In this article (and the next few), we will discuss tensor products of modules over a ring. Here is a motivating example of tensor products.
If and are real vector spaces, then is the vector space with basis over . Hence it follows that .
Also is the complex vector space of all where . Thus .
These properties will be proven as we cover the theory.
Throughout this article, A is a fixed ring and all modules and linear maps are over A.
Given modules M, N, P, an A-bilinear map is a function
- for each , the function is A-linear;
- for each , the function is A-linear.
For example over the base ring , the Euclidean inner product
is bilinear. It is important to distinguish between a bilinear and a linear . For example, in a bilinear map, we must have for all while this is hardly ever true for a linear map.
Philosophically, we want the tensor product of M and N to be the module that classifies all bilinear maps from .
The tensor product of modules M, N comprises of:
where is a module, is a bilinear map, such that for any pair
where Q is a module and is bilinear, there is a unique linear map such that .
1. Prove that the tensor product is unique up to unique isomorphism.
2. For our initial example of , let X be the space of all with a, b real. Now take the bilinear
Compute the resulting .
To prove that the tensor product exists for any A-modules M and N, let be the free A-module generated by as a set. The standard basis of F is denoted by for . Now let P be the quotient of F by the submodule generated by elements of the form:
over all . The map is defined by ; by definition of P this is A-bilinear.
The pair forms the tensor product of M and N.
Let Q be any module and be any bilinear map. We first define a linear by taking . This map factors through P because:
- by bilinearity of we have and thus takes to 0; Similarly it also takes all elements of the form , and to zero.
Hence we obtain a linear which takes . We have
and so .
To prove f is unique, since we must have . But the set of generates F so this proves uniqueness. ♦
Let us set some notation. For the given , any and gives us , denoted by . The universal property of tensor product thus says the following.
- Any A-bilinear induces a unique linear which sends .
From the explicit construction of the tensor product, we also have:
The module is generated by over all and .
Every element of is thus a sum of various ; but in general not every element is of the form . Counter-examples should be easy to find if you take in the beginning of this article.
Prove functoriality: if and are linear maps, they induce a linear satisfying . We denote this map by .
In other words, we get a covariant functor
Even if f is injective, is not injective in general. Thus the functor does not take injective maps to injective maps. An example will be given at the end of this article.
Now we prove some basic properties of the tensor product.
For any module M, , where .
The isomorphism is natural. To be exact, both sides are functorial in M and we get a natural isomorphism between the functors.
Let us construct A-linear maps
To define f we take the map , . This is easily checked to be A-bilinear so it induces an satisfying .
Defining g is easy: set , which is A-linear. It remains to show the maps are mutually inverse. Let , .
- We have due to bilinearity. Since the pure tensors generate we have .
- Also so too. ♦
For any modules M, N, P, we have
Again, this isomorphism is functorial in M, N and P.
Fix and consider the map which takes . This is an A-bilinear map so it induces a linear
Now the map which takes is A-bilinear so it induces a linear
Similarly, we can define a linear which takes . Thus takes elements of the form back to themselves; since these generate the whole module we have . Similarly . ♦
For any modules M, N, we have
Consider , the multiplication-by-2 map. Taking gives us , still multiplication-by-2 but now it is the zero map. Hence tensor products do not take injective maps to injective maps in general.