If $R$ is a ring without divisors of $0$ (not necessarily commutative), is it true that for every $x\neq 0$, $xR=Rx$?

Asked Oct 10 '23 at 22:50

Active Oct 10 '23 at 23:08

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Let $R$ be a ring (not necessarily commutative), is it true that for every $x\neq 0$, $xR=Rx$?.

What I've tried so far was:

Let $y\in Rx$. Then $y=rx$ for every $r\neq 0,~r\in R$. Then $xrx\in xR$.

$\therefore yr\in xR.$

I'm not sure, but couldn't find a counterxample.

edited Oct 10 '23 at 23:08

asked Oct 10 '23 at 22:50

Fabrizio Gambelín

2,006

I don't follow your argument. Why would $rxr \in xR$? – Sammy Black Oct 10 '23 at 23:04
Sorry, I edited it now. – Fabrizio Gambelín Oct 10 '23 at 23:08
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Try taking $R$ to be the ring of two-by-two matrices over some field, $x=\pmatrix{1&0\cr 0&0}$, and compare $xR$ and $Rx$? – paul garrett Oct 10 '23 at 23:15
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How about the example of $\mathbb{Z}\langle a, b \rangle$ where $aR$ is the elements where every "monomial" with nonzero coefficient starts with $a$, whereas $Ra$ is instead the elements where every "monomial" with nonzero coefficient ends with $a$? So for example, $ab \in aR$ but $ab \notin Ra$. – Daniel Schepler Oct 10 '23 at 23:18
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@paulgarrett The ring of matrices has zero divisors. – absolution-ever73 Oct 10 '23 at 23:19
I am not following your reasoning—what do you mean by $y=rx$ for all $r\in R$ nonzero? Can you please post your definitions of the sets $xR$ and $Rx$? – absolution-ever73 Oct 10 '23 at 23:22
@absolution-ever73, true, that ring of matrices has zero-divisors, but avoiding that didn't seem to be a requirement...? – paul garrett Oct 10 '23 at 23:49
$xR={y\in\mathbb{R}:~\exists r\text{ such that}y=xr}$. $Rx$ is defined the same way. – Fabrizio Gambelín Oct 11 '23 at 00:25
Sorry, but what is $\mathbb{Z}\langle a,b\rangle$? – Fabrizio Gambelín Oct 11 '23 at 00:26
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$\mathbb{Z}\langle a,b\rangle$ is like a polynomial ring, but the variables $a$ and $b$ are not assumed to commute with each other. So $ab\neq ba$. – Arturo Magidin Oct 11 '23 at 00:51
Great, thank you all very much!! – Fabrizio Gambelín Oct 11 '23 at 00:52

If $R$ is a ring without divisors of $0$ (not necessarily commutative), is it true that for every $x\neq 0$, $xR=Rx$?

0 Answers0