Abstract Nonsense

Crushing one theorem at a time

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


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

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Once again, this concretely means that if M is an (A,R)-bimodule and N a (R,B)-bimodule then an (A,B) tensor product of M by N over R is an (A,B) module L and an R-biadditive (A,B)-map f:M\times N\to L which has the property such that given any other (A,B)-bimodule K and an R-biadditive (A,B)-map g:M\times N\to K there exists a unique (A,B)-map \phi:L\to K such that \phi\circ f=g.

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For example, the previous construction of a tensor product as an abelian group corresponded to (\mathbb{Z},\mathbb{Z})-tensor products over R.

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Once again, we have by mere definition the following important theorem:

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Theorem: Let M be a (A,R)-bimodule and N a (R,B)-bimodule. Then, if f:M\times N\to L and g:M\times N\to K are two (A,B)-tensor products over R, of M by N then there exists a unique (A,B)-isomorphism \phi:L\to K such that \phi\circ f=g.

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Unfortunately, similar to before there is absolutely no indication that (A,B)-tensor products over a ring should exist in general. Strangely enough, we have the following theorem that brings together all of the theory:

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Theorem: Let M be an (A,R)-bimodule and N an (R,B)-bimodule. Then, the multiplications a(m\otimes n)=(am)\otimes n and (m\otimes n)b=m\otimes (nb) extend to give a (A,B)-bimodule structure on M\otimes_R N, and moreover \otimes:M\times N\to M\otimes_R N is then a (A,B)-tensor product of M by N over R.

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Proof: We begin by showing that there really is a (necessarily unique) A-multiplication on M\otimes_R N for which a(m\otimes n)=(am)\otimes n. Indeed, consider the map f_a:M\times N\to M\otimes_R N given by f_a(m,n)=(am,n). We note then that for any  r\in R, m,m'\in M and n,n'\in N we have that

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\begin{aligned}f_a(mr+m',n+n') &=(a(mr+m'))\otimes(n+n')\\ &=(amr+am')\otimes(n+n')\\ &=(amr)\otimes n+(amr)\otimes n'+(am')\otimes n+(am')\otimes n'\\ &=(am)\otimes (rn)+(am)\otimes(rn')+(am')\otimes n+(am')\otimes n'\\ &= f_a(m,rn)+f_a(m,rn')+f_a(m',n)+f_a(m',n')\end{aligned}

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so that f_a is R-biadditive. Thus, by the universal property of tensor products we get a group homomorphism \widetilde{f}_a:M\otimes_R N\to M\otimes_R N such that \widetilde{f}_a(m\otimes n)=(am)\otimes n. We then define the multiplication map \mu:A\times(M\otimes_R N)\to M\otimes_R N by \mu(a,x)=\widetilde{f}_a(x) and denote, as per usual, \mu(a,x) as just ax. We note though that since

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\begin{aligned}\widetilde{f}_{a+a'}(m\otimes n) &=((a+a')m)\otimes n\ \&=(am+a'm)\otimes n\\ &=(am)\otimes n+(a'm)\otimes n\\ &=\widetilde{f}_a(m)+\widetilde{f}_{a'}(m)\end{aligned}

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and maps on M\otimes_R N are determined by their values on elements of the form m\otimes n we may conclude that \widetilde{f}_{a+a'}=\widetilde{f}_a+\widetilde{f}_{a'}. Similarly, one can check that on elements of the form \widetilde{f}_{aa'}=\widetilde{f}_{a}\circ\widetilde{f}_{a'} and \widetilde{f}_1=\text{id}_{M\otimes_R N} and so they are equal on all of M\otimes_R N. We then proceed to verify the desired axioms

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\begin{aligned}&(a+a')x=\widetilde{f}_{a+a'}(x)=\widetilde{f}_a(x)+\widetilde{f}_{a'}(x)=ax+a'x\\ & a(m+m')=\widetilde{f}_a(x+x')=\widetilde{f}_a(x)+\widetilde{f}_{a'}(x)=ax+ax'\\ & (aa')x=\widetilde{f}_{aa'}(x)=\widetilde{f}_a(\widetilde{f}_{a'}(x))=a(a'x)\\ &1x=\widetilde{f}_1(x)=\text{id}_{M\otimes_R N}(x)=x\end{aligned}

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Thus, we see that we have defined a left A-multiplication (a,x)\mapsto ax on M\otimes_R N such that a(m\otimes n)=(am)\otimes n as desired. The exact same process shows that we have a right B-multiplication (x,b)\mapsto xb on M\otimes_R N such that (m\otimes n)b=m\otimes (nb). To prove that everything defines a (A,B)-bimodule structure on M\otimes_R N it suffices to show that a(xb)=(ax)b for all a\in A, b\in B, and x\in X. It of course suffices to do this on simple tensors:

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a((m\otimes n)b)=a(m\otimes (nb))=(am)\otimes(nb)=((am)\otimes n)b=(a(m\otimes n))b

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thus this really is a (A,B)-bimodule structure on M\otimes_R N.

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Let’s now verify that \otimes:M\times N\to M\otimes_R N really is an (A,B)-tensor product of M by N over R. To do this let’s first check that \otimes is an R-biadditive (A,B)-map. The fact that it’s R-biadditive follows from previous note, and so it suffices to show that it is (A,B)-biadditive. To do this we merely note that

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\otimes (am,nb)=(am)\otimes (nb)=a(m\otimes n)b=a\otimes(m,n)b

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and so the fact that it’s an (A,B)-map follows.

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Let’s now show that it’s initial in the desired category. To this end suppose that we have some  (A,B)-bimodule L and an R-biadditive (A,B)-map f:M\times N\to L. Since f is R-biadditive we are guaranteed, by construction, a group homomorphism \phi:M\otimes_R N\to L such that \phi\circ \otimes=f, it thus remains to show that \phi is an (A,B)-map. To do this let \displaystyle \sum_i m_i\otimes n_i\in M\otimes_R N be arbitrary and a\in A, b\in B we see then that

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\displaystyle \begin{aligned}\phi\left(a\left(\sum_i m_i\otimes n_i\right)b\right) &=\phi\left(\sum_i (am_i)\otimes (n_ib)\right)\\ &=\sum_i \phi((am_i)\otimes (n_ib))=\sum_i f((am_i,n_ib))\\ &=\sum_i f(a(m_i,n_i)b))=\sum_i af(m_i,n_i)b\\ &=\sum_i a \phi(m_i\otimes n_i)b\\ &=a\left(\sum_i \phi(m_i\otimes n_i)\right)b\\ &=a\phi\left(\sum_i m_i\otimes n_i\right)b\end{aligned}

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Thus, \phi is an (A,B)-map M\otimes_R N\to L such that \phi\circ \otimes=f. Since the uniqueness is guaranteed by the construction of M\otimes_R N the fact that \otimes:M\times N\to M\otimes_R N is an (A,B)-tensor product over R 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 3, 2012 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , , , ,

1 Comment »

  1. […] The Construction of the Tensor Product of Modules (Pt. IV) Point of Post: This is a continuation of this post. […]

    Pingback by The Construction of the Tensor Product of Modules (Pt. IV) « Abstract Nonsense | January 3, 2012 | Reply


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