Abstract Nonsense

Crushing one theorem at a time

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.

[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 | , , , , , , , , , , , ,

2 Comments »

  1. […] Basic Properties of Tensor Products (Pt. II) Point of Post: This is a continuation of this post. […]

    Pingback by Basic Properties of Tensor Products (Pt. II) « Abstract Nonsense | January 5, 2012 | Reply

  2. […] example of a flat -module is for we know that is naturally isomorphic to the forgetful functor which is definitively […]

    Pingback by Flat Modules (Pt. I) « Abstract Nonsense | May 4, 2012 | Reply


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