Abstract Nonsense

Basic Properties of the Tensor Product (Pt. I)

Point of Post: In this post we prove some of the basic properties of tensor products: commutativity (in some cases), associativity, identity (i.e. that $R$ acts as an identity for tensor product), etc.

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Motivation

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This is just the standard post when we define a new operation, the checking that the basic properties of the operation hold: associativity, commutativity, etc.

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The Basics

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Theorem (Identity): Let $R$ be a ring and $M$ a left $R$-module. Then, if we consider $R$ as an $(R,R)$-bimodule in the usual way so that $R\otimes_R M$ is considered as a left $R$-module. Then, $R\otimes_R M\cong M$. Similarly, if $M$ is a right $R$-module then $M\otimes_R R\cong M$ as right $R$-modules. Moreover, these isomorphisms are natural.

Proof: We do the case when $M$ is a left $R$-module because the other case is done similarly. So, define $f:R\times M\to M$ by $(r,m)\mapsto rm$. Note then that

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\begin{aligned}&f(r+r',m)=(r+r')m=rm+r'm=f(r,m)+f(r',m)\\ &f(r,m+m')=r(m+m')=rm+rm'=f(r,m)+f(r,m')\\& f(rs,m)=(rs)m=r(sm)=f(r,sm)\\ &f(sr,m)=(sr)m=s(rm)=sf(r,m)\end{aligned}

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so that $f$ is an $R$-biadditve $R$-map. Thus, by the universal characterization of tensor products $f$ lifts to a map $\widetilde{f}:R\otimes_R M\to M$ such that $\widetilde{f}(r\otimes m)=rm$.  We now claim that $\widetilde{f}$ is an isomorphism. Indeed, to see that $\widetilde{f}$ is surjective we merely note that for any $m\in M$ one has that $\widetilde{f}(1\otimes m)=m$. Now, to see that $\widetilde{f}$ is an injection we merely note that if $\displaystyle \widetilde{f}\left(\sum_i r_i\otimes m_i\right)=\sum_i r_i m_i=0$ then

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$\displaystyle \sum_i r_i\otimes m_i=\sum_i 1\otimes r_i m_i=1\otimes\left(\sum_i r_im_i\right)=1\otimes 0=0$

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and so $\ker\widetilde{f}$ is trivial. Thus, $\widetilde{f}$ is an isomorphism as desired.

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Now, what we now want to prove is that the tensor functor $R\otimes_R\bullet$ is naturally equivalent to the identity functor on $R\text{-}\mathbf{Mod}$. To prove this we let $\eta=\{\eta_M\}$ (where $\eta_M:R\otimes_R M\to M$ is the above isomorphism) and merely have to check that if $M$ and $N$ are two left $R$-modules and $f:M\to N$ is an $R$ map then the following diagram commutes

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$\begin{matrix}R\otimes_R M & \overset{\eta_M}{\longrightarrow} & M\\ ^{1_R\otimes f}\big\downarrow & & \big\downarrow ^{f}\\ R\otimes N & \underset{\eta_N}{\longrightarrow} & N\end{matrix}$

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but since this is equivalent to checking two maps out of a tensor product are equal, we must only check this commutativity is true on simple tensors. But, $f(\eta_M(r\otimes m))=f(rm)=rf(m)$ and $\eta_N((1_R\otimes f)(r\otimes m))=\eta(r\otimes f(m))=rf(m)$ from where the conclusion follows. $\blacksquare$

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Theorem(Commutativity): Let $M$ be a $(A,R)$-bimodule and $N$ a $(R,A)$-bimodule where $A$ is commutative. Then $M$ is a $(R^\text{op},A)$-bimodule and $N$ a $(A,R^\text{op})$-bimodule in a natural way. Then,

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$M\otimes_R N\cong N\otimes_{R^\text{op}}M$

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as $(A,A)$-bimodules such that $m\otimes n\mapsto n\otimes m$. Moreover, this isomorphism is natural.

Proof: Define $f:M\times N\to N\otimes_{R^\text{op}}M$ by $(m,n)\mapsto n\otimes m$. To see that this is an $R$-biadditive $(A,A)$-bimap we note that

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\begin{aligned}&f(m,n+n')=(n+n')\otimes m=n\otimes m+n'\otimes m=f(m,n)+f(m,n')\\ & f(m+m',n)=n\otimes (m+m')=n\otimes m+n\otimes m'=f(m,n)+f(m,n')\\ &f(mr,n)=n\otimes mr=n\otimes rm=nr\otimes m=rn\otimes m=f(m,rn)\\ & f(am,na')=(na')\otimes (am)=(a'n)\otimes (ma)=a'(n\otimes m)a=a(n\otimes m)a'=af(m,n)a'\end{aligned}

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and so $f$ is lifted to a map $\widetilde{f}:M\otimes_R N\to N\otimes_{R^\text{op}} M$ such that $m\otimes n\mapsto n\otimes m$. Now, we could repeat the same procedure to produce a map $q:N\otimes_{R^\text{op}}M\to M\otimes_R N$ such that $n\otimes m\mapsto m\otimes n$. It’s easy then to see that $\widetilde{f}$ and $q$ are mutual inverses and so $\widetilde{f}$ is an isomorphism.

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Now we wish to show that the two functors sending $(M,N)\mapsto M\otimes_R N$ and $(M,N)\mapsto N\otimes_{R^\text{op}}M$ are naturally isomorphic. This amounts to showing that if $\eta=\{\eta_{(M,N)}\}$ is the set of isomorphisms $\eta_{(M,N)}:M\otimes_R N\to N\otimes_{R^\text{op}}M$ and if $(M,N)\xrightarrow{(f,g)}(M',N')$ is a morphism in the product category $(A,R)\text{-}\mathbf{Mod}\times (R,A)\text{-}\mathbf{Mod}$ then the follwoing diagram commutes

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$\begin{matrix}M\otimes_R N & \overset{\eta_{(M,N)}}{\longrightarrow} & N\otimes_{R^\text{op}}M\\ ^{f\otimes g}\big\downarrow & & \big\downarrow^{g\otimes f}\\ M'\otimes_R N' & \underset{\eta_{(M',N')}}{\longrightarrow} & N'\otimes_{R^\text{op}} M'\end{matrix}$

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But, as always, since we are dealing with maps out of tensor products it suffices to check this is true on simple tensors. But, this is easy since

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$(g\otimes f)(\eta_{(M,N)}(m\otimes n))=(g\otimes f)(n\otimes m)=g(n)\otimes f(m)$

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and

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

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The conclusion follows. $\blacksquare$

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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.