The Unapologetic Mathematician

Mathematics for the interested outsider

Free Algebras

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

Before we can invoke our theorem to cook up free K-algebras, we need to verify the hypotheses. First of all, K\mathbf{-mod} is symmetric. Well, remember that the tensor product is defined so that K-bilinear functions from A\times B to C are in bijection with K-linear functions from A\otimes B to C. So we’ll define the function T(a,b)=b\otimes a. Now there is a unique function \tau:A\otimes B\rightarrow B\otimes A which sends a\otimes b to b\otimes a. Naturality and symmetry are straightforward from here.

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

Finally, we need to see that K\mathbf{-mod} has countable coproducts. But the direct sum of modules gives us our coproducts (but not products, since our index set is infinite). Then since K\mathbf{-mod} 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 K-algebra on a K-module A is \bigoplus\limits_n(A^{\otimes n}). 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 \mathbb{Z}-modules and rings are \mathbb{Z}-algebras.

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August 3, 2007 - Posted by | Category theory

1 Comment »

  1. [...] 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 | Reply


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