Rings from Sets

Hi everyone!  This is my first post and it is more of an aside than a true post.  In this post I will briefly discuss a few methods on how one creates a ring out of a set X.  Forgive me for being so brief, but this far from my area of math so I am less knowledgeable on the subject; however I found this rather interesting.

First we begin with the rather “stupid” method.  If X is finite we consider the bijection to \mathbb{Z}/n.  If X is countable we consider the bijection to \mathbb{Z}.   This is rather uninteresting.

We can also construct the free group on the set X, here we denote this F_X.  We can then look at the group ring \mathbb{Q}F_X.

Next, (forgive any abusive notation here) we can consider \mathbb{Q}[x]: the set of all polynomials with variables from X where

\mathbb{Q}[x]=\frac{\mathbb{Q}F_X}{(\{x_ix_j-x_jx_i\}, \ x_1^{-1}-id_{F_X})}.

We can also construct the power set \mathcal{P}(X) and form a ring with addition being the symmetric difference and multiplication being the intersection.

Finally we can create E_X, the exterior algebra on X which is generated by things of the form a_i[x_i] where x_i\in X and a_i\in \mathbb{Q} with multiplication defined by the wedge product: [x_i]\wedge[x_j] = [x_i\wedge x_j] with the relationship [x_i\wedge x_j] = -[x_j\wedge x_i] and [x_i\wedge x_i] = 0.  Thus a[x_i]\wedge b[x_j] = ab[x_i\wedge x_j].  This will have basis \mathcal{P}(X) when viewed as a vector space if X is finite.

This entry was posted in Asides. Bookmark the permalink.

2 Responses to Rings from Sets

  1. JCummings says:

    Yay Nicki!! Now we just need Corey and we’ll have the whole gang! This post is cool. I especially like the symmetric difference way.

  2. Z Norwood says:

    Good one, Nicki!

    Hotovy: it’s possible countable choice is all that’s necessary to prove that the group ring \mathbb{Q} F_X — which can be made commutative by taking F_X to be the free abelian group on X — has the same cardinality as X (provided X is infinite). The problem is that amorphous sets can appear in the absence of countable choice, and it’s much harder to compute the cardinality of things related to them.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s