1
$\begingroup$

I was computing the left annihilators of the elements $x$ of a ring $R$ with $1_R$ (denoted by Ann$_R(x)$) and encountered the following scenario:

For any $a\in R$, $$\text{Ann}_R(1_R-a)=\{r\in R: r(1_R-a)=0\}=\{r\in R: r1_R-ra=0\}=\{r\in R: r=ra\}.$$ From this observation, I feel like concluding that $$\text{Ann}_R(1_R-a)=Ra.$$

My question is, should I conclude that that $$\text{Ann}_R(1_R-a)=Ra?$$ Does it make sense?

$\endgroup$
  • 1
    $\begingroup$ You can certainly conclude that $\mathrm{Ann}_R(1_R-a)\subset Ra$, but the other inclusion does not necessarily hold. For example : any $a\neq1_R$ in a domain $R$. $\endgroup$ – Segipp Aug 9 '18 at 11:38
  • $\begingroup$ Yes: the strategy of thinking of extremes is apropos here. In a domain, the only annihilator for a nozero element is $\{0\}$. $\endgroup$ – rschwieb Aug 9 '18 at 13:14
2
$\begingroup$

No, all you can conclude is that $r$ is a left-annihilator of $1-a$ if and only if $r=ra$.

That doesn't imply that for all $r\in R$, the element $ra$ is a left-annihilator of $1-a$.

For example, if $a\ne a^2$, the element $1a=a$ is not a left-annihilator of $1-a$.

$\endgroup$
  • $\begingroup$ Thank you! But I am not understanding the implication of $r\in \text{Ann}_R(1-a)$ iff $r=ra$, because I don't see how $r(1-a)=ra(1-a)=ra-ra^2$ goes to zero if $a\neq a^2$! $\endgroup$ – Sulayman Aug 9 '18 at 16:08
  • $\begingroup$ If $r$ is a left-annihilator of $1-a$, then $r(1-a)=0$, hence $r=ra$. Conversely, if $r=ra$, then $r(1-a)=0$, hence $r$ is a left-annihilator of $1-a$. $\endgroup$ – quasi Aug 9 '18 at 16:13
  • $\begingroup$ Thus, if $a\ne a^2$, then $a(1-a)\ne 0$, so $a$ is not a left-annihilator of $1-a$. $\endgroup$ – quasi Aug 9 '18 at 16:21
  • $\begingroup$ Exactly! my problem is the converse, you say if $r=ra$, then $r(1−a)=0$. But all I see is that $r(1−a)=ra(1-a)=ra-ra^2\neq0$ $\endgroup$ – Sulayman Aug 9 '18 at 16:27
  • $\begingroup$ $$r=ra\iff r-ra=0\iff r1-ra=0\iff r(1-a)=0$$ $\endgroup$ – quasi Aug 9 '18 at 16:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.