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 . 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 is finite we consider the bijection to . If is countable we consider the bijection to . This is rather uninteresting.

We can also construct the free group on the set , here we denote this . We can then look at the group ring .

Next, (forgive any abusive notation here) we can consider : the set of all polynomials with variables from where

.

We can also construct the power set and form a ring with addition being the symmetric difference and multiplication being the intersection.

Finally we can create , the exterior algebra on which is generated by things of the form where and with multiplication defined by the wedge product: with the relationship and . Thus . This will have basis when viewed as a vector space if is finite.

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.

Good one, Nicki!

Hotovy: it’s possible countable choice is all that’s necessary to prove that the group ring — which can be made commutative by taking to be the free abelian group on — has the same cardinality as (provided 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.