## 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.

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.