Given $R:=\mathcal M_2(\mathbb{R})$, the set of $2\times 2$ matrices, define the multiplicative set $$S = \{E_{1,1}\}$$ where $E_{1,1}$ is the matrix with $a_{1,1}=1$ and zero else where. Show that the localization $R_S = 0$.
Can you give me some hints? I only learned the commutative localization, and the only reference that I could find about noncommutative localization is from here, and it seems very complicated.
The localisation of the ring $R$ at the element $s \in R$ is the ring of sums of 'words' in the language $R \cup \{x\}$ modulo all relations that come from $R$ and the relations $xs=sx=1$ and $1x=x1=x$. Elements look like the following sort of thing. . . .
$xax^2b$
$x^3 + axbxc$
$abxcx^2dx + b$
. . . where $a,b,c$ and $d$ are elements of $R$.
Modulo our equivalences we have $xax^2b = xax^2bsx$ for example.
In case $s$ has some nice properties, the localisation has nice properties. For example the element $s$ is invertible and the localisation is a nontrivial ring (Most authors only define the localisation when $s$ is this nice).
You want to show the ring defined above is in fact the zero ring for your example. You can do this using the equivalences. It is enough to show each 'word' is equivelent to zero. That can be done using how your $s$ is a right and a left zero divisor, and performing some symbol manipulation.
Edit: It is even enough to prove the subring $R \subset R_S$ is the zero ring.
Denote $R:= \mathcal M_2(\mathbb{R})$, suppose we have the $S$-inverting map $f: R \rightarrow R_S$, then note $$f\left(\begin{bmatrix} a & b \\ 0 & 0 \end{bmatrix}\right) = f\left(\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) = f(\left(\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\right) f\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right)$$ is a unit if we pick some invertible matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
Similarly, $f\left( \begin{bmatrix} a & 0 \\ c & 0 \end{bmatrix}\right)$ is a unit.
If we start with the invertible matrix $ \begin{bmatrix} 1 & -1 \\ 1 & 0 \end{bmatrix}$, we will get
$$ f\left(\begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix}\right)f\left(\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}\right) = f\left( \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} \right) = 0,$$ but the left side is a product of units, this implies $1 = 0$ in $R_S$ thus, $R_S = 0$.