Tensor Product as a Universal Object (Category Theory & Module Theory)

Introduction

It is quite often to see direct sum or direct product of groups, modules, vector spaces. Indeed, for modules over a ring , direct products are also direct products of -modules as well. On the other hand, the direct sum is a coproduct in the category of -modules.

But what about tensor products? It is some different kind of product but how? Is it related to direct product? How do we write a tensor product down? We need to solve this question but it is not a good idea to dig into numeric works.

The category of bilinear or even -multilinear maps

From now on, let be a commutative ring, and are -modules. Mainly we work on and , i.e. and . For -multilinear one, simply replace with and with . The only difference is the change of symbols.

The bilinear maps of determines a category, say or we simply write . For an object in this category we have as a bilinear map and as a -module of course. For two objects and , we define the morphism between them as a linear function making the following diagram commutative:

morphism-in-BL

This indeed makes a category. If we define the morphisms from to by (for simplicity we omit and since they are already determined by and ) we see the composition satisfy all axioms for a category:

CAT 1 Two sets and are disjoint unless and , in which case they are equal. If but for example, for any , we have , hence . Other cases can be verified in the same fashion.

CAT 2 The existence of identity morphism. For any , we simply take the identity map . For , we see . For , we see .

CAT 3 The law of composition is associative when defined.

There we have a category. But what about the tensor product? It is defined to be initial (or universally repelling) object in this category. Let's denote this object by .

For any , we have a unique morphism (which is a module homomorphism as well) . For and , we write . We call the existence of the universal property of .

The tensor product is unique up to isomorphism. That is, if both and are tensor products, then in the sense of module isomorphism. Indeed, let and be the unique morphisms respectively, we see , , and therefore Hence is the identity of and is the identity of . This gives .

What do we get so far? For any modules that is connected to with a bilinear map, the tensor product of and , is always able to be connected to that module with a unique module homomorphism. What if there are more than one tensor products? Never mind. All tensor products are isomorphic.

But wait, does this definition make sense? Does this product even exist? How can we study the tensor product of two modules if we cannot even write it down? So far we are only working on arrows, and we don't know what is happening inside an module. It is not a good idea to waste our time on 'nonsenses'. We can look into it in an natural way. Indeed, if we can find a module satisfying the property we want, then we are done, since this can represent the tensor product under any circumstances. Again, all tensor products of and are isomorphic.

A natural way to define the tensor product

Let be the free module generated by the set of all tuples where and , and be the submodule generated by tuples of the following types: First we have a inclusion map and the canonical map . We claim that is exactly what we want. But before that, we need to explain why we define such a .

The reason is quite simple: We want to make sure that is bilinear. For example, we have due to our construction of (other relations follow in the same manner). This can be verified group-theoretically. Note but Hence we get the identity we want. For this reason we can write Sometimes to avoid confusion people may also write if both and are -modules. But before that we have to verify that this is indeed the tensor product. To verify this, all we need is the universal property of free modules.

tensor-product-universal

By the universal property of , for any , we have a induced map making the diagram inside commutative. However, for elements in , we see takes value , since is a bilinear map already. We finish our work by taking . This is the map induced by , following the property of factor module.

Trivial tensor product

For coprime integers , we have where means that the module only contains and is considered as a module over for . This suggests that, the tensor product of two modules is not necessarily 'bigger' than its components. Let's see why this is trivial.

Note that for and , we have since, for example, for and . If you have trouble understanding why , just note that the submodule in our construction contains elements generated by already.

By Bézout's identity, for any , we see there are and such that , and therefore Hence the tensor product is trivial. This example gives us a lot of inspiration. For example, what if and are not necessarily coprime, say ? By Bézout's identity still we have This inspires us to study the connection between and . By the universal property, for the bilinear map defined by (there should be no difficulty to verify that is well-defined), there exists a unique morphism such that Next we show that it has a natural inverse defined by Taking , we show that , that is, we need to show that By Bézout's identity, there exists some such that . Hence , which gives since So is well-defined. Next we show that this is the inverse. Firstly Secondly, Hence and we can say If are coprime, then , hence is trivial. More interestingly, . But this elegant identity raised other questions. First of all, , which implies Further, for , we have , which gives hence Hence for modules of the form , we see the tensor product operation is associative and commutative up to isomorphism. Does this hold for all modules? The universal property answers this question affirmatively. From now on we will be keep using the universal property. Make sure that you have got the point already.

Tensor product as a binary operation

Let be -modules, then there exists a unique isomorphism for , , .

Proof. Consider the map where . Since is bilinear, we see is bilinear for all . Hence by the universal property there exists a unique map of the tensor product: Next we have the map which is bilinear as well. Again by the universal property we have a unique map This is indeed the isomorphism we want. The reverse is obtained by reversing the process. For the bilinear map we get a unique map Then from the bilinear map we get the unique map, which is actually the reverse of : Hence the two tensor products are isomorphic.

Let and be -modules, then there exists a unique isomorphism where and .

Proof. The map is bilinear and gives us a unique map given by . Symmetrically, the map gives us a unique map which is the inverse of .

Therefore, we may view the set of all -modules as a commutative semigroup with the binary operation .

Maps between tensor products

Consider commutative diagram:

tensor-prouct

Where are some module-homomorphism. What do we want here? On the left hand, we see sends to , which is quite natural. The question is, is there a natural map sending to ? This is what we want from the right hand. We know exists, since we have a bilinear map by . So for , we have as what we want.

But in this graph has more interesting properties. First of all, if an , both and are identity maps, then we see is the identity as well. Next, consider the following chain We can make it a double chain:

tensor-double-chain

It is obvious that , which also gives Hence we can say is functorial. Sometimes for simplicity we also write or simply , as it sends to . Indeed it can be viewed as a map