## Free Algebras

Let’s work an explicit example from start to finish to illustrate these free monoid objects a little further. Consider the category of modules over the commutative ring , with tensor product over as its monoidal structure. We know that monoid objects here are -algebras with units.

Before we can invoke our theorem to cook up free -algebras, we need to verify the hypotheses. First of all, is symmetric. Well, remember that the tensor product is defined so that -bilinear functions from to are in bijection with -linear functions from to . So we’ll define the function . Now there is a unique function which sends to . Naturality and symmetry are straightforward from here.

Now we need to know that is closed. Again, this goes back to the definition of tensor products. The set consists of -linear functions from to , which correspond to -bilinear functions from to . Now we can use th same argument we did for sets to see such a function as a -linear function from to the -module . Remember here that every modyle over is both a left and a right module because is commutative. That is, we have a bijection . Naturality is easy to check, so we conclude that is indeed closed.

Finally, we need to see that has countable coproducts. But the direct sum of modules gives us our coproducts (but not products, since our index set is infinite). Then since is closed the tensor product preserves all of these coproducts.

At last, the machinery of our free monoid object theorem creaks to life and says that the free -algebra on a -module is . And we see that this is exactly how we constructed the free ring on an abelian group! In fact, that’s a special case of this construction because abelian groups are -modules and rings are -algebras.

[…] is exactly the free algebra on a vector space, and it’s just like we built the free ring on an abelian group. If we […]

Pingback by Tensor and Symmetric Algebras « The Unapologetic Mathematician | October 26, 2009 |