Let be a ring. The set of all polynomials in the variable with coefficients in the ring is denoted by . An example of an element in the set would be .

Some of you would know that is a module over !

Now, let’s look at the polynomial rings and - they’re essentially the same except one consists of polynomials in the variable , and the other in . How are the two related to the polynomial ring ?

To make things easier, let’s fix to be our favourite ring . If I pick an element in , say , and an element from , say , and multiply the two, then I get the polynomial , which is polynomial in AND , i.e. an element in .

OK - so multiplying an element in with an element in seems to give us an element in . So maybe we can define a multiplication map: which sends to .

Can we dare to hope that is a bijection (or maybe an isomorphism of -modules)?

Well, if you’ve had any experience in solving differential equations, and therein a technique called , you’d know this map wouldn’t be surjective. Indeed, not every polynomial in two variables and can be split into two polynomials of one variable each - one in and the other in . For instance, can’t be split into the product of a polynomial in and one in .

Is it at least injective? Well, the product of and is the same as the product of and , meaning that the pairs and map to the same element under , meaning isn’t even injective.

So is neither injective nor surjective - very disappointing for a map so promising. Is there ANYTHING nice about ?

Let’s look at what stopped from being a bijection.

In addition, thanks to the distributive property of polynomials multiplication, we have , and a similar “linearity” in the second component. This, along with the condition described above in Equation make the map into what’s called an - map. These properties are what make VERY special as we shall see now!

Suppose we had another bilinear map , where is any -module, or even just an abelian group. Let us try to define a group homomorphism using .

As observed above, is generated by polynomials of the form , and so it is sufficient to describe where these generators are mapped to under to completely determine the map. We can therefore define . The product is of course equal to , and therefore we have .

The properties of noted above make sure that this map is well-defined i.e. even though the definition of depends on how you split a polynomial - say - as a product, it won’t send the same element in to two different elements in based on how you split it. For instance, if you write as , then would send it to , and if you split it as , then would send it to the element in . But since and are bilinear, these two elements in are equal, so there’s no confusion (check this)!

This gives rise to the following commutative diagram (i.e. if you pick an element in , it’ll land on the same element in no matter which path you take) -

Since this holds for choice of abelian group with a map , the above property of is called a . This group becomes special because it turns bilinear maps from into group homomorphisms from . We call the of and and denote it by (the subscript refers to the fact that it is elements that can be moved around between the two components in the map as in Equation ) .

We are now (hopefully) motivated enough to define the tensor product of two objects- say modules over a commutative ring -

Let and be two modules over . The tensor product of and over is an abelian group with an -bilinear map which satisfies the following universal property - for any abelian group and an -bilinear map , there exists a unique group homomorphism such that the following diagram commutes:

The whole discussion started with the observation that the product of a polynomial in with a polynomial in gives a polynomial in two variables and i.e. an element in the set . But we could have also taken the SUM of and and still ended up with an element in . Could this have been used to get to the tensor product? Or does the sum map give us some other universal property?