I cant understand free algebra or free ring.

Question

I want to write as a set any free algebra or ring.For instance,
$K[t]/(t^{2})=${$ g(t)+(t^{2}):g(t) \in K[t]$}
$\hspace{1.9cm}=${$at+b+(t^{2}):a,b \in K$}
$(t^{2})$={$f(t)t^{2}:f(t)\in K[t]$}
As above I can write this ring as a set,but
Let k be any ring,and {$x_{i}:i \in I$} be a system of independent,noncommuting indeterminates over k.Then we can form the ''free k-ring'' generated by {$x_{i}:i \in I$},which we denote by
$R=k<x_{i}:i\in I>$
Example:Let $k$ and $R$ as above,and let $F=${$f_{j}:j\in J$}$\subset R$.Writing $(F)$ for ideal generated by $F$ in $R$.
My question is $R=\mathbb{R}<x,y>=?$ and,if $F=${$x^{2}+1,y^{2}+1,xy+yx$},then $R/(F)=?$
I want to write instead of $?$ a set like a above example $K[t]$.

Answer 1

In general, this kind of problem is rather difficult, but in this case it's tractable. First the elements of $R$ are linear combination of non-commutative monomials, like $x^6yx^2xyx$ for instance. So you can see it as the monoid algebra $k[M]$ where $M$is the free monoid on $x$ and $y$ (so words in the two letters $x$ and $y$).

You add that $x$ and $y$ anti-commute, so any monomial like above will be equal in the quotient to $\pm x^ny^m$, by making $x$ and $y$ anti-commute as much as necessery to group all $x$ and all $y$ together.

Then you also add the condition that $x^2 =y^2 = -1$, so any $x^n$ will be equal to either $\pm 1$ or $\pm x$ in the quotient. The same thing happens for $y^n$. Then $x^ny^m\in \{\pm 1,\pm x, \pm y, \pm xy\}$.

In the end, $R/(F)$ is a free $k$-module of rank $4$ : its element are of the (unique) form $a+bx+cy+dxy$ with $a,b,c,d\in k$, and the multiplication is given by the relations $x^2 = y^2 = -1$ and $xy=-yx$.

If you know about quaternions this is exactly the algebra of quaternions $(-1,-1)_k$ over $k$. So for instance if $k = \mathbb{R}$ it is the usual Hamilton quaternions $\mathbb{H}$, and if $k = \mathbb{C}$ it is $M_2(\mathbb{C})$.