Tensor products of abelian groups
Often enough we’re going to see the following situation. There are three abelian groups — , , and — and a function that is linear in each variable. What does this mean?
Well, “linear” just means “preserves addition” as in a homomorphism, but I don’t mean that is a homomorphism from to . That would mean the following equation held:
Instead, I want to say that if we fix one variable, the remaining function of the other variable is a homomorphism. That is
we call such a function “bilinear”.
The tensor product is a construction we use to replace bilinear functions by linear functions. It is defined by yet another universal property: a tensor product of and is an abelian group and a bilinear function so that for every other bilinear function there is a unique linear function so that . Like all objects defined by universal properties, the tensor product is automatically unique up to isomorphism if it exists.
So here’s how to construct one. I claim that has a presentation by generators and relations as an abelian group. For generators take all elements of , and for relations take all the elements and for all and in and and in .
By the properties of such presentations, any function of the generators — of — defines a linear function on if and only if it satisfies the relations. That is, if we apply a function to each relation and get every time, then defines a unique function on the presented group. So what does that look like here?
So a bilinear function gives rise to a linear function , just as we want.
Usually we’ll write the tensor product of and as , and the required bilinear function as .
Now, why am I throwing this out now? Because we’ve already seen one example. It’s a bit trivial now, but catching it while it’s small will help see the relation to other things later. The distributive law for a ring says that multiplication is a bilinear function of the underlying abelian group! That is, we can view a ring as an abelian group equipped with a linear map for multiplication.