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?

  • 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

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 $1-a$.

  • $\begingroup$ 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

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