# Abstract Nonsense

## 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.