The Unapologetic Mathematician

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.

Like this:

Posted by John Armstrong | 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 | October 26, 2009 | Reply

« Previous | Next »

About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 40, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9179376363754272, "perplexity_flag": "head"}