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 Post subject: An interesting resultPosted: Mon Jan 18, 2016 4:07 am
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Joined: Tue Nov 10, 2015 8:25 pm
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Let $\displaystyle R$ be a non zero associative ring such that

$\displaystyle \forall a \in \left( R \smallsetminus \{ 0_{R} \} \right) \; \exists \, ! \, x \in R \, : \, a = axa$

Show that $\displaystyle R$ has a unit $\displaystyle 1_{R}$ and that $\displaystyle R$ is a division ring.

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 Post subject: Re: An interesting resultPosted: Mon Jan 18, 2016 4:08 am

Joined: Sat Dec 12, 2015 11:19 am
Posts: 9
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 $\displaystyle{aaxa=axaa\Rightarrow aaxa-axaa=0\Rightarrow a\left ( ax-xa \right )a=0}$

since $\displaystyle{R}$ is an integral domain we take that $\displaystyle{ax-xa=0\Rightarrow ax=xa}$

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 $\displaystyle{\left ( 1 \right ),\left ( 2 \right )\Rightarrow crsr=ckcr\Rightarrow crsr-ckcr=0\Rightarrow c\left ( rs-kc \right )r=0\Rightarrow rs-kc=0\Rightarrow rs=kc=e=rs=sr \left (3 \right )}$

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.

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