and the left $R$-module structure of $N$, leaving no structure left. But, if $M$ or $N$ has some additional module structure then this is “left untouched” by the tensor, allowing us to imbue the tensor product with this structure. So, enough with the strange analogies. For rings $R$ and $S$ we define an $(R,S)$bimodule to be an abelian group $M$ along with a left $R$-module structure $(r,m)\mapsto rm$ and a right $S$-modules structure $(m,s)\mapsto ms$ such that $r(ms)=(rm)s$. If $R=S$ We often called $(R,S)$-bimodules, just $R$-bimodules.  Note that this associativity axiom allows us to unambiguously write $rms$. We can consider then the category $(R,S)\text{-}\mathbf{Mod}$ of all $(R,S)$-bimodules, where a morphism $M\to N$ is a set map $M\to N$ which is both a left $R$-map and a right $S$-map (i.e. $f(rms)=rf(m)s$), called an $(R,S)$-map. The set of all $(R,S)$-maps $M\to N$ is denoted $\text{Hom}_{(R,S)}(M,N)$.  It’s not hard to see that $(R,S)\text{-}\mathbf{Mod}$ is a preadditive category with the usual sum of functions and the zero object being the zero $(R,S)$-bimodule. Obviously the composition of $(R,S)$-maps is an $(R,S)$-map.

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Suppose that $M$ is an $(A,R)$-bimodule and $N$ an $(R,B)$-bimodule. . Now, suppose that we are given a $(A,B)$-bimodule $L$ we define a map $f:M\times N\to L$ to be an $R$-biadditive $(A,B)$-bimap (yes, I know, it’s kind of ugly) to be an $R$-biadditive map such that $f(am,nb)=af(m,n)b$. We can form a category out of such maps, namely we define $\mathbf{BiAdd}_R(M,N;(A,B)\text{-}\mathbf{Mod})$ to be the set of all $R$-biadditive $(A,B)$-bimaps $f:M\times N\to L$ where $L$ is an $(A,B)$-bimodule, and a morphism from $f:M\times N\to L$ to $g:M\times N\to K$ to be an $(A,B)$-map $\phi:L\to K$ such that $\phi\circ f=g$. This is readily verified to be a category, with an isomorphism $\phi$ from $f:M\times N\to L$ to $g:M\times N\to K$ to be an $(A,B)$-isomorphism $\phi:L\to K$ such that $\phi\circ f=g$.

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We then define a $(A,B)$-tensor product over $R$ to be an initial object of $\mathbf{BiAdd}_R\left(M,N;(A,B)\text{-}\mathbf{Mod}\right)$.

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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.

[5] Grillet, Pierre A. Abstract Algebra. New York: Springer, 2007. Print.