Let \(\displaystyle{r_{1},r_{2}\in R : r_{1}r_{2}= 0}\) ,
if \(\displaystyle{r_{1}\neq 0}\) then there exists unique \(x_{1} : r_{1}= r_{1}x_{1}r_{1}\)
since \(\displaystyle{r_{1}\left (x_{1}+r_{2} \right )r_{1}=r_{1}x_{1}r_{1}+\left (r_{1}r_{2} \right )r_{1}=r_{1}+0=r_{1}}\) and \(\displaystyle{x_{1}}\) is unique we take that
\(\displaystyle{x_{1}=x_{1}+r_{2}\Rightarrow r_{2}=0}\)
if \(\displaystyle{r_{2}\neq 0}\) then, if \(\displaystyle{r_{1}\neq0}\) from the fact that \(\displaystyle{r_{1}r_{2}= 0}\) we take that \(\displaystyle{r_{2}=0}\)
so \(\displaystyle{r_{1}=0}\) . Thus R is an integral domain.
For each \(\displaystyle{a\in R : a\neq 0}\) there is unique \(\displaystyle{x}\) such that \(\displaystyle{a=axa}\) multiplicating by both left and right with \(\displaystyle{a}\) we take
\(\displaystyle{aa=aaxa}\) and \(\displaystyle{aa=axaa}\) so \(\displaystyle{aaxa=axaa\Rightarrow aaxaaxaa=0\Rightarrow a\left ( axxa \right )a=0}\)
since \(\displaystyle{R}\) is an integral domain we take that \(\displaystyle{axxa=0\Rightarrow ax=xa}\)
Let \(\displaystyle{c\in R,c\neq 0}\) a specific element of \(\displaystyle{R}\) then \(\displaystyle{c=ckc}\) for a unique \(\displaystyle{k\in R}\) setting
\(\displaystyle{kc=ck=e}\) and by taking a random \(\displaystyle{r\in R , r\neq 0}\) we have that
\(\displaystyle{r=rsr}\) for a unique \(\displaystyle{s\in R\Rightarrow cr=crsr \left ( 1 \right )}\) and since \(\displaystyle{c=ckc\Rightarrow cr=ckcr \left ( 2 \right )}\) .
From \(\displaystyle{\left ( 1 \right ),\left ( 2 \right )\Rightarrow crsr=ckcr\Rightarrow crsrckcr=0\Rightarrow c\left ( rskc \right )r=0\Rightarrow rskc=0\Rightarrow rs=kc=e=rs=sr \left (3 \right )}\)
so \(\displaystyle{re=r\left ( sr \right )=r=\left ( rs \right )r=er}\) ,obviously \(\displaystyle{0e=e0==0}\) so for each \(\displaystyle{r\in R , re=r=er}\) so \(\displaystyle{e}\) is the unit of \(\displaystyle{R}\).
Finally equations \(\displaystyle{\left (3 \right )}\) shows that for each \(\displaystyle{r\in R}\) such that \(\displaystyle{r\neq 0}\) there is \(\displaystyle{r^{1}=s\in R}\).So \(\displaystyle{R}\) is a division ring.
