# The Unapologetic Mathematician

## Tensor products of modules

The notion of a tensor product also extends to modules, but the generalization is not quite as straightforward as it was for direct sums.

We start with a ring $R$ and consider a right $R$-module $A_R$ and a left $R$-modules ${}_RB$. We consider functions $f:A\times B\rightarrow X$ which take an element of each module and return an element of some abelian group $X$. Such a function is called “middle linear” if

• $f(a_1+a_2,b)=f(a_1,b)+f(a_2,b)$
• $f(a,b_1+b_2)=f(a,b_1)+f(a,b_2)$
• $f(a\cdot r,b)=f(a,r\cdot b)$

That is, if it is bilinear as a function of abelian groups, and if we can pull the action of $R$ from the first argument to the second and back without changing the value of the function. This condition may seem a little artificial, but I’ll motivate it a bit more later.

The tensor product $A\otimes_RB$ is the abelian group for the universal middle linear function, just as the tensor product for abelian groups was the abelian group for the universal bilinear function. We show that such an object exists by a similar construction. Take the free abelian group generated by all elements of $A\times B$ and write a generator as $a\otimes b$. Then impose the relations $(a_1+a_2)\otimes b$, $a\otimes(b_1+b_2)$, and $(a\cdot r)\otimes b=a\otimes(r\cdot B)$. Then given any middle linear map $f:A\times B\rightarrow X$ there is a unique homomorphism of abelian groups $\bar{f}:A\otimes B\rightarrow X$ so that $f(a,b)=\bar{f}(a\otimes b)$.

In general, the tensor product over $R$ is just an abelian group — it eats $R$-module structures like $\hom$ does. That said, like $\hom$ plays well with extra module structures, so does tensor product. If ${}_SA_R$ is a right $R$-module and a left $S$-module, then $A\otimes_RB$ carries the structure of a left $S$-module. Indeed we can define $s\cdot(a\otimes b)=(s\cdot a)\otimes b$ and check that this action respects the relations we imposed. Similarly, if $A$ has an additional right $S$-module structure commuting with the action of $R$ then $A\otimes_RB$ is a right $S$-module, and the same goes for extra modules structures on $B$. Unlike in the case of $\hom$, no structures get “flipped over” in this process.

If $R$ is a commutative ring, then every module is both a left and a right $R$-module. Thus, the same is true of $A\otimes_RB$ — the tensor product eats the left module structure on $A$ and the right module structure on $B$, but leaves the other two structures.

Now for an example that should motivate the idea of a middle-linear map. Let $A_R$, ${}_SB_R$, and $C_R$ be right $R$-modules with an extra left $S$-module structure on $B$. Then $\hom(A,B)$ is a left $S$-module and $\hom(B,C)$ is a right $S$-module. We consider $f\in\hom(A,B)$ and $g\in\hom(B,C)$ and calculate the two composites
$\left[g\circ (s\cdot f)\right](a)=g(\left[s\cdot f\right](a))=g(s\cdot(f(a)))$
$\left[(g\cdot s)\circ f\right](a)=\left[g\cdot s\right](f(a))=g(s\cdot(f(a)))$
so we can use the $S$ action on either factor. This means that composition of these homomorphisms is a middle-linear function, and so defines a linear function $\hom(B,C)\otimes_S\hom(A,B)\rightarrow\hom(A,C)$.

April 30, 2007 - Posted by | Ring theory

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