# Tensor Products

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.

**Example**

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*.

Definition.Given modules M, N, P, an A-

bilinear mapis a functionsuch that:

- 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 .

Definition.The

tensor productof 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 .

**Exercise A**

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 .

# Construction

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.

Lemma.The pair forms the tensor product of M and N.

**Proof**

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. ♦

**Note**

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:

Corollary 1.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.

**Exercise B**

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.

# Basic Properties

Now we prove some basic properties of the tensor product.

Proposition 1.For any module M, , where .

**Note**

The isomorphism is *natural*. To be exact, both sides are functorial in *M* and we get a natural isomorphism between the functors.

**Proof**

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. ♦

Proposition 2.For any modules M, N, P, we have

**Note**

Again, this isomorphism is functorial in *M*, *N* and *P*.

**Proof**

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 . ♦

Proposition 3.For any modules M, N, we have

**Proof**

Exercise. ♦

**Example**

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.*

In the statement ” Thus the functor – \otimes_A N does not take submodules to submodules. ” did you mean ” does not take injective maps to injective maps” ?

Agreed: that would be more accurate.