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An interesting result

Linear Algebra, Algebraic structures (Groups, Rings, Modules, etc), Galois theory, Homological Algebra
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Tsakanikas Nickos
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An interesting result


Post by Tsakanikas Nickos » Mon Jan 18, 2016 4:07 am

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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Re: An interesting result


Post by Alkesk » Mon Jan 18, 2016 4:08 am

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

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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