# 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 $\hom(B,C)\otimes_S\hom(A,B)\rightarrow\hom(A,C)$.

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Posted by John Armstrong | Ring theory

## 6 Comments	»

1. [...] and in particular a zero object — the trivial -dimensional vector space . It also has a tensor product, which makes this a monoidal category, using the one-dimensional vector space itself as monoidal [...]

Pingback by | May 19, 2008 | Reply

2. [...] Since we’re looking at vector spaces, which are special kinds of modules, we know that has a tensor product structure. Let’s see what this means when we pick [...]

Pingback by | May 23, 2008 | Reply

3. [...] Remember that the structure is just a bilinear multiplication, which is just a linear map from the tensor square to . And we know that the basis for a tensor product consists of pairs of basis elements. So we [...]

Pingback by | July 28, 2008 | Reply

4. [...] concepts tied to linear algebra, and before we get into that we need to revisit an old topic: tensor powers and the subspaces of symmetric and antisymmetric tensors. Specifically, how do all of these [...]

Pingback by | October 22, 2009 | Reply

5. [...] finite groups and , and we have (left) representations of each one: and . It turns out that the tensor product naturally carries a representation of the product group . Equivalently, it carries a [...]

Pingback by | November 4, 2010 | Reply

6. [...] the tensor product of vector spaces is old hat by now, as is using the dual space . We’ll put them together by [...]

Pingback by | July 6, 2011 | Reply

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